Applications of optimization
with Xpress-MP

Revised translation from the French language edition of:
Programmation linéaire
by Christelle Guéret, Christian Prins, Marc Sevaux
c 2000 Editions Eyrolles, Paris, France.

Translated and revised by Susanne Heipcke

Published by Dash Optimization Ltd.
www.dashoptimization.com

Published by:
Dash Optimization Ltd.
Blisworth House
Blisworth
Northants
NN7 3BX
United Kingdom

c 2000 Editions Eyrolles, Paris, France
Revised translation first published 2002

Translation and editing of this book has been supported by the EC Framework 5 project LISCOS(contract G1RD-1999-00034).

Cover design: James Atkins Design, www.jades.co.uk
Manufacturing coordinator: Software Logistics, www.softwarelogistics.com

ISBN: 0-9543503-0-8

Last update 19 January, 2006

About the authors
Christelle Guéret is an Associate Professor in the Automatic Control and Production Systems department
at the ‘Ecole des Mines de Nantes’ (France). She received her Ph.D. degree in 1997 at the ‘Universitéde Technologie de Compiègne’ and defended her Habilitation in December 2004 at the University of
Nantes in the IRCCyN laboratory. Her teaching courses include graph algorithms, scheduling and linear
programming. Her main research areas concern optimization problems in production and logistics, andmore specifically scheduling and vehicle routing problems.
Susanne Heipcke worked for BASF-AG (Germany) before joining Dash Optimization in 1998. Her Ph.D.
research (awarded in 1999 by the University of Buckingham) focused on the solution of large-scale industrial problems by a combination of constraint programming and mixed integer programming. Morerecently she has worked on various aspects of modeling, including the development of teaching material
for Mosel, and interfaces to different types of solvers and solution methods. Since 2001 she has participated in teaching the course on mathematical modeling in the OR master program at the University
Aix-Marseille II.
Christian Prins received his Ph.D. and Habilitation in computer science from Pierre and Marie Curie University (Paris VI). He is now a Professor in the Industrial Systems Engineering Department at the Universityof Technology of Troyes (UTT, France). His research interest include combinatorial optimization, vehicle
routing and transportation, scheduling and algorithmic tools for optimization (e.g. metaheuristics andreusable software components). He is a member of the editorial board of Computers and Operations
Research.
Marc Sevaux is a Professor at the University of South-Britanny in Lorient (France). He received his Ph.D.
degree in 1998 at the Pierre and Marie Curie University (Paris VI) and defended his Habilitation in July
2004 at the University of Valenciennes in the LAMIH laboratory. He is primarily interested in solving combinatorial optimization problems, particularly planning, scheduling and routing problems. Marc Sevaux isthe coordinator of the EUropean chapter on MEtaheuristics (EU/ME). He is an associate editor of Journalof Heuristics and INFOR.

Contents
Foreword

1

Preliminaries
What you need to know before reading this book . . . . . . . . . . . . . . . . . . . . . . . . . . .
Symbols and conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4
4
4

I

6

Developing Linear and Integer Programming models

1 What is modeling? Why use models?
1.1 The chess set problem: description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 A first formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Solving the chess set problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Building the model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 The results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.3 Divisibility again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.4 Unboundedness and infeasibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Diagnosing infeasibility and unboundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 The benefits of modeling and optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Data in models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7
7
7
9
11
11
12
13
15
15
16
17
18

2 Typical LP model constructs
2.1 Simple upper and lower bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Flow constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Simple resource constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Material balance constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Quality requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Accounting constraints, non-constraining ‘constraints’ . . . . . . . . . . . . . . . . . . . . . .
2.7 Blending constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9 Soft constraints and ‘panic variables’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10 Objective functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10.1 Minimax objective functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10.2 Ratio objective functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19
19
20
21
22
24
25
25
26
27
28
28
28

3 Integer Programming models
3.1 IP modeling objects: ‘global entities’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 IP solving: the ideas behind Branch and Bound . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Modeling with binary variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Do/don’t do decisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2 Logical conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2.1 Choice among several possibilities . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2.2 Simple implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2.3 Implications with three variables . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2.4 Generalized implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.3 Products of binary variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.4 Dichotomies: either/or constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Binary variables ‘do everything’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.1 General integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.2 Partial integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.3 Semi-continuous variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30
30
31
33
33
34
34
34
35
36
37
38
38
39
39
39

i

Applications of optimization with Xpress-MP

Contents

3.4.4 Special Ordered Sets of type 1 (SOS1) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.5 Special Ordered Sets of type 2 (SOS2) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Connecting real variables to binary variables . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 Modeling fixed costs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.3 Partial integers again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.4 Price breaks and economies of scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.5 The product of a binary and a real variable . . . . . . . . . . . . . . . . . . . . . . . .
3.6 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39
40
42
42
42
43
43
44
45

4 Quadratic Programming
4.1 Revenue optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Portfolio optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

46
46
46
47

II

Application examples

48

Classification of the example problems

49

5 The basics of Xpress-MP
5.1 Introductory example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Using Xpress-Mosel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Using Xpress-IVE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Modeling with Mosel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 The burglar problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.2 Reading data from text files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.3 Reserved words . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55
55
56
56
57
57
60
61

6 Mining and process industries
6.1 Production of alloys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Animal food production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Refinery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Cane sugar production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Opencast mining . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6 Production of electricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

62
62
62
63
65
65
66
67
68
68
69
71
73
73
73
74
74
75
75
76
77
77
77
78
79
80

7 Scheduling problems
7.1 Construction of a stadium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.1 Model formulation for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.2 Implementation of question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.3 Results for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.4 Model formulation for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.5 Implementation of question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.6 Results for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Flow-shop scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81
81
81
83
83
83
84
86
86

ii

Applications of optimization with Xpress-MP

7.3

7.4

7.5

7.6

7.7

7.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
7.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
7.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Job Shop Scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
7.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
7.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
7.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
Sequencing jobs on a bottleneck machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
7.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
7.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
7.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Paint production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
7.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
7.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
7.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
Assembly line balancing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
7.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
7.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
7.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

8 Planning problems
103
8.1 Planning the production of bicycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
8.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
8.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
8.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
8.2 Production of drinking glasses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
8.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
8.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
8.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
8.3 Material Requirement Planning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
8.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
8.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
8.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
8.4 Planning the production of electronic components . . . . . . . . . . . . . . . . . . . . . . . . 112
8.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
8.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
8.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
8.5 Planning the production of fiberglass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
8.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
8.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
8.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
8.6 Assignment of production batches to machines . . . . . . . . . . . . . . . . . . . . . . . . . . 117
8.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
8.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
8.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
8.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
9 Loading and cutting problems
121
9.1 Wagon load balancing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
9.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
9.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
9.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
9.2 Barge loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
9.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
9.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
9.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
9.3 Tank loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
9.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
9.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
9.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
9.4 Backing up files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
9.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
9.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

Contents

iii

Applications of optimization with Xpress-MP

9.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
9.5 Cutting sheet metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
9.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
9.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
9.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
9.6 Cutting steel bars for desk legs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
9.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
9.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
9.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
9.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
10 Ground transport
139
10.1 Car rental . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
10.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
10.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
10.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
10.2 Choosing the mode of transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
10.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
10.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
10.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
10.2.4 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
10.3 Depot location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
10.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
10.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
10.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
10.4 Heating oil delivery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
10.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
10.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
10.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
10.5 Combining different modes of transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
10.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
10.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
10.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
10.6 Fleet planning for vans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
10.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
10.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
10.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
10.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
11 Air transport156
11.1 Flight connections at a hub . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
11.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
11.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
11.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
11.2 Composing flight crews . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
11.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
11.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
11.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
11.3 Scheduling flight landings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
11.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
11.3.2 Generalization to arbitrary types of time windows . . . . . . . . . . . . . . . . . . . . 162
11.3.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
11.3.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
11.4 Airline hub location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
11.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
11.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
11.4.3 Revised formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
11.4.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
11.5 Planning a flight tour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
11.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
11.5.2 Implementation and results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
11.6 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
12 Telecommunication problems

Contents

173

iv

Applications of optimization with Xpress-MP

12.1 Network reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
12.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
12.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
12.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
12.2 Dimensioning of a mobile phone network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
12.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
12.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
12.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
12.3 Routing telephone calls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
12.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
12.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
12.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
12.4 Construction of a cabled network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
12.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
12.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
12.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
12.5 Scheduling of telecommunications via satellite . . . . . . . . . . . . . . . . . . . . . . . . . . 185
12.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
12.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
12.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
12.6 Location of GSM transmitters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
12.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
12.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
12.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
12.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
13 Economics and finance194
13.1 Choice of loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
13.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
13.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
13.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
13.2 Publicity campaign . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
13.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
13.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
13.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
13.3 Portfolio selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
13.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
13.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
13.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
13.4 Financing an early retirement scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
13.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
13.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
13.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
13.5 Family budget . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
13.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
13.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
13.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
13.6 Choice of expansion projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
13.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
13.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
13.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
13.7 Mean variance portfolio selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
13.7.1 Model formulation for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
13.7.2 Implementation for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
13.7.3 Results for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
13.7.4 Model formulation for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
13.7.5 Implementation for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
13.7.6 Results for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
13.7.7 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
13.8 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
14 Timetabling and personnel planning210
14.1 Assigning personnel to machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
14.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

Contentsv

Applications of optimization with Xpress-MP

14.1.1.1 Parallel machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
14.1.1.2 Machines working in series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
14.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
14.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
14.2 Scheduling nurses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
14.2.1 Model formulation for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
14.2.2 Implementation of question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
14.2.3 Results for question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
14.2.4 Model formulation for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
14.2.5 Implementation of question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
14.2.6 Results for question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
14.3 Establishing a college timetable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
14.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
14.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
14.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
14.4 Exam scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
14.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
14.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
14.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
14.5 Production planning with personnel assignment . . . . . . . . . . . . . . . . . . . . . . . . . 221
14.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
14.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
14.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
14.6 Planning the personnel at a construction site . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
14.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
14.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
14.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
14.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
15 Local authorities and public services
228
15.1 Water conveyance / water supply management . . . . . . . . . . . . . . . . . . . . . . . . . . 228
15.1.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
15.1.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
15.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
15.2 CCTV surveillance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
15.2.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
15.2.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
15.2.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
15.3 Rigging elections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
15.3.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
15.3.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
15.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
15.4 Gritting roads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
15.4.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
15.4.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
15.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
15.5 Location of income tax offices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
15.5.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
15.5.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
15.5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
15.6 Efficiency of hospitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
15.6.1 Model formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
15.6.1.1 General idea of the DEA method . . . . . . . . . . . . . . . . . . . . . . . . . 244
15.6.1.2 Modeling our problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
15.6.2 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
15.6.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
15.7 References and further material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

Index

Contents

253

vi

Applications of optimization with Xpress-MP

Foreword
In the past, many students have become acquainted with optimization through a course on Linear Programming. Until very recently computers were too expensive and optimization software only available toindustrial groups and research centers with powerful machines. Courses on Linear Programming therefore were usually restricted to solving problems in two variables graphically and applying the famous
simplex algorithm in tableau form to tiny problems. The size of the problems thus solved by hand rarely
exceeded 6 to 8 variables. The practice of modeling was commonly neglected due to a lack of time andthe unavailability of software to verify the models.
Nowadays Linear Programming is still frequently taught from a very academic perspective. Of course, inthe majority of course programs Linear Programming appears together with other classics of Operations
Research such as the PERT method for project planning. But with the exception of a few students inapplied mathematics who have learned to deal with applications, the others do not retain much: they
have no precise idea of the possible applications, they do not know that now there are readily available
software tools and most importantly, they have not been taught how to model.
The idea of this book was born based on this observation. With PCs becoming ever more powerful andthe rapid progress of the numerical methods for optimization the market of Linear/Mixed Integer Programming software is changing. Contrary to the first products (subroutine libraries reserved for specialistuse), the more recent generations of software do not require any programming. These tools are easy
to use thanks to their user-friendly interface and a more and more natural syntax that is close to themathematical formulae.
In different study programs (sciences as well as business/economics) a significant increase in the students’interest and ‘productivity’ has been observed when such tools are used. With such an approach, evenstudents without any mathematical background are well able to model and solve real problems of considerable size, whilst they would be lost in a purely theory-oriented course about optimization methods.
These students experience the satisfaction of mastering powerful tools and learn to appreciate the fieldof Linear/Mixed Integer Programming. Later, at work, they may recall the potential of these tools andbased on the experience they have gained, may use them where it is necessary. What else should one aim
for to spread the use of Mathematical Programming?
This book has been written for students of science and business/economics and also for decision makers,
professionals, and technical personnel who would like to refresh their knowledge of Linear/Mixed Integer
Programming and more importantly, to apply it in their activities.
To focus on modeling and on the application examples, the theoretical aspects of Linear/Mixed Integer
Programming are only briefly mentioned in this book, without giving any detailed explanations. Ample references are provided and commented on to point the interested reader to where to find more
information about the different topics.
To avoid a pre-digested textbook approach, the examples chosen for this book describe real situations.
Although simplified compared to the real world so as to remain accessible for beginners, most examples
are sufficiently complicated not to be solvable by hand on a piece of paper and they all require a nontrivial modeling phase.
The software used for the implementation and problem solving, Xpress-MP, may be downloaded fromhttp://www.dashoptimization.com/applications_book.html
free of charge in a version limited in size (but sufficient to solve the problems in this book). So thereader does not have to worry about how to solve the problems and is free to concentrate his effortson the most noble task that nevertheless requires progressive and methodical training: the modeling.
The opportunity to validate a model immediately using a software tool is gratifying: results are obtained
immediately, modeling errors are discovered faster, and it is possible to perform various simulation runs
by interactively modifying the problem parameters.
It is not the aim of this book to turn the reader into an Xpress-MP specialist at the expense of the model-

1

Applications of optimization with Xpress-MP

ing work. Every problem is therefore first modeled in mathematical form without making any reference
to the implementation with the modeling software. The translation of the model into a program is given
afterwards, accompanied by an explanation of any specific language features used, and followed by thepresentation and discussion of the results. The mathematical model and the results are obviously independent of the software. All Xpress-MP model and data files are available from the Dash Optimization
website and can be used to verify the results immediately. It would of course be possible to implement
the models in a different way.
To avoid establishing a heterogeneous catalogue of applications, the collection has been structured stressing the large variety of application areas of Linear/Mixed Integer Programming. The introductory chapters about modeling and the use of Xpress-MP are followed by ten chapters with applications. Every
chapter is dedicated to a different application area and may be read independently of the rest. It always
starts with a general description of the topic, followed by a choice of (on the average) six problems. A
summary section at the end of every chapter provides additional material and references.
Chapter 6 describes blending problems or the separation of components that are usually encountered inmining and process industries. These problems only rarely use integer variables and are the simplest andbest suited for beginners. The two following chapters are dedicated to two other important groups ofindustrial applications, namely scheduling (Chapter 7) and production planning problems (Chapter 8).
In many applications, it is required to fill a limited space (boxes, holds of ships, computer disks) with
objects or, on the contrary, cut material to extract patterns whilst minimizing the trim loss. This type ofproblem is the subject of Chapter 9 about loading and cutting.
Flow problems in the widest sense give rise to a large number of interesting optimization problems.
This problem type is discussed across three representative application areas: ground transport (Chapter10), air transport (Chapter 11), and the burgeoning world of telecommunications (Chapter 12). Some
problems may be shifted between these three chapters, but other applications are more specific, like thetours driven by a vehicle in ground transport, the problems of flight connections in air transport, and allthat concerns dimensioning and reliability of networks in telecommunications.
Since this book is not aimed exclusively at scientists, industrialists, and logisticians, more recent and lesswell known application areas of Mathematical Programming have also been included. Chapter 13 is dedicated to problems in economics and finance. The subjects of Chapter 14 are timetabling and personnel
planning problems. Mathematical Programming may equally be of use to local authorities, administrations, and public services in general: Chapter 15 describes six examples.
The classification of the problems has been a subject of discussion: theoreticians usually prefer a classification based on the theoretical model type. For instance the so-called flow models are well known classics
that concern several problems in this book (water conveyance, ground transport, telecommunications).
Preference has been given to a classification by application area to allow a professional or a student ofa given subject to identify himself with at least one entire chapter. Nevertheless, the other classification
has not been omitted: the additions at the end of every chapter indicate the applications of the same
theoretical type in other chapters of the book. An overview table at the beginning of the examples partalso lists all models with their theoretical types.
The bibliography entries are all grouped at the end of the book because certain references are citedin several chapters. However, the section ‘References and further material’ at the end of every chapter
comments on the references that are useful to learn more about a certain topic. The references havebeen chosen carefully to provide a blend of basic articles and works, sometimes already quite old, andvery recent articles that illustrate the rapid progress of the discipline.
The Xpress-MP software may be seen as a product of the progress achieved in optimization in recent
years. The first software tools in the 60s and 70s required the user to enter the problem in numerical form
via matrices that are of little intuition, difficult to re-read and to modify. Starting in the 80s, algebraic
modeling languages have been made available that enable users to write linear programs in a form
close to the algebraic notation. An algebraic modeling language allows the user to write generic models
using indexed variables and data arrays. This means that it is possible to separate the model from thedata (stored in separate files) and to make it independent of the size of the problem. Such a high level
language focuses the user’s attention on the modeling, reduces the possible sources of errors, accelerates
model development and facilitates the understanding and modifications to the model at a later time.
The modeling and solving environment Xpress-Mosel [CH02] used in this book belongs to a new generation of optimization tools: it provides all the functionality of an algebraic modeling language andit also gives access to the problem solving. Through its programming facilities it becomes possible toimplement solution heuristics and data pre- and post-treatment in the same environment as the model
itself: all application examples (to a different extent) use this facility to formulate and solve problems
and display the results. An interesting feature for more advanced uses is Mosel’s modular design throughForeword

2

Applications of optimization with Xpress-MP

which, for instance, access to other solver or solution algorithm types or different data sources becomes
possible. This feature is not used in this book, since we employ a single solver, the Xpress-Optimizer for
Linear/Mixed Integer and Quadratic Programming.
For working with Mosel models under Microsoft Windows the graphical user interface Xpress-IVE may
be used: it provides debugging support, makes accessible additional information about the problem andthe solution process, and allows the user to display his results graphically. As a default, on all supported
operating systems, Mosel works with a command line interpreter based on standard text files.
Only a few other books have been published with an aim similar to the present one: a book by Williams
provides twenty case studies, in the third edition [Wil93] accompanied by implementations with XpressMP. The case studies are well suited for student project work, requiring several days of work. Unfortunately, this book only covers relatively few application areas. Another example is the book by Schrage[Sch97] about the software Lindo. This work contains more applications, but classified by theoreticalmodels. The problems are implemented directly with the software, without using a modeler.
Compared to the previously cited works, the present book contains a larger number of applications (sixty)
structured by application areas. Although all models are implemented with Xpress-MP, every problem
is first modeled with a mathematical syntax which makes it possible to use the models with different
implementations. Finally, many models are re-usable for related problems or large-size instances due to a
generic way of modeling (maximum flow, minimum cost flow, assignment, transport, traveling salesman
problem etc.).
Updates to this book will be made available at
http://www.dashoptimization.com/applications_book.html
and you are encouraged to look there regularly.
The partial funding of the work on this book by the EC Framework 5 project LISCOS (contract G1RD-199900034) is gratefully acknowledged.
We would specially like to thank Yves Colombani for conceiving, designing and implementing Mosel. Hisvision has made Mosel the powerful and open software that it is today.

Foreword

3

Applications of optimization with Xpress-MP

Preliminaries
What you need to know before reading this book
Before reading this book you should be comfortable with the use of symbols such as x or y to represent
unknown quantities, and the use of this sort of variable in simple linear equations and inequalities, forexample:x+y ≤6
Experience of a basic course in Mathematical or Linear Programming is worthwhile, but is not essential.
Similarly some familiarity with the use of computers would be helpful.
For all but the simplest models you should also be familiar with the idea of summing over a range ofvariables. For example, if producej is used to represent the number of cars produced on production line jthen the total number of cars produced on all N production lines can be written as:NX

producej

j=1

This says ‘sum the output from each production line producej over all production lines j from j = 1 to
j = N’.
If our target is to produce at least 1000 cars in total then we would write the inequality:
NX

producej ≥ 1000

j=1

We often also use a set notation for the sums. Assuming that LINES is the set of production lines {1, . . , N},
we may write equivalently:X
producej ≥ 1000j∈LINESThis may be read ‘sum the output from each production line producej over all production lines j in theset LINES’.
Other common mathematical symbols that are used in the text are IN (the set of non-negative integer
numbers {0, 1, 2, . . . }), ∩ and ∪ (intersection and union of sets), ∧ and ∨ (logical ‘and’ and ‘or’), the allquantifier ∀ (read ‘for all’), and ∃ (read ‘exists’).
Computer based modeling languages, and in particular the language we use, Mosel, closely mimic themathematical notation an analyst uses to describe a problem. So provided you are happy using the above
mathematical notation the step to using a modeling language will be straightforward.

Symbols and conventions
We have used the following conventions within this book:• Mathematical objects are presented in italics.
• Examples of commands, models and their output are printed in a Courier font. Filenames are
given in lower case Courier.
• Decision variables have lower case names; in the application examples these usually are verbs (suchas buy, make).

4

Applications of optimization with Xpress-MP

• Constraint names start with an upper case letter, followed by mostly lower case (e.g. Profit,
TotalCost).
• Data (arrays and sets) and constants are written entirely with upper case (e.g. DEM, JOBS, PRICE).
• The vertical bar symbol | is found on many keyboards as a vertical line with a small gap in the
middle, but often confusingly displays on-screen without the small gap. In the UNIX world it isreferred to as the pipe symbol. (Note that this symbol is not the same as the character sometimes
used to draw boxes on a PC screen.) In ASCII, the | symbol is 7C in hexadecimal, 124 in decimal.

Preliminaries

5

Applications of optimization with Xpress-MP

I. Developing Linear andInteger Programming models

Chapter 1

What is modeling?
Why use models?
Rather than starting with a theoretical overview of what modeling is, and why it is useful, we shall lookat a problem facing a very small manufacturer, and how we might go about solving the problem. When
we have gained some practical modeling experience, we shall come back to see what the benefits ofmodeling have been.
Section 1.1 gives a word description of the problem and its translation into a mathematical form. We thendefine in Section 1.2 the notion of a Linear Program and show how the example problem fits this form. In
Section 1.3 the problem is implemented with Mosel and solved with the Xpress-Optimizer. The discussionof the results comprises a graphical representation of the solution information that may be obtained fromthe solver. Important issues in modeling and solving linear problems are infeasibility and unboundedness(Section 1.4). The chapter closes with reflections on the benefits of modeling and optimization (Section
1.5) and the importance of the data (Section 1.6).

1.1

The chess set problem: description
A small joinery makes two different sizes of boxwood chess sets. The small set requires 3 hours of machining on a lathe, and the large set requires 2 hours. There are four lathes with skilled operators whoeach work a 40 hour week, so we have 160 lathe-hours per week. The small chess set requires 1 kg of
boxwood, and the large set requires 3 kg. Unfortunately, boxwood is scarce and only 200 kg per week
can be obtained.
When sold, each of the large chess sets yields a profit of $20, and one of the small chess set has a profitof $5.
The problem is to decide how many sets of each kind should be made each week so as to maximize profit.

1.1.1

A first formulation
Within limits, the joinery can vary the number of large and small chess sets produced: there are thus two
decision variables in our model, one decision variable per product. What we want to do is to find thebest (i.e. optimal) values of these decision variables, where by ’best’ we mean that we get the largest
profit. We shall give these variables abbreviated names:
xs :
xl :

the number of small chess sets to makethe number of large chess sets to makeThe number of large and small chess sets we should produce to achieve the maximum contribution to
profit is determined by the optimization process. In other words, we look to the optimizer to tell us thebest values of xs, and xl.
The values which xs and xl can take will always be constrained by some physical or technological limits.
One of the main tasks in building a model is to write down in a formal manner the exact constraints thatdefine how the system can behave. In our case we note that the joinery has a maximum of 160 hours ofmachine time available per week. Three hours are needed to produce each small chess set, and two hours
are needed to produce each large set. So if in the week we are planning to make xs small chess sets and7

Applications of optimization with Xpress-MP

xl large chess sets, then in total the number of hours of machine time we are planning to use is:
3 · xs + 2 · xl
where the 3 · xs comes from the time making small sets, and the 2 · xl from the time machining large sets.
Note that we have already made some assumptions here. Firstly we have assumed that the lathe-hours tomachine xs small sets is exactly xs times the lathe-hours required to machine one small set. This probably
will not be exactly true in practice — one tends to get faster at doing something the more one does it, so
it will probably take a slightly smaller amount of time to machine the 2nd and subsequent sets than thefirst set. But it is unlikely that this will be a very important effect in our small joinery.
The second assumption we have made is much more likely to be inaccurate. We have assumed that thetime for making small and large sets is the sum of the times for the sets. We have not allowed for any
changeover time: resetting the lathes, cleaning up, getting different size tools etc. In some situations,the time that we lose in changeovers can be very large compared with the time we actually spend doing
constructive work and then we have to resort to more complex modeling. But for the moment, we shallassume that the changeover times are negligible.
Our first constraint is:
3 · xs + 2 · xl ≤ 160 (lathe-hours)
which says ‘the amount of time we are planning to use must be less than or equal to the amount of time
available’, or equivalently ‘we cannot plan to use more of the resource (time) than we have available’.
The allowable combinations of small and large chess sets are restricted to those that do not exceed thelathe-hours available.
In addition, only 200 kg of boxwood is available each week. Since small sets use 1 kg for every set made,against 3 kg needed to make a large set, a second constraint is:
1 · xs + 3 · xl ≤ 200 (kg of boxwood)
where the left hand side of the inequality is the amount of boxwood we are planning to use and theright hand side is the amount available.
The joinery cannot produce a negative number of chess sets, so two further non-negativity constraintsare:xs ≥ 0xl ≥ 0In a similar way, we can write down an expression for the total profit. Recall that for each of the large
chess sets we make and sell we get a profit of $20, and one of the small chess set gives us a profit of $5.
Assuming that we can sell all the chess sets we make (and note that this may not always be a reasonableassumption) the total profit is the sum of the individual profits from making and selling the xs small sets
and the xl large sets, i.e.
Profit = 5 · xs + 20 · xl
Profit is the objective function, a linear function which is to be optimized, that is, maximized. In this case
it involves all of the decision variables but sometimes it involves just a subset of the decision variables.
Note that Profit may be looked at as a dependent variable, since it is a function of the decision variables.
In maximization problems the objective function usually represents profit, turnover, output, sales, market share, employment levels or other ‘good things’. In minimization problems the objective function
describes things like total costs, disruption to services due to breakdowns, or other less desirable process
outcomes.
Consider some possible values for xs, and xl (see Table 1.1). The aim of the joinery is to maximize profit,
but we cannot select any combination of xs and xl that uses more of any of the resources than we haveavailable. If we do plan to use more of a resource than is available, we say that the plan violates theconstraint, and the plan is infeasible if one or more constraints is violated. If no constraints are violated,
the plan is feasible. The column labeled ‘OK?’ in the table tells us if the plan is feasible. Plans C and E are
infeasible.
In terms of profit, plan H looks good. But is it the best plan? Is there a plan that we have not considered that gives us profit greater than 1320? To answer this question we must move to the notion of
optimization.

What is modeling? Why use models?

8

Applications of optimization with Xpress-MP

Table 1.1: Values for xs and xl

1.2

xs

xl

Lathe-hours

Boxwood

OK?

Profit

Notes

A
B
C

0
10
-10

0
10
10

0
50
-10

0
40
20

Yes
Yes
No

0
250
150

D

53

0

159

53

Yes

265

E
F

50
25

20
30

190
135

110
115

No
Yes

650
725

G
H

12
0

62
66

160
130

198
198

Yes
Yes

1300
1320

Unprofitable!
We won’t get rich doing this.
Planning to make a negative
number of small sets.
Uses all the lathe-hours. Thereis spare boxwood.
Uses too many lathe-hours.
There are spare lathe-hours
and spare boxwood.
Uses all the resources
Looks good. There are spareresources.

Linear Programming
We have just built a model for the decision process that the joinery owner has to make. We have isolated
the decisions he has to make (how many of each type of chess set to manufacture), and taken his objectiveof maximizing profit. The constraints acting on the decision variables have been analyzed. We have given
names to his variables and then written down the constraints and the objective function in terms of thesevariable names.
At the same time as doing this we have made, explicitly or implicitly, various assumptions. The explicit
assumptions that we noted were:• For each size of chess set, manufacturing time was proportional to the number of sets made.
• There was no down-time because of changeovers between sizes of sets.
• We could sell all the chess sets we made.
But we made many implicit assumptions too. For instance, we assumed that no lathe will ever break or
get jammed; that all the lathe operators will turn up for work every day; that we never find any flaws in
the boxwood that lead to some being unusable or a chess set being unacceptable; that we never have to
discount the sale price (and hence the per unit profit) to get an order. And so on. We have even avoided
a discussion of what is the worth of a fraction of a chess set — is it a meaningless concept, or can we justcarry the fraction that we have made over into next week’s production?
All mathematical models necessarily contain some degree of simplification of the real world that we areattempting to describe. Some assumptions and simplifications seem eminently reasonable (for instance,that we can get the total profit by summing the contributions of the individual profits from the two sizes);
others may in some circumstances be very hopeful (no changeover time lost when we swap between
sizes); whilst others may just be cavalier (all the lathe operators will arrive for work the day after theWorld Cup finals).
Modeling is an art, not a precise science. Different modelers will make different assumptions, and comeup with different models of more or less precision, and certainly of different sizes, having different numbers of decision variables. And at the same time as doing the modeling, the modeler has to be thinking
about whether he will be able to solve the resulting model, that is find the maximum or minimum valueof the objective function and the values to be given to the decision variables to achieve that value.
It turns out that many models can be cast in the form of Linear Programming models, and it is fortunatethat Linear Programming (LP) models of very large size can be solved in reasonable time on relatively
inexpensive computers. It is not the purpose of this book to discuss the algorithms that are used to solve
LP problems in any depth, but it is safe to assume that problems with tens of thousands of variables andconstraints can be solved with ease. So if you can produce a model of your real-world situation, without
too many wild assumptions, in the form of an LP then you know you can get a solution.
So we next need to see what a Linear Programming problem consists of. To do so, we first introduce thenotion of a linear expression. A linear expression is a sum of the following form
A1 · x1 + A2 · x2 + A3 · x3 +. . . +AN · xN

What is modeling? Why use models?

9

Applications of optimization with Xpress-MP

which in mathematical notation is usually written asNX

Aj · xj

j=1

where A1 , . . . , AN are constants and x1 , . . . , xN are decision variables. So for instance, if we have variables
x, makeP and makeQ
2 · x − 3 · makeP + 4 · makeQ
is a linear expression, but
2 · x · makeP − 3 · makeP + 4 · makeQ
is not, as the first term contains the product of two variables.
Next, we introduce the notion of linear inequalities and linear equations. For any linear expressionPN
j=1

Aj · xj and any constant B, the inequalities
NX

Aj · xj ≤ B and

N
X

j=1

Aj · xj ≥ B

j=1

are linear inequalities, and the equation
NX

Aj · xj = B

j=1

is a linear equation. So for example our lathe-hours constraint
3 · xs + 2 · xl ≤ 160
is a linear inequality, but
2 · xs · xl + 3 · xs = 200
is not a linear equation because of the first term, which is a product of two decision variables.
Now, if we have decision variables
x1 , x2 , x3 , . . . , xN ,a linear expression
C1 · x1 + C2 · x2 + C3 · x3 +. . . +CN · xN
and a number of linear inequalities and linear equations
Ai1 · x1 + Ai2 · x2 + Ai3 · x3 +. . . +AiN · xN ≤ Bi for i = 1, . . . , M1
Ai1 · x1 + Ai2 · x2 + Ai3 · x3 +. . . +AiN · xN = Bi for i = M1 + 1, . . . , M2
Ai1 · x1 + Ai2 · x2 + Ai3 · x3 +. . . +AiN · xN ≥ Bi for i = M2 + 1, . . . , M3 ,then a Linear Programming problem is tomaximize or minimizeN
X

Cj · xj (the objective function)i=jsubject to the constraints

NX

Aij · xj ≤ Bi for i = 1, . . . , M1

j=1
N
X

Aij · xj = Bi for i = M1 + 1, . . . , M2

j=1
N
X

Aij · xj ≥ Bi for i = M2 + 1, . . . , M3

j=1

and

xj ≥ 0 for each j = 1, . . . , N (non-negativity constraints)

The Bi are often called the right hand sides (RHS).
So, for instance, the chess set model is a linear program as it has variables xs and xl and is tomaximize

5 · xs + 20 · xl

subject to

1 · xs + 3 · xl ≤ 200 (kg of boxwood)
3 · xs + 2 · xl ≤ 160 (lathe-hours)
xs ≥ 0xl ≥ 0What is modeling? Why use models?

10

Applications of optimization with Xpress-MP

We have stressed the linearity condition, where the objective function and all of the constraints must be
linear in the decision variables. But there are two further properties that we must have, Divisibility andDeterminism.
Divisibility means that in an acceptable solution any values of the decision variables are allowed within
the restrictions imposed by the linear constraints. In particular, we are not constrained to accept only
whole number (integer) values for some or all of the decision variables. We have already alluded to this,when we remarked on the debate as to whether a fraction of a chess set is worth something (or, more
precisely, whether a fraction f of a chess set is worth exactly f times the worth of a whole chess set).
The final requirement we have of an LP problem is that it is deterministic — all the coefficients in theconstraints and the objective function are known exactly. Determinism is sometimes a very strong assumption, particularly if we are building planning models which extend some way into the future. Is it
reasonable, for example, to assume that it will always take 7.5 days for the oil tanker to reach our refineryfrom the Gulf? Or that all of the lathe operators will arrive for work every day next week?
Problems where we must consider the variability in objective function coefficients, right hand sides or
coefficients in the constraints are Stochastic Programming problems. We do not consider them in thisbook, because they are difficult to deal with, not because they are of little practical interest. In fact, it
can be argued that all planning models have stochastic elements, and we will later demonstrate somemethods for dealing with uncertainty in an LP framework.

1.3

Solving the chess set problem

1.3.1

Building the model
The Chess Set problem can be solved easily using Mosel. We will discus Mosel much more in the next
chapter, and then throughout the book, but give the Mosel model here so you can see the solution to
the problem we have modeled.
The first stage is to get the model we have just developed into Mosel. Here it is.
Table 1.2: Chess Set Model
model Chess
uses "mmxprs"
declarations
xs, xl: mpvar
end-declarations
Profit:= 5*xs + 20*xl
Boxwood:= 1*xs + 3*xl <=Lathe:=
3*xs + 2*xl <=! We use the Xpress-MP Optimizer

Notes
1
2

! These are the decision variables

3
4

200
160

! kg of boxwood! Lathehours

maximize(Profit)

5

writeln("LP Solution:")writeln(" Objective: ", getobjval)
writeln("Make ", getsol(xs), " small sets")
writeln("Make ", getsol(xl), " large sets")
end-model

6

7

When we were building our mathematical model we used the notation that items in italics (for example,xs) were the mathematical variables. The corresponding Mosel decision variables have the same name in
non-italic courier font (for example, xs).
Once the model has been defined, the objective function can be optimized and the optimal number ofchess sets to make is obtained.
Notice that the character ‘*’ is used to denote multiplication of a decision variable by the constant associated with it in the constraint. Blanks are not significant and the modeling language distinguishes
between upper and lower case, so X1 would be recognized as different from x1.
By default, Mosel assumes that all variables are constrained to be non-negative unless it is informed
otherwise. There is therefore no need to specify non-negativity constraints on variables. It is possible totell Mosel that variables are not non-negative: we shall see how to do this later.

What is modeling? Why use models?

11

Applications of optimization with Xpress-MP

Here are some notes on the perhaps non-obvious parts of the model. They follow the numbering in themodel listing 1.2. All of the points will be elaborated later, so do not be overly concerned if you do notunderstand some point now.
1 and 7. We give the model a name. The end-model statement terminates the model.
2. We tell Mosel we shall be using the Xpress-MP Optimizer to maximize the problem.
3. We declare the ‘make’ variables to be decision variables, Mosel’s mpvar variables.
4. The objective function and constraints are given here.
5. Here we tell Mosel to maximize the objective function Profit. 6. This part writes out the solution
values. getobjval returns the optimal value of the objective function, while getsol(x) returns the
optimal value of x.

1.3.2

The results
We shall see later how to run the Mosel model. Here is the output.
LP Solution:
Objective: 1333.33
Make 0 small sets
Make 66.6667 large sets

Considering the original model:
maximize

5 · xs + 20 · xl

subject to

1 · xs + 3 · xl ≤ 200 (kg of boxwood)
3 · xs + 2 · xl ≤ 160 (lathe-hours)
xs ≥ 0xl ≥ 0we can check the answer. Mosel tells us that xl is the only non-zero variable, with value 66.6667. The
profit is indeed 20 · 66. 6667 = 1333. 33. We can easily see that the amount of boxwood we use is 3 ·66. 6667 = 200, exactly what we have available; and the lathe-hours we use are 2 · 66. 6667 = 133. 3333,
66.6667 fewer than are available; and we satisfy the non-negativity constraints. Note that we do not use
all the lathe-hours, but we do use all the boxwood.
We may represent the solution to this problem graphically as in Figure 1.1. The grey shaded area is thefeasible region. The values of the objective function Profit are marked with dotted lines.

xl

120

optimal solution
80

Profit=200

40
Profit0=160

Profit

0

=120

40

80

120
Profit
=

160
400

Lathe

200
Profit
=

0

240
800

xs
Boxwood

Figure 1.1: Solution of the Chess Set problem

If we want to analyze the solution to this problem further, we may wish to obtain the constraint activities
and their dual values (shadow prices), and also the reduced costs of the decision variables. We might add
some more writing as follows to the Mosel model:

What is modeling? Why use models?

12

Applications of optimization with Xpress-MP

writeln("Activities/Dual Values")
writeln(" Lathe: ", getact(Lathe)," / ", getdual(Lathe))writeln(" Boxwood: ", getact(Boxwood)," / ", getdual(Boxwood))
writeln("Reduced Costs")
writeln(" xs: ", getrcost(xs))
writeln(" xl: ", getrcost(xl))

The output then becomes
LP Solution:
Objective: 1333.33
Make 0 small sets
Make 66.6667 large sets
Activities/Dual Values
Lathe: 133.333 / 0
Boxwood: 200 / 6.66667
Reduced Costs
xs: -1.66667
xl: 0

The activity of a constraint is the evaluation of its left hand side (that is, the sum of all terms involving
decision variables). The optimal solution only uses 133.333 lathe-hours, but all 200 kg of boxwood thatare available. As the 160 available lathe-hours come from 4 machines that operate 40 hours per week,
we could reduce the weekly working time to 35 hours (resulting in a total of 4 · 35 = 140 hours) withoutchanging anything in the solution (Figure 1.2).

xl

120

optimal solution
80

Profit=200

040Profit=160

Profit

0

=120

40

80

120
Profit160

200
Profit

=400

Lathe

0

240
=800

xs
Boxwood

Figure 1.2: Reduction of lathe-hours

If we modify the limit on the availability of boxwood, this will have an immediate effect on the solution:
the dual value (also called shadow price) of 6.66667 for the boxwood constraint tells us that one additional unit (i.e. 1 kg) of boxwood will lead to an increase of the objective function by 6.66667, giving a
total profit of $1340. Graphically, this may be represented as follows (Figure 1.3).
The reduced cost values of the variables indicate the cost of producing one (additional) unit of every typeof chess set: if we decided to produce one small chess set this would cost us 1.66667 more (in terms ofresource usage) than what its sale contributes to the total profit. Or put otherwise, we need to increase
the sales price of xs by 1.66667 to make small chess sets as profitable as large ones. Figure 1.4 shows theeffect on the solution when we increase the price of xs to 6.66667 (its break-even point): any point on
the highlighted edge of the feasible region now leads to the same objective value of $1333.33, so wemight choose to produce a small number of small chess sets with the same total profit.

1.3.3

Divisibility again
In the present example we are faced with the Divisibility problem mentioned earlier. In reality accepting
fractional answers may be practical since a chess set that is not completed in one week’s production sched-

What is modeling? Why use models?

13

Applications of optimization with Xpress-MP

xl

120

optimal solution
80

Profit

=200

40

0

Profit=160

0Profit

=120

40

80

120
Profit

200

160

240

Profit

=400

Lathe

0

=800

xs
Boxwood

Figure 1.3: Additional unit of wood

xl

120

optimal solutions
80

40

Prof

it=2

Prof

000

it=1

40

80
Lathe

120

Prof

it=4
0

0

160
Prof
it=8
0

0

200
Prof

600
240

it=1

200

xs
Boxwood

Figure 1.4: Increasing the price of xs to its break-even point

What is modeling? Why use models?

14

Applications of optimization with Xpress-MP

ule may be completed the following week. However, some optimization problems have variables thatcannot be fractional in this way. Such problems require the techniques of Mixed Integer Programming,
which we cover later (Chapter 3).

1.3.4

Unboundedness and infeasibility
Although we have found an optimal solution to the problem given, this may not always be possible. If
the constraints were insufficient it might be possible to improve the objective function indefinitely. In
this case, the problem is said to be unbounded. This usually means that a relevant constraint has been
omitted from the problem or that the data are in error, rather than indicating an infinite source of profit.
Conversely, if the constraints are too restrictive it may not be possible to find any values for the decisionvariables that satisfy all constraints. In this case, the problem is said to be infeasible. This would be thecase in the chess set problem if an order of 100 large chess sets had to be made each week.

1.4

Diagnosing infeasibility and unboundedness
One of the most common, and unwelcome, things that happens when one is building LP models is when
the solver says that the solution is unbounded or infeasible, both of which mean that we have done
something wrong in our modeling or that the data we are feeding to the model are wrong in some way.
Let us deal with unboundedness first. Unboundedness in a maximization problem means that we canincrease the profit without limit, thereby ultimately getting infinite profit. Rather than immediately
retiring, buying a yacht and going off to the sunshine, we should pause for a moment and consider
that infinite profits do not arise in practice — we have done something wrong. Unboundedness usuallysprings from having forgotten some constraint or having put in an item which has a cost associated with
it with a profitable coefficient in the objective function. This mistake should usually be very easy to spot
as the LP solver will tell you which decision variable can be increased indefinitely and so theoreticallygives an infinite profit. All you have to do is to look at that particular variable and see what in the realworld is stopping it from increasing indefinitely. You then have to model the forgotten constraint, andtry solving again. In practice it is usually fairly easy to diagnose unboundedness using this technique.
Unfortunately, it is not usually as easy to diagnose infeasibility, the situation where there is no set ofvalues for the decision variables that satisfy all the constraints simultaneously. Unlike unboundedness
it is perfectly possible for infeasibility to arise in practice in a planning situation where one is making
assumptions about resource availabilities, market demands etc.
If one is optimizing an existing system, which by observation we know has a feasible solution, then we
have an easier job to diagnose the infeasibility. Take the values of the decision variables that we areusing right now and put them into the linear programming problem. An easy way to do this is to put thevariables in as fixed variables. Then see which of the constraints are being violated. These constraints areprime candidates for having been wrongly modeled in some way. For example, if we know that today
we can produce certain numbers of 60 watt and 100 watt light bulbs but our model tells us that we haveviolated a manufacturing constraint then we have wrongly modeled that constraint and it is in that areaof the model where we need to concentrate our efforts.
If we do not have a set of values that we know we can realize then we are faced with a much tougher
problem. When seeing an infeasible problem, naive users often say ‘which constraint is wrong?’ orequivalently ‘which constraint is causing the infeasibility?’ There is no answer to this question otherthan to tell the naive user that it is the combination of constraints that is causing the infeasibility, or, inother words, that the combination of the constraints is contradictory. Consider the following contrived
problem:
make1 ≤ 2
make2 ≤ 3
make3 ≤ 4
make1 + make2 + make3 ≥ 10
It is obvious that these constraints are contradictory (add up the first three of them and compare withthe last) but without some knowledge of what the model refers to it is impossible to see which one (oreven all four) of these constraints is ‘wrong’. All we know is that these four constraints taken together
are contradictory.
It turns out to be useful in practice to be able to isolate from a large model sets of constraints thattogether are contradictory but such that if you removed any one of them from the set the remainder

What is modeling? Why use models?

15

Applications of optimization with Xpress-MP

are not contradictory. These are the so called Irreducible Infeasible Sets (IIS) which good LP solvers canprovide for infeasible problems. It is beyond the scope of this chapter to talk about how to apply IISs and
we refer you to the Xpress-MP manuals for a full discussion.
It has to be acknowledged that diagnosing infeasibilities is the hardest job in modeling and even themost experienced analyst sighs when faced with the ‘Problem is Infeasible’ message from an optimization
system. Since infeasibilities can arise from bad modeling or bad data there is no obvious first place tolook. Experienced modelers of course know that the data are the most likely things to be in error but
even experienced modelers do make mistakes from time to time.
There is one small crumb of comfort — infeasibility has nothing to do with the objective function and is
only influenced by the constraints.
Here are a few things to look for when you are facing an infeasible problem:• Have I got the right sense of a constraint? (For instance, I may have put in a ≥ constraint when it
should be a ≤ one; or an equality when an inequality would do.)• Have I got the right flow direction for a flow variable (see Section 2.2)? (It is easy to get the sign
wrong in flow balance constraints.)• Have I got a non zero value for a resource availability? (Often missing data shows up us a zero
value and if this turns into a zero right hand side for a resource availability we frequently get an
infeasibility.)

1.5

The benefits of modeling and optimization
Having just discussed rather technical points it is now time to reflect on modeling and optimization andthe benefits they bring to an organization. But first we will discuss how the somewhat idealized processes
we have considered so far actually occur in practice.
Building a model, solving it and then implementing the ‘answers’ is not generally a linear process. As
we have hinted above we often make mistakes in our modeling which are usually only detected by
the optimization process, where we could get answers that were patently wrong (e.g. unbounded orinfeasible) or that do not accord with our intuition. If this happens we are forced to reflect further about
the model and go into an iterative process of model refinement, re-solution and further analyses of the
optimum solution. During this process it is quite likely that we will add extra constraints, perhaps remove
constraints that we were mislead into adding, correct erroneous data or even be forced to collect new
data that we had previously not considered necessary.

Analysis
Description

Modeling

Deployment
Solving

Figure 1.5: Simplified scheme of a typical optimization project

The problems of data accuracy and reconciliation are frequently the most difficult to tackle. By its very
nature a model that attempts to capture the influence of constraints coming from different parts of theorganization will require the use of data that similarly have come from parts of the organization that
may not normally communicate with each other. These data may have been collected in different ways,
to different standards and indeed assumptions. The severe process of optimization, which by its nature
will seek out extreme solutions if they exist, is very testing of the accuracy of data. Though one will try
to use data that already exist in the organization one is sometimes faced with the problem of collecting
primary data where existing data are just not adequate.
Though we have posed the discussion above as a set of problems, a more reflective view of the process
of modeling and optimization shows that the thoughtful and critical analysis it requires is in fact one ofits major benefits. The integrative nature of a model requires that we have data of a consistent standard
and consistent accuracy, ideals which any organization should strive for anyway.
Of course, the most obvious benefit of modeling and optimization is that we come up with a recommended answer which in some sense that we have defined gives us the best way of running the system

What is modeling? Why use models?

16

Applications of optimization with Xpress-MP

we are modeling. We may choose to reject the solution proffered and indeed there may be very goodreasons for so doing. For instance a plan we are suggesting for tomorrow may differ radically in theset of tasks that are suggested but have almost the same objective function value as that we would getfrom doing exactly the same as we are doing today. In this case it makes perfect sense to forego a small
amount of profit for the sake of continuity. The important thing, however, about a quantitative modelis that we know how much profit we are foregoing and we can make a rational and informed decision,
not just one based on hunch or intuition. We can get an idea of the sensitivity of our solution to theinevitable changes and variations in the data, which leads on to the idea of robust solutions, solutions
which will in some sense remain ‘good’ as the input data change.
The final major benefit of modeling is that it enables us to understand the organization or part of theorganization or system that we are studying in a more quantitative and demonstrably rational way, hopefully devoid of organizational politics. This understanding may lead us to consider ways of working ordifferent processes that we have not considered before, which again might lead to considerable increasein profitability or efficiency. The boundaries of the model may then increase as we consider an enhanced
system or start to consider the system evolving over time.
One thing that is certain about models: as they develop and are worked upon they grow in size. At theadvent of the personal computer, a Linear Programming model with 1000 constraints and perhaps 2000
decision variables was considered big, and in fact would probably not be solved on a PC. Now such amodel would be considered to be small and models of one hundred to a thousand times that size areregularly solved. Not all of this size increase has come from expansions of the boundary of the typicalmodel, most of it has come from the ability to solve models that have been disaggregated by considering
subsets of machinery, or products, time periods, customer zones, labor categories etc.
Models will continue to increase in size but it has to be admitted that the benefits of models do not growin proportion to the model’s size – smaller models that capture the essence of the decision problem maygive greater insights than bloated models where every factor is disaggregated to the most minute detail.

1.6

Data in models
The primary focus of this book is on the construction of clear, accurate and maintainable models, but weshould pause to discuss the data that go into the model.
One of the important characteristics of models is that they often bring together different aspects of theorganization in an attempt to achieve an overall optimal solution. A consequence of this is that a model
will often require data from many different sources, and there is a need for the accuracy of these data
sources to be approximately comparable. There is little point in having some technological data accurate
to 6 digits when some of the cost data have been derived from arbitrary accounting principles. In fact
one of the major obstacles to successful modeling is the belief in many organizations that accounting
data are accurate, whereas frequently much cost accounting data are derived from very questionableassumptions.
Experienced modelers tend to be very distrustful of data in the initial stages of a project. By its very
nature, optimization will find extreme solutions, and it is very testing of the accuracy of data. In practice
it often turns out that the original data have to be rejected and minor studies initiated with the objectiveof getting data of better quality.
There is however a danger of over-stressing the need for accuracy in data when we should equally bethinking about the internal consistency of the data. As a perhaps extreme example, it would not affectour optimal policy (i.e. the values of the optimal decision variables) if all costs in a model were consistently
underestimated by 50%. The value of the objective function of course would be wrong but the actions
we propose would be correct. The problem comes when some sources of data are precise whilst others
are consistently biased in some way.
It is always best, if possible,• to use data in their original format, rather than in some aggregated or disaggregated form, and• to obtain data directly from the owner of the data, that is the person who is responsible for collecting the data and maintaining their accuracy.
Such data will almost invariably be kept in a spreadsheet or a database system and so any high quality
modeling system will have links to the popular spreadsheets and databases. That way the data can beextracted from the primary source, alleviating the danger of using out of date information.
It is hard to make other than sweeping generalizations about the role of data gathering in LP modeling. Sometimes one is very fortunate and the data that the modeler requires are immediately available;

What is modeling? Why use models?

17

Applications of optimization with Xpress-MP

sometimes data of the required accuracy are just not present in the organization and the whole LP project
has to be abandoned as the costs and time involved in gathering the primary data cannot be justified.
Fortunately the latter is a rare event but the authors have found that most modelers consistently underestimate the time involved in acquiring good data.

1.7

References and further material
Linear Programming is often dealt with in the form of a few chapters in general books about Operations
Research, such as the one by Winston [Win94]. Other books are entirely dedicated to this topic; let us citehere the works by Chvàtal [Chv83], Schrijver [Sch86], and Bazaraa and Jarvis [BJ90]. The topic of Linear
Programming is also discussed in general books on optimization such as [Lue84] and [SN96].
The Simplex method that is commonly used for solving linear problems was introduced by Dantzig in
the 1940s; his monograph on Linear Programming [Dan63] is re-edited regularly. A more recent method
for solving linear problems is the so-called Interior Point or Newton-Barrier algorithm (original article in[Kar84], see [ARVK89] for a conceptually simpler form). The book by Arbel [Arb93] comes with a collectionof software; a textbook on Linear Programming based on the Interior Point method is [Roo97]. Interior
Point algorithms are available as part of commercial software packages such as Xpress-MP.
As mentioned at different places, especially in the following Chapter 2, commercial LP solvers usually do
not work with a problem in the form it is given but transform it using so-called presolving techniques tosimplify the problem and to reduce its size (see for instance [AA95]). At the end of the solution algorithm
the transformation is reversed (postsolving) to provide the user with the solution to his original problem.

What is modeling? Why use models?

18

Applications of optimization with Xpress-MP

Chapter 2

Typical LP model constructs
We move now to a more disciplined approach to modeling. Though we have noted that modeling is anart, this does not mean that we should treat each modeling exercise in an ad hoc way. Rather, we needto learn from our previous modeling experience so as to increase our productivity and skill in the next
modeling work we do.
A natural way to approach any new modeling exercise is to try to understand the constraints in terms of
constraint types that we have seen before and so know how to model. If we can do a disaggregation of
the overall set of constraints into individual constraints that we know how to model, then we can feelquite confident that we will be able to complete the whole model.
So we shall look at some typical constraint types. We will first give various ‘word descriptions’ of constraints, that is, phrases in which the constraint might be articulated by the person responsible for thesystem we are modeling. We will then give an algebraic formulation of the constraint, and see how
it might be generalized. The presentation below is given in terms of the following constraint types:bounds (Section 2.1), flow constraints (Section 2.2), simple resource constraints (Section 2.3), materialbalance constraints (Section 2.4), quality requirements (Section 2.5), accounting constraints (Section 2.6),
blending constraints (Section 2.7), modes (Section 2.8), soft constraints (Section 2.9), and as a special formof non-constrained ‘constraints’, objective functions (Section 2.10),
It should be noted that these constraint types are somewhat arbitrary — another presentation could havecategorized them in different ways.

2.1

Simple upper and lower bounds
A) ‘Marketing tells me that we cannot sell more than 100 units of Product 4 this period.’B) ‘I’m committed to sending at least 20 tonnes of Blend 6 to Boston tomorrow.’C) ‘I have to send exactly 30 tonnes of Blend 5 to Paris next period.’D) ‘Electricity can flow in either direction along the Channel Cable.’
In A, one of the decision variables will be the amount of Product 4 that we sell, and suppose this isvariable sell4 . The words say that there is an upper limit or upper bound of 100 on the single variable
sell4 , i.e.
sell4 ≤ 100
In case B, we have a restriction in the opposite direction. If the decision variable is, say, sendb6,Boston , thenwe have a lower limit or lower bound on the single variable,
sendb6,Boston ≥ 20
Note that this is more restrictive than the normal non-negativity constraint
sendb6,Boston ≥ 0
We can see case C, where the decision variable might be called sendb5,Paris , as an example of a variablebeing fixed, or equivalently that its lower and upper bounds are identical, i.e.
sendb5,Paris = 30
orsendb5,Paris ≥ 30 and sendb5,Paris ≤ 30

19

Applications of optimization with Xpress-MP

One might ask ‘why have sendb5,Paris as a decision variable at all, as it is fixed and there is no decision to
make about its value?’. If every day we always sent exactly 30 tonnes to Paris then it might be legitimate
to remove this from the decision process, remembering to account for the resources that are used in theblend, the transport capacity that is used up in moving the blend, etc. But of course there will inevitably
come a day when Paris needs more of Blend 5, and we have to spend time recalculating the resource
availabilities, transport capacity etc., net of the new Parisian demand. We will get things wrong unlesssomeone remembers to do the recalculation; and of course forgetting to redo the sums is likely when
we are having to rejig our production plan because of the extra demands from other cities in France (itis Bastille Day tomorrow). What we gain by removing one variable from the model is not worth the risk
arising from later inflexibility.
The last case, D, expresses the fact that sometimes a variable is not bound by the normal non-negativity
constraint, and in this situation it is said to be free. Since, as we have mentioned before, all decisionvariables are by default assumed to be non-negative, we have to explicitly tell our modeling softwarethat the decision variable here (which we shall call eflow), denoting the flow from Britain to France
where a positive value means that electricity is flowing from Britain to France and a negative valuemeans that it is flowing in the reverse direction is not non-negative−∞ ≤ eflow ≤ +∞In Mosel, this is indicated by the notationeflow is_free

Free variables can be replaced by normal non-negative variables, e.g.
eflow = eflowBF − eflowFB
where eflowBF (eflowFB ) is the non-negative flow from Britain to France (respectively, France to Britain),but since optimization software can handle free variables directly there may be no reason to do thistransformation. If, however, there is some asymmetry in the flows (perhaps a flow from Britain to France
is taxed in some way by the French government, whereas a flow in the other direction is not taxed), theeflowBF and eflowFB variables may be required anyway by other parts of the model, so the substitution isharmless, or even necessary.

2.2

Flow constraints
Flow constraints arise where one has some sort of divisible item like electricity, water, chemical fluids,
traffic etc. which can be divided into several different streams, or alternatively streams can come together.
Some word formulations might beA) ‘I have got a tank with 1000 liters in it and 3 customers C1, C2 and C3 to supply.’B) ‘I buy in disk drives from 3 suppliers, S1, S2 and S3. Next month I want to have at least 5000 disk drives
arriving in total.’C) ‘I have 2 water supplies into my factory, S1 and S2. I get charged by the water supplier for the amountof water that enters my site. I lose 1% of the water coming in from S1 by leakage and 2% of that from
S2. My factory needs 100,000 gallons of water a day.’Case A is where there is an outflow from a single source, whilst case B is where we want to model a total
inflow. Case C is a little more complicated, as we have to model losses.
In case A we have flow variables supply1 , supply2 and supply3 , being the amounts we supply to the threecustomers. The total amount supplied is the sum of these three decision variables i.e. supply1 + supply2 +supply3 , and this total must be less than or equal to the 1000 liters we have available. Thus we havesupply1 + supply2 + supply3 ≤ 1000
(providing we have defined the units of supply to be liters).
Case B is very similar. The obvious decision variables are buy1 , buy2 and buy3 , and the total amount webuy is buy1 + buy2 + buy3 . Our requirement is for at least 5000 drives in total, so the constraint isbuy1 + buy2 + buy3 ≥ 5000
If we wanted exactly 5000 drives, then the constraint would bebuy1 + buy2 + buy3 = 5000

Typical LP model constructs

20

Applications of optimization with Xpress-MP

Case C is more challenging. ‘I have 2 water supplies into my factory, S1 and S2. I get charged by thewater supplier for the amount of water that enters my site. I lose 1% of the water coming in from S1 byleakage and 2% of that from S2. My factory needs 100,000 gallons of water a day.’
There are several different ways we might approach the modeling. One way is to define decision variables
as follows: buy1 and buy2 are the amounts we buy from supplier 1 and 2 respectively, and get1 and get2
are the amounts that we actually get (after the losses). If we measure the decision variables in thousandsof gallons, then we have the requirement constraint
get1 + get2 = 100
and we can model the losses as follows (loss equations):
get1 = 0. 99 · buy1
get2 = 0. 98 · buy2
since, for example, we only receive 98% of the water from supplier 2 that we have to pay for. Then theobjective function would, among other entries, have terms PRICE1 · buy1 + PRICE2 · buy2 , where PRICE1 andPRICE2 are the prices per thousand gallons of water from the two suppliers.
We have the option of using the loss equations to substitute for buy1 and buy2 (or, alternatively, get1 andget2 ), and doing so would reduce the number of equations and variables by 2. But as we have discussed
before, it almost certainly is not worth the effort, and the LP solver should be clever enough to do thework for us automatically anyway if it thinks it will reduce the solution time.

2.3

Simple resource constraints
We have already seen examples of this sort of constraint in the chess set problem where, to remind you,
we had limited amounts of lathe-hours and boxwood, and we had to create a production plan that usedneither too many lathe-hours nor too much boxwood.
If we generalize what we mean by a resource, all linear programs can be formulated as maximizing some
objective function subject to not using any more of any resource than we have available. But it is notparticularly helpful to think of everything as a resource, so we will give more word formulations here
which involve real resources.
A) ‘I can only get hold of 10,000 connectors a month. Each Industry PC I make needs 8 connectors andeach Home PC I make requires 5 connectors.’
B) ‘Each tonne of chemical C1 uses 50 grams of a rare catalyst and 1 kg of a certain fine chemical, whereas
each tonne of chemical C2 requires 130 grams of the catalyst and 1.5 kg of the fine chemical. I can only
afford to buy 10 kilograms of catalyst per month and I can only acquire 200 kg of the fine chemical eachmonth.’C) ‘If I do one unit of activity i it takes up COSTi of my disposable income. My disposable income next
week is $150.’
If we consider just one scarce resource then a resource constraint can be characterized by the activities
that potentially use up some of that resource, the level at which we are planning to conduct those
activities, the availability of the resource and, finally, the amount of the resource that is used per unit ofeach activity.
These latter numbers, the resources used per unit of activity, are called technological coefficients. because
they are determined by the technology that we are using. Typically, when we have several products which
are each made from several resources, we will have a technological coefficient matrix which gives thenumber of units of each resource required to make one unit of the product. It would be presented in
the form of Table 2.1, where, for example, the entry 0.8528 means that to make 1 unit of Product 3 werequire 0.8528 units of Resource 3. We have to be careful to use consistent and sensible units for how wemeasure products and resources.
Case A where we stated that ‘I can only get hold of 10,000 connectors a month. Each Industry PC I makeneeds 8 connectors and each Home PC I make requires 5 connectors’ has a simple technological matrix:
If we have decision variables makeI and makeH , being the number of Industrial and Home computers
we plan to make, then the number of connectors we plan to use is the number used making industrial
computers plus the number used making Home computers. If we make 1 Industrial computer then weuse 8 connectors, so if we make makeI Industrial computers we use 8 · makeI connectors; and if we make1 Home computer then we use 5 connectors, so if we make makeH Home computers we use 5 · makeH

Typical LP model constructs

21

Applications of optimization with Xpress-MP

Table 2.1: Technological coefficient matrix
Resource 1
Product 1
Product 2
Product 3...
Product P

Resource 2

1.28235
0.95474
12.33312

Resource 3...

Resource R
0.24
1.83616.128

0.8528etc.

11.2322

5.208

Table 2.2: Simple technological coefficient matrix
Connector
Industry PC
Home PC8
5connectors. So in total if we make makeI Industrial computers and makeH Home computers we will use
8 · makeI + 5 · makeH connectors. This total must not be greater than the number available i.e.
8 · makeI + 5 · makeH ≤ 10000
In case B we said ‘Each tonne of chemical C1 uses 50 grams of a rare catalyst and 1 kg of a certain finechemical, whereas each tonne of chemical C2 requires 130 grams of the catalyst and 1.5 kg of the finechemical. I can only afford to buy 10 kilograms of catalyst per month and I can only acquire 200 kg of thefine chemical each month.’ So we have two products and two resources.
Table 2.3: Resource use
Catalyst

Fine Chemical

50
130

1.0
1.5

C1
C2

We measure the catalyst in grams, and the fine chemical in kg. Defining decision variable make1 andmake2 to be the number of tonnes of the two chemicals (C1 and C2) we plan to make, and using anargument similar to the one in case A), we get two constraints, one for the catalyst and one for the finechemical.
50 · make1 + 130 · make2 ≤ 1000 (catalyst)
1. 0 · make1 + 1. 5 · make2 ≤ 200 (fine chemical)
We have been careful to keep the units of measurement consistent.
In general, we have one constraint for each resource, relating the amount we are planning to use to theamount available.
Case C is now hardly challenging. ‘If I do one unit of activity i it takes up COSTi of my disposable income.
My disposable income next week is $150.’ The decision variables are doacti , the amount of activity i that
I plan to do. Making the linearity assumptionthat if I do doacti units of activity i it costs me COSTi ·P
doacti , my total planned expenditure is i COSTi · doacti and this can be no more than the money I have(assuming credit is not an option for me!). Then the constraint is

X

COSTi · doacti ≤ 150

i

2.4

Material balance constraints
Many systems have constraints which can be expressed in words as ‘what goes out must in total equal
what comes in’. In other words, there can be no loss of mass when some particular process takes place. If
the process is some sort of material flow and the thing that is flowing is incompressible, then sometimes
the constraint is expressed in terms of conservation of volume.
Pictorially if we have a process with inflow variables flowini where i ranges from 1 to NIN and outflows
flowoutj where j ranges from 1 to NOUT , and an inflow L from the outside world, we might have

Typical LP model constructs

22

Applications of optimization with Xpress-MP

Lflowin3
flowout1
flowin2

flowout2
flowout3

flowin1

Figure 2.1: Material flow balance in a node

If the flow variables are measured in terms of mass or weight then we will have
L+

MIN
X

NOUT

flowini =i=1

X

flowoutj

j=1

A particular form of material balance constraint occurs when we are accounting for the flow of materials
between time periods in a multi-time period model, that is, a model where we represent time as aseries of intervals, not concerning ourselves with the details of when during an interval events happen.
Suppose that we have a simple factory that only makes one product and that we are trying to decide the
manufacturing levels (maket ) for the next NT periods, i.e. t ranges from 1 to NT. Another set of decisionvariables are sellt , the amount of the product that we decide to sell in time period t. If we have thepossibility of storing product from one time period to the next time period we can introduce a further
set of decision variables storet which are the amounts of product to have in stock (inventory) at the endof time period t.
Let us consider the material balance in time period t. In words we can say ‘the stock at the end of time
period t is equal to the stock at the beginning of time period t, plus what we make, minus what we sell.’We are faced with a slight problem now — we do not know the decision variable for the stock at the
beginning of time period t but we can see that, assuming there is no loss of stock, the stock at the
beginning of time period t is the same at the end of time period t -1. So in different words, the previous
statement can be phrased as ‘the stock at the end of time period t is equal to the stock at the end of time
period t − 1, plus what we make in period t, minus what we sell in period t’. In algebraic form this canbe written
stockt = stockt−1 + maket − sellt
This is deceptively simple but it contains one flaw: there is no time period preceding time period 1, so
there is no stock0 variable. We need a special constraint that relates to time period 1 only, and says inwords ‘the stock at the end of time period 1 is equal to the opening inventory SINIT, plus what we makein time period 1, minus what we sell in time period 1’, or in algebra
stock1 = SINIT + make1 − sell1
Other material balance constraints arise when we are talking about raw materials that we buy in froman outside supplier and use in our production process. A word description might be, ‘for a particular raw
material, the stock of the raw material at the end of time period t is equal to the stock at the end oftime period t − 1, plus the amount of the raw material that we buy in time period t, minus the amountof raw material that we use in making the products during time period t’. Then considering one raw
material we might have decision variables rbuyt , rstockt and ruset , so the constraints for all but the firsttime period are
rstockt = rstockt−1 + rbuyt − ruset
We have to remember again that there is a special constraint for time period 1 as we already know the
opening stock level in that period: it is what we have in stock right now.
It is also likely that the ruset variables will be related to the decision variables in another part of the model
which represent how much product we are going to make. If we have products indexed by p = 1, . . . , NP
with technological coefficients REQpr , the number of units of raw material r required by 1 unit of productp, and decision variables makept denoting the amount of product p that we make in time period t, we
can immediately write down an expression for the amount of raw material we use in time period t (ruset )Typical LP model constructs

23

Applications of optimization with Xpress-MP

asruset =NP
X

REQpr · makept

p=1(see Section 2.3 ‘Simple resource constraints’ above for an explanation of this).
Since we have just written down an equation for ruset we might consider substituting for it in the constraint in which it occurs. This would generally not be a particularly good thing to do, but not particularly
bad either. It would reduce the number of variables a little but we are probably going to have to use
these variables somewhere else in the model anyway and so the substitution would have to take place everywhere the ruset variables appeared in the model. The model would certainly be less comprehensible,and consequently harder to maintain.
A new form of material balance equations of the multi-period type occurs where we have fixed demands
for our product or products in the NT time periods. In other words, the selling decision variables (sellt in
the example above) are fixed. We have the choice of expressing this as a set of simple equality constraints
with sellt variables e.g.
sellt = MUSTSELLt
or by substituting the sellt variables by MUSTSELLt everywhere they occur. This time the decision as towhether to substitute for the sellt variables is probably a matter of choice — some modelers will eliminate
variables while others might put in the simple equality constraint arguing that the sellt variables willprobably at some stage in the future be of interest in themselves.

2.5

Quality requirements
Here are some word statementsA) ‘I cannot have more than 0.02% of sulphur in this gasoline.’B) ‘There must be at least 3% of protein in this dried apricot mixture.’C) ‘This foodstuff must have no less than 5% fat but no more than 10%.’
These sorts of quality requirements frequently occur when we are blending together various raw materials with differing properties to create a final product. Typically each of the raw materials will comewith a set of properties, for instance in foodstuffs it might be the fat content or the percentage of carbohydrates; in the petrochemical industry it might be sulphur content or the octane number. For somecharacteristics of the final product we require that quality specifications are adhered to. For ‘bad’ things
these are usually expressed as a maximum percentages; for desirable properties they will be expressed asa minimum percentage and other properties may have to lie between specified lower and upper bounds(perhaps so that the product has stability or other characteristics that we desire). An extreme form of
the latter case is where we have to have an exact percentage of a particular characteristic in the final
product.
Consider a number NR of raw materials and just one characteristic that we are concerned about in the
final product. Suppose that one unit of raw material r contains CONTr units of the characteristic (forinstance CONT3 might be 0.1 if raw material number 3 is 10% fat and the characteristic we are considering
is fat). As usual with LP we assume that we know the CONT data precisely.
In many manufacturing processes it is reasonable to assume that the characteristic in the resulting end
product comes from blending the raw materials linearly. For example if we blend together one kg of agasoline with 0.01% sulphur and 1 kg of gasoline with 0.03% sulphur, then we will get 2 kg of the mixture
with 0.02% sulphur. Whether this assumption of linearity is correct very much depends upon the physics
or chemistry of the blending process and in some industries it certainly does not hold. However, making
the assumption of linearity, we can see that the proportion of the characteristic in the final product isgiven bytotal mass of constituentsp = proportion =total mass of blend
If the decision variables are ruser for the amount of raw material r to use, then we have

PNRp=

Cr · ruser

r=1
P
NR

r=1

ruser

and we can apply the desired inequality (or equality) to this proportion. For instance, p ≤ 0. 1 (i.e. 10%)
becomesPNR
C · ruser
r=1 r≤ 0. 1
P
NR
ruser
r=1

Typical LP model constructs

24

Applications of optimization with Xpress-MP

At first sight this does not seem like a linear inequality and indeed it is not as it is a constraint applying
to the ratio of two linear expressions. However, by cross multiplying we can get the constraint into the
form
NR
X

Cr · ruser ≤ 0. 1

r=1

NR
X

ruser

r=1

Note: we have to be careful when multiplying inequalities. The direction of the inequality is preserved
if we multiply by a non-negative amount, but is reversed if we multiply by a negative amount. Here, we
are multiplying both sides of the inequality by the sum of a set of non-negative variables, so the directionof the inequality is maintained.
If we collect together two terms for each decision variable ruser we get the linear inequality
NRX(Cr − 0. 1) · ruser ≤ 0

r=1

2.6

Accounting constraints, non-constraining ‘constraints’
A typical way of building up a model is to construct it in quasi-independent pieces. So, for instance, wemight have a constraint expressed as ‘the total cost of operating region 1 plus the cost of operating region
2,... plus the cost of operating region 16 must be less than $10M’, and then later on in the model wehave 16 different equations which capture the cost of operating each of the regions. If we had decisionvariables totalcost and rcosti this would be expressed as
totalcost =16
X

rcostii=1where rcosti = some_expressioni for i = 1 to 16.
Obviously it is possible to eliminate the rcosti variables (the cost of operating region i) by substituting
some_expressioni into the equation defining totalcost but it is very likely that we will want the costs ofoperating different regions to appear in some report, so we might as well let the LP system calculate thevalues of these variables for us. We call the rcosti variables accounting variables and the constraints thatdefine them accounting constraints.
Whether you choose to use the LP system to do the accounting for you, or do them afterwards when
producing reports, is a matter of taste. In the past, when LP systems were limited and one had to squeeze
the problem into as few constraints as possible, it was obviously a good idea to eliminate accounting
variables and then to calculate their values from the optimum values of the variables contributing tothem. Now, where a few hundred extra constraints are of no consequence it is better to leave accounting
variables and constraints in the model.
There is however one subtlety that we must be aware of: it is possible that an accounting variable does
not necessarily have to be non-negative. Suppose we had variables revenuei and expenditurei which were
the revenue and expenditure of plant i. If we were to define accounting variables profiti and accounting
constraints
profiti = revenuei − expendi
to pick up the values of the profiti variables then we must remember that it is possible for a plant tohave negative profitability and so the profiti variables must be declared as free variables (unless for somereason we have to run each plant in a profitable mode).
Forgetting that decision variables are by default non-negative is a common cause of error with accounting variables. However, if one just wants the LP system to do accounting for you, it is possible to add
extra unconstrained linear expressions to a model, letting the LP system work out the value of the linear
expression for you. It used to be the case that adding lots of these expressions could slow down the
LP solution process, but modern solvers recognize such constraints and discard them during the solution
process, only reinstating them when the rest of the LP has been solved.

2.7

Blending constraints
A very common use of LP modeling is to deal with the situation where we have a set of inputs each withcertain percentages of different characteristics and we want to blend these inputs together to get a final

Typical LP model constructs

25

Applications of optimization with Xpress-MP

product with certain desired characteristics. Blending typically falls into one of two major categories:
either we use fixed recipes where the proportion of different inputs is determined in advance (rather likethe recipes you see in cookery books which specify the exact weights of the constitutes required to makea certain weight of cake) or we have variable recipes where it is up to us to decide, perhaps within limits,
the proportions of the inputs that we are going to blend together.
An example of a variable recipe might be in the manufacture of animal feed stuffs which can be madefrom a wide variety of different raw materials, not necessarily always mixed in the same proportions. Aninteresting example of recipes which pretend to be fixed but in fact are variable are those found in books
that tell you how to mix cocktails. They usually pretend that the proportions of gin, tequila, vermouth
etc are fixed, but of course anyone who has been in the position of having restricted stocks of any ofthese raw materials knows that within limits the recipes are very variable.
Suppose we have 3 raw materials and the constraint in words is ‘the ratios of raw material 1, raw material
2 and raw material 3 are 6:3:1’. If we have decision variables raw1 , raw2 and raw3 representing the weightof the materials in the blend, then the total weight in the blend is raw1 + raw2 + raw3 and we have theconstraints
raw1
6=
raw1 + raw2 + raw3 10
raw2
3=
raw1 + raw2 + raw3 10
raw3
1=
raw1 + raw2 + raw3 10
These are ratio constraints, not linear equations, but as before we can convert them into linear equations
by cross multiplying to get(1 − 0. 6) · raw1 = 0. 6 · raw2 + 0. 6 · raw3 , i. e.
0. 4 · raw1 = 0. 6 · raw2 + 0. 6 · raw3
The cross multiplication preserves the direction of the inequality since the denominator is always nonnegative.
Similarly we get
0. 7 · raw2 = 0. 3 · raw1 + 0. 3 · raw3 , and
0. 9 · raw3 = 0. 1 · raw1 + 0. 1 · raw2
One of these equations is redundant as it is implied by the other two but it does no harm to put all threeequations into the model. In fact it is probably a good idea to put all three equations down as inevitably
at some time in the future we will have a fourth raw material and if we try to be too clever in eliminating
redundant constraints we will forget that we have omitted a previously redundant equation, and makea mistake that will be hard to detect.

2.8

Modes
A generalization of blending constraints is where we have M inputs and N outputs which must be in strict
proportions. As an example, this might be expressed as ‘My 3 inputs have to be used in the ratio of 2 :1.6 : 2.2 and they produce outputs of 1.6 of output 1 and 1.7 of output 2’. So, for example we might put
in 1 kg of input 1, 0.8 kg of input 2, and 1.1 kg of input 3, and get out 0.8 kg of output 1 and 0.85 kg of
output 2.
It is easy to see that if we have just one output then this is a simple fixed ratio blending example.
Such M-input/N-output constraints often arise where we have a plant that we can operate in different
ways (modes), and the ratios differ for different modes. At any point in time, the plant can only be in
one mode.
Consider a very simple example, where we have 3 inputs, 2 outputs and 3 possible operating modes. We
present the operating characteristics in the tables below, where we have defined a unit of a mode as 1hour i.e. we have shown the kg of each input used, and output produced, by the plant.
The decision variables are the number of hours the plant spends in each mode m, say usemodem . Thenthe usage of Input 1, for instance, is 20 · usemode1 + 22 · usemode2 + 24 · usemode3 and the production of
output 2 is 17 · usemode1 + 20 · usemode2 + 24 · usemode3 .

Typical LP model constructs

26

Applications of optimization with Xpress-MP

Table 2.4: Fixed ratio blending example
Quantities used/produced
Mode 1 Mode 2 Mode 3
Input 1
Input 2
Input 3

20
16
22

22
14
21

24
13
18

Output 1
Output 2

16
17

15
20

14
24

In any given planning period we will also have a constraint on the total number of hours the plant spends
in all the modes. For instance, if we are doing a weekly plan (168 hours), then we will have the constraint
that
usemode1 + usemode2 + usemode3 ≤ 168
We are assuming that we can work at most 168 hours in the week, and that we do not lose any timechanging from one mode to another. The latter assumption may be totally unrealistic for some plants,
and if we cannot assume that change-over times are negligible then we have to use an Integer Programming formulation.

2.9

Soft constraints and ‘panic variables’
The constraint that we have just modeled, that we can run a plant for at most 168 hours, is an example
of a hard constraint. It is impossible to get more than 168 hours into a week. Other examples of hardconstraints are ones that relate to physical, chemical or engineering properties, such as a boiler’s capacity,
or a reaction rate. Other hard constraints come from accounting or definitional constraints (for instance,
profit = revenue-cost is a hard constraint, as it is just a definition really).
But there are other, more ‘economic’ constraints where in reality if we have to violate a constraint we
can do so at a cost. If there is a scarce resource, the economic environment invariably sets up a market inthat resource, and we can go to the market if we really do need to acquire more of that resource. (Thisis somewhat of a generalization as a particular resource may actually be so specific to us that we are theonly people who can make it to the standards or in the locality or to the timescale we require.)
Consider again the statement of case B of Section 2.3 ‘Simple resource constraints’ above: We said ‘Eachtonne of chemical C1 uses 50 grams of a rare catalyst and 1 kg of a certain fine chemical, whereas eachtonne of chemical C2 requires 130 grams of the catalyst and 1.5 kg of the fine chemical. I can only afford
to buy 10 kilograms of catalyst per month and I can only acquire 200 kg of the fine chemical each month.’
The key part of this phrasing is the last sentence, where we say we can only afford so much of the catalyst,
and we can only acquire 200 kg of the fine chemical. We expressed the fine chemical constraint as
1. 0 · make1 + 1. 5 · make2 ≤ 200 (fine chemical)
i.e. as a hard constraint. But suppose that if we were willing to pay P per kg we could get more of thefine chemical. Then we could introduce an extra decision variable buyFC to represent the number of
kg of fine chemical we bought at this price. If we were building a profit maximizing LP then we would
have to add a term −P · buyFC to the objective function, accounting for the money we pay for the finechemical, but now the constraint becomes
1. 0 · make1 + 1. 5 · make2 ≤ 200 + buyFC (fine chemical, with buy-in)
We now have a soft constraint, i.e. we can violate the original constraint but at a penalty to our objectivefunction. The buyFC variable can be considered a panic variable: it is a variable which is there in case wecannot manage with our normal resources and have to resort to the market to satisfy our needs.
The way we have introduced panic variables is not the way they are most commonly used in modeling.
We have mentioned before that getting an infeasible solution to an LP is very difficult to diagnose, andgood modelers have learned the art of what we might call ‘defensive modeling’. One of the worst things
that can happen to a model when it has been deployed to end-users is that it gives an infeasible solution
message to the poor end-user, who then does not know what to do. Defensive modelers try to avoid
this by adding panic variables to constraints that might possibly be violated, at the same time trying to
get estimates of how much it will realistically cost to violate the constraint. Often the end-user can help
in this, as they will know what can be done to get extra resources — for instance, if it is a question of
shutting down the plant or renting a plane to fly in the scarce resource, you may well decide to rent theplane.

Typical LP model constructs

27

Applications of optimization with Xpress-MP

Panic variables do not just apply to resource availability constraints; they can also be added to demand
constraints. For instance, we might have a constraint that a particular customer C has to have a total
supply from our depots of 100 tonnes, modeled asX

flowdC ≥ 100d∈DEPOTS

If, however, it is just not possible to supply him then we may undersupply. There may be a contractual
cost to doing this, in which case we know what the direct cost will be, or we might have some measureof the cost in terms of loss of goodwill (admittedly these costs are hard to estimate, and can be very
subjective).
If a panic variable is non-zero in an optimal solution, then we can infer one of two things: either we have
put in too low a penalty cost, or we are truly being forced to use the panic variable to achieve feasibility.
Rerunning the model with higher penalty costs will enable us to distinguish these cases. But in either case
we have an implementable solution, and we are not left with an end-user staring at a computer screen
wondering what on earth to do.

2.10

Objective functions
An objective function is not a constraint, but since it consists of a linear expression usually involving lotsof variables its construction has many similarities to the process of building constraints. In fact, now wehave so extensively covered modeling constraints there is little new to say about modeling an objectivefunction.
The objective is usually cost (when we minimize) or profit (when we maximize). Most LP systems minimize
by default and the industry MPS standard for presenting an LP problem to an optimizer does not specify
whether the objective is to be minimized or maximized, so everyone has at one time or another selected
the wrong sense for optimization, and come up with a silly answer. One feels foolish, but it is usually
obvious that a mistake has been made.
Problems do exist where there is no objective function, and the LP optimizer is being used just to obtaina feasible solution, or to answer the question in a design study as to whether the constraints are too
restrictive. But such problems are rare.

2.10.1Minimax objective functions
Two apparently non-linear objective functions can be converted into LP forms by tricks. The first is where
we wish to:minimize max(t1 , t2 , t3 , . . . , tN )i.e. minimize the maximum of a set of decision variables. We can model this by introducing a newdecision variable, say s, and thenminimizessubject tos ≥ t1s ≥ t2s ≥ t3...
s ≥ tNYou can see that s has to be no less than each of the t’s, and the minimization objective will force it down
to take the value of the largest t. Unfortunately, the same trick cannot be applied where we want tominimize min(t1 , t2 , t3 , . . . , tN ) or maximize max(t1 , t2 , t3 , . . . , tN ).

2.10.2

Ratio objective functions
The second trick enables us to deal with a ratio objective function of the form
minimizeP
N · xjj j
Obj = Pjsubject to LP constraints

X

Dj · xj

Aij · xj ∼ Rij

Typical LP model constructs

28

Applications of optimization with Xpress-MP

where ∼ is one of ≤, ≥ or =.
Define new variables d = P 1
j

Dj ·xjand yj = xj · d (i.e. xj = yj / d). Then the objective becomesP= d · j Nj · xjP
= j Nj · d · xjP
= j Nj · yj

Obj

The definition of d can be linearized by cross multiplying:1=X

Dj · d · xj =X

Dj · y jjj

and into the normal LP constraints we can substitute for xj thus:X

Aij · xj ∼ Rijor (multiplied by d)d·X

Aij · xj ∼ d · Ri

j

or

X

Aij · d · xj ∼ d · Ri

ji.e. orX

Aij · yj ∼ d · RijThe resulting LP has variables d and yj . When it has been solved, the optimal values of the original xj
variables can be recovered from the definition xj = yj / d.
Note that we are in trouble if variable d can become 0. We also have to taken care if d is always negative,
as the signs of the inequalities must be reversed when we multiply by d above.

Typical LP model constructs

29

Applications of optimization with Xpress-MP

Chapter 3

Integer Programming models
Though many systems can accurately be modeled as Linear Programs, there are situations where discontinuities are at the very core of the decision making problem. There are three major areas where non-linear
facilities are required• where entities must inherently be selected from a discrete set of possibilities;• in modeling logical conditions; and• in finding the global optimum over functions.
Modeling languages let you model these non-linearities and discontinuities using a range of discrete
entities and then a Mixed Integer Programming (MIP) optimizer can be used to find the overall (global)
optimum of the problem. Usually the underlying structure is that of a Linear Program, but optimization
may be used successfully when the non-linearities are separable into functions of just a few variables.
Note that in this book we use the terms Integer Programming (IP) and Mixed Integer Programming interchangeably. Strictly speaking MIP models have both discrete entities and the continuous variables, andpure IP models only have discrete entities. But it seems pedantic to distinguish these two cases unless thedistinction is important to our modeling or optimization; and it is not.
The first Section 3.1 of this chapter introduces the different types of IP modeling objects (also referred toas ‘global entities’). The following Section 3.2 describes the branch and bound method for solving MIP
problems. Binary variables merit special attention: they may be used to formulate do/don’t do decisions,
implications and other logical expressions, and even products between several binary variables (Section
3.3). As shown in Section 3.4, it would indeed be possible to reduce the whole set of IP modeling objects
to binary variables — but this at the expense of efficiency. In combination with real variables, binary
variables can be used, among others, to express discontinuities (Section 3.5).

3.1

IP modeling objects: ‘global entities’
In principle, all you need in building MIP models are continuous variables and binary variables, wherebinary variables are decision variables that can take either the value 0 or the value 1. They are often
called ‘0/1 variables’ or ‘do/don’t do’ variables. But it is convenient to extend the set of modeling entities
to embrace objects that frequently occur in practice.
• Integer variables: an integer variable is a variable that must take only an integer (whole number)value.
• Partial integer variables: a partial integer variable is a variable that must take an integer value if it
is less than a certain user-specified limit L, or any value above that limit.
• Semi-continuous (SC) variables: an SC variable is a decision variable that can take either the value
0, or a value between some user specified lower limit L and upper limit U. SCs help model situations
where if a variable is to be used at all, it has to be used at some minimum level.
• Semi-continuous integer variables (SI): decision variables that can take either the value 0, or an
integer value between some lower limit and upper limit. SIs help model situations where if avariable is to be used at all, it has to be used at some minimum level, and has to be integer.
• Special Ordered Sets of type 1 (SOS1 or S1): an ordered set of non-negative variables at most oneof which can take a non-zero value.

30

Applications of optimization with Xpress-MP

• Special Ordered Sets of type 2 (SOS2 or S2): an ordered set of non-negative variables, of which at
most two can be non-zero, and if two are non-zero these must be consecutive in their ordering.
Why might some of these variables and sets, which collectively we call global entities, be used? We shallgive a brief overview, and later we shall expand on these examples.
Integer variables occur naturally, being most frequently found where the underlying decision variable
really has to take on a whole number value for the optimal solution to make sense. If we are modelingthe number of trains that leave a railway station in a half-hour interval, then the variable representing
this number must take on an integer value — it makes no sense to say that 11.23 trains are to leave.
The idea of partial integers arose from a study where car engines in short supply were being allocatedto different regions. Some of the regions were large, and took many engines, so it was permissible toround a solution saying the region should take 1654.19 engines to the integer value 1654. But for aregions taking only a few engines, it was not acceptable to round 6.32 engines to 6 engines. It is up to
the decision maker to decide the limit L below which it is not acceptable to round a fractional answer toa nearby integer. The modeling system allows the decision maker to specify a different L for each partial
integer variable, if so desired. So partial integers provide some computational advantages in problems
where it is acceptable to round the LP solution to an integer if the optimal value of a decision variable isquite large, but unacceptable if it is small.
Semi-continuous variables are useful where, if some variable is to be used at all, its value must be no lessthan some minimum amount. For example, in a blending problem we might be constrained by practicalreasons either to use none of a component or at least 20 kg of it. This type of constraint occurs veryfrequently in practical problems.
Semi-continuous integer variables often arise in shift planning modeling. If we are going to set up amachine, then we have to run it for a whole number of shifts, and in any case for at least some minimum
number of shifts. For example, it might be possible only to run the machine for 0, or 12, 13, 14, ... shifts,
with the case of running it for 1, 2, ..., 11 shifts making no economic sense.
Special Ordered Sets of type 1 are often used in modeling choice problems, where we have to select at
most one thing from a set of items. The choice may be from such sets as: the time period in which to starta job; one of a finite set of possible sizes for building a factory; which machine type to process a part on,
etc.
Special Ordered Sets of type 2 are typically used to model non-linear functions of a variable. They are
the natural extension of the concepts of Separable Programming, but when embedded in a Branch andBound code (see below Section 3.2) enable truly global optima to be found, and not just local optima.
(A local optimum is a point where all the nearest neighbors are worse than it, but where we have no
guarantee that there is not a better point some way away. A global optimum is a point which we know
to be the best. In the Himalayas the summit of K2 is a local maximum height, whereas the summit ofEverest is the global maximum height.)
Theoretically, models that can be built with any of the entities we have listed above can equally well bemodeled solely with binary variables. The reason why modern IP systems have some or all of the extra
entities is that they often provide significant computational savings in computer time and storage whentrying to solve the resulting model. Most books and courses on Integer Programming do not emphasize
this point adequately. We have found that careful use of the non-binary global entities often yields very
considerable reductions in solution times over ones that just use binary variables.

3.2

IP solving: the ideas behind Branch and Bound
Most commercial Integer Programming optimizers, including that in Xpress-MP, use a Branch and Bound(B&B) search as the basis of the algorithm which locates and proves the optimal solution. It is not the
purpose of this book to go into great detail about the ideas behind the solution algorithms, but tounderstand why the modeling entities we have introduced are useful it is necessary to give here a short
explanation of the ideas behind Branch and Bound. The presentation is intended to give you the flavorof what a B&B search does, and is not supposed to be an exact description.
The first idea underpinning B&B is relaxation. If we have global entities in the problem, we temporarily
forget about their discreteness and solve the resulting problem (the relaxed problem) as an LP. We may
be lucky, and the solution values that should be discrete happen by accident to take allowable values.
The binaries are all 0 or 1, the integer variables take integer values etc. If this fortuitous event occurs,
then we have got the integer solution, and the B&B search is over.
If not, we have at least one global entity which is not satisfied, and the solution is said to be integer

Integer Programming models

31

Applications of optimization with Xpress-MP

infeasible. A binary or integer variable is taking a fractional value, a partial integer is below its criticallimit and is fractional, or a Special Ordered Set condition is not satisfied. So we have to do something
to move towards integrality. We introduce the notion of separation, which is most easily described in
the case of an integer variable. Suppose one particular integer variable’s (say, variable v) optimal value
in the LP solution is 12.3. Then the obvious remark that in a solution either v ≤ 12 or v ≥ 13 leads
us to separate the initial problem into two sub-problems, one with the added constraint v ≤ 12 (callthis the down branch) and the other with the added constraint v ≥ 13 (call this the up branch). If we
can solve both sub-problems, then the better of the two solutions gives us the solution to the original
problem. In other words, we have separated the original problem into two subproblems. We note thatsince each sub-problem is more constrained than the original problem, the solutions to the sub-problems
will certainly be no better than the solution to the original problem.
What do we do now? We have replaced solving the original problem by the task of solving two subproblems. So for both of the sub-problems we again apply the idea of relaxation, i.e. drop the discreteness conditions and solve as LP problems.
At first sight, this seems like nonsense — we have replaced solving one problem by having to solve two
sub-problems. However, there is a good argument why the two sub-problems should naturally be more‘integer’ than the original problem, where by ‘integer’ we mean ‘more likely to have an integer solution
to the LP relaxation’. In the LP solution to the original problem variable v wants to take the value 12.3,
but in the two sub-problems we force it away from that value. So in the LP solution of the down branch
sub-problem we will have v = 12 and in the LP solution of the up branch we will have v = 13. In bothcases one of the previously fractional variables is taking an integer solution, so in some very vague way
we can see that the sub-problems are naturally more ‘integer’ than the original problem.
However, the LP relaxation solutions to the sub-problems may still be integer infeasible, but now weknow what to do. We apply the idea of separation again, selecting an integer infeasible variable.
So the B&B process consists of successively separating on (branching on) variables which are not satisfying
their discreteness condition, thereby creating two new sub-problems every time we separate. We canrepresent this as a tree, drawn upside-down in the usual computer science manner.
Obj = 112v = 12.3

1v

Obj = 115
w = 7.6
w

7

12

v 13

2

4
w

8

w

2

Obj = 118
w = 2.4
w

3

3

7

6

5

Obj = 128
(cut off)

infeasible

Obj = 120

Obj = 120(cut off)

Figure 3.1: Branch and Bound tree

The nodes of the tree represent the solution of the LP relaxation. Beside each node we have written thevalue of the objective function of the LP relaxation at the node. Note that we assume we have beensolving a minimization problem, so the values of the objective functions get worse (bigger) as we go
down a branch of the tree, as we are continuously adding new constraints. If the LP solution at a node isinteger feasible, we stop exploring down from that node, otherwise we separate on one of the integer
infeasible variables, creating two descendant nodes. If the LP at a node is infeasible, we cannot continue
from that node as adding constraints to an already infeasible problem will only make it more infeasible,and no descendant can possible give us a feasible integer solution.
If in the course of the search we find an integer solution, we may be able to exploit this fact through thenotion of fathoming. In the figure above, we find an integer solution with value 120 at node 6. Looking
at node 3 we observe that adding constraints will only make its descendants have objective function
values worse than 128, but we already have a solution with value 120. So we do not need to look at any
descendants of node 3: it has been fathomed or cut-off. Similarly, node 5 is fathomed. And node 7 isfathomed by the fact that it is infeasible.

Integer Programming models

32

Applications of optimization with Xpress-MP

If we are fortunate and find an integer solution early in the search, then you can see that a very large
fraction of the possible nodes may be fathomed, and we will not have to search the entire tree. But if
there are N discrete variables, in theory we might have to generate nearly 2N nodes, and thus have tosolve that many LP problems, a formidable task.
As you can see, there are three main decisions to make when doing a B&B search
1. which of the outstanding nodes to select for solving by LP (the node selection strategy)
2. which of the fractional integer variables to select for separation (the variable selection strategy)
3. which direction to branch on first when you have separated (the branching direction strategy)
Though any way of making these decisions will work in theory, since B&B is guaranteed to find the optimal integer solution, given long enough time, the practical effectiveness of a B&B algorithm depends
crucially on making the decisions correctly. In many cases it is best to leave the choice to the IP solver;
however there is some opportunity for the modeler to influence the solution process by passing knowledge of the semantics of the model to the ‘out-of-the-box’ solver: often it is helpful to assign branchingpriorities to decision variables if the values of certain (sets of) variables have a strong impact on the restof the solution (for instance, if we need to decide whether to open a factory and how much to produce
there in different time periods, the decision on the opening should usually be made first). The modeler
may also indicate preferred branching directions for single variables (‘up’ or ‘down’ branch first) if he hassome intuition of the value a variable is likely to take in the solution.
Like other solvers, Xpress-MP has a set of default strategies which have been found to work well on mostproblems. However, the user should note carefully that sophistication in modeling is important in large
scale MIP work. Theoretical developments in the last two decades have lead to insights as to how best
to model IP problems, and how to choose between alternative formulations. We cannot over-stress thepoint that a good formulation can often speed an IP search by orders of magnitude.

3.3

Modeling with binary variables
We have seen that a rich modeling environment supplies more than just binary variables, and we shalllater see examples of using these extended entities, but we start the discussion of IP modeling by showing
the use of binary variables.

3.3.1

Do/don’t do decisions
The most common use of binary variables is when we are modeling activities that either have to be donein their entirety or not done at all. For instance, we might want to model whether we drive a particular
truck between city A and city B on a particular day. In this case we either do drive the truck or we do not
drive the truck at all: there are just two options. Typically we model the option of doing something by
having a binary variable being equal to 1, and the option of not doing the thing with the binary variablebeing equal to 0. Thus we just have two choices, of which we must select one.
Associated with the activity and with its binary variable, there may be some other properties. Suppose
for instance we are considering what we do with the three trucks that we own. Truck 1 has a capacity of
20 tonnes, truck 2 has a capacity of 25 tonnes and truck 3 has a capacity of 30 tonnes. If on a particular
day we decide to send truck 1 to a customer then we either send the entire truck or we do not send theentire truck. So we might have a decision variable (a binary variable) send1 taking the value 1 if we send
truck 1 out or 0 if we do not send truck 1. Similarly we will have another binary variable send2 which is
1 if truck 2 goes out and 0 otherwise; and variable send3 taking on the value 1 if truck 3 goes out, and 0
otherwise.
Elsewhere in the model we will probably be interested in the total tonnage that goes out on that particular day and we can see that this is
20 · send1 + 25 · send2 + 30 · send3
because if truck 1 goes out (send1 = 1) it takes 20 tonnes, if truck 2 goes out it independently takes 25
tonnes, and if truck 3 goes out it carries 30 tonnes, so the total is 20 · send1 + 25 · send2 + 30 · send3 .
Suppose that we have a decision variable called out which gives the total tonnage leaving our depot onthat particular day. Then we have the equationout = 20 · send1 + 25 · send2 + 30 · send3

Integer Programming models

33

Applications of optimization with Xpress-MP

3.3.2

Logical conditions
To show many uses of binary variables we will take as an example a set of projects that we may or may
not decide to do. We shall call the projects rather unimaginatively A, B, C, D, E, F, G and H and with eachof these projects we will associate a decision variable (a binary variable) which is 1 if we decide to do theproject and 0 if we decide not to do the project. We call the corresponding variables a, b, c, d, e, f, g andh. So decision variable a taking on the value 1 means that we do project A, whilst a taking on the valueof 0 means that we do not do project A. We are now going to express some constraints in words and see
how they can be modeled using these binary variables.

3.3.2.1 Choice among several possibilities
The first constraint that we might impose is ‘we must choose no more than one project to do’. We caneasily see that this can be expressed by the constraint
a+b+c+d+e+f +g+h≤1
Why is this true? Think of selecting one project, for instance project C. If c is 1 then the constraint
immediately says that a, b, d, e, f, g and h must be 0 because there is no other setting for the values ofthese variables that will satisfy that constraint. So if c is 1 then all the others are 0 and, as there is nothing
special about C, we can see immediately that the constraint means that we can only do one project.
If the constraint was that we could do no more than three projects then obviously all we have to do in
the constraint above is to replace the 1 by 3 and then we can easily see that in fact we can have no morethan three of the 0-1 variables being equal to 1. It is possible to have just two of them or one of them oreven none of them being equal to 1 but we can certainly not have four or more being 1 and satisfy theconstraint at the same time.
Now suppose that the constraint is that we must choose exactly two of the projects. Then the constraint
we can use to model this isa+b+c+d+e+f +g+h=2
Since the binary variables can take only the values 0 or 1, we must have exactly two of them being 1 andall the rest of them being 0, and that is the only way that we can satisfy that constraint.

3.3.2.2 Simple implications
Now consider an entirely different sort of constraint. Suppose that ‘if we do project A then we must do
project B’. How do we model this? Like a lot of mathematics, it is a question of learning a trick that has
been discovered. Consider the following constraintb≥a
To show that this formulation is correct, we consider all the possible combinations of settings (there are
four of them) of a and b. First, what happens if we do not do project A. If we also do not do project B
then the constraint is satisfied (0 = b ≥ a = 0) and our word constraint is satisfied. If, on the other hand,
we do do project B, then 1 = b ≥ a = 0, and the constraint is again satisfied. Now consider what happens
if we actually do project A, i.e. a = 1. Then the constraint is violated if b = 0 and is satisfied (0 is not ≥ 1)
if b = 1, in other words we do project B. The last case to consider is if we do not do project A, i.e. a = 0
and we do project B, i.e. b = 1. Then the constraint is satisfied (1 ≥ 0) and the word constraint is indeed
satisfied. We can lay these constraints out in a table. There are only four possible conditions: {do A, do
B}, {do A, do not do B}, {do not do A, do not do B}, {do not do A, do B}, and we can see that the illegal
one is ruled out by the algebraic constraint.
Table 3.1: Evaluation of a binary implication constraint
b≥a

a=0

a=1

b=0
b=1

Yes
Yes

No
Yes

The next word constraint we consider is ‘if we do project A then we must not do project B’. How might
we set about modeling this? The way to think of it is to notice that the property of notdoing B can bemodeled very easily when we already have a binary variable b representing doing B. We invent a newvariableb=1−bInteger Programming models

34

Applications of optimization with Xpress-MP

where b represents the doing of project B i.e., the project ‘not doing B’. If b = 1 then b = 0 (in otherwords, if we do project B then we do not do ‘not B’) whereas if b = 0 then b = 1 (if we do not do project
B then we do project B). This is very convenient and b is called the complement of b. We can use this trick
frequently. Just above we learned how to model ‘if we do A then we must do B’ and now we are trying
to model ‘if we do A then we must not do B’, i.e. ‘if we do A then we must do B’. As ‘if we do A then wemust do B’ was modeled by b ≥ a we can immediately see that the constraint ‘if we do A then we must
do B’ can be obtained by replacing b by b in the constraint, in other words b ≥ a. Replacing b by 1 − b
we get
1−b≥a
or
1≥a+b
or alternatively
a+b≤1
Now that we have obtained this constraint, it is quite obvious. What it says is that if we do project A (a =1) then b must be 0. This is exactly what we wanted to model. The point of the somewhat long-windedargument we showed above, however, is that we have used the result from the first logical constraintthat we wanted to model, plus the fact that we have now introduced the notion of the complement of
the project, to build up a newer and more complicated constraint from these two primitive concepts. This
is what we will do frequently in what follows.
We see an example of this by trying to model the word constraint ‘if we do not do A then we must do
B’, in other word, ‘if not A then B’. We can go back immediately to our first logical constraint ‘if we do A
then we must do B’ which was modeled as b ≥ a. Now we are actually replacing A by not A, so we cansee immediately that our constraint isb≥1−awhich isa+b≥1
Again this constraint is obvious now that we have got to it. If we do not do A then a = 0, so b ≥ 1 andsince the maximum value of b is 1 then this immediately means that b = 1. Again we have just taken our
knowledge of how to model ‘if A then B’ and the notion of the complement of a variable to be able tobuild up a more complex constraint.
The next constraint to consider is ‘if we do project A we must do project B, and if we do project B wemust do project A’. We have seen that the first is modeled as b ≥ a and the second as a ≥ b. Combining
these two constraints we get
a=b
in other words projects A and B are selected or rejected together, which is exactly what we expressed in
our word constraint.

3.3.2.3 Implications with three variables
The next constraint we consider is ‘if we do project A then we must do project B and project C’. The firstthing to note is that this is in fact two constraints. One, the first, is ‘if we do A then we must do project
B’ and the second constraint is ‘if we do A then we must do project C’. With this observation we can seethat the word constraint can be modeled by the two inequalitiesb ≥ a and c ≥ aso that if a = 1, then both b = 1 and c = 1.
Another constraint might be ‘if we do project A then we must do project B or project C’. It is like theprevious constraint, except we now have an ‘or’ in place of the ‘and’. The constraintb+c ≥amodels this correctly. To see this, consider the following situation. If a is 0 then b + c can be anything, andso b and c are not constrained. If a = 1, then one or both of b and c must be 1.
We may also try to model the inverse situation: ‘if we do Project B or project C then we must do A’. Thisis again a case that may be formulated as two separate constraints: ‘if we do B then we must do A’ and‘if we do C then we must do A’, giving rise to the following two inequalitiesa ≥ b and a ≥ cInteger Programming models

35

Applications of optimization with Xpress-MP

so that if either b = 1 or c = 1, then we necessarily have a = 1.
A harder constraint to model is the following ‘if we do both B and C then we must do A’. How might wemodel this? One way to think about it is to express it in the following way: ‘if we do both B and C thenwe must not do not-A’, or, ‘we can do at most two of B, C or not-A’ which we would model asb + c + (1 − a) ≤ 2
or in other words
b+c−a≤1or perhaps more conventionallya≥b+c−1
Looking at this last inequality, we can see that there is no effect on a when b and c are 0, or when justone of b and c is 1, but a does have to be ≥ 1 when both b and c are 1. A binary variable having to begreater than or equal to 1 means that the binary variable has to be precisely 1.

3.3.2.4 Generalized implications
Generalizing the previous we now might try to model ‘if we do two or more of B, C, D or E then we must
do A’, and our first guess to this might be the constrainta≥b+c+d+e−1
Certainly if, say, b and c are both equal to 1 then we have a ≥ 1 and so a must equal 1, but the problem
comes when three of the variables are equal to 1, say b, c and d. Then the constraint says that a ≥ 3 − 1,
i.e. a ≥ 2, which is impossible as a is a binary variable. So we have to modify the constraint as followsa≥1· (b + c + d + e − 1)
3

The biggest value that the expression inside the parentheses can take is 3, if b = c = d = e = 1. The 31
in front of the parenthesis means that in the worst case a must be ≥ 1 (so a is equal to 1). But we mustverify that the constraint is true if, say, just b = c = 1 (and d and e are equal to 0). In this case we havethat a ≥ 13 · (1 + 1 + 0 + 0 − 1) i.e. a ≥ 13 . But since a can only take on the values 0 or 1 then the constrainta ≥ 31 means that a must be 1. This is exactly what we want.
We can generalize this to modeling the statement ‘if we do M or more of N projects (B, C, D, ...) then wemust do project A’ by the constrainta≥b + c + d+. . . −M + 1
N−M+1

So far we have only given one way of modeling each of these constraints. We return to the constraint ‘if
we do B or C then we must do A’ which we modeled as two constraints a ≥ b and a ≥ c. It is possible tomodel this with just one constraint if we take this as a special case of the ‘M or more from N’ constraint
we have just modeled. ‘If we do B or C then we must do A’ is the same as ‘if we do M or more of N
projects (B, C, D, ...) then we must do project A’ with M = 1 and N = 2. So the constraint isb+c−M+1i.e.
N−M+1b+c−1+1
a≥
i.e.
2−1+1
1
a ≥ · (b + c)2a≥

So this single constraint is exactly the same in terms of binary variables as the two constraints which we
produced before. Which of these two representations is better? In fact the representation in terms oftwo constraints is better. But both of the two are correct and both will give a correct answer if put into
an Integer Programming system. It is just that the first pair of constraints will in general give a solution
more rapidly.
More and more complicated constraints can be built up from the primitive ideas we have explored so far.
Since these more complicated constraints do not occur very frequently in actual practical modeling weshall not explore them further. Table 3.2 summarizes the formulations of logical conditions we have seenin the different paragraphs of Section 3.3.2.

Integer Programming models

36

Applications of optimization with Xpress-MP

Table 3.2: Formulation of logical conditions using binary variables
At most one of A, B,...,H
Exactly two of A, B,...,H
If A then B
Not B
If A then not B
If not A then B
If A then B, and if B then A
If A then B and C
If A then B or C
If B or C then A

a+b+c+d+e+f +g+h≤1
a+b+c+d+e+f +g+h=2
b≥a
b=1−b
a+b≤1
a+b≥1
a=b
b ≥ a and c ≥ ab+c ≥aa ≥ b and a ≥ cor alternatively: a ≥ 21 · (b + c)a≥b+c−1a ≥ 31 · (b + c + d + e − 1)a ≥ b+c+d+...−M+1
N−M+1

If B and C then A
If two or more of B, C, D or E then A
If M or more of N projects (B, C, D, ...) then A

3.3.3

Products of binary variables
We move on to modeling the product of binary variables. Suppose we have three binary variables b1 , b2
and b3 and we want to model the equation
b3 = b1 · b2
This is not a linear equation since it involves the product of two variables, so we have to express it insome linear form. There is a trick and it is another one of these tricks that one just has to learn. Consider
the following set of three inequalities
b3 ≤ b1b3 ≤ b2b3 ≥ b1 + b2 − 1Then we claim that this represents the product expression that we wish to model. To see we construct
the following Table 3.3.
Table 3.3: Product of two binaries
b1

b2

b3

b3 = b1 · b2 ?

b3 ≤ b1 ?

b3 ≤ b2 ?

b3 ≥ b1 + b2 − 1?

0
0
0
0
1
1
1
1

0
0
1
1
0
0
1
1

0
1
0
1
0
1
0
1

Yes
No
Yes
No
Yes
No
No
Yes

Yes
No
Yes
No
Yes
Yes
Yes
Yes

Yes
No
Yes
Yes
Yes
No
Yes
Yes

Yes
Yes
Yes
Yes
Yes
Yes
No
Yes

We can see that the column headed b3 = b1 · b2 ? is true if and only if we have a ‘Yes’ in the three columns
b3 ≤ b1 ?, b3 ≤ b2 ? and b3 ≥ b1 + b2 − 1?, so the three linear equations do exactly represent and are true
at exactly the same time as the product is true.
This is a particularly long winded way of demonstrating the equivalence of the product term and thethree linear equations and in fact now we have got it it is actually quite easy to see why these threeinequalities are correct. Since b3 is b1 multiplied by something that is less than or equal to 1, b3 will
always be less than or equal to b1 and by a similar argument b3 will always be less than or equal to b2 .
The only further case we have to consider is when both b1 and b2 are equal to 1 and then we have toforce b3 to be equal to 1. This is done by the constraint b3 ≥ b1 + b2 − 1 which is non restrictive if only
one or none of b1 and b2 are 1 but forces b3 to be 1 when b1 = b2 = 1.
Looking at the constraint this way immediately enables us to model for instanceb4 = b1 · b2 · b3 ,in other words the product of three variables, as the four constraints
b4 ≤ b1

Integer Programming models

37

Applications of optimization with Xpress-MPb4 ≤ b2
b4 ≤ b3
b4 ≥ b1 + b2 + b3 − 2
If any of b1 , b2 and b3 are 0 then b4 must be 0 but if b1 , b2 and b3 are 1 then b4 must be greater than orequal to 3 − 2, i.e. b4 must be greater than or equal to 1 so b4 must be 1.

3.3.4

Dichotomies: either/or constraints
All the constraints we have seen so far have had to be satisfied simultaneously, but sometimes we need tomodel that either Constraint 1 or Constraint 2 has to be satisfied, not necessarily both of them. Consider
the problem:
minimize

Z = x1 + x2

subject tox1 ≥ 0, x2 ≥ 0 andEither 2 · x1 + x2 ≥ 6 (Constraint 1) or x1 + 2 · x2 ≥ 7 (Constraint 2)

Since we have just two variables, we can graph the feasible region, and we have done this in the figure
below where the grey shaded area is the feasible region.

x2

6

4

L

7
Z=

c2

5

Z=

2

3
Z=
Z=
1

G
2

4

8

6

x1

c1
Figure 3.2: Example of either/or constraints

Point L (x1 = 0, x2 = 3. 5, Z = 3. 5) is a local optimum. Point G (x1 = 3, x2 = 3, Z = 3) is the global minimum.
We can model this with one additional binary variable b.
2 · x1 + x2 ≥ 6 · b
x1 + 2 · x2 ≥ 7 · (1 − b)
To show that this is true, we just have to consider the two cases:
b=0:
b=1:

3.4

2 · x1 + x2 + x3 ≥ 0
2 · x1 + x2 + x3 ≥ 6

x1 + 2 · x2 + 3 · x3 ≥ 7
x1 + 2 · x2 + 3 · x3 ≥ 0Constraint 2 is satisfied
Constraint 1 is satisfied

Binary variables ‘do everything’All global entities (general integers, partial integers, semi-continuous variables, and both sorts of Special
Ordered Sets) can be expressed in terms of binary variables. However, as shown by the examples in thissection, it is usually preferable to use the specific global entities.

Integer Programming models

38

Applications of optimization with Xpress-MP

3.4.1

General integers
Consider an integer variable v which must take a value between 0 and 10. We could replace this integer
variable with four binary variables b1 , b2 , b3 , and b4 everywhere in the model using the expressionv = b1 + 2 · b2 + 4 · b3 + 8 · b4
remembering that we also have to have the constraint
b1 + 2 · b2 + 4 · b3 + 8 · b4 ≤ 10
This is the binary expansion of v. It has the major disadvantage that we now have four global entities
rather than just one global entity. Furthermore, a particular integer value of v, for example 6, can beachieved by many fractional settings of the binary variables. For instance, v = 6 can be achieved by{b1 = b2 = b3 = 0, b4 = 0. 75} or {b1 = 0, b2 = 1, b3 = 1, b4 = 0}, and there are a host of other possibilities.
Thus even though v is integer it can be achieved by fractional values of the binary variables and thesefractional values would force our integer programming code to do more branching.
Since all commercial integer programming codes allow one to use general integer variables, and with thedisadvantage of expanding integers in terms of binary variables, it is usually silly to use binary variables
where we really have an underlying integer.

3.4.2

Partial integers
We will postpone the demonstration of how to model partial integer variables in terms of binaries until
later (Section 3.5.3), when we have discussed linking binary variables and real variables.

3.4.3

Semi-continuous variables
Suppose we have a semi-continuous variable s which can take on either the value 0 or any value between
some lower limit L and some upper limit U. If we introduce a binary variable b then we can represent thesemi-continuity of s by the following pair of constraints
L·b≤s
s≤U·b
We shall see why this is true. There are two cases to consider. Either b = 0, in which case s is constrained tobe 0, or b = 1, in which case s is constrained to lie between L and U. These are exactly the two conditions
that we want to impose.
Why are semi-continuous variables useful? It seems a very small penalty just to have to replace a semicontinuous variable by one extra binary variable. The answer is that the semi-continuous property may
be satisfied by the variable s, but if we introduce a binary variable the integrality of that binary variable
may not be satisfied. Consider the case where s either has to take on the value 0 or it has to take on areal value in the range between 5 and 10. Suppose that we solve the LP relaxation and we get a value
for s of 7.5 and we get a value of b of 0.75. This certainly satisfies the two inequalities but the branch
and bound search would note that the binary variable was not satisfied (it is not either 0 or 1) and so wewould have to go on branching and bounding. But the underlying semi-continuity of s is satisfied; s does
indeed lie between its lower bound 5 and its upper bound 10 and so we would not have to branch any
further on the s. Thus not having semi-continuous variables in the modeling language or in the optimizer
can lead to more branching and bounding in the branch and bound search, and indeed if you need touse semi-continuous variables you should look for an optimizer that has these built into it.

3.4.4

Special Ordered Sets of type 1 (SOS1)
SOS1s are a set of variables, at most one of which can take a strictly positive value, all others being at 0.
They most frequently apply where a set of variables are actually 0-1 variables: in other words, we haveto choose one from a set of possibilities. These might arise for instance where we are deciding on whatsize of factory to build, when we have a set of options, perhaps small, medium, large or no factory at all,and we have to chose one and only one size.
To take a small concrete example, suppose that we are considering building a factory and we have options
S for a small factory, M for a medium sized factory, B for a big factory and N which is no factory at all.
We can model this by introducing four binary variables bS , bM , bB and bN and model the fact that we canonly choose one of these options by the constraintbN + bS + bM + bB = 1

Integer Programming models

39

Applications of optimization with Xpress-MP

Clearly since these are binary variables adding up to 1, only one of them can be 1 and one of them mustbe 1. So the SOS1 property applies.
If that was all there was to special ordered sets then it would not be much of an invention. But infact there is rather more that can be said because we have forgotten about the word ordered; in thedefinition. Suppose that the capacity of the various options for the factories are Small 10,000 tonnes,
Medium 20,000 tonnes, Big 38,000 tonnes and of course the ‘No factory’ has 0 tonnes capacity. One of
the things we will be interested in is the selected capacity of the factory that we will be building, andsomewhere in the model we would have a decision variable size and an equation
size = 0 · bN + 10000 · bS + 20000 · bM + 38000 · bB .
We can use the coefficients in this equation (which we call the reference row) to order the variables. The
order emerging is bN , bS , bM , bB where we have ordered them by their coefficients in the Reference Row.
Now suppose that we have solved the linear programming relaxation and got the following solution:bN = 0. 5, bS = 0, bM = 0, bB = 0. 5
indicating that we are going to build half of the 0 sized factory and half of the big factory (remember
these are the values of the LP relaxation where we have relaxed the integrality of the binary variables).
bN = 0.5, bB = 0.5

1
bN 0 and bS 0
(size 20,000)

bM 0 and bB 0
(size 10,000)

2nFigure 3.3: Possible SOS branching tree

The branch and bound search will look at that solution and say that there are only two fractional variables
bN and bB and so it will branch on one or other of those variables. But there is a lot more information in
the variables than the fact that they are fractional or not fractional. We are being told by the LP solutionthat the tonnage that we require from that factory is a half of the biggest one: in other words 19,000
tonnes. Now consider branching on, say, bN . Either we are forcing the factory to have zero capacity
if we branched on bN up to 1, or we are forcing the factory to be open. It would be much better tohave some branching scheme that understood and tried to exploit the fact that the indicated capacity is
about 19,000 tonnes. This is precisely what the special ordered set property indicates to the branch andbound tree, and the branching occurs on sets of variables rather than on the individual variables which
form the Special Ordered Set. The details of precisely how to branch on sets of variables rather than
individual variables are outside the scope of this chapter. It is, however, an observed fact that modelingthese selection constraints in terms of Special Ordered Sets generally gives a faster branch and boundsearch. Again as with the case of general integer variables, the solution we get will be exactly the same
whichever way we model. The ‘only’ benefit of using Special Ordered Sets is that the search procedure
will generally be noticeably faster.

3.4.5

Special Ordered Sets of type 2 (SOS2)
SOS2s are generally used for modeling piecewise approximations of functions of a single variable. As an
example look at the figure below
We wish to represent f as a function of x and we are prepared to consider four line segments between five points. The five points are (R1 , F1 ), (R2 , F2 ), (R3 , F3 ), (R4 , F4 ) and (R5 , F5 ) and associated witheach point i is a weight variable yi . We have a binary variable bi associated with each of the intervals(R1 , R2 ), (R2 , R3 ), (R3 , R4 ) and (R4 , R5 ). bi is 1 if the value of x lies between Ri and Ri+1 .
Then
x=

5
X

Ri · yi (reference row)

i=1

Integer Programming models

40

Applications of optimization with Xpress-MP

f

F2

b1

F4

b2

b3

F1

b4

F3
F5

weights:

R1

R2

R3

R4

R5

y

y

y

y

y

1

2

3

4

x

5

Figure 3.4: Function approximation with SOS2

f=

5
X

Fi · yii=1
5X

yi = 1 (convexity row)

i=1

The SOS2 property of having at most two non-zero yi , and if there are two non-zero then they must be
adjacent, means that we are always on the piece-wise linear function. If we did not have the adjacency
condition, then by having, say, y1 = y3 = 0. 5 (all the others 0) we could ‘cheat’ at x = R2 as the function
value is not F2 but (F1 + F3 ) / 2, considerably lower (see Figure 3.5).

f

F2
F4

b1

b2

b3

F1

b4

F3
F5

R1

R2

R3

R4

R5

x

Figure 3.5: Function approximation with SOS2: adjacency condition

Consider the constraints
y1 ≤ b1y2 ≤ b1 + b2
y3 ≤ b2 + b3
y4 ≤ b3 + b4
y5 ≤ b4
b1 + b2 + b3 + b4 = 1
The last constraint ensures that one and only one bi is 1 in an acceptable solution. For instance, if b3 is 1,
the constraints on the yi ensure that y1 = y2 = y5 = 0, leaving only the adjacent y3 and y4 to be (possibly)
non-zero. The other case to consider is when one of the ‘end’ bi is 1, say b1 . Then y3 = y4 = y5 = 0, and y1
and y2 , which are adjacent, can be non-zero. So the constraints ensure the SOS2 property, at the expenseof introducing N − 1 additional binary variables if there are N points.

Integer Programming models

41

Applications of optimization with Xpress-MP

The same remarks that applied to a reduction in the search times with SOS1 applies to SOS2. It is foundin practice that branching on subsets of the variables within a SOS2, rather than on the individual binary
variables, is generally beneficial.

3.5

Connecting real variables to binary variables

3.5.1

Modeling fixed costs
The first case we look at is where we want to model fixed costs, that is, where we incur a fixed cost K if aparticular real variable x is strictly greater than 0. This fixed cost K might represent the cost of setting up
a piece of machinery — we do not incur the fixed cost if we do not set up the piece of machinery but if
the output level x of the machine is anything different from zero then we have to incur the fixed cost.
The graph of the cost against the output level x might be represented in the figure below.

Cost
C

K

XMAX

x

Figure 3.6: Combining fixed and variable cost

Here x represents the throughput of the piece of equipment and the cost consists of two parts. If thethroughput is zero then the cost is zero, whereas if the throughput takes a value which is different from0 then the cost is Cost = K + C · x, where C is the per unit cost of output x.
We introduce a binary variable b which is equal to zero if x is equal to zero and equal to 1 if x is strictly
greater than zero. Then the cost equals b · K + C · x, but we have to ensure that if b is zero then x is zero.
We can do this by introducing the constraintx ≤ Xmax · b
where Xmax is the largest value that x can take.
We shall see why this is correct. If b = 0 then x is constrained to be less than or equal to zero and so x iszero, whereas if b = 1 we just have the constraint that x ≤ Xmax , i.e. that x is not more than its maximum
value (which might be for instance the maximum output of that particular piece of equipment).

3.5.2

Counting
A frequent use of binary variables in modeling is where we wish to be able to count whether a realvariable x is different from 0. Often we want to have a constraint that only a certain number of realvariables are strictly greater than 0.
An example might be where we have a whole variety of different ingredients that could go into a blend
and we have a constraint that perhaps at most 10 of these ingredients can actually go into the blend. If
the quantity of item i going into the blend is xi and we require no more than 10 of these to be non zero,for each item i we introduce a binary variable bi and have constraints of the formxi ≤ VOL · bi
where VOL is the volume of the blend to be made. These constraints ensure that if a particular xi isgreater than zero then the corresponding bi must be equal to 1. Then the constraint that says that no
more than 10 of the ingredients must be non zero could be expressed as
10X

bi ≤ 10i=1Integer Programming models

42

Applications of optimization with Xpress-MP

3.5.3

Partial integers again
Earlier we postponed the discussion of how partial integer variables can be expressed in terms of binary
variables. Recall that a partial integer is a variable which has to be integral in any acceptable solution if
the value is less than some bound, and otherwise can be real. The natural way to express a partial integer
p is
p=v+x
where variable v is an integer that must be less than or equal to U (i.e. 0 ≤ v ≤ U) and x is a real variable.
We have already seen that the integer variable v can be expressed in terms of several binary variables
using the binary expansion. Now we have to express the fact that if v is strictly less than U then x must be
zero. Suppose that we know some value Xmax which x is guaranteed not to exceed, i.e. an upper boundfor x. Then we can introduce a new binary variable b and the following constraints
x ≤ Xmax · bb ≤ v / U (or, equivalently, v ≥ U · b )To show that this formulation is valid, consider the two possible values of b. First, if b = 0 then x ≤ 0, i.e.
x is equal to 0 and v ≥ 0. If b = 1 then x ≤ Xmax and v ≥ U, which combined with the fact that v ≤ Umeans v = U. So if b = 1, v is at its maximum, and we can then start having non-zero amounts of x.
As was the case with general integer variables, it seems profligate to express partial integers in terms ofbinary variables. Firstly, we have to express the integer part in terms of binary variables and then we haveto introduce a new binary variable b to model the partial integrality of variable p. Furthermore, as withsemi-continuous variables, introducing the extra binary variable will never make the branch and boundtree search quicker and may often make the tree search worse.
Consider the following example. Suppose that U = 10 and that the LP solution gives the optimum valueof v to be 3 and x to be 0, but the binary variable b is at 0.3. We can see that this satisfies all theinequalities we have given but that the binary variable b is fractional and we will have to branch on b
to satisfy integrality. But v is feasible as far as partial integrality is concerned, so we have a case wherepartial integer variables enable us to say that we are integer feasible but the introduction of the binary
variable results in it not being integer feasible and we have to branch further. In models with many partial
integer variables this can lead to a great expansion of the branch and bound tree and a lengthening of
the search.

3.5.4

Price breaks and economies of scale
Consider Figure 3.7 which represents the situation where we are buying a certain number of items andwe get discounts on all items that we buy if the quantity we buy lies in certain price bands.

Cost

COST3
COST2

COST1

B1

B2

B3

x

Figure 3.7: All items discountIn the figure we see that if we buy a number of items between 0 and B1 then for each item we pay anamount COST1 , whereas if the quantity we buy lies between B1 and B2 we pay an amount COST2 for each
item. Finally if we buy an amount between B2 and B3 then we pay an amount COST3 for each item. For
the sake of concreteness we assume that if we order exactly B1 items then we get the unit price COST2 ,whereas if we order just a little less than B1 then we would get the unit price COST1 , and similarly for all
the other price breaks. This sort of pricing is called all item discount pricing.
To model these item price breaks we should use binary variables b1 , b2 and b3 , where bi is 1 if we havebought any items at a unit cost of COSTi .

Integer Programming models

43

Applications of optimization with Xpress-MP

Now introduce the real decision variables x1 , x2 and x3 which represent the number of items bought at
price COST1 , COST2 and COST3 respectively. Note that we cannot buy any items at price COST2 until wehave bought the maximum number of items at the higher price COST1 , otherwise the solution would beeasy as we buy all items at the least expensive unit price!
The total amount x that we buy is given byx = x1 + x2 + x3
We claim that the following set of constraints models the all-item discounts.
B1 · b2 ≤ x1 ≤ B1 · b1

(3.5.1)(B2 − B1 ) · b3 ≤ x2 ≤ (B2 − B1 ) · b2

(3.5.2)x3 ≤ (B3 − B2 ) · b3

(3.5.3)b1 ≥ b2 ≥ b3

(3.5.4)Equations (3.5.1) say that if we have bought any in the second price range (b1 = b2 = 1) then x1 is fixed toB1 . Equations (3.5.2) ensure that if b2 = 0, then x2 = 0, whereas if b2 = 1 then x2 is only constrained by the
maximum amount that can be bought at the price COST2 . Equation (3.5.3) ensures that x3 = 0 if b3 = 0,
and that if b3 = 1 then we cannot buy more than the maximum number at price COST3 . Equations (3.5.4)
ensure that we can only buy at a lower price if we have bought at all the higher prices.

Cost

COST3

COST2

COST1

B1

B2

B3

x

Figure 3.8: Incremental price breaks

Now consider Figure 3.8 which represents the situation where we are buying a certain number of items
and we get discounts incrementally. The unit cost for items between 0 and B1 is C1 , whereas items
between B1 and B2 cost C2 each, and items between B2 and B3 cost C3 each.
We can model this using Special Ordered Sets of type 2 (SOS2). At the points 0, B1 , B2 and B3 we introduce
real valued decision variables wi (i = 0, 1, 2, 3). We also define cost break points CBPi that correspond to
the total cost of buying quantities 0, B1 , B2 and B3 . So
CBP0 = 0
CBP1 = C1 · B1
CBP2 = CBP1 + C2 · (B2 − B1 )CBP3 = CBP2 + C3 · (B3 − B2 )We then have
w0 + w1 + w2 + w3 = 1
TotalCost = 0 · w0 + CBP1 · w1 + CBP2 · w2 + CBP3 · w3
x = 0 · w0 + B1 · w1 + B2 · w2 + B3 · w3
and the wi form a SOS2 with reference row coefficients given by the coefficients in the definition of x.
For a solution to be valid, at most two of the wi can be non-zero, and if there are two non-zero theymust be contiguous, thus defining one of the line segments.

3.5.5

The product of a binary and a real variable
With two real variables, x and y, and a binary variable b, we want to model
y =b·xInteger Programming models

44

Applications of optimization with Xpress-MP

Suppose we have some upper bound U on the value of x. Then consider the following constraints
y≤x(3.5.5)y ≥ x − U · (1 − b)(3.5.6)y ≤U·b

(3.5.7)

If b = 0, then (3.5.7) means that y = 0. If b = 1, then (3.5.7) is harmless, and we have y ≤ x from (3.5.5)
and y ≥ x from (3.5.6), i.e. y = x, which is what is desired.

3.6

References and further material
Most of the books cited in the ‘References’ section of Chapter 1 also contain a part on Integer Programming, for instance the one by Schrijver [Sch86]. An excellent text on Integer Programming is themonograph by Wolsey [Wol98]. The book on combinatorial optimization by Papadimitriou and Steiglitz[PS98] also deals with this topic. Syslo et al. [SDK83] provide Pascal code for some algorithms.
The idea of Branch and Bound stems from Land and Doig [LD60]. For certain huge problems this algorithm
may be combined with column generation, leading to Branch and Price algorithms, a description of thistechnique and its main application areas is given in [BJN+ 98].
The notion of Special Ordered Sets was introduced by Beale et al. ([BT70] and [BF76]). For semi-continuous
variables see Ashford and Daniel [AD93].
Similarly to presolving for linear problems, commercial MIP software usually performs some work to obtain an improved problem formulation before starting the branch and bound search, and sometimes also
at the nodes of the search tree. The techniques that are used may be summarized as integer preprocessing and cut generation. Integer preprocessing tries to reduce the size of a problem and tighten bounds
based on integrality considerations; an overview on standard techniques is given in [Sav94]. Cut generation techniques generate additional constraints (valid inequalities, cutting planes, or in general cuts)that tighten the problem formulation: these cuts do not change the solution of the MIP but draw theLP relaxation closer to the MIP. Research related to cuts is a very active field (e.g. [GP01] and [MW01]).
Cuts are generated based on the identification of certain mathematical structures, but the user may also
‘help’ by specifying so-called model cuts (see the example in Section 10.4).
As mentioned in Section 3.2, to solve large applications efficiently it may be worthwhile experimenting
with the different search strategy options provided by a solver. A computational study of search strategies
is described in [LS98]. Integer Programming algorithms have made rapid progress in recent years [JNS00]:
problems with several thousand discrete variables may now be solved to optimality. However, certainclasses of problems still remain extremely difficult, like the example given by [CD99] for which instanceswith more than forty variables cannot be solved with commercial solvers.
An entirely different method for representing and solving problems with discrete variables is Constraint
Programming (CP) [VH98]. An introduction to CP is given in the textbook by Marriott and Stuckey [MS98].
Many concepts in (M)IP have their correspondence in CP and vice versa; Heipcke [Hei99b] compares andexplains the most common terms of both fields. With CP it is often relatively easy to find feasible solution(s) to a problem but proving optimality is a much more difficult task (there is no real correspondence
to the bounding information provided by the LP relaxation in MIP). A very active field of research is theintegration of both approaches to exploit their complementarity and also to be able to work with continuous variables in the context of Constraint Programming (see for instance [BDB95], [RWH99], [BK98],[Hei99a]).

Integer Programming models

45

Applications of optimization with Xpress-MP

Chapter 4

Quadratic Programming
Quadratic Programming (QP) is the name given to the problem of finding the minimum (or maximum)of a quadratic function of the decision variables subject to linear equality or inequality constraints. Notethat QP problems do not allow quadratic terms in the constraints, only in the objective function. QP
problems occur much less frequently than LP or MIP problems, so we shall not discuss them in much
depth.
Mixed Integer Quadratic Programs (MIQP) are QPs where some or all of the decision variables have totake on discrete values; in other words, they are MIPs with a quadratic objective function. The discrete
constraints can be any of the global objects that we have seen above (binaries, integers, etc.)
Since QPs are just LPs with quadratic objective functions, the only thing special to note is modeling theobjective function. We now give two typical examples.

4.1

Revenue optimization
If a company wishes to plan the amount q of a product to be sold, and at the same time determine theunit price p at which it is to be sold, then the objective function will contain a quadratic term p · q, whichis the revenue (price · quantity) from selling the product.

4.2

Portfolio optimization
Portfolio optimization is the classic example of QP. Given various opportunities (assets) in which an investor can place his money, the problem is to minimize the risk of achieving a given level of return. The
problem explicitly takes into account the fact that overall risk is not just the sum of the riskiness of theindividual assets, as their future performances are highly likely to be correlated: shares tend to go up anddown together, and shares in a particular sector tend to be highly correlated. The past variability of ashare’s price movements can be considered a measure of its risk.
We shall take a small example. Suppose we have B
C 1 million to invest in three stocks. Let Ri be the randomvariable representing the annual return on B
C 1 invested in stock i. Analysis of historical data has lead us
to estimate the returns as E(R1 ) = 0. 09, E(R2 ) = 0. 07, and E(R3 ) = 0. 06 where E(x) denotes the expectationof x. The variance of the annual returns have been estimated as var(R1 ) = 0. 20, var(R2 ) = 0. 07, andvar(R3 ) = 0. 15, and the covariances as cov(R1 , R2 ) = 0. 03, cov(R1 , R3 ) = 0. 04, cov(R2 , R3 ) = 0. 05.
If xi is the number of (millions of) Euros invested in stock i, then the annual return is
x1 · R1 + x2 · R2 + x3 · R3
and the expected annual return is
x1 · E(R1 ) + x2 · E(R2 ) + x3 · E(R3 )If we are seeking a return of at least 7.5%, we must have
0. 09 · x1 + 0. 07 · x2 + 0. 06 · x3 ≥ 0. 075
and the constraint that says we spend all the money
x1 + x2 + x3 = 1

46

Applications of optimization with Xpress-MP

The variance of the portfolio, which we want to minimize, isvar(x1 · R1 + x2 · R2 + x3 · R3 )=var(x1 · R1 ) + var(x2 · R2 ) + var(x3 · R3 )+2 · cov(x1 · R1 , x2 · R2 ) + 2 · cov(x1 · R1 , x3 · R3 )+2 · cov(x2 · R2 , x3 · R3 )=x12 · var(R1 ) + x22 · var(R2 ) + x32 · var(R3 )+2 · x1 · x2 · cov(R1 , R2 ) + 2 · x1 · x3 · cov(R1 , R3 )+2 · x2 · x3 · cov(R2 , R3 )=0. 20 · x12 + 0. 07 · x22 + 0. 15 · x32+0. 06 · x1 · x2 + 0. 08 · x1 · x3 + 0. 10 · x2 · x3

The usual non-negativity constraints apply to the xi . So we have a QP: the objective function has quadratic
terms (there are no linear terms, which are allowable in a general QP), and we have linear (in)equalities.

4.3

References and further material
Many standard works on Mathematical Programming or Operations Research also include a section on
Quadratic Programming. A good general reference for Quadratic Programming is Fletcher [Fle87]. Beale
has published a book entirely dedicated to Quadratic Programming [Bea59].
QP problems occur much less frequently in practice than LP or MIP problems, so we shall just give one
application example (Section 13.7).

Quadratic Programming

47

Applications of optimization with Xpress-MP

II. Application examples

Classification of
the example problems
The application examples in this part are grouped by application area into ten chapters:• Chapter 6: Mining and process industries• Chapter 7: Scheduling problems• Chapter 8: Planning problems• Chapter 9: Loading and cutting problems• Chapter 10: Ground transport• Chapter 11: Air transport• Chapter 12: Telecommunication problems• Chapter 13: Economics and finance• Chapter 14: Timetabling and personnel planning• Chapter 15: Local authorities and public services
The following tables give an overview of the examples, including their classification (theoretical problemtype), a rating between * (easy) and ***** (difficult), and a list of their modeling specifics and any Mosel
features used in the implementation that go beyond the basics explained in Chapter 5. The entry ‘NN’ in
the classification field means that the problem does not correspond to any standard type: the problem
may be a mixture of standard types, or have application-specific features for which no classification isavailable.
The reference numbers of the problems in the tables indicate the names given to the Mosel implementations of the models. For instance, the Mosel file for problem A-1 Production of alloys is a1alloy.mosand the data for this example is contained in the file a1alloy.dat. All application examples belonging
to one chapter start with the same letter, the number corresponds to the section number in this chapter.

49

Applications of optimization with Xpress-MP

Table 4.1: Classification in order of appearance (Chapters 6 and 7)
Difficulty

Features

A-1 Production of alloys
Blending problem

Problem name and type*

formulation of blending constraints; datawith numerical indices, solution printout,
if-then, getsol

A-2 Animal food production
Blending problem*

formulation of blending constraints; datawith string indices, as, formatted solution
printout, use of getsol with linear expressions, strfmt

A-3 Refinery
Blending problem**

formulation of blending constraints; sparse
data with string indices, dynamic initialization, dynamic arrays, finalize, create,
union of sets*ceil, is_binary**

encoding of arcs, solving LP-relaxation only**

inline if, is_integer***

2 problems; selection with ‘|’, sparse/dense
format, naming and redefining constraints,
subroutine: procedure for solution printing,forward declaration****

alternative formulation using SOS1***

formulating disjunctions (BigM); dynamicarray, range, exists, forall-do

A-4 Cane sugar production
Minimum cost flow (in a bipartite
graph)
A-5 Opencast mining
Minimum cost flow
A-6 Production of electricity
Dispatch problem
B-1 Construction of a stadium
Project scheduling (Method of Potentials)

B-2 Flow shop scheduling
Flow shop scheduling
B-3 Job shop scheduling
Job shop scheduling

B-4 Sequencing jobs on a bottleneck machine
Single machine scheduling***
B-5 Paint production
Asymmetric Traveling Salesman
Problem (TSP)
B-6 Assembly line balancing
Assembly line balancing

Classification of the example problems

3 different objectives; subroutine: procedure
for solution printing, if-then***solution printing, repeat-until, cast to
integer, selection with ‘|’, round**encoding of arcs, range

50

Applications of optimization with Xpress-MP

Table 4.2: Classification in order of appearance (Chapters 8 and 9)
Problem name and type

Difficulty

C-1 Planning the production of bicycles
Production planning (single product)
C-2 Production of drinking glasses
Multi-item production planningFeatures***modeling inventory
forall-dobalance;inline**modeling stock balance constraints; inline if,
index value 0

if,

C-3 Material requirement planning
Material requirement planning**working with index (sub)sets, dynamic initial(MRP)
ization, finalize, create, as
C-4 Planning the production of electronic components
Multi-item production planning**modeling stock balance constraints; inline if
C-5 Planning the production of fiberglass
Production planning with time***representation of multi-period production as
dependent production cost
flow; encoding of arcs, exists, create,
isodd, getlast, inline if
C-6 Assignment of production batches to machines
Generalized assignment problem*
D-1 Wagon load balancing
Nonpreemptive schedulingparallel machines****

heuristic solution requiring sorting algorithm,
formulation of maximin objective; nested
subroutines: function returning heuristic solution value and sorting procedure,ceil, getsize, if-then, break, exit,
all loop types (forall-do, repeat-until,
while-do), setparam, cutoff value

D-2 Barge loading
Knapsack problem**

incremental problem definition with 3 different objectives, procedure for solution printing

D-3 Tank loading
Loading problem***

2 objectives; data preprocessing, as, dynamic
creation of variables, procedure for solution
printing, if-then-else

D-4 Backing up files
Bin-packing problem**

2 versions of mathematical model, symmetry
breaking; data preprocessing, ceil, range

D-5 Cutting sheet metal
Covering problem
D-6 Cutting steel bars for desk legs
Cutting-stock problem

Classification of the example problems

on

*
**

51

set operation(s) on range sets,
integer (data as set contents)

set ofApplications of optimization with Xpress-MP

Table 4.3: Classification in order of appearance (Chapters 10 and 11)
Problem name and type

Difficulty

Features

E-1 Car rental
Transport problem***

data preprocessing, set operations, sqrt andˆ2, if-then-elif

E-2 Choosing the mode of transport
Minimum cost flow**

formulation with extra nodes for modes oftransport; encoding of arcs, finalize, unionof sets, nodes labeled with strings

E-3 Depot location
Facility location problem***

modeling flows as fractions, definition ofmodel cuts

E-4 Heating oil delivery
Vehicle routing problem (VRP)****

elimination of inadmissible subtours, cuts; selection with ‘|’, definition of model cuts

E-5 Combining different modes of transport
NN***

E-6 Fleet planning for vans
NN***

F-1 Flight connections at a hub
Assignment problem
F-2 Composing flight crews
Bipartite matching

F-3 Scheduling flight landings
Scheduling problem with time
windows
F-4 Airline hub location
Hub location problem

F-5 Planning a flight tour
Symmetric travelingproblemClassification of the example problems

salesmanmodeling implications, weak and strong formulation of bounding constraints; triple indicesmaxlist, minlist, max, min*****

2 problems, data preprocessing, incremental definition of data array, encoding ofarcs, logical or (cumulative version) and and,procedure for printing solution, forall-do,max, finalize***

generalization of model to arbitrary time
windows; calculation of specific BigM,
forall-do***

quadruple indices; improved (re)formulation(first model not usable with student version),
union of index (range) sets*****loop over problem solving, TSP subtour elimination algorithm; procedure for generating additional constraints, recursive subroutine calls, working with sets, forall-do,
repeat-until, getsize, not

52

Applications of optimization with Xpress-MP

Table 4.4: Classification in order of appearance (Chapters 12 and 13)
Problem name and typeG-1 Network reliability
Maximum flow with unitary capacities

Difficulty

Features***encoding of arcs, range, exists, create,algorithm for printing paths, forall-do,
while-do, roundG-2 Dimensioning of a mobile phone networkNN**
if-then, exit
G-3 Routing telephone calls
Multi-commodity network flow***
encoding of paths, finalize, getsize
problem
G-4 Construction of a cabled network
Minimum weight spanning tree***
formulation of constraints to exclude subcyproblem
cles
G-5 Scheduling of telecommunications via satellite
Preemptive open shop scheduling ***** data preprocessing, algorithm for preemptive
scheduling that involves looping over optimization, “Gantt chart” printing
G-6 Location of GSM transmitters
Covering problem*
modeling an equivalence; sparse data format
H-1 Choice of loansNN
H-2 Publicity campaign
NN
H-3 Portfolio selectionNN*

calculation of net present value*

forall-do**

sets of integers, second formulation withsemi-continuous, parameters

H-4 Financing an early retirement schemeNN**
H-5 Family budgetNN**

inline if, selection with ‘|’
formulation of monthly balance constraints
including different payment frequencies; as,mod, inline if, selection with ‘|’

H-6 Choice of expansion projects
NN**experiment with solutions: solve LP problem
explicitly, “round” some almost integer variable and re-solve

H-7 Mean variance portfolio selection
Quadratic Programming problem***

parameters, forall-do, min, max, loop overproblem solving

Classification of the example problems

53

Applications of optimization with Xpress-MP

Table 4.5: Classification in order of appearance (Chapters 14 and 15)
Difficulty

Features

I-1 Assigning personnel to machines
Assignment problem

Problem name and type****

formulation of maximin objective; heuristic solution + 2 different problems (incremental definition) solved, working with sets,
while-do, forall-do, negative index values

I-2 Scheduling nurses
NN***

2 problems, using mod to formulate cyclic
schedules; forall-do, set of integer

I-3 Establishing a college timetableNN***

many specific constraints, tricky (pseudo) objective function; finalize

I-4 Exam scheduleNN**
symmetry breaking, no objective
I-5 Production planning with personnel assignmentNN***
2 problems, defined incrementally with partial re-definition of constraints (named constraints), exists, create, dynamic arrayI-6 Planning the personnel at a construction site
NN**
formulation of balance constraints using inline if
J-1 Water conveyance / water supply management
Maximum flow problem**
encoding of arcs, finalize, selection with ‘|’
J-2 CCTV surveillance
Maximum vertex cover problem**
encoding of network, exists
J-3 Rigging elections
Partitioning problem**** algorithm for data preprocessing; file inclusion, 3 nested/recursive procedures, working with sets, if-then, forall-do, exists,
finalize
J-4 Gritting roads
Directed Chinese postman prob- **** algorithm for finding Eulerian path/graph forlemprinting; encoding of arcs, dynamic array,
exists, 2 functions implementing Euleriancircuit algorithm, round, getsize, break,while-do, if-then-else
J-5 Location of income tax offices
p-median problem**** modeling an implication, all-pairs shortest
path algorithm (Floyd-Warshall); dynamicarray, exists, procedure for shortest path
algorithm, forall-do, if-then, selection
with ‘|’J-6 Efficiency of hospitals
Data Envelopment Analysis (DEA)***
description of DEA method; loop over problem solving with complete re-definition ofproblem every time, finalize, naming anddeclaring constraints

Classification of the example problems

54

Applications of optimization with Xpress-MP

Chapter 5

The basics of Xpress-MP
All problems in this book are formulated using the Xpress-Mosel (for short, Mosel) Language. To run thesemodels, the user has the choice between using the Mosel Command Line Interpreter or the graphical userinterface Xpress-IVE. To solve the optimization problems in this book we use the Xpress-Optimizer linear
and mixed integer solver by accessing it from the Mosel Language.
All software with reference manuals and the complete set of examples discussed in this book can be
downloaded fromhttp://www.dashoptimization.com/applications_book.html
In the first section of this chapter we show how to execute the small production planning problem from
Chapter 1 using the Mosel Command Line Interpreter or Xpress-IVE. Section 5.2 introduces the basics ofthe Mosel Language.

5.1

Introductory example
A small joinery makes two different sizes of boxwood chess sets. The small set requires 3 hours of machining on a lathe, and the large set requires 2 hours. There are four lathes with skilled operators whoeach work a 40 hour week, so we have 160 lathe-hours per week. The small chess set requires 1 kg of
boxwood, and the large set requires 3 kg. Unfortunately, boxwood is scarce and only 200 kg per week
can be obtained. When sold, each of the large chess sets yields a profit of $20, and one of the small chessset has a profit of $5. The problem is to decide how many sets of each kind should be made each week
to so as to maximize profit.
In Chapter 1 the transformation of this description into a mathematical model was discussed in somedetail and will therefore not be repeated here. An implementation of the model has also already been
given (see Section 1.3). We assume that the following Mosel model has been entered into a text file
named chess.mos (mos is the standard file extension expected by Mosel):model Chess
uses "mmxprs"! We shall use Xpress-Optimizer

declarations
xs, xl: mpvar
end-declarations
Profit:= 5*xs + 20*xl
Boxwood:= 1*xs + 3*xl <=Lathe:=
3*xs + 2*xl <=maximize(Profit)! Decision variables: produced quantities

200
160

! Objective function! kg of boxwood! Lathehours! Solve the problem

writeln("LP Solution:")! Solution printing
writeln(" Objective: ", getobjval)
writeln("Make ", getsol(xs), " small sets")
writeln("Make ", getsol(xl), " large sets")
end-model

55

Applications of optimization with Xpress-MP

5.1.1

Using Xpress-Mosel
Mosel is an advanced modeling and solving language and environment, where optimization problems
can be specified and solved with the utmost precision and clarity. The modeling component of Mosel
provides an easy to use yet powerful language for describing optimization problems. Through its modular
architecture, Mosel provides access to data in different formats (including spreadsheets and databases)
and gives access to a variety of solvers, which can find optimal or near-optimal solutions to a problem.
Mosel is provided either as a standalone program (the Mosel Command Line Interpreter used in this book)or in the form of libraries that make it possible to embed a model into a larger application written in a
programming language.
To run the model we have entered into the file chess.mos, we start Mosel at the command prompt, andtype the following sequence of commands
mosel
exec chess
quit

which will start Mosel, compile the model and (if no syntax error has been detected) run the model,
and then quit Mosel. We will see output something like that below, where we have highlighted Mosel’s
output in bold face.
mosel** Xpress-Mosel **(c) Copyright Dash Associates 1998-2002> exec chess
LP Solution:
Objective: 1333.33
Make 0 small sets
Make 66.6667 large sets
Returned value: 0> quitExiting.

The same steps may be done immediately from the command line:
mosel -c "exec chess"

The -c option is followed by a list of commands enclosed in double quotes.
If after having started Mosel you type a command that is not recognized by the Mosel Command LineInterpreter (for instance: h), Mosel displays the full list of commands (or the possible completions to valid
commands) with short explanations.
The different options that may be used from the operating system’s command line can be obtained bytyping mosel -h.
The distribution of Mosel contains several modules that add extra functionality to the language. A fulllist of the functionality of a module can be obtained by using the exam command; for instance to see
what is provided by the Xpress-Optimizer module mmxprs:
mosel -c "exam mmxprs"

For a complete description of the Mosel Language and the Mosel Command Line Interpreter, the reader
is referred to the Mosel Reference Manual, available at
http://www.dashoptimization.com/applications_book.html
From the same address, individual manuals for the Mosel modules can also be downloaded.

5.1.2

Using Xpress-IVE
Xpress-IVE, sometimes called just IVE, is the Xpress Interactive Visual Environment, a complete modeling and optimization development environment running under Microsoft Windows. It presents Mosel
in an easy-to-use Graphical User Interface (GUI), with a built-in text editor. IVE can be used for the development, management and execution of multiple models and is ideal for developing and debugging
prototype models.
To execute the model file chess.mos you need to carry out the following steps.

The basics of Xpress-MP

56

Applications of optimization with Xpress-MP

• Start up IVE.
• Open the model file by choosing File > Open. The model source is then displayed in the central
window (the IVE Editor).
• Click the Run button (green triangle) or alternatively, choose Build > Run. The resulting screen
display is shown in Figure 5.1.
The Build pane at the bottom of the workspace is automatically displayed when compilation starts. If
syntax errors are found in the model, they are displayed here, with details of the line and character
position where the error was detected and a description of the problem, if available. Clicking on theerror takes the user to the offending line.
When a model is run, the Output/Input pane at the right hand side of the workspace window is selected
to display program output. Any output generated by the model is sent to this window. IVE will also provide graphical representations of how the solution is obtained, which are generated by default whenevera problem is optimized. The right hand window contains a number of panes for this purpose, dependent
on the type of problem solved and the particular algorithm used. IVE also allows the user to draw graphs
by embedding subroutines in Mosel models (see the documentation on the website for further detail).
IVE makes all information about the solution available through the Entities pane in the left hand window.
By expanding the list of decision variables in this pane and hovering over one with the mouse pointer, its
solution and reduced cost are displayed. Dual and slack values for constraints may also be obtained.

Figure 5.1: Xpress-IVE display after running model chess.mos

5.2

Modeling with Mosel
Let us now consider a second, slightly larger model which represents the problem faced by a burglar. With
the help of this model we shall explain the basic features of the Mosel language that are used repeatedly
in the implementation of the example problems in the following chapters.

5.2.1

The burglar problem
A burglar sees eight items, of different values and weights. He wants to take the items of greatest total
value whose total weight is not more than the maximum WTMAX he can carry.
We introduce binary variables takei for all i in the set of all items (ITEMS) to represent the decision
whether item i is taken or not. takei has the value 1 if item i is taken and 0 otherwise. Furthermore, let

The basics of Xpress-MP

57

Applications of optimization with Xpress-MP

VALUEi be the value of item i and WEIGHTi its weight. A mathematical formulation of the problem isthen given by:
maximizeX

VALUEi · takei (maximize the total value)i∈ITEMSX

WEIGHTi · takei ≤ WTMAX (weight restriction)i∈ITEMS∀i ∈ ITEMS : takei ∈ {0, 1}This problem is an example of a knapsack problem. It may be implemented with Mosel as follows:model "Burglar 1"
uses "mmxprs"
declarations
ITEMS = 1..8
WTMAX = 102! Index range for items! Maximum weight allowed

VALUE: array(ITEMS) of real
WEIGHT: array(ITEMS) of real

! Value of items! Weight of items

take: array(ITEMS) of mpvar
end-declarations! 1 if we take item i; 0 otherwise! Item:
1
2
3
4
5
6
7
8
VALUE := [15, 100, 90, 60, 40, 15, 10, 1]
WEIGHT:= [ 2, 20, 20, 30, 40, 30, 60, 10]! Objective: maximize total value
MaxVal:= sum(i in ITEMS) VALUE(i)*take(i)! Weight restriction
sum(i in ITEMS) WEIGHT(i)*take(i) <= WTMAX! All variables are 0/1
forall(i in ITEMS) take(i) is_binary
maximize(MaxVal)! Solve the MIP-problem! Print out the solution
writeln("Solution:\n Objective: ", getobjval)
forall(i in ITEMS) writeln(" take(", i, "): ", getsol(take(i)))
end-model

When running this model we get the following output:
Solution:Objective: 280
take(1): 1
take(2): 1
take(3): 1
take(4): 1
take(5): 0
take(6): 1
take(7): 0
take(8): 0

The structure of this model and Mosel models in general is the following:• Model: Every Mosel program starts with the keyword model, followed by a name, and terminateswith end-model.
• Declarations: All objects must be declared in a declarations block, unless they are defined unambiguously through an assignment (e.g. i:=1 defines i as an integer and assigns it the value 1;
in our example the objective function MaxVal is defined by assigning it a linear expression).
There may be several such declarations blocks at different places in a model.
• Problem definition: Typically, a model starts with the specification of the data (here: assignment ofvalues to VALUE and WEIGHT), followed by the statement of the problem (here: definition of theobjective function MaxVal, definition of one inequality constraint, and restricting the variables tobe binaries)

The basics of Xpress-MP

58

Applications of optimization with Xpress-MP

• Solving: With the procedure maximize, we call Xpress-Optimizer to maximize the objective function MaxVal. Since there is no ‘default’ solver in Mosel, we specify that Xpress-Optimizer is to be
used with the statement uses "mmxprs" at the beginning of the program.
• Output printing: The last two lines print out the value of the optimal solution and the solution
values for the decision variables.
• Line breaks: It is possible to place several statements on a single line, separating them by semicolons (like x1 <= 4; x2 >= 7). Conversely, since there are no special ‘line end’ or continuation
characters, every line of a statement that continues over several lines must end with an operator(+, >= etc.) or characters like , that make it obvious that the statement is not terminated.
• Comments: As shown in the example, the symbol ! signifies the start of a comment, which continues to the end of the line. Comments over multiple lines start with (! and terminate with!).
We shall now explain certain features used in this model in more detail:• Ranges and sets:
ITEMS = 1..8

defines a range set, that is, a set of consecutive integers from 1 to 8. This range is used as an index
set for the data arrays (VALUE and WEIGHT) and for the array of decision variables take.
Instead of a using numerical indices, we could, for instance, have defined ITEMS as a set of strings
by replacing the current definition ITEMS = 1..8 with the following definition (without making
any other changes to the model):ITEMS = {"camera", "necklace", "vase", "picture", "tv", "video","chest", "brick"}! Index set for items• Arrays:
VALUE: array(ITEMS) of real

defines a one-dimensional array of real values indexed by the range ITEMS.
Multi-dimensional arrays are declared in the obvious way e.g.
VAL3: array(ITEMS, 1..20, ITEMS) of realdeclares a 3-dimensional real array. Arrays of decision variables (type mpvar) are declared likewise,
as shown in our example.
All objects (scalars and arrays) declared in Mosel are always initialized with a default value:
real, integer: 0
boolean: false
string: ’’ (i.e. the empty string)
The values of data arrays may either be assigned in the model as we show in the example orinitialized from file (see Section 5.2.2).
• Summations:
MaxVal:= sum(i in Items) VALUE(i)*x(i)defines a linear expression called MaxVal as the sum

X

VALUEi · xii∈Items• Simple Looping:
forall(i in ITEMS) take(i) is_binary

illustrates looping over all values in an index range. Recall that the index range ITEMS is 1, ..., 8, so
the statement says that take(1), take(2), ..., take(8) are all binary variables.
There is another example of the use of forall at the penultimate line of the model when writingout all the solution values.
Other types of loops are used in some of the application examples (see the classification tables atthe beginning of Part II).
• Integer Programming variable types:
To make an mpvar variable, say variable xbinvar, into a binary (0/1) variable, we just have to say
xbinvar is_binary

To make an mpvar variable an integer variable, i.e. one that can only take on integral values in a
MIP problem, we would havexintvar is_integer

The basics of Xpress-MP

59

Applications of optimization with Xpress-MP

5.2.2

Reading data from text files
The following example illustrates how data may be read into tables from text files. In the Burglar problem
instead of having the item data embedded in the model file we have the data in a file. We might havethe following Mosel model in a file burglar2.mos.
model "Burglar 2"
uses "mmxprs"
declarations
ITEMS: set of string
WTMAX = 102
VALUE: array(ITEMS) of real
WEIGHT: array(ITEMS) of real
end-declarations! Set of items! Maximum weight allowed! Value of items! Weight of items

initializations from ’burglar.dat’
VALUE
WEIGHT
end-initializations
declarations
take: array(ITEMS) of mpvar
end-declarations! 1 if we take item i; 0 otherwise! Objective: maximize total value
MaxVal:= sum(i in ITEMS) VALUE(i)*take(i)! Weight restriction
sum(i in ITEMS) WEIGHT(i)*take(i) <= WTMAX! All variables are 0/1
forall(i in ITEMS) take(i) is_binary
maximize(MaxVal)! Solve the MIP-problem! Print out the solution
writeln("Solution:\n Objective: ", getobjval)
forall(i in ITEMS) writeln(" take(", i, "): ", getsol(take(i)))
end-model

The file burglar.dat contains
VALUE: [("camera") 15 ("necklace") 100 ("vase") 90 ("picture") 60("tv") 40 ("video") 15 ("chest") 10 ("brick") 1]
WEIGHT:[("camera") 2 ("necklace") 20 ("vase") 20 ("picture") 30("tv") 40 ("video") 30 ("chest") 60 ("brick") 10]

The initializations block tells Mosel where to get data from to initialize sets and arrays. The orderof the data items in the file does not have to be the same as that in the initializations block. Notethat the contents of the set ITEMS is defined indirectly through the index values of the arrays VALUE andWEIGHT. We only declare the variables once the data has been initialized and hence, the set ITEMS is
known.
In the application examples, where appropriate, we show how to work with dynamic arrays of data anddecision variables (see the classification tables at the beginning of Part II).
The data may also be given in the form of a single record, say, KNAPSACK. The initialization then takes
the following form:
initializations from ’burglar2.dat’[VALUE, WEIGHT] as ’KNAPSACK’
end-initializations

and the data file burglar2.dat has the following contents:
KNAPSACK: [ ("camera")[ 15 2]("necklace") [100 20]

The basics of Xpress-MP

60

Applications of optimization with Xpress-MP

("vase")("picture")("tv")("video")("chest")("brick")[ 90 20][ 60 30][ 40 40][ 15 30][ 10 60][ 1 10] ]In the examples of this book we always read data from text files. However, with Mosel it is also possible toread and write data from/to other sources (such as spreadsheets and databases) or input data in memory.
For further information, the reader is referred to the documentation on the website.

5.2.3

Reserved words
The following words are reserved in Mosel. The upper case versions are also reserved (i.e. AND and andare keywords but not And). Do not use them in a model except with their built-in meaning.
and, array, as
boolean, breakcasedeclarations, div, do, dynamic
elif, else, end
false, forall, forward, from, function
if, in, include, initialisations, initializations, integer, inter,
is_binary, is_continuous, is_free, is_integer, is_partint, is_semcont,is_semint, is_sos1, is_sos2
linctrmax, min, mod, model, mpvar
next, notof, options, orparameters, procedure, public, prod
range, real, repeat
set, string, sumthen, to, true
union, until, uses
while

The basics of Xpress-MP

61

Applications of optimization with Xpress-MP

Chapter 6

Mining and process industries
This chapter is about fairly simple linear programming problems where one tries to mix or extract ingredients from raw materials subject to quality constraints with the objective of minimizing the total
cost. These blending or product mix problems are typical for process industries that uses big quantities
of product in an almost continuous manner: refineries, chemical, metallurgical, and farm-produce industries. These problems are relatively simple because they deal with fractional quantities, so Mixed Integer
Programming is not required.
We are going to study three problems of this type: the production of alloys in the metallurgical industry(Section 6.1), animal food production (Section 6.2), and refining of petrol (Section 6.3). These problems
concern the primary sector, that is, economical activities related to the production of raw materials like
agriculture or mining. In addition to the blending problems, this chapter also presents some other problems that belong to this industrial sector: the treatment of cane sugar lots that degrade quickly throughfermentation (Section 6.4), the exploitation of an opencast mine (Section 6.5), and the planning of electricity production by a set of power generators (Section 6.6).

6.1

Production of alloys
The company Steel has received an order for 500 tonnes of steel to be used in shipbuilding. This steel
must have the following characteristics (‘grades’).
Table 6.1: Characteristics of steel ordered
Chemical element

Minimum grade

Maximum grade

Carbon (C)Copper (Cu)Manganese (Mn)

2
0.4
1.2

3
0.6
1.65

The company has seven different raw materials in stock that may be used for the production of this steel.
Table 6.2 lists the grades, available amounts and prices for all raw materials.
Table 6.2: Raw material grades, availabilities, and prices
Raw material

C%

Cu %

Mn %

Availability in t

Cost in B
C/tIron alloy 1
Iron alloy 2
Iron alloy 3
Copper alloy 1
Copper alloy 2
Aluminum alloy 1
Aluminum alloy 2

2.5
3
0
0
0
0
0

0
0
0.3
90
96
0.4
0.6

1.3
0.8
0
0
4
1.2
0

400
300
600
500
200
300
250

200
250
150
220
240
200
165

The objective is to determine the composition of the steel that minimizes the production cost.

6.1.1

Model formulation
We use RAW to represent the set of raw materials and COMP for the set of components (chemical elements) that are relevant for the grade requirements. We want to determine the quantity user of every

62

Applications of optimization with Xpress-MP

raw material r that is used to produce the quantity produce of steel to satisfy the given demand DEM.
We denote by Prc the percentage of the chemical element c in raw material r and COSTr the buying priceper kg of r. Furthermore, the minimum and maximum grades PMINc and PMAXc are given for every
component c. We obtain the following mathematical model:Xminimize

COSTr · user

(6.1.1)user(6.1.2)r∈RAW

produce =Xr∈RAW∀c ∈ COMP :X

Prc · user ≥ PMINc · produce(6.1.3)

Prc · user ≤ PMAXC · produce(6.1.4)r∈RAW∀c ∈ COMP :

X
r∈RAW

∀r ∈ RAW : user ≤ AVAILr(6.1.5)produce ≥ DEM(6.1.6)∀r ∈ RAW : user ≥ 0, produce ≥ 0(6.1.7)The objective, given by (6.1.1), is to minimize the total cost that is calculated as the sum of the raw
material prices times the quantity used. The constraint (6.1.2) indicates that the resulting product weightis equal to the sum of the weights of the raw materials used for its production.
The constraints (6.1.3) and (6.1.4) impose the limits on the grade of the final product. They have beenobtained based on the observation that if the steel contains user tonnes of a raw material that contains
Prc percent of a chemical element c, then the quantity of this element in the steel is given by the sum

X

Prc · userfr∈RAW

and the percentage of the element c in the final product is given by the ratio

P
r∈RAW

Prc · user

produce
Adding the lower bound PMINc to express the constraint on the minimum grade of c results in the following relation:P
P · userfr∈RAW rc
≥ PMINc
produceThis relation is non-linear but it can be transformed by multiplying it with produce – and we finally obtainthe linear constraint (6.1.3). The constraint (6.1.4) is derived following the same scheme.
The constraints (6.1.5) make sure that only the available quantities of raw material are used. The constraint (6.1.6) guarantees that the produced amount of steel satisfies the demand. The last set of constraints (6.1.7) establishes the non-negativity condition for all variables of the problem.

6.1.2

Implementation
The algebraic model translates into the following Mosel program. The correspondence between the two
is easy to see. In this model, just like for most others in this book, we read in all data from a separate file.
Note that whilst we simply number the raw materials in the model formulation, we read in their names
after solving the problem to obtain a more readable solution output.
model "A-1 Production of Alloys"
uses "mmxprs"
declarations
COMP = 1..3
RAW = 1..7

! Components (chemical elements)! Raw materials (alloys)

P: array(RAW,COMP) of real
! Composition of raw materials (in percent)
PMIN,PMAX: array(COMP) of real ! Min. & max. requirements for components
AVAIL: array(RAW) of real
! Raw material availabilities
COST: array(RAW) of real
! Raw material costs per tonne
DEM: real! Amount of steel to produce
use: array(RAW) of mpvar

Mining and process industries! Quantity of raw mat. used

63

Applications of optimization with Xpress-MP

produce: mpvar
end-declarations! Quantity of steel produced

initializations from ’a1alloy.dat’
P PMIN PMAX AVAIL COST DEM
end-initializations! Objective function
Cost:= sum(r in RAW) COST(r)*use(r)! Quantity of steel produced = sum of raw material usedproduce = sum(r in RAW) use(r)! Guarantee min. and max. percentages of every chemical element
forall(c in COMP) dosum(r in RAW) P(r,c)*use(r) >= PMIN(c)*producesum(r in RAW) P(r,c)*use(r) <= PMAX(c)*produce
end-do! Use raw materials within their limit of availability
forall(r in RAW) use(r) <= AVAIL(r)! Satisfy the demand
produce >= DEM! Solve the problem
minimize(Cost)! Solution printing
declarations
NAMES: array(RAW) of string
end-declarations
initializations from ’a1alloy.dat’
NAMES
end-initializations! Get the names of the alloys

writeln("Total cost: ", getobjval)
writeln("Amount of steel produced: ", getsol(produce))
writeln("Alloys used:")forall(r in RAW)if(getsol(use(r))>0) thenwrite(NAMES(r), ": ", getsol(use(r))," ")end-ifwrite("\nPercentages (C, Cu, Mn): ")forall(c in COMP)
write( getsol(sum(r in RAW) P(r,c)*use(r))/getsol(produce), "%writeln")end-model

The data file a1alloy.dat has the following contents:
P: [2.5 0
1.3
3
0
0.8
0
0.3 0
0
90
0
0
96
4
0
0.4 1.20
0.6 0]! Raw material composition

PMIN: [2 0.4 1.2]
PMAX: [3 0.6 1.65]! Min. & max. requirements

AVAIL: [400 300 600 500 200 300 250]
COST: [200 250 150 220 240 200 165]

! Raw material availabilities! Raw material costs

DEM: 500! Amount of steel to produce

NAMES: ["iron 1" "iron 2" "iron 3" "copper 1" "copper 2""aluminum 1" "aluminum 2"]

The solution printout introduces the if-then statement of Mosel, it may also be used in its forms

Mining and process industries

64

Applications of optimization with Xpress-MP

if-then-else or if-then-elif-then-else. Another observation concerning the solution printout
is that the getsol function may be applied not only to variables but also to linear expressions.

6.1.3

Results
The required 500 tonnes of steel are produced with 400 tonnes of iron alloy 1, 39.776 tonnes of iron alloy
3, 2.761 tonnes of copper alloy 2, and 57.462 tonnes of aluminum alloy 1. The percentages of Carbon,
Copper, and Manganese are 2%, 0.6%, and 1.2% respectively. The total cost of production is B
C 98121.60.

6.2

Animal food production
The company CowFood produces food for farm animals that is sold in two forms: powder and granules.
The raw materials used for the production of the food are: oat, maize and molasses. The raw materials(with the exception of molasses) first need to be ground, and then all raw materials that will form a
product are blended. In the last step of the production process the product mix is either transformed togranules or sieved to obtain food in the form of powder.

MolassesOat
Maize

Granulating

GranulesSieving

PowderBlending

Grinding

Figure 6.1: Animal food production process

Every food product needs to fulfill certain nutritional requirements. The percentages of proteins, lipids
and fibers contained in the raw materials and the required percentages in the final products are listed inTable 6.3.
Table 6.3: Contents of nutritional components in percent
Raw material

Proteins

Lipids

Fiber

Oat
Maize
Molasses

13.6
4.1
5

7.1
2.4
0.3

7
3.7
25

Required contents

≥ 9.5

≥2

≤6

There are limits on the availability of raw materials. Table 6.4 displays the amount of raw material that isavailable every day and the respective prices.
Table 6.4: Raw material availabilities and prices
Raw material

Available amount in kg

Cost in B
C/kg

11900
23500
750

0.13
0.17
0.12

Oat
Maize
Molasses

The cost of the different production steps are given in the following table.
Table 6.5: Production costs in B
C /kg
Grinding

Blending

Granulating

Sieving

0.25

0.05

0.42

0.17

With a daily demand of nine tonnes of granules and twelve tonnes of powder, which quantities of raw
materials are required and how should they be blended to minimize the total cost?

Mining and process industries

65

Applications of optimization with Xpress-MP

6.2.1Model formulation
This model is very similar to the preceding except that we now want to produce two products and thatinstead of the minimum and maximum grades we have to fulfill certain nutritional requirements. Let
FOOD = {1, 2} (1 = granules, 2 = powder) be the set of food product types that are produced, RAW
the set of raw materials, and COMP = {1, 2, 3} the set of nutritional components (with 1 = protein, 2 =lipids, 3 = fiber). We further use COSTr to denote the price per kg of raw material r, PCOSTp for the costof production process p in B
C /kg, REQc for the required content of nutritional component c, Prc for thecontent of c in raw material r, AVAILr for the maximum available quantity of raw material r, and DEMf
for the daily demand of food product f.
With variable userf representing the quantity of raw material r used for the production of food type f
and producef the amount of food f produced, we obtain the following model:
minimize

X

X

COSTr · userf +r∈RAW f∈FOOD

X

+

X

X

X

r∈RAW

f∈FOOD
r6=molasses

PCOSTblending · userf +r∈RAW f∈FOOD

X

+

X

PCOSTgrinding · userf

PCOSTgranulating · user1r∈RAW(6.2.1)

PCOSTsieving · user2r∈RAW∀f ∈ FOOD :X

userf = producef(6.2.2)r∈RAW∀f ∈ FOOD, c ∈ 1. . 2 :X

Prc · userf ≥ REQc · producef(6.2.3)r∈RAW∀f ∈ FOOD :X

Pr3 · userf ≤ REQ3 · producef(6.2.4)userf ≤ AVAILr(6.2.5)r∈RAW∀r ∈ RAW :X
f∈FOOD∀f ∈ FOOD : producef ≥ DEMf

(6.2.6)∀r ∈ RAW, f ∈ FOOD : userf ≥ 0, ∀f ∈ FOOD : producef ≥ 0(6.2.7)The objective function (6.2.1) that is to be minimized is the sum of all costs. The first term is the price paid
for the raw materials, the following terms correspond to the production costs of the different process
steps. Grinding is applied to all raw materials except molasses, blending to all raw material. Granules(f = 1) are obtained through granulating and powder (f = 2) by sieving the blended product. The
constraint (6.2.2) expresses the fact that the produced quantity of every food type corresponds to thesum of the raw material used for its production. Through the constraints (6.2.3) and (6.2.4) we obtain therequired content of nutritional components. They are based on the same reasoning that we saw for the‘grades’ in the previous example. The percentage for the content of a nutritional component c in a foodproduct f is given by the following ratio:

P
r∈RAW

Prc · userf

producef(6.2.8)This percentage must be at least REQc for the first two components (protein, lipid) and at most REQc forthe third one (fiber). For the first two, this results in a relation of type (6.2.9), for the third in the relation(6.2.10).

Pr∈RAW

Prc · userf

producef
P
P · userfr∈RAW rc
producef≥ REQc(6.2.9)≤ REQc(6.2.10)

Both relations are non-linear since the variables producef appear in the denominator, but as before it ispossible to transform them into linear constraints by multiplying them by producef . Doing so, we obtainthe constraints (6.2.3) and (6.2.4).
The constraints (6.2.5) guarantee that the quantities of raw material used remain within their limits ofavailability. The constraints (6.2.6) serve to satisfy the demand. And the last set of constraints (6.2.7)
establishes that all variables are non-negative.

Mining and process industries

66

Applications of optimization with Xpress-MP

6.2.2

Implementation
Below follows the implementation of this problem with Mosel. Note that in this implementation some of
the indexing sets are of numerical type and others are sets of strings. Any indexing set may be assigned
its contents at its declaration (a constant set), like FOOD and COMP, or it may be created dynamically asis the case for for the (unnamed) indexing set of PCOST that is filled in when this array is read from file.
model "A-2 Animal Food Production"
uses "mmxprs"
declarations
FOOD = 1..2
! Food types
COMP = 1..3
! Nutritional components
RAW = {"oat", "maize", "molasses"}! Raw materials
P: array(RAW,COMP) of real
! Composition of raw materials (in percent)
REQ: array(COMP) of real
! Nutritional requirements
AVAIL: array(RAW) of real
! Raw material availabilities
COST: array(RAW) of real
! Raw material prices
PCOST: array(set of string) of real ! Cost of processing operations
DEM: array(FOOD) of real
! Demands for food types
use: array(RAW,FOOD) of mpvar
produce: array(FOOD) of mpvar
end-declarations! Quantity of raw mat. used for a food type! Quantity of food produced

initializations from ’a2food.dat’
P REQ PCOST DEM[AVAIL, COST] as ’RAWMAT’
end-initializations! Objective function
Cost:= sum(r in RAW,f in FOOD) COST(r)*use(r,f) +sum(r in RAW,f in FOOD|r<>"molasses") PCOST("grinding")*use(r,f) +sum(r in RAW,f in FOOD) PCOST("blending")*use(r,f) +sum(r in RAW) PCOST("granulating")*use(r,1) +sum(r in RAW) PCOST("sieving")*use(r,2)! Quantity of food produced corresponds to raw material used
forall(f in FOOD) sum(r in RAW) use(r,f) = produce(f)! Fulfill nutritional requirements
forall(f in FOOD,c in 1..2)sum(r in RAW) P(r,c)*use(r,f) >= REQ(c)*produce(f)forall(f in FOOD) sum(r in RAW) P(r,3)*use(r,f) <= REQ(3)*produce(f)! Use raw materials within their limit of availability
forall(r in RAW) sum(f in FOOD) use(r,f) <= AVAIL(r)! Satisfy demands
forall(f in FOOD) produce(f) >= DEM(f)! Solve the problem
minimize(Cost)! Solution printing
writeln("Total cost: ", getobjval)
write("Food type"); forall(r in RAW) write(strfmt(r,9))
writeln(" protein lipid
fiber")
forall(f in FOOD) do
write(strfmt(f,-9))
forall(r in RAW) write(strfmt(getsol(use(r,f)),9,2))
forall(c in COMP) write("",
strfmt(getsol(sum(r in RAW) P(r,c)*use(r,f))/getsol(produce(f)),3,2),"%")
writeln
end-doend-model

The data file a2food.dat has the following contents. Whereas strings in Mosel model files must always
be surrounded by single or double quotes, in data files this is only required if a string contains blanks orother non-alphanumeric characters, or starts with a numerical value.

Mining and process industries

67

Applications of optimization with Xpress-MP

Similarly to the way the raw material availabilities and prices are listed in Table 6.4, the data for the arrays
AVAIL and COST are given in a single record labeled RAWMAT. In the initializations block, the namesof the data arrays that are to be read from a single record need to be surrounded by [ and ], followedby the keyword as and the label (name) of the record.
P: [(oat 1)
13.6 7.1 7(maize 1)
4.1 2.4 3.7(molasses 1) 5
0.3 25 ]! Composition of raw materials

REQ: [9.5 2 6]

! Nutritional requirements

RAWMAT: [(oat)[11900 0.13](maize)[23500 0.17](molasses) [750 0.12]]! Raw material availabilities and prices! Cost of processing operations
PCOST: [(grinding) 0.25 (blending) 0.05 (granulating) 0.42 (sieving) 0.17]
DEM: [9000 12000]! Demands for food types

In this example we again show how the solution may be printed after the problem has been solved. The
output is printed in table format with the help of strfmt that for a given string (first parameter) reserves
the indicated space (second parameter) or for a number the total space and the number of digits after
the decimal point (third parameter). If the second parameter has a negative value, the output is printed
left justified.
Since there is only little variation in the solution printing for the large majority of examples, it will beomitted in the listings from now on.

6.2.3

Results
The minimum cost for producing the demanded nine tonnes of granule and twelve tonnes of powder is
B
C 15086.80. The composition of the food products is displayed in Table 6.6.
Table 6.6: Optimal composition of food products

6.3

Food type

Oat

Maize

MolassesProtein

Lipid

Fiber

GranulePowder5098.566798.073719.534959.37181.91
242.55

9.50%
9.50%

5.02%
5.02%

6.00%
6.00%

RefineryA refinery produces butane, petrol, diesel oil, and heating oil from two crudes. Four types of operations
are necessary to obtain these products: separation, conversion, upgrading, and blending.
The separation phase consists of distilling the raw product into, among others, butane, naphtha, gasoil,
and a residue. The residue subsequently undergoes a conversion phase (catalytic cracking) to obtainlighter products. The different products that come out of the distillation are purified (desulfurizationor sweetening) or upgraded by a reforming operation that augments their octane value. Finally, toobtain the products that will be sold, the refinery blends several of the intermediate products in order tofulfill the prescribed characteristics of the commercial products. The following drawing gives a simplified
overview on the production processes in this refinery.
After the distillation, crude 1 gives 3% butane, 15% naphtha, 40% gasoil, and 15% residue. Crude 2
results in 5% butane, 20% naphtha, 35% gasoil, and 10% residue. The reforming of the naphtha gives
15% butane and 85% of reformate (reformed naphtha). The catalytic cracking of the residue results in
40% of cracked naphtha and 35% of cracked gasoil (note that these percentages do not add up to 100%
because the process also produces 15% of gas, 5% coke and another type of residue that are not taken
into consideration in our example). The petrol is produced with three ingredients: reformed naphtha(reformate), butane, and cracked naphtha. The diesel oil is obtained by blending sweetened gasoil,
cracked gasoil, and cracked naphtha. The heating oil may contain gasoil and cracked naphtha without
any restrictions on their proportions.
Certain conditions on the quality of the petrol and diesel oil are imposed by law. There are three important characteristics for petrol: the octane value, vapor pressure and volatility. The octane value is a

Mining and process industries

68

Applications of optimization with Xpress-MP

ButaneButaneButaneNaphthaReforming

Reformate

CrudesPetrol blending

Petrol

HO blending

Heating oil

Diesel blending

Diesel oil

Crk. naphtha

ResidueCat. cracking

CrackedgasoilGasoilDesulfurization
Sweetened gasoil

Distillation
Figure 6.2: Simplified representation of a refinery

measure of the anti-knock power in the motor. The vapor pressure is a measure of the risk of explosion
during storage, especially with hot weather. The volatility is a measure for how easy the motor is started
during cold weather. Finally, the maximum sulfur content of the diesel oil is imposed by antipollution
specifications. The following table summarizes the required characteristics of the final products and thecomposition of the intermediate ones. Fields are left empty if no particular limit applies. We work withthe assumption that all these characteristics blend linearly by weight (in reality, this is only true for thesulfur contents).
Table 6.7: Characteristics of intermediate and final products
Characteristic

ButaneReformate

CrackednaphthaCrackedgasoilDesulfurizedgasoilPetrolDieseloil

Octane value
Vapor pressure
Volatility

120
60
105

100
2.6
3

74
4.1
12

–
–
–

–
–
–

≥ 94
≤ 12.7≥ 17–––

Sulfur (in %)

–

–

0.12

0.76

0.03

–

≤ 0.05In the next month the refinery needs to produce 20,000 tonnes of butane, 40,000 tonnes of petrol, 30,000
tonnes of diesel oil, and 42,000 tonnes of heating oil. 250,000 tonnes of crude 1 and 500,000 tonnes ofcrude 2 are available. The monthly capacity of the reformer are 30,000 tonnes, for the desulfurization
40,000 tonnes and for the cracking 50,000 tonnes. The cost of processing is based on the use of fuel andcatalysts for the different operations: the costs of distillation, reforming, desulfurization, and cracking
are B
C 2.10, B
C 4.18, B
C 2.04 and B
C 0.60 per tonne respectively.

6.3.1

Model formulation
Let producebutane , producepetrol , producediesel , and produceheating be the quantities to produce of butane, petrol, diesel oil, and heating oil. These four products form the set FINAL of final products.
We need to determine the composition of each of these products. We shall use the following variables for the intermediate products: producepetbutane , producereformate , producepetcrknaphtha for the quantities of butane, reformate and cracked naphtha used in the production of petrol (the three products
are grouped in the set IPETROL); similarly, producedslgasoil , producedslcrknaphtha , and producedslcrkgasoil forthe quantities of sweetened gasoil, cracked naphtha, and cracked gasoil used in the production ofdiesel oil (set IDIESEL) and producehogasoil , producehocrknaphtha , and producehocrkgasoil for the quantities ofsweetened gasoil, cracked naphtha, and cracked gasoil blended to heating oil (set IHO). We also define production variables producep for the quantities of intermediate products resulting from the different production processes, namely for the sets IDIST = {distbutane, naphtha, residue, gasoil} (products resulting from distillation), IREF = {refbutane, reformate} (products resulting from reforming), andICRACK = {crknaphtha, crkgasoil} (products resulting from cracking). All final and intermediate products
are regrouped in the set ALLPRODS = FINAL ∪ IDIST ∪ IREF ∪ ICRACK ∪ IPETROL ∪ IHO ∪ IDIESEL.
In addition to the production variables, we define variables usec for the quantity of crude c ∈ CRUDES
used in the production.
Our objective is to minimize the production costs, that is, the cost of distilling the crudes plus the cost ofreforming, desulfurizing or cracking the intermediate products resulting from the distillation. Hence theMining and process industries

69

Applications of optimization with Xpress-MP

objective function is:Xminimize

COSTc · usec +c∈CRUDESX

COSTp · producep(6.3.1)p∈IDIST

The following constraints (6.3.2) establish restrictions on the maximum quantities of every component of
the raw materials, where DISTcp stands for the contents of the intermediate product p of the crude c.

X∀p ∈ IDIST : producep ≤DISTcp · usec

(6.3.2)c∈CRUDES

For instance, the total quantity of naphtha going into the reformer (producenaphthaP ) cannot exceed the
DISTc,naphtha · usec .
maximum quantity of naphtha that is obtained from the distillation of the crudesc∈CRUDES
Similar relations are given for the reforming (6.3.3) and cracking (6.3.4) processes (in both cases, there
is only a single input product) where REFp denotes the percentage composition of naphtha and CRACKp
the percentage composition of the residue:∀p ∈ IREF : producep ≤ REFp · producenaphtha(6.3.3)∀p ∈ ICRACK : producep ≤ CRACKp · produceresidue(6.3.4)The products coming out of the cracking and desulfurization processes may be used for several of the final
products. We therefore have the following three equations. Cracked naphtha is used in the productionof petrol, heating oil, and diesel oil (6.3.5). Cracked gasoil (6.3.6) and sweetened gasoil (6.3.7) are bothused for heating oil and diesel oil.
producecrknaphtha = producepetcrknaphtha + producehocrknaphtha + producedslcrknaphtha(6.3.5)producecrkgasoil = producehocrkgasoil + producedslcrkgasoil(6.3.6)

producegasoil = producehogasoil + producedslgasoil(6.3.7)

The following four constraints (6.3.8) – (6.3.11) state how the quantities of the final products are obtained
from the intermediate products: butane is produced by distillation and reforming and used in the production of petrol (6.3.8). The quantities of the three other final products (petrol, diesel oil, and heatingoil) result from blending as the sum of the intermediate product quantities used in their production.
producebutane = producedistbutane + producerefbutane − producepetbutane

X

producepetrol =(6.3.8)producep(6.3.9)producep(6.3.10)producep(6.3.11)p∈IPETROL

X

producediesel =p∈IDIESEL

X

produceheating =p∈IHO

Furthermore, we have a set of constraints to obey the legal restrictions on the composition of petrol anddiesel oil. For instance, the octane value of petrol must be better than 94. This constraint translates tothe following relation (with OCTp the octane value of the components):

Pp∈IPETROL

OCTp · producep

producepetrol≥ 94(6.3.12)This constraint is non-linear since it contains a variable in the denominator. To turn it into a linear
constraints, we simply multiply everything by producepetrol . We then obtain the following constraint(6.3.13):
X
OCTp · producep ≥ 94 · producepetrol(6.3.13)p∈IPETROL

The following constraints (6.3.14) – (6.3.16) establish the corresponding relations for the vapor pressureand volatility of petrol and the sulfur contents of diesel oil.

X

VAPp · producep ≤ 12. 7 · producepetrol(6.3.14)

VOLp · producep ≥ 17 · producepetrol(6.3.15)

SULFp · producep ≤ 0. 05 · producediesel(6.3.16)p∈IPETROLXp∈IPETROLXp∈IDIESEL

Mining and process industries

70

Applications of optimization with Xpress-MP

We also need constraints that establish the limits on production capacities for reforming (6.3.17), desulfurization (6.3.18) and cracking (6.3.19). The amount usec of every crude c that is used in production islimited by the available amount AVAILc . The constraints (6.3.21) guarantee that the demand DEMp for
all final products is satisfied. To this system we add the non-negativity condition for all variables (6.3.22)
and (6.3.23).
producenp ≤ 30000(6.3.17)

producegd + producegh ≤ 50000(6.3.18)

producerp + producerd + producerh ≤ 40000(6.3.19)∀c ∈ CRUDES : usec ≤ AVAILc(6.3.20)∀p ∈ FINAL : producep ≥ DEMp

(6.3.21)∀p ∈ ALLPRODS : producep ≥ 0(6.3.22)∀c ∈ CRUDES : usec ≥ 0(6.3.23)

The lines (6.3.1) to (6.3.11), and (6.3.13) to (6.3.23) define the LP model.
Compared to the two other blending problems in Sections 6.1 and 6.2 the formulation of this problem
looks quite ‘messy’: this is typical for production situations where we have a relatively small number of
products going through a large number of stages that are related in complicated ways. There are manydifferent types of constraints, and often lots of exceptions and special cases.

6.3.2

Implementation
The translation of the model in the previous section into the following formulation with Mosel is straightforward.
Contrary to the implementations of the previous examples, this time we do not define the index sets forthe data arrays directly in the model but initialize the contents of the data arrays (and the corresponding
index sets) dynamically from file. The only set defined directly in the model is IHO since it does not appear as an index set for any data array – its definition could also be read from file. The set of all products
ALLPRODS is defined as the union of all final and intermediate products.
model "A-3 Refinery planning"uses "mmxprs"
declarations
CRUDES: set of string! Set of crudes
ALLPRODS: set of string! Intermediate and final products
FINAL: set of string! Final products
IDIST: set of string! Products obtained by distillation
IREF: set of string! Products obtained by reforming
ICRACK: set of string! Products obtained by cracking
IPETROL: set of string! Interm. products for petrol
IDIESEL: set of string! Interm. products for diesel
IHO={"hogasoil", "hocrknaphtha", "hocrkgasoil"}! Interm. products for heating oil
DEM: array(FINAL) of real
! Min. production
COST: array(set of string) of real
! Production costs
AVAIL: array(CRUDES) of real
! Crude availability
OCT, VAP, VOL: array(IPETROL) of real
! Octane, vapor pressure, and! volatility values
SULF: array(IDIESEL) of real
! Sulfur contents
DIST: array(CRUDES,IDIST) of real
! Composition of crudes (in %)REF: array(IREF) of real
! Results of reforming (in %)CRACK: array(ICRACK) of real
! Results of cracking (in %)end-declarations
initializations from ’a3refine.dat’
DEM COST OCT VAP VOL SULF AVAIL DIST REF CRACK
end-initializations
finalize(FINAL); finalize(CRUDES); finalize(IPETROL); finalize(IDIESEL)
finalize(IDIST); finalize(IREF); finalize(ICRACK)ALLPRODS:= FINAL+IDIST+IREF+ICRACK+IPETROL+IHO+IDIESEL
declarations
use: array(CRUDES) of mpvar
produce: array(ALLPRODS) of mpvar

Mining and process industries! Quantities used
! Quantities produced

71

Applications of optimization with Xpress-MP

end-declarations! Objective function
Cost:= sum(c in CRUDES) COST(c)*use(c) + sum(p in IDIST) COST(p)*produce(p)! Relations intermediate products resulting of distillation - raw materials
forall(p in IDIST) produce(p) <= sum(c in CRUDES) DIST(c,p)*use(c)! Relations between intermediate products! Reforming:
forall(p in IREF) produce(p) <= REF(p)*produce("naphtha")! Cracking:forall(p in ICRACK) produce(p) <= CRACK(p)*produce("residue")
produce("crknaphtha") >= produce("petcrknaphtha") +produce("hocrknaphtha") + produce("dslcrknaphtha")
produce("crkgasoil") >= produce("hocrkgasoil") + produce("dslcrkgasoil")! Desulfurization:
produce("gasoil") >= produce("hogasoil") + produce("dslgasoil")! Relations final products - intermediate products
produce("butane") = produce("distbutane") + produce("refbutane") produce("petbutane")produce("petrol") = sum(p in IPETROL) produce(p)produce("diesel") = sum(p in IDIESEL) produce(p)produce("heating") = sum(p in IHO) produce(p)! Properties of petrol
sum(p in IPETROL) OCT(p)*produce(p) >= 94*produce("petrol")sum(p in IPETROL) VAP(p)*produce(p) <= 12.7*produce("petrol")sum(p in IPETROL) VOL(p)*produce(p) >= 17*produce("petrol")! Limit on sulfur in diesel oilsum(p in IDIESEL) SULF(p)*produce(p) <= 0.05*produce("diesel")! Crude availabilities
forall(c in CRUDES) use(c) <= AVAIL(c)! Production capacities
produce("naphtha") <= 30000
produce("gasoil") <= 50000
produce("residue") <= 40000! Reformer! Desulfurization! Cracker! Satisfy demands
forall(p in FINAL) produce(p) >= DEM(p)! Solve the problem
minimize(Cost)
end-model

The dynamic initialization of the index sets necessitates a special treatment of the arrays of variables use
and produce.
If we declared these arrays at the same time as the data arrays, this would result in empty, dynamic arrays
since the index sets are not known before the data has been read from file. Whilst the entries of dynamicarrays of types other than mpvar are created when the entry is addressed, this is not the case for arraysof variables. There are two possibilities for handling this situation:• In the implementation shown above we define the variables in a separate declarations block
after the data has been read from file. This block is preceded by a series of finalize statements.
With the help of the finalize procedure, the (dynamic) index sets are turned into constant sets,that is, their contents cannot be modified any more. As a consequence, all arrays with these sets astheir indices and that are declared after this point are created as arrays of fixed size.
• Alternatively, we may declare the array of type mpvar together with the data arrays and create therequired variables explicitly by adding the following lines after the data has been read from fileand the two sets ALLPROD and CRUDES have been finalized:
forall(p in ALLPROD) create(produce(p))
forall(c in CRUDES) create(use(c))

In larger applications, typically most data (and the corresponding index sets) are initialized dynamicallyso that the variable definition needs to be done with one of the two methods described above.

Mining and process industries

72

Applications of optimization with Xpress-MP

To keep the example implementations in this book as easy as possible, we usually define the index sets asconstants directly in the model so that the definition of variables does not require any special care.

6.3.3

Results
The total production cost is B
C 1,175,400. We produce 40,000 tonnes of petrol from 6,500 tonnes of butane,
25,500 tonnes of reformate, and 8,000 tonnes of cracked naphtha. With this composition, we obtain an
octane value of 98.05, the vapor pressure is 12.23, and the volatility of the petrol is 21.38. The required
30,000 tonnes of diesel oil are produced with 30,000 tonnes of sweetened gasoil, which gives a sulfur
content of 0.03%. The 42,000 tonnes of heating oil are composed of 20,000 tonnes of sweetened gasoil,
8,000 tonnes of cracked naphtha, and 14,000 tonnes of cracked gasoil. In addition to this production wealso have the required 20,000 tonnes of butane.

6.4

Cane sugar production
The harvest of cane sugar in Australia is highly mechanized. The sugar cane is immediately transported
to a sugar house in wagons that run on a network of small rail tracks. The sugar content of a wagon load
depends on the field it has been harvested from and on the maturity of the sugar cane. Once harvested,
the sugar content decreases rapidly through fermentation and the wagon load will entirely lose its value
after a certain time. At this moment, eleven wagons all loaded with the same quantity have arrived at
the sugar house. They have been examined to find out the hourly loss and the remaining life span (inhours) of every wagon, these data are summarized in the following table.
Table 6.8: Properties of the lots of cane sugar
Lot

1

2

3

4

5

6

7

8

9

10

11

Loss (kg/h)Life span (h)

43
8

26
8

37
2

28
8

13
4

54
8

62
8

49
8

19
8

28
8

30
8

Every lot may be processed by any of the three, fully equivalent production lines of the sugar house.
The processing of a lot takes two hours. It must be finished at the latest at the end of the life span of
the wagon load. The manager of the sugar house wishes to determine a production schedule for thecurrently available lots that minimizes the total loss of sugar.

6.4.1

Model formulation
This problem also concerns the primary sector but it is different from the blending problems we haveseen so far in this chapter.
Let WAGONS = {1, . . . , NW} be the set of wagons, NL the number of production lines and DUR the
duration of the production process for every lot. The hourly loss for every wagon w is given by LOSSw
and its life span by LIFEw . We observe that in an optimal solution the production lines need to workwithout any break – otherwise we could reduce the loss in sugar by advancing the start of the lot that
follows the break. This means that the completion time of every lot is of the form s · DUR, with s > 0 and
integer. The maximum value of s is the number of time slots (of length DUR) that the sugar house will
work, namely NS = ceil ( NW / NL ), where ceil stands for ‘rounded to the next larger integer’. If NW / NL
is an integer, every line will process exactly NS lots. Otherwise, some lines will process NS − 1 lots, but atleast one line processes NS lots. In all cases, the length of the optimal schedule is NS · DUR hours. We call
SLOTS = {1, . . . , NS} the set of time slots.
Every lot needs to be assigned to a time slot. We define binary variables processws that take the value 1 if
and only if lot w is assigned to slot s (constraints (6.4.1)). Every lot needs to be assigned to a slot (6.4.2),
and any slot may take up to NL lots because there are NL parallel lines (6.4.3).
∀w ∈ WAGONS, s ∈ SLOTS : processws ∈ {0, 1}∀w ∈ WAGONS :X(6.4.1)processws = 1(6.4.2)

processws ≤ NL

(6.4.3)s∈SLOTS∀s ∈ SLOTS :X
w∈WAGONS

The life span of the sugar lots is given in hours and not in periods of DUR hours. The maximum slot
number for a wagon load w is therefore LIFEw / DUR. The constraints (6.4.4) establish this bound for

Mining and process industries

73

Applications of optimization with Xpress-MP

every lot w. Note that the number of the time slot is expressed through a sum of decision variables: dueto the constraints (6.4.2) only the index of the slot s for which processws = 1 will be counted for lot w.

X∀w ∈ WAGONS :s · processws ≤ LIFEw / DUR(6.4.4)s∈SLOTSThe loss of sugar per wagon load w and time slot s is s · DUR · LOSSw . The objective function (total loss ofsugar) is therefore given by the following expression (6.4.5):

X

minimizeX

s · DUR · LOSSw · processws(6.4.5)w∈WAGONS s∈SLOTSThe lines (6.4.1) to (6.4.5) define the model for our problem. This linear problem is a minimum cost flow
problem of a particular type. It is related to the transportation problems (like the car rental problem
in Chapter 10). For minimum cost flow problems, the optimal LP solution calculated by the simplex
algorithm is integer feasible. We could therefore replace the integrality constraints in our model by
non-negativity constraints.

6.4.2

Implementation
The following model is the Mosel implementation of the linear problem. It uses the function ceil tocalculate the maximum number of time slots.
model "A-4 Cane sugar production"
uses "mmxprs"
declarations
NW = 11
NL = 3
WAGONS = 1..NW
SLOTS = 1..ceil(NW/NL)
LOSS: array(WAGONS) of real
LIFE: array(WAGONS) of real
DUR: integer! Number of wagon loads of sugar! Number of production lines! Time slots for production! Loss in kg/hour! Remaining time per lot! Duration of the production

process: array(WAGONS,SLOTS) of mpvar! 1 if wagon processed in slot,! 0 otherwise

end-declarations
initializations from ’a4sugar.dat’
LOSS LIFE DUR
end-initializations! Objective function
TotalLoss:= sum(w in WAGONS, s in SLOTS) s*DUR*LOSS(w)*process(w,s)! Assignment
forall(w in WAGONS) sum(s in SLOTS) process(w,s) = 1! Wagon loads per time slot
forall(s in SLOTS) sum(w in WAGONS) process(w,s) <= NL! Limit on raw product life
forall(w in WAGONS) sum(s in SLOTS) s*process(w,s) <= LIFE(w)/DUR
forall(w in WAGONS, s in SLOTS) process(w,s) is_binary! Solve the problem
minimize(TotalLoss)

6.4.3

Results
If we solve the problem without the constraints limiting the life span of the wagon loads we obtain aminimum loss of 1518 kg of sugar. In this case the lots are processed in the order of decreasing loss. With
the complete set of constraints we obtain a total loss of 1620 kg of sugar. The corresponding schedule oflots is shown in the following table (there are several equivalent solutions).

Mining and process industries

74

Applications of optimization with Xpress-MP

Table 6.9: Optimal schedule for the cane sugar lots

6.5

Slot 1

Slot 2

Slot 3

Slot 4lot 3 (74 kg)
lot 6 (108 kg)
lot 7 (124 kg)

lot 1 (172 kg)
lot 5 (52 kg)
lot 8 (196 kg)

lot 9 (114 kg)
lot 10 (168 kg)
lot 11 (180 kg)

lot 2 (208 kg)
lot 4 (224 kg)Opencast mining
An opencast uranium mine is being prospected. Based on the results of some test drillings the mine has
been subdivided into exploitation units called blocks. The pit needs to be terraced to allow the trucks todrive down to its bottom. The uranium deposit extends from east to west. The pit is limited in the west
by a village and in the east by a group of mountains. Taking into account these constraints, 18 blocks of
10,000 tonnes on three levels have been identified (Figure 6.3). To extract a block, three blocks of thelevel above it need to be extracted: the block immediately on top of it, and also, due to the constraints
on the slope, the blocks to the right and to the left.

VillageMountainsLevel 1
Level 2

1

2

3

4

5

6

7

9

10

11

12

13

14

15

16

17

18

Level 38

Figure 6.3: Length cut through the opencast mine

It costs B
C 100 per tonne to extract a block of level 1, B
C 200 per tonne for a block of level 2, and B
C 300 pertonne for a block of level 3, with the exception of the hatched blocks that are formed of a very hard rockrich in quartz and cost B
C 1000 per ton. The only blocks that contain uranium are those displayed in a grayshade (1, 7, 10, 12, 17, 18). Their market value is 200, 300, 500, 200, 1000, and B
C 1200/tonne respectively.
Block 18, although rich in ore, is made of the same hard rock as the other hatched blocks. Which blocks
should be extracted to maximize the total benefit?

6.5.1

Model formulation
Let BLOCKS be the set of blocks, VALUEb the value per tonne of a block b, and COSTb the cost per tonneof extracting it. The benefit of extracting block b per tonne is then given by VALUEb −COSTb . The possible
extraction sequences of blocks can be represented by a directed graph G = (BLOCKS, ARCS), where ARCS
stands for the set of arcs between blocks. An arc (b, a) means that block b can only be extracted if blocka has already been taken. For instance, block 16 gives rise to three arcs in G: (16,10), (16,11), and (16,12)
and blocks 10, 11, and 12 again to three arcs each, as shown in the following graph.
To decide which blocks are extracted we introduce binary variables extractb that take the value 1 if andonly if block b is extracted. We obtain the following, very compact problem with 0-1 variables:
maximizeX(VALUEb − COSTb ) · extractb(6.5.1)b∈BLOCKS∀(b, a) ∈ ARCS : extractb ≤ extracta(6.5.2)∀b ∈ BLOCKS : extractb ∈ {0, 1}(6.5.3)The objective function (6.5.1) that is to be maximized is the sum of the benefits from the blocks thatare extracted. The constraints (6.5.2) make sure that the blocks are extracted in the right order: if b is
extracted (extractb = 1) then a also needs to be extracted (extracta = 1). The constraints (6.5.3) define thebinary variables.

Mining and process industries

75

Applications of optimization with Xpress-MP

2

16

10

3

11

4

12

5
6

Figure 6.4: Precedence graph for extraction of block 16

It would be possible to replace the constraints (6.5.2) by their aggregate form (6.5.4). In large problems that also include many other variables and constraints this formulation may help reduce the total
number of constraints. This reformulation provides a weaker, i.e. relaxed, formulation than the original
constraints. In our example this has the disadvantage that whilst for the original formulation the solutionof the linear relaxation is already integer, this is no longer true for the relaxed formulation.
∀b ∈ BLOCKS, ∃(b, a) ∈ ARCS : 3 · extractb ≤X

extracta(6.5.4)a∈BLOCKS∃(b,a)∈ARCS

6.5.2

Implementation
The following model implements the LP of the previous section with Mosel. The graph is coded as a
two-dimensional array that for every block of levels 2 and 3 contains its three predecessors in the level
above it. For instance for block 16, it contains the line (16 1) 10 11 12 which is the short form of (161) 10 (16 2) 11 (16 3) 12.
model "A-5 Opencast mining"uses "mmxprs"
declarations
BLOCKS = 1..18! Set of blocks
LEVEL23: set of integer! Blocks in levels 2 and 3
COST: array(BLOCKS) of real
! Exploitation cost of blocks
VALUE: array(BLOCKS) of real
! Value of blocks
ARC: array(LEVEL23,1..3) of integer ! Arcs indicating order of extraction
extract: array(BLOCKS) of mpvar
end-declarations! 1 if block b is extracted

initializations from ’a5mine.dat’
COST VALUE ARC
end-initializations! Objective: maximize total profit
Profit:= sum(b in BLOCKS) (VALUE(b)-COST(b))* extract(b)! Extraction orderforall(b in LEVEL23)
forall(i in 1..3) extract(b) <= extract(ARC(b,i))
forall(b in BLOCKS) extract(b) is_binary! Solve the problem
maximize(Profit)
end-model

As mentioned earlier, for the model formulation we are using in the implementation the solution to
the linear problem is integer. As a general rule, it may be helpful not to declare any special types of

Mining and process industries

76

Applications of optimization with Xpress-MPvariables at the first attempt at solving a problem: the problem studied may have a specific structure
for which the integrality constraints are redundant (this is the case for the present example) or for acertain data set all variables may be integral in the optimal LP solution. Switching integrality conditions
off without removing them from the model can also be obtained by adding a parameter to the call tothe optimization algorithm. For instance in this example
maximize(XPRS_LIN,Profit)

Another reason not to establish immediately all integrality constraints may be size limitations in the
optimization software that is used (the student version of Xpress-Optimizer is limited to 800 variables +constraints out of which at most 500 may be MIP variables or SOS sets). A problem that defines too many
integer variables cannot be executed, even if the optimal integer solution could be obtained by solving
the LP.

6.5.3

Results
The maximal profit is B
C 4,000,000. Blocks 1 to 7, 10 to 13, and 17 are extracted. Blocks 8, 9, 14 to 16, and18 remain in the pit.

6.6

Production of electricity
Power generators of four different types are available to satisfy the daily electricity demands (in megawatts) summarized in the following table. We consider a sliding time horizon: the period 10pm-12am of
day d is followed by the period 0am-6am of day d + 1.
Table 6.10: Daily electricity demands (in MW)
Period0am–6am

6am–9am

9am–12pm

12pm–2pm

2pm–6pm

6pm–10pm

10pm–12am

Demand

12000

32000

25000

36000

25000

30000

18000

The power generators of the same type have a maximum capacity and may be connected to the network
starting from a certain minimal power output. They have a start-up cost, a fixed hourly cost for working
at minimal power, and an hourly cost per additional megawatt for anything beyond the minimal output.
These data are given in the following table.
Table 6.11: Description of power generators

Type 1
Type 2Type 3
Type 4

Available
numberMin. output
in MW

Max. capacity
in MW

Fix cost
B
C/hAdd. MW costB
C/hStart-up
cost

10
4
8
3

750
1000
1200
1800

1750
1500
2000
3500

2250
1800
3750
4800

2.7
2.2
1.8
3.8

5000
1600
2400
1200

A power generator can only be started or stopped at the beginning of a time period. As opposed to thestart, stopping a power plant does not cost anything. At any moment, the working power generators
must be able to cope with an increase by 20% of the demand forecast. Which power generators should
be used in every period in order to minimize the total daily cost?

6.6.1

Model formulation
Let TIME = {1, . . . , NT} be the set of time periods per day and TYPES the set of generator types. For agiven time period t, DEMt denotes the electricity demand in the network and LENt the length (number ofhours) of the period. For any power generator type p, PMINp and PMAXp are the minimum and maximum
capacity respectively, and AVAILp is the available number of generators. The start-up cost is CSTARTp , thehourly cost for working at minimum level is CMINp , and the cost per additional MW hour is CADDp .
To formulate this model and account for the different types of cost we need three sets of variables. startpt
is the integer number of power generators of type p that start working at the beginning of time periodt (6.6.1). workpt is the number of power generators of type p that are working in time period t. This
integer number is bounded by the available number of generators AVAILp , constraints (6.6.2). paddpt
represents the additional production beyond the minimum output level of generators of type p in time

Mining and process industries

77

Applications of optimization with Xpress-MP

period t.
∀p ∈ TYPES, t ∈ TIME : startpt ∈ IN

(6.6.1)∀p ∈ TYPES, t ∈ TIME : workpt ∈ {0, 1, 2, . . . , AVAILp }(6.6.2)∀p ∈ TYPES, t ∈ TIME : paddpt ≥ 0(6.6.3)For a power generator of type p the additional production is limited by PMAXp − PMINp . For every type
and time period we can therefore link the additional output with the number of working generators:∀p ∈ TYPES, t ∈ TIME : paddpt ≤ (PMAXp − PMINp ) · workpt(6.6.4)

To satisfy the demand in every period we write the constraints (6.6.5). The first sum is the total basis
output by the generators of different types. The second sum is the total additional output beyond theminimum level by all power generators.

X∀t ∈ TIME :

PMINp · workpt + paddpt ≥ DEMt

(6.6.5)p∈TYPES

Without starting any new generators, those that are working must be able to produce 20% more thanplanned:X∀t ∈ TIME :PMAXp · workpt ≥ 1. 20 · DEMt
(6.6.6)p∈TYPESIf no generators were stopped, the number of power generators starting to work in a period t would
correspond to the difference between the numbers of generators working in periods t and t − 1. Since
generators may be stopped (not explicitly counted as there is no impact on the cost), the number ofgenerators starting to work is at least the value of this difference (6.6.7). The constraints (6.6.8) are forthe particular case of the transition from the last period of a day to the first period of the following day.
∀p ∈ TYPES, t ∈ {2, . . . , NT} : startpt ≥ workpt − workp,t−1(6.6.7)∀p ∈ TYPES : startp1 ≥ workp1 − workp,NT(6.6.8)The daily cost (6.6.9) comprises the start-up cost, the cost for working at the minimum level and the cost
for any additional production. These last two costs are proportional to the length of the time periods.
minimizeX

X

CSTARTp · startpt + LENt · CMINp · workpt + CADDp · paddpt

(6.6.9)p∈TYPES t∈TIME

The lines (6.6.1) to (6.6.9) form the final mixed integer model.

6.6.2

Implementation
The algebraic model translates into the following Mosel model. The constraints (6.6.7) and (6.6.8) are
obtained with a single line, using the inline if function of Mosel.
Note further that we do not define variables start as integer since they result from a difference of integer
variables work through constraints (6.6.7) and therefore automatically take integer values.
model "A-6 Electricity production"
uses "mmxprs"
declarations
NT = 7
TIME = 1..NT
TYPES = 1..4

! Time periods! Power generator types

LEN, DEM: array(TIME) of integer! Length and demand of time periods
PMIN,PMAX: array(TYPES) of integer ! Min. & max output of a generator type
CSTART: array(TYPES) of integer! Start-up cost of a generator
CMIN: array(TYPES) of integer! Hourly cost of gen. at min. output
CADD: array(TYPES) of real
! Cost/hour/MW of prod. above min. level
AVAIL: array(TYPES) of integer! Number of generators per typestart: array(TYPES,TIME) of mpvar
work: array(TYPES,TIME) of mpvar
padd: array(TYPES,TIME) of mpvar

Mining and process industries! No. of gen.s started in a period! No. of gen.s working during a period! Production above min. output level

78

Applications of optimization with Xpress-MP

end-declarations
initializations from ’a6electr.dat’
LEN DEM PMIN PMAX CSTART CMIN CADD AVAIL
end-initializations! Objective function: total daily cost
Cost:= sum(p in TYPES, t in TIME) (CSTART(p)*start(p,t) +LEN(t)*(CMIN(p)*work(p,t) + CADD(p)*padd(p,t)))! Number of generators started per period and per type
forall(p in TYPES, t in TIME)
start(p,t) >= work(p,t) - if(t>1, work(p,t-1), work(p,NT))! Limit on power production above minimum level
forall(p in TYPES, t in TIME) padd(p,t) <= (PMAX(p)-PMIN(p))*work(p,t)! Satisfy demands
forall(t in TIME) sum(p in TYPES) (PMIN(p)*work(p,t) + padd(p,t)) >= DEM(t)! Security reserve of 20%
forall(t in TIME) sum(p in TYPES) PMAX(p)*work(p,t) >= 1.20*DEM(t)! Limit number of available generators; numbers of generators are integer
forall(p in TYPES, t in TIME) dowork(p,t) <= AVAIL(p)work(p,t) is_integer
end-do! Solve the problem
minimize(Cost)
end-model

6.6.3

Results
The optimal LP solution to this problem is a cost of B
C 985164.3. The optimal integer solution has a cost ofB
C 1,465,810. For every time period and generator type the following table lists the total output and thepart of the production that is above the minimum level.
Table 6.12: Plan of power generator use
Type

0am–6am 6am–9am 9am–12pm 12pm–2pm 2pm–6pm 6pm–10pm 10pm–12am

1 No. used
Tot. output
Add. output

3
2250
0

4
4600
1600

4
3000
0

7
8600
3350

3
2250
0

3
2600
350

3
2250
0

2 No. used
Tot. output
Add. output

4
5750
1750

4
6000
2000

4
4200
200

4
6000
2000

4
4950
950

4
6000
2000

4
5950
1950

3 No. used
Tot. output
Add. output

2
4800
1600

8
16000
6400

8
16000
6400

8
16000
6400

8
16000
6400

8
16000
6400

4
8000
3200

4 No. used
Tot. output
Add. output

0
0
0

3
5400
0

1
1800
0

3
5400
0

1
1800
0

3
5400
0

1
1800
0

By looking at this table we may deduce when power generators are started and stopped: one generatorof type 1 starts at the beginning of period 2, and three at the beginning of period 4, four generators stop
after period 4. The four generators of type 2 work continuously. Six generators of type 3 start working inperiod 2, four generators of this type are stopped after period 6 and two more at the end of every day.
All three generators of type 4 start at the beginning of period 2, two of which are switched off in thenext period, then switched on, switched off, and switched on again and at the end of every day all three
are stopped.

Mining and process industries

79

Applications of optimization with Xpress-MP

6.7

References and further material
Blending problems are historically the first civil applications of Linear Programming. References for the
first three problems in this chapter: Glen describes a problem of mixing food for cattle [Gle80], Sutton
and Coates deal with blending in the iron and steel industry [SC81], and McColl presents applications of
LP in the petrol industry [McC69].
The problem of the sugar house is a member of the big family of flow problems, dealt with in detail inthe book by Ahuja, Magnanti and Orlin [AMO93] which is a reference for this topic. The opencast mining
problem boils down to calculating the maximum weight closure of a graph, a problem that can equally
be solved by a flow algorithm (Chapter 19 of the book by Ahuja et al.). An example where blocks arearranged in a three-dimensional way is described by Williams [Wil93].
The electricity generation problem has been studied by Garver [Gar63]. In this family of dispatchingproblems, the aim is to satisfy at the least cost the demand of each period, with costs for activating
and deactivating resources (see for example [Gar00]). The main difference from the production planningproblems in Chapter 8 lies in the costs and the fact that the power plants cannot be used below a certain
minimum output level. The variation in power supply may even be very narrow for water barrages whichusually leads to combining them with thermal or nuclear power plants to follow the demand more easily[SM74].

Mining and process industries

80

Applications of optimization with Xpress-MP

Chapter 7

Scheduling problems
Scheduling problems are an important class of problems in Operations Research. They consist of distributing the execution of a set of tasks over time subject to various constraints (sequence of tasks, duedates, limited resource availability), with the objective to optimize a criterion like the total duration, thenumber of tasks that finish late etc. These problems may come from many different fields: project management, industrial production, telecommunications, information systems, transport, timetabling. With
the exception of the well known case of project scheduling, these problems are difficult to solve to optimality if they grow large. Nevertheless, instances of smaller size have become solvable by LP, due to thepower of current software.
This chapter only deals with project and production scheduling problems. The Section 7.1 concerns theconstruction of a stadium, a typical case of project scheduling in civil engineering. The next three problems are about the management of workshops: Section 7.2 presents a workshop organized in lines (flowshop), that all products are run through in the same order. In the job-shop of Section 7.3 every producthas a different processing order on the machines of the workshop. Section 7.4 deals with the case of a
critical machine, the bottleneck of a workshop. The problem in Section 7.5 consists of distributing theworkload of an amplifier assembly line to maximize the throughput.
This book also discusses scheduling problems in other domains: in air transport (Chapter 11), telecommunications (Chapter 12) and for establishing a timetable (Chapter 14).

7.1

Construction of a stadium
A town council wishes to construct a small stadium in order to improve the services provided to the people
living in the district. After the invitation to tender, a local construction company is awarded the contract
and wishes to complete the task within the shortest possible time. All the major tasks are listed in thefollowing table. The durations are expressed in weeks. Some tasks can only start after the completion ofcertain other tasks. The last two columns of the table refer to question 2 which we shall see later.
Question 1: Which is the earliest possible date of completing the construction?
Question 2: The town council would like the project to terminate earlier than the time announced by thebuilder (answer to question 1). To obtain this, the council is prepared to pay a bonus of B
C 30K for every
week the work finishes early. The builder needs to employ additional workers and rent more equipment
to cut down on the total time. In the preceding table he has summarized the maximum number of weeks
he can save per task (column "Max. reduct.") and the associated additional cost per week. When will the
project be completed if the builder wishes to maximize his profit?

7.1.1

Model formulation for question 1
This problem is a classical project scheduling problem. We add a fictitious task with 0 duration that
corresponds to the end of the project. We thus consider the set of tasks TASKS = {1, . . . , N} where N is
the fictitious end task. Let DURi be the duration of task i. To establish the precedences between tasks, weuse a precedence graph G = (TASKS, ARCS) where ARCS stands for the set of arcs (an arc (i, j) symbolizes
that task i precedes task j). It is easy to construct such a graph starting from the lists of predecessors in
Table 7.1. The fictitious task follows all tasks that have no successor.
We introduce variables starti to represent the earliest start time of tasks i. The only constraints that are
given are the precedences. A task j may only start if all its predecessors have finished, which translates

81

Applications of optimization with Xpress-MP

Table 7.1: Data for stadium construction
Task

Description

Duration

Predecessors

Max.
reduct.

Add. cost per
week (in 1000 B
C)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

Installing the construction site
TerracingConstructing the foundations
Access roads and other networks
Erecting the basement
Main floor
Dividing up the changing rooms
Electrifying the terraces
Constructing the roof
Lighting of the stadium
Installing the terraces
Sealing the roof
Finishing the changing rooms
Constructing the ticket officeSecondary access roads
Means of signalling
Lawn and sport accessories
Handing over the building

2
16
9
8
10
6
2
2
9
5
3
2
1
7
4
3
9
1

none
1
2
2
3
4,5
4
6
4,6
4
6
9
7
2
4,14
8,11,14
12
17

0
3
1
2
2
1
1
0
2
1
1
0
0
2
2
1
3
0

–
30
26
12
17
15
8
–
42
21
18
–
–
22
12
6
16
–

7

13

4
10
9
1

2

3

5

6

12

17

18

F

8
11

16

14
15
Figure 7.1: Precedence graph of construction tasks

Scheduling problems

82

Applications of optimization with Xpress-MP

into constraints (7.1.1): if there is an arc between i and j, then the completion time of i (starti + DURi )must not be larger than the start time of j.
∀(i, j) ∈ ARCS : starti + DURi ≤ startj(7.1.1)The objective is to minimize the completion time of the project, that is the start time of the last, fictitious
task N. We complete the mathematical model with the following lines:

7.1.2

minimize startN(7.1.2)∀i ∈ TASKS : starti ≥ 0(7.1.3)

Implementation of question 1
The translation of the mathematical model into Mosel is straightforward. The set of arcs is implemented
as a two-dimensional binary matrix ARC, an element ARCij = 1 if and only if the arc (i, j) exists. At its
definition, we do not fix the index sets of this array and use the keyword range instead, which resultsin the definition of a dynamic array for which only those entries exist that are read in from the data
file. The latter only contains the list of defined arcs (using an array format that is usually referred to assparse format as opposed to the dense format where all entries must be defined). Note that the objectivefunction of this problem is simply a single variable (the fictitious end task).
model "B-1 Stadium construction (First part)"
uses "mmxprs"
declarations
N = 19TASKS=1..N
ARC: array(range,range) of real
DUR: array(TASKS) of real
start: array(TASKS) of mpvar
end-declarations! Number of tasks in the project! (last = fictitious end task)! Matrix of the adjacency graph! Duration of tasks! Start times of the tasks

initializations from ’b1stadium.dat’
ARC DUR
end-initializations! Precedence relations between tasks
forall(i,j in TASKS | ARC(i,j)=1) start(i) + DUR(i) <= start(j)! Solve the first problem: minimize the total duration
minimize(start(N))
end-model

7.1.3

Results for question 1
The execution of the program gives us a duration of 64 weeks to complete the construction of the stadium, which correspond to 1 year and 3 months. A possible solution is represented in Figure 7.2 (there
are several possible plans, since for instance task 10 may start any time in the interval [26,59] without
having any effect on the total completion time). In this bar chart every task is represented by a rectangle
the length of which is proportional to the duration of the task. The vertical position of the tasks has no
signification. The shaded areas in the rectangles representing the tasks correspond to the maximum timeby which the task may be shortened (see question 2).

7.1.4

Model formulation for question 2
This second problem is called scheduling with project crashing. To reduce the total duration of the
project, we need to take into account the result of the preceding optimization run. We define variables
savei that correspond to the number of weeks that we wish to save for every task i. The column ‘Max.
reduct.’ in Table 7.1 gives an upper bound on these variables savei . We call MAXWi this maximum
reduction (in weeks). The constraints (7.1.4) must be satisfied for all tasks i except the last, fictitious taskN.
∀i ∈ TASKS\{N} : savei ≤ MAXWi(7.1.4)

Scheduling problems

83

Applications of optimization with Xpress-MP

14
4
1

2

0

10

13
7

3
20

5
30

11 16

15

8

10
9

6
40

12
50

17
60 64
18

Figure 7.2: Solution to question 1

For the last task, the variable saveN represents the number of weeks the project finishes earlier than thesolution obj1 calculated in answer to question 1. The new completion time of the project startN must be
equal to the previous completion time minus the advance saveN , which leads to the constraint (7.1.5).
startN = obj1 − saveN(7.1.5)The constraints (7.1.1) need to be modified to take into account the new variables savei . The new completion time of a task is equal to its start, plus its duration, minus the savings, or starti + DURi − savei . We
rename these constraints to (7.1.6).
∀(i, j) ∈ ARCS : starti + DURi − savei ≤ startj(7.1.6)The objective has also been redefined by the second question. We now want to maximize the builder’sprofit. For every week finished early, he receives a bonus of BONUS kB
C. In exchange, the savings in time
for a task i costs COSTi kB
C (column ‘Add. cost per week’ of Table 7.1). The new objective function will belabelled (7.1.7).
X(7.1.7)
maximize BONUS · saveN −COSTi · saveii∈TASKS\{N}

The new mathematical model consists of relations (7.1.6), (7.1.7), (7.1.3), (7.1.4), and (7.1.5), and thenon-negativity conditions for variables savei .

7.1.5

Implementation of question 2
It is possible to implement the model for question 2 as an addition to the model for question 1 shown
above, all in a single Mosel program. This makes it possible to retrieve the solution value of the firstproblem and use it in the definition of the second problem without the need for any interaction.
Since we need to make some additions to the first model, we repeat the complete example. The additions
to the previous implementation of the first part are the following:• A solution printing subroutine: we want to print the solution to the problem twice (that is, thestart times assigned to tasks). To avoid having to repeat the same lines of code several times wedefine a subroutine that is called whenever we want to print a solution. The third line of themodel now contains the declaration of the procedure print_sol that is defined at the end of the
program. This declaration (using the keyword forward) is required if a user-defined subroutine iscalled before it is defined. A subroutine has a similar structure to a model in Mosel; we shall see
more examples of subroutines in the following chapters.
• The precedence constraints are named: Mosel defines models incrementally. If we simply add the
new definition of the precedence constraints in the second part of the model, we also keep the previous definition of these constraints. By naming the constraints and re-assigning them we override
the first definition.
• A real number obj1 is used to store the solution value of the first problem.
After solving the second problem, we print the solution value of this problem using the function getsol(we could equally have used getobjval, this is just to make clear which is the constraint that defines
the current objective function). If it is applied to a constraint, the function getsol returns its evaluation
with the current solution (note that a constraint is stored in the form variableterms − RHSvalue which isthe expression that is evaluated by the function).

Scheduling problems

84

Applications of optimization with Xpress-MP

model "B-1 Stadium construction (Complete model)"uses "mmxprs"
forward procedure print_sol
declarations
N = 19TASKS=1..N
ARC: array(range,range) of real
DUR: array(TASKS) of real
start: array(TASKS) of mpvar
obj1: real
end-declarations! Number of tasks in the project! (last = fictitious end task)! Matrix of the adjacency graph! Duration of tasks! Start times of tasks! Solution of the first problem

initializations from ’b1stadium.dat’
ARC DUR
end-initializations! Precedence relations between tasks
forall(i,j in TASKS | ARC(i,j)=1)
Prec(i,j):= start(i) + DUR(i) <= start(j)! Solve the first problem: minimize the total duration
minimize(start(N))
obj1:=getobjval! Solution printing
print_sol! **** Extend the problem ****
declarations
BONUS: integer
MAXW: array(TASKS) of real
COST: array(TASKS) of real
save: array(TASKS) of mpvar
end-declarations! Bonus per week finished earlier! Max. reduction of tasks (in weeks)! Cost of reducing tasks by a week
! Number of weeks finished early

initializations from ’b1stadium.dat’
MAXW BONUS COST
end-initializations! Second objective function
Profit:= BONUS*save(N) - sum(i in 1..N-1) COST(i)*save(i)! Redefine precedence relations between tasks
forall(i,j in TASKS | ARC(i,j)=1)
Prec(i,j):= start(j) - start(i) + save(i) >= DUR(i)! Total duration
start(N) + save(N) = obj1
! Limit on number of weeks that may be saved
forall(i in 1..N-1) save(i) <= MAXW(i)! Solve the second problem: maximize the total profit
maximize(Profit)! Solution printing
writeln("Total profit: ", getsol(Profit))print_sol!----------------------------------------------------------------procedure print_sol
writeln("Total duration: ", getsol(start(N)), " weeks")
forall(i in 1..N-1)
write(strfmt(i,2), ": ", strfmt(getsol(start(i)),-3),
if(i mod 9 = 0,"\n",""))
writeln
end-procedure
end-model

Scheduling problems

85

Applications of optimization with Xpress-MP

7.1.6

Results for question 2
The solution of the complete problem shows that it is possible to complete the project 10 weeks earlier with a total profit of B
C 87k for the builder. The new start times are shown in the schedule that isrepresented below (note that this is again one of several possible plans).

13
7
14

15

11 16

4

10

8

5

1

2

3

0

10

20

18

6

9

30

40

12

17
50 54

Figure 7.3: Solution to question 2

7.2

Flow-shop schedulingA workshop that produces metal pipes on demand for the automobile industry has three machines forbending the pipes, soldering the fastenings, and assembling the links. The workshop has to produce
six pieces, for which the durations of the processing steps (in minutes) are given in the following table.
Every workpiece first goes to bending, then to soldering, and finally to assembly of the links. Oncestarted, any operations must be carried out without interruption, but the workpieces may wait between
the machines.
Table 7.2: Processing durations in minutes
Workpiece

1

2

3

4

5

6

Bending
Soldering
Assembly

3
5
5

6
4
2

3
2
4

5
4
6

5
4
3

7
5
6

Every machine only processes one piece at a time. A workpiece may not overtake any other by passing
onto the following machine. This means that if at the beginning a sequence of the workpieces is established, they will be processed on every machine in exactly this order. Which is the sequence of workpieces
that minimizes the total time for completing all pieces?

7.2.1

Model formulation
From now on, we work with MACH = {1, . . . , NM} the set of machines and JOBS = {1, . . . , NJ} the setof pieces (jobs) to produce. The processing duration of a piece j on a machine m is given by DURm,j .
Every workpiece has to run through the machines 1, . . . , NM in exactly this order, without being able toovertake other pieces. We can therefore define a schedule by the initial order of the workpieces. The
duration of the schedule is given as the time instant when the machine NM finishes the last job.
The sequence of the jobs can be defined with the help of binary variables rankjk that are 1 if and only
if piece j has the rank (position) k in the starting sequence (7.2.1). The set of starting positions RANKS is
the same as the set of jobs JOBS since every job needs to be assigned a rank (7.2.2), and every rank mustbe occupied by one job only (7.2.3). These constraints are typical of assignment problems (see also theassignment of personnel to workstations in Chapter 14).
∀j ∈ JOBS, k ∈ RANKS : rankjk ∈ {0, 1}∀k ∈ RANKS :X(7.2.1)

rankjk = 1(7.2.2)

rankjk = 1(7.2.3)j∈JOBSX∀j ∈ JOBS :k∈RANKS

Scheduling problems

86

Applications of optimization with Xpress-MP

In this problem it is relatively difficult to calculate the starting or completion times of the operations
from the ranks. To obtain these values, we introduce two additional sets of variables emptymk and waitmk(7.2.4) – (7.2.5). The variables emptymk (with m in MACH and k in 1, . . . , NJ − 1) denote the time between
the processing of the jobs with rank k and k + 1 on a machine m, i.e. the time that the machine m isidle after the termination of the workpiece with rank k. A variable waitmk (with m in 1, . . . , NM and k in
RANKS) is the waiting time for the job with rank k between its processing on machines m and m + 1.
The workpieces can be processed without any pause on the first machine that does not have to wait
for any preceding machines, so the variables empty1k are therefore always 0 (7.2.6). Similarly, the first
workpiece in the sequence can pass through all machines without any waiting times, which means thevariables waitm1 can also be fixed to 0 (7.2.7).
∀m ∈ MACH, k = 1, . . . , NJ − 1 : emptymk ≥ 0(7.2.4)∀m = 1, . . . , NM − 1, k ∈ RANKS : waitmk ≥ 0(7.2.5)∀k = 1, . . . , NJ − 1 : empty1k = 0(7.2.6)∀m = 1, . . . , NM − 1 : waitm1 = 0(7.2.7)To simplify the formulation of the following constraints, we introduce the notation durmk for the processing duration of the job of rank k on machine m, defined by the relations (7.2.8). Through the constraints(7.2.3) only a single variable rankjk takes the value 1 in this sum, and only the duration of the corresponding workpiece will be counted. Using this notation and the variables emptymk we are now able to write
down the objective function (7.2.9).
∀m ∈ MACH, k ∈ RANKS : durmk =X(7.2.8)

DURmj · rankjk

j∈JOBSNM−1

minimizeX

NJ−1

durm1 +

X

m=1

NM−1

emptyNM,k =

X X

NJ−1

DURmj · rankj1 +m=1 j∈JOBS

k=1

X

emptyNM,k(7.2.9)k=1

The first sum results in the point of time when the last machine (machine NM) starts working: this is thetotal time required for the first job on all preceding machines (1, . . . , NM − 1). The second sum is thetotal duration of idle times between processing operations on the last machine. The following graphic
represents these quantities. Normally, we also needed to count the durations of all operations on the lastmachine, but this is a constant and may be omitted.

Machine 1

dur 11dur 21

Machine 2

...

...

dur NM,1 emptyNM,1 dur NM,2

Machine NM... emptyNM,NJ-1 dur NM,NJ

Figure 7.4: Relation between production and idle time

The tricky bit that is now left is to link the variables waitmk and emptymk to the variables rankjk . To do so,
we introduce dnextmk , the time between the completion of job k on machine m and the start of job k + 1
on machine m + 1.
emptymk

Machine m

dur m,k+1

dur mkwait mk

Machine m+1

dur m+1,kwait m,k+1

emptym+1,k

dur m+1,k+1

Figure 7.5: Calculation of the total duration

Scheduling problems

87

Applications of optimization with Xpress-MP

With the help of Figure 7.5, it is easy to see that the equations (7.2.10) hold. These can be developed tothe constraints (7.2.11).
∀m = 1, . . . , NM − 1, k = 1, . . . , NJ − 1 :dnextmk = emptymk + durm,k+1 + waitm,k+1 = waitmk + durm+1,k + emptym+1,k(7.2.10)∀m = 1, . . . , NM − 1, k = 1, . . . , NJ − 1 :emptymk +X

DURmj · rankj,k+1 + waitm,k+1

j∈JOBS= waitmk +X

DURm+1,j · rankjk + emptym+1,k(7.2.11)j∈JOBS

The resulting model is given through the lines (7.2.1) to (7.2.7), (7.2.9), and (7.2.11). This is a mixed-integer
problem since only the variables rankjk are constrained to be integer (more exactly, binary).

7.2.2

Implementation
The following Mosel program implements the mathematical model of the previous section. In order toextract the resulting schedule more easily, variables startmk have been added that indicate the start of
the job with rank k on machine m. These variables are linked to the rankjk variables via an additional setof constraints.
model "B-2 Flow shop"
uses "mmxprs"
declarations
NM = 3
NJ = 6
MACH = 1..NMRANKS = 1..NJ
JOBS = 1..NJ! Number of machines! Number of jobs

DUR: array(MACH,JOBS) of integer! Durations of jobs on machines

rank: array(JOBS,RANKS) of mpvar! 1 if job j has rank k, 0 otherwise
empty: array(MACH,1..NJ-1) of mpvar ! Space between jobs of ranks k and k+1
wait: array(1..NM-1,RANKS) of mpvar ! Waiting time between machines m! and m+1 for job of rank kstart: array(MACH,RANKS) of mpvar! Start of job rank k on machine m! (optional)
end-declarations
initializations from ’b2flowshop.dat’DUR
end-initializations! Objective: total waiting time (= time before first job + times between! jobs) on the last machine
TotWait:= sum(m in 1..NM-1,j in JOBS) (DUR(m,j)*rank(j,1)) +sum(k in 1..NJ-1) empty(NM,k)! Every position gets a jobs
forall(k in RANKS) sum(j in JOBS) rank(j,k) = 1! Every job is assigned a rank
forall(j in JOBS) sum(k in RANKS) rank(j,k) = 1! Relations between the end of job rank k on machine m and start of job on! machine m+1
forall(m in 1..NM-1,k in 1..NJ-1)
empty(m,k) + sum(j in JOBS) DUR(m,j)*rank(j,k+1) + wait(m,k+1) =wait(m,k) + sum(j in JOBS) DUR(m+1,j)*rank(j,k) + empty(m+1,k)! Calculation of start times (to facilitate the interpretation of results)
forall(m in MACH, k in RANKS)
start(m,k) = sum(u in 1..m-1,j in JOBS) DUR(u,j)*rank(j,1) +sum(p in 1..k-1,j in JOBS) DUR(m,j)*rank(j,p) +sum(p in 1..k-1) empty(m,p)! First machine has no idle times

Scheduling problems

88

Applications of optimization with Xpress-MP

forall(k in 1..NJ-1) empty(1,k) = 0! First job has no waiting times
forall(m in 1..NM-1) wait(m,1) = 0
forall(j in JOBS, k in RANKS) rank(j,k) is_binary! Solve the problem
minimize(TotWait)
end-model

It may be noted that it is possible to use an alternative formulation with Special Ordered Sets of type 1(SOS1) instead of constraining variables rankjk to be binary. The corresponding line in the program above
needs to be replaced by one of the following two lines:
forall(j in JOBS) sum(k in RANKS) k*rank(j,k) is_sos1

orforall(k in RANKS) sum(j in JOBS) j*rank(j,k) is_sos1

In the first case, we define a SOS1 for every job, where the rank number is used as weight coefficient in
the set; in the second case, a SOS1 is established for every rank position using the job numbers as weight
coefficients. Exactly one variable of a SOS1 takes a value greater than 0, and due to the constraints (7.2.2)and (7.2.3) this value will be 1.

7.2.3

Results
The minimum waiting time calculated for the last machine is 9. With the variables startmk we can deducea total duration of 35. A schedule (there are several possibilities) that has this minimum duration is shownin the following table.
Table 7.3: Optimal schedule of workpieces

7.3

Rank number

1

2

3

4

5

6

Workpiece number
Start on machine 1:
Start on machine 2:
Start on machine 3:

3
0
3
5

1
3
6
11

4
6
11
16

6
11
18
23

5
18
23
29

2
23
29
33

Job Shop Scheduling
A company has received an order for three types of wallpapers: one (paper 1) has a blue background withyellow patterns, another (paper 2) a green background and blue and yellow patterns, and the last (paper3) has a yellow background with blue and green patterns. Every paper type is produced as a continuous
roll of paper that passes through several machines, each printing a different color. The order in which
the papers are run through the machines depends on the design of the paper: for paper 1 first the blue
background and then the yellow pattern is printed. After the green background for paper 2, first theblue and then the yellow patterns are printed. The printing of paper 3 starts with the yellow background,followed by the blue and then the green patterns.
Paper 2Paper 1

Paper 3Paper 1

Green

Paper 2Blue

Paper 3Paper 2

Yellow

Paper 3Paper 1

Paper 3Paper 2

Figure 7.6: Production flows through printing machines

The processing times differ depending on the surface that needs to be printed. The times (in minutes) for
applying every color of the three paper types are given in the following table.

Scheduling problems

89

Applications of optimization with Xpress-MP

Table 7.4: Times required for applying every color
MachineColor

Paper 1

Paper 2Paper 3

1
2
3

Blue
Green
Yellow

45
10

20
10
34

12
17
28

Knowing that every machine can only process one wallpaper at a time and that a paper cannot be processed by several machines simultaneously, how should the paper printing be scheduled on the machines
in order to finish the order as early as possible?

7.3.1

Model formulation
Let JOBS be the set of jobs (paper types) and MACH the set of machines. We are going to use the variables
startmj for the start of job j on machine m (supposing the the schedule starts at time 0). We write DURmj
for the processing duration of job j on machine m. The variable finish indicates the completion time ofthe entire schedule. Since we want to minimize the completion time, the objective function is simply(7.3.1).
minimize finish(7.3.1)The schedule terminates when all paper types are completed, that is, when the last operation for every
types finishes. The completion time of the schedule therefore needs to satisfy the following constraints(7.3.2) – (7.3.4).
finish ≥ start31 + DUR31(7.3.2)finish ≥ start32 + DUR32(7.3.3)finish ≥ start23 + DUR23(7.3.4)The constraints between processing operations are of two types: the so-called conjunctive constraints
represent the precedences between the operations for a single paper type, and the disjunctive constraints
express the fact that a machine can only execute a single operation at a time.
To start, we consider the conjunctive constraints. Paper 1 first goes onto the machine printing the blue
color (machine 1), and then onto the one printing yellow (machine 3). This means that the processingof paper 1 on machine 1 needs to be finished when the processing on machine 3 starts, and hence, theconstraint (7.3.5) needs to hold.
start11 + DUR11 ≤ start31(7.3.5)
Similarly, paper 2 is first processed on machine 2, then on machine 1 and then on machine 3, which leads
to the constraints (7.3.6) and (7.3.7).
start22 + DUR22 ≤ start12(7.3.6)

start12 + DUR12 ≤ start32(7.3.7)For paper 3, that is processed in the order machine 3, machine 1, machine 2, we obtain the following
constraints (7.3.8) and (7.3.9).
start33 + DUR33 ≤ start13(7.3.8)start13 + DUR13 ≤ start23(7.3.9)We now still need to model the disjunctions. Machine 1 has to process all three wallpaper types. Sincea machine can only perform a single operation at any time, either paper 1 is printed before paper 2 or
paper 2 is printed before paper 1. This is expressed by the following:
start11 + DUR11 ≤ start12 ∨ start12 + DUR12 ≤ start11
These two mutually exclusive constraints can be written as the constraints (7.3.10) and (7.3.11), where M
is a large positive number and y1 is a binary variable that takes the value 1 if paper 1 comes before paper
2 on machine 1, and 0 otherwise.

Scheduling problems

start11 + DUR11 ≤ start12 + M · (1 − y1 )(7.3.10)start12 + DUR12 ≤ start11 + M · y1

(7.3.11)

90

Applications of optimization with Xpress-MP

Let us see why this claim is correct. If paper 1 comes before paper 2 on machine 1, then y1 has the value
1 and we obtain the constraints start11 + DUR11 ≤ start12 and start12 + DUR12 ≤ start11 + M. The first
constraint requires that the processing of paper 2 on machine 1 takes places after the printing of paper 1
is terminated. Whatever value start11 and start12 may take, the second constraint is automatically fulfilled
due to the large value M. If, on the contrary, paper 2 is processed before paper 1 on machine 1, then y1
has the value 0 and we obtain the constraints start11 + DUR11 ≤ start12 + M and start12 + DUR12 ≤ start11 .
This time, the first constraint is automatically fulfilled and the second guarantees that the processing ofwallpaper 2 is terminated before the start of paper 1.
Proceeding in this way, we can translate all pairs of disjunctions between operations on the same machineby defining a binary variable yd for every disjunction (d = 1, . . . , ND). We thus obtain the following
constraints (7.3.12) – (7.3.23):
start11 + DUR11 ≤ start13 + M · (1 − y2 )(7.3.12)

start13 + DUR13 ≤ start11 + M · y2

(7.3.13)start12 + DUR12 ≤ start13 + M · (1 − y3 )(7.3.14)

start13 + DUR13 ≤ start12 + M · y3

(7.3.15)

start22 + DUR22 ≤ start23 + M · (1 − y4 )(7.3.16)

start23 + DUR23 ≤ start22 + M · y4

(7.3.17)

start31 + DUR31 ≤ start32 + M · (1 − y5 )(7.3.18)

start32 + DUR32 ≤ start31 + M · y5

(7.3.19)

start31 + DUR31 ≤ start33 + M · (1 − y6 )(7.3.20)

start33 + DUR33 ≤ start31 + M · y6

(7.3.21)start32 + DUR32 ≤ start33 + M · (1 − y7 )(7.3.22)start33 + DUR33 ≤ start32 + M · y7

(7.3.23)

The constraints (7.3.12) – (7.3.15) establish the rest of the disjunctions on machine 1, constraints (7.3.16)
and (7.3.17) the disjunction of papers 2 and 3 on machine 2, and the constraints (7.3.18) – (7.3.23) the
disjunctions between the processing of the three papers on machine 3.
To avoid numerical instabilities, the value of M should not be chosen too large. It is, for instance, possible
to use as its value an upper bound UB on the schedule determined through some heuristic. In this casethe following constraint (7.3.24) needs to be added to the mathematical model, which may also help toreduce the number of nodes explored by the tree search.
(7.3.24)

finish ≤ UB

In our small example, we are going to use as the value for M the sum of all processing times as this gives
a rough but safe upper bound.
To complete the formulation of the model, we need to add the non-negativity constraints for the start
time variables (7.3.25) and constrain the yd to be binaries (7.3.26).
∀m ∈ MACH, j ∈ JOBS : startmj ≥ 0(7.3.25)∀d ∈ {1, . . . , ND} : yd ∈ {0, 1}(7.3.26)

By renumbering the operations (tasks) using a single subscript (resulting in the set TASKS) we obtain amore general model formulation. The precedence relations and disjunctions can be represented as a
disjunctive graph G = (TASKS, ARCS, DISJS) where ARCS is the set of arcs (precedences) and DISJS the setof disjunctions. There is an arc (i, j) between two tasks i and j if i is the immediate predecessor of j. A
disjunction [i, j] indicates that tasks i and j are disjoint. With these conventions we obtain the following
model (7.3.27) – (7.3.32).
minimize finish(7.3.27)∀j ∈ TASKS : startj + DURj ≤ finish(7.3.28)∀(i, j) ∈ ARCS : starti + DURi ≤ startj(7.3.29)∀[i, j] ∈ DISJ : starti + DURi ≤ startj + M · yij ∧ startj + DURj ≤ starti + M · (1 − yij )(7.3.30)∀j ∈ TASKS : startj ≥ 0(7.3.31)∀[i, j] ∈ DISJ : yij ∈ {0, 1}(7.3.32)

The objective function remains the same as before. The constraints (7.3.2) – (7.3.4) are generalized by
the constraints (7.3.28): the completion time of the schedule is greater than or equal to the completion

Scheduling problems

91

Applications of optimization with Xpress-MP

time of the last operation for every paper time, and hence to the completion times of all operations. The
constraints (7.3.29) state the conjunctions (precedences) and (7.3.30) the disjunctions. The last two setsof constraints are the non-negativity conditions for the start time variables (7.3.31) and the constraints
turning the disjunction variables yij into binaries.

7.3.2

Implementation
The following Mosel program implements the mathematical model given in lines (7.3.27) – (7.3.32) of theprevious section, that is, it uses a single index set TASKS for numbering the operations instead of a double
machine-paper type index for start time variables and durations. In the set of examples on the book’s
website a second implementation is provided that uses the double subscript of the first mathematicalmodel (lines (7.3.1) – (7.3.25) of the previous section).
The sequence of operations for every paper type is given in the form of a list of precedences: an entry inthe table ARCij is defined if task i immediately precedes task j. The disjunctions are input in a similar way:
an entry in the table DISJij is defined if and only if tasks i and j are processed on the same machine – andare therefore in disjunction.
model "B-3 Job shop"
uses "mmxprs"
declarations
TASKS=1..8! Set of tasks (operations)

DUR: array(TASKS) of integer! Durations of jobs on machines
ARC: dynamic array(TASKS,TASKS) of integer! Precedence graph
DISJ: dynamic array(TASKS,TASKS) of integer ! Disjunctions between jobs
start: array(TASKS) of mpvar
finish: mpvar
y: array(range) of mpvar
end-declarations! Start times of tasks! Schedule completion time
! Disjunction variables

initializations from ’b3jobshop.dat’
DUR ARC DISJ
end-initializations
BIGM:= sum(j in TASKS) DUR(j)! Some (sufficiently) large value! Precedence constraints
forall(j in TASKS) finish >= start(j)+DUR(j)forall(i,j in TASKS | exists(ARC(i,j)) ) start(i)+DUR(i) <= start(j)! Disjunctions
d:=1
forall(i,j in TASKS | i<j and exists(DISJ(i,j)) ) do
create(y(d))y(d) is_binary
start(i)+DUR(i) <= start(j)+BIGM*y(d)start(j)+DUR(j) <= start(i)+BIGM*(1-y(d))
d+=1
end-do! Bound on latest completion timefinish <= BIGM
! Solve the problem: minimize latest completion timeminimize(finish)
end-model

The implementation of this model uses some features of the Mosel language that have not yet been
introduced:
In the declarations block the reader may have noted the keywords dynamic array and range. Bothindicate that the corresponding arrays are defined as dynamic arrays. range indicates an unknown index
set that is a consecutive sequence of integers (which may include 0 or negative values). We do not haveto enumerate the array of variables y so that there is no need to name its index set. We could proceed
in the same way for the arrays ARC and DISJ, but since we know that their index sets are subsets of
TASKS (which is also used later on for enumerating the array entries) we use this name for the index sets.

Scheduling problems

92

Applications of optimization with Xpress-MP

However, the set TASKS is defined as a constant set directly at its declaration, which means that any arraywith this index will be created with a fixed size. For both arrays, ARC and DISJ only very few entries aredefined, and it is therefore preferable to force them to be initialized dynamically with the data read fromfile – this is done with the keyword dynamic.
In the forall loops enumerating the precedence arcs and the disjunctions, we use the function exists
to test whether an array entry is defined. For arrays with few defined entries (sparse arrays) this is a very
efficient way of enumerating these entries. Note that for an efficient use of exists the index sets in theloop must be the same as those used in the definition of the array that the condition bears on.
Another new feature is the forall-do loop used in the formulation of the disjunctions: so far we have
only seen the inline version that loops over a single statement. The full version needs to be used if theloop applies to a block of program statements.

7.3.3

Results
After solving this problem we obtain a total completion time of 97. The start and completion times forall jobs are given in the following table (there are several possible solutions).
Table 7.5: Starting times of operations
Blue
Green
Yellow

Paper 1

Paper 2Paper 342 – 87–
87 – 97

10 – 30
0 – 1030 – 64

30 – 42
42 – 59
0 – 28This schedule may be represented as a bar chart, also called a Gantt chart, that has the time as its horizontal axis. Every task (job) is represented by a rectangle, the length of which is proportional to its duration.
All operations on a single machine are placed in the same line.

Paper 2Blue

Paper 3

Green Paper 2
Yellow

Paper 3Paper 3

10

Paper 120

Paper 2

30

40

50

Paper 1

60

70

80

90

100

Figure 7.7: Gantt chart

7.4

Sequencing jobs on a bottleneck machine
In workshops it frequently happens that a single machine determines the throughput of the entire production (for example, a machine of which only one is available, the slowest machine in a production
line, etc.). This machine is called the critical machine or the bottleneck. In such a case it is important toschedule the tasks that need to be performed by this machine in the best possible way.
The aim of the problem is to provide a simple model for scheduling operations on a single machine
and that may be used with different objective functions. We shall see here how to minimize the total
processing time, the average processing time, and the total tardiness.
A set of tasks (or jobs) is to be processed on a single machine. The execution of tasks is non-preemptive(that is, an operation may not be interrupted before its completion). For every task i its release date andduration are given. For the last optimization criterion (total tardiness), a due date (latest completiontime) is also required to measure the tardiness, that is, the amount of time by which the completion ofjobs exceeds their respective due dates. The following table lists the data for our problem.
What is the optimal value for each of the objectives: minimizing the total duration of the schedule (=makespan), the mean processing time or the total tardiness?

Scheduling problems

93

Applications of optimization with Xpress-MP

Table 7.6: Task time windows and durations

7.4.1

Job

1

2

3

4

5

6

7

Release date
DurationDue date

2
5
10

5
6
21

4
8
15

0
4
10

0
2
5

8
4
15

9
2
22

Model formulation
We are going to deal with the different objective functions in sequence, but the body of the model willremain the same. For writing a model that corresponds to the three objectives at a time, we are going touse binary variables rankjk (j, k ∈ JOBS = {1, . . . , NJ}) that have the value 1 if job j has the position (rank)
k and 0 in all other cases. There can be only one job per position k and every job j takes a single position.
This leads to constraints (7.4.1) and (7.4.2).

X∀k ∈ JOBS :rankjk = 1(7.4.1)

rankjk = 1(7.4.2)j∈JOBSX∀j ∈ JOBS :k∈JOBSP

The processing time for the job in position k is given by the sum j∈JOBS DURj ·rankjk (where DURj denotes
the duration given in the table in the previous section). Only for the job j in position k does the variablerankjk have the value 1. By multiplying with the duration of job j, we are able to retrieve the durationof the job at position k. Using this technique we can write the constraints of our problem. If startk is thestart time of the job at position k, this value must be at least as great as the release date (given as RELj )of the job assigned to this position. This gives the constraints (7.4.3):∀k ∈ JOBS : startk ≥X(7.4.3)

RELj · rankjk

j∈JOBS

Another constraint used by all models specifies that two jobs cannot be processed simultaneously. The
job in position k + 1 must start after the job in position k has finished, hence the constraints (7.4.4).

X∀k ∈ {1, . . . , NJ − 1} : startk+1 ≥ startk +DURj · rankjk(7.4.4)j∈JOBS

Objective 1: The first objective is to minimize the makespan (completion time of the schedule), or equivalently, to minimize the completion time of the last job (job with rank k). The line (7.4.5) models thisobjective. The complete model is then given by the following:
minimize startNJ +X(7.4.5)

DURj · rankj,NJ

j∈JOBS∀k ∈ JOBS :X

rankjk = 1(7.4.6)

rankjk = 1(7.4.7)X(7.4.8)j∈JOBS∀j ∈ JOBS :Xk∈JOBS∀k ∈ JOBS : startk ≥RELj · rankjk

j∈JOBS∀k ∈ {1, . . . , NJ − 1} : startk+1 ≥ startk +X

DURj · rankjk(7.4.9)j∈JOBS∀k ∈ JOBS : startk ≥ 0(7.4.10)∀j, k ∈ JOBS : rankjk ∈ {0, 1}(7.4.11)

Objective 2: For minimizing the average processing time, we introduce additional variables compk (representing the completion times of the job in position k) to simplify the notation. We add the following
two sets of constraints to the problem to obtain these completion times:∀k ∈ JOBS : compk = startk +X

DURj · rankjk(7.4.12)j∈JOBS∀k ∈ JOBS : compk ≥ 0

Scheduling problems

94(7.4.13)

Applications of optimization with Xpress-MP

The new objective (7.4.14) consists of minimizing the average processing time, or equivalently, minimizing
the sum of the job completion times.
X
minimize
compk(7.4.14)k∈JOBS

The complete model for the second objective consists of lines (7.4.14) and (7.4.6) to (7.4.13).
Objective 3: If we now aim to minimize the total tardiness, we again introduce new variables – this time
to measure the amount of time that jobs finish after their due date. We write latek for the variable that
corresponds to the tardiness of the job with rank k. Its value is the difference between the completiontime of a job j and its due date DUEj . If the job finishes before its due date, the value must be 0. We thus
obtain the constraints (7.4.15) and (7.4.16).
∀k ∈ JOBS : latek = max(0, compk −X

DUEj · rankjk )(7.4.15)j∈JOBS∀k ∈ JOBS : latek ≥ 0(7.4.16)

The new objective function (7.4.17) minimizes the total tardiness of all jobs:
minimizeX(7.4.17)latekk∈JOBS

The new model is given by the constraints (7.4.17), (7.4.6) to (7.4.13), and (7.4.15) to (7.4.16).

7.4.2

Implementation
The Mosel implementation below solves the same problem three times, each time with a different objective, and prints the resulting solutions. To simplify the representation, we use the completion time
variables already in the formulation of the first objective. For the calculation of the tardiness of jobs, this
implementation uses the fact that by default all variables are non-negative. That means that if the constraint that calculates the difference between the completion time and the due date results in a negativelower bound for latek , then the Xpress-Optimizer will set this variable to 0.
model "B-4 Sequencing"uses "mmxprs"
forward procedure print_sol(obj:integer)
declarations
NJ = 7
JOBS=1..NJ! Number of jobs

REL: array(JOBS) of integer
DUR: array(JOBS) of integer
DUE: array(JOBS) of integer! Release dates of jobs
! Durations of jobs
! Due dates of jobs

rank: array(JOBS,JOBS) of mpvar
start: array(JOBS) of mpvar
comp: array(JOBS) of mpvar
late: array(JOBS) of mpvar
finish: mpvar
end-declarations! =1 if job j at position k! Start time of job at position k! Completion time of job at position k! Tardiness of job at position k! Completion time of the entire schedule

initializations from ’b4seq.dat’
DUR REL DUE
end-initializations! One job per position and one position per job
forall(k in JOBS) sum(j in JOBS) rank(j,k) = 1
forall(j in JOBS) sum(k in JOBS) rank(j,k) = 1! Sequence of jobs
forall(k in 1..NJ-1)
start(k+1) >= start(k) + sum(j in JOBS) DUR(j)*rank(j,k)! Start times
forall(k in JOBS) start(k) >= sum(j in JOBS) REL(j)*rank(j,k)! Completion times

Scheduling problems

95

Applications of optimization with Xpress-MP

forall(k in JOBS) comp(k) = start(k) + sum(j in JOBS) DUR(j)*rank(j,k)
forall(j,k in JOBS) rank(j,k) is_binary! Objective function 1: minimize latest completion time
forall(k in JOBS) finish >= comp(k)minimize(finish)
print_sol(1)! Objective function 2: minimize average completion timeminimize(sum(k in JOBS) comp(k))
print_sol(2)! Objective function 3: minimize total tardinessforall(k in JOBS) late(k) >= comp(k) - sum(j in JOBS) DUE(j)*rank(j,k)
minimize(sum(k in JOBS) late(k))
print_sol(3)!----------------------------------------------------------------! Solution printing
procedure print_sol(obj:integer)
writeln("Objective ", obj, ": ", getobjval,
if(obj>1, " completion time: " + getsol(finish), "") )write("\t")forall(k in JOBS) write(strfmt(getsol(sum(j in JOBS) j*rank(j,k)),4))
write("\nStart\t")forall(k in JOBS) write(strfmt(getsol(start(k)),4))
write("\nEnd\t")forall(k in JOBS) write(strfmt(getsol(comp(k)),4))
write("\nDue\t")forall(k in JOBS) write(strfmt(getsol(sum(j in JOBS) DUE(j)*rank(j,k)),4))if(obj=3) thenwrite("\nLate\t")forall(k in JOBS) write(strfmt(getsol(late(k)),4))
end-ifwriteln
end-procedure
end-model

Similarly to the implementation of the first problem in this chapter, we define a procedure for printing
out the solution that is called after every optimization run. In this example, we want to introduce somevariations in the way the solutions are printed depending on the optimization criterion, we therefore
pass the number of the latter as parameter to the subroutine. For selecting the information that is to beprinted we use two different versions of the if statement: the inline if and if-then that includes a
block of statements.
The reader may have been wondering why we did not use the probably more obvious pair start − end
for naming the variables in this example: end is a keyword of the Mosel language (see Section 5.2.3),
which means that neither end nor END may be redefined by a Mosel program. It is possible though, touse versions combining lower and upper case letters, like End, but to prevent any possible confusion we
do not recommend their use.

7.4.3

Results
The minimum makespan of the schedule is 31, the minimum sum of completion times is 103 (which gives
an average of 103 / 7 = 14. 71). A schedule with this objective value is 4 → 5 → 2 → 6 → 7 → 3 → 1. If
we compare the completion times with the due dates we see that the jobs 1, 2, 3, and 6 finish late (witha total tardiness of 21). The minimum tardiness is 18. A schedule with this tardiness is 5 → 1 → 4 → 6 →2 → 7 → 3 where jobs 4 and 7 finish one time unit late and job 3 is late by 16, it terminates at time 31
instead of being ready at its due date 15. This schedule has an average completion time of 15.71.

7.5

Paint production
As a part of its weekly production a paint company produces five batches of paints, always the same, forsome big clients who have a stable demand. Every paint batch is produced in a single production process,all in the same blender that needs to be cleaned between two batches. The durations of blending paint

Scheduling problems

96

Applications of optimization with Xpress-MP

batches 1 to 5 are respectively 40, 35, 45, 32, and 50 minutes. The cleaning times depend on the colors
and the paint types. For example, a long cleaning period is required if an oil-based paint is produced
after a water-based paint, or to produce white paint after a dark color. The times are given in minutes in
the following table CLEAN where CLEANij denotes the cleaning time between batch i and batch j.
Table 7.7: Matrix of cleaning times
1
2
3
4
5

1

2

3

4

5

0
5
13
9
3

11
0
15
13
7

7
13
0
5
7

13
15
23
0
7

11
15
11
3
0

Since the company also has other activities, it wishes to deal with this weekly production in the shortest
possible time (blending and cleaning). Which is the corresponding order of paint batches? The order
will be applied every week, so the cleaning time between the last batch of one week and the first of thefollowing week needs to be counted for the total duration of cleaning.

7.5.1

Model formulation
We may try to model this problem as an assignment problem like the assignment of personnel to workposts (Chapter 14) and the problem of flight connections (Chapter 11). Let JOBS = {1, . . . , NJ} be the setof batches to produce, DURj the processing time for batch j, and CLEANij the cleaning time between theconsecutive batches i and j. We introduce decision variables succij that take the value 1 if and only ifbatch j succeeds batch i. This means that we need to decide which successor is assigned to every batch i.
We thus find the classical formulation of an assignment problem.
minimize

X

X(DURi + CLEANij ) · succij(7.5.1)i∈JOBS j∈JOBS,j6=i∀i ∈ JOBS :X

succij = 1(7.5.2)

succij = 1(7.5.3)j∈JOBS,j6=i∀j ∈ JOBS :Xi∈JOBS,i6=j∀i, j ∈ JOBS, i 6= j : succij ∈ {0, 1}(7.5.4)Note that we do not use any variables succij with i = j since a batch may not appear more than once inthe production cycle. It would be possible to include these variables and penalize them through a large
value in the diagonal of CLEAN. The constraints (7.5.2) and (7.5.3) guarantee that every batch has a single
successor and a single predecessor.
The objective function (7.5.1) sums up the processing time of i and the cleaning time between i and j forevery pair of batches (i, j). Due to the constraints (7.5.2) and (7.5.3), for every batch i these durations will
only be counted for the batch j that immediately succeeds i (that is, the j for which succij = 1 holds).
Unfortunately, this model does not guarantee that the solution forms a single cycle. Solving it indeed
results in a total duration of 239 with an invalid solution that contains two sub-cycles 1 → 3 → 2 → 1
and 4 → 5 → 4. In Linear Programming, the constraints that force a set of assignment variables to forma single cycle are tricky. A first possibility is to add the constraints (7.5.5) to our model.
∀S ⊆ {2, . . . , NJ} :X

succij ≤ |S| − 1(7.5.5)(i,j)∈S

If a solution contains a sub-cycle with a subset S of batches, the sum of the succij at 1 in S has the samevalue as the number of batches in S. For example, there are three batches (1, 2, and 3) in the first sub-cycleof the previous solution. By imposing a smaller cardinality, the constraints (7.5.5) force the sequence ofbatches to ‘enter’ and ‘leave’ the set S. These constraints concern all subsets S that do not contain the firstbatch (or any other fixed batch), because otherwise even the single cycle we wish to generate would be
excluded. This formulation generates an exponential number of constraints (2NJ−1 ). We therefore choosea weaker formulation with a real variable yj per batch and NJ · (NJ − 1) constraints (7.5.7).

Scheduling problems∀j ∈ JOBS : yj ≥ 0(7.5.6)∀i ∈ JOBS, ∀j = 2, . . . , NJ, i 6= j : yj ≥ yi + 1 − NJ · (1 − succij )(7.5.7)

97

Applications of optimization with Xpress-MP

To understand these constraints, suppose the solution of the assignment problem consists of several subcycles and we choose a sub-cycle that does not contain the batch 1, such as 4 → 5 → 4. The constraints(7.5.7) with succij = 1 for this sub-cycle are:y5 ≥ y4 + 1
y4 ≥ y5 + 1
By combining these constraints, we obtain y4 − 1 ≥ y5 ≥ y4 + 1 or −1 ≥ 1, a contradiction. A solutionthat is divided into sub-cycles is therefore not feasible for the assignment problem with the additional
constraints (7.5.7). If, on the contrary, the cycle is not subdivided, then values for yj exist that fulfill theconstraints. This is for instance the case if yj denotes the rank of the batch j in the cycle, choosing batch
1 as the starting position (with y1 = 1). For the variables succij at 1, the constraints result in yj ≥ yi + 1
and are therefore satisfied. For the 0-valued variables succij they evaluate to yi − yj ≤ NJ − 1 and are alsosatisfied since the yj take values between 1 and NJ, and hence their difference does not exceed NJ − 1.
To summarize, the mathematical model for our paint problem is the assignment problem in lines (7.5.1)
to (7.5.4) with the additional constraints in (7.5.6) and (7.5.7).

7.5.2

Implementation
The transformation of the mathematical model into a Mosel program is fairly straightforward.
model "B-5 Paint production"
uses "mmxprs"
declarations
NJ = 5
JOBS=1..NJ! Number of paint batches (=jobs)

DUR: array(JOBS) of integer! Durations of jobs
CLEAN: array(JOBS,JOBS) of integer ! Cleaning times between jobs
succ: array(JOBS,JOBS) of mpvar
y: array(JOBS) of mpvar
end-declarations! =1 if batch i is followed by batch j,! =0 otherwise! Variables for excluding subtours

initializations from ’b5paint.dat’
DUR CLEAN
end-initializations! Objective: minimize the duration of a production cycle
CycleTime:= sum(i,j in JOBS | i<>j) (DUR(i)+CLEAN(i,j))*succ(i,j)! One successor and one predecessor per batch
forall(i in JOBS) sum(j in JOBS | i<>j) succ(i,j) = 1
forall(j in JOBS) sum(i in JOBS | i<>j) succ(i,j) = 1! Exclude subtours
forall(i in JOBS, j in 2..NJ | i<>j) y(j) >= y(i) + 1 - NJ * (1 - succ(i,j))
forall(i,j in JOBS | i<>j) succ(i,j) is_binary! Solve the problem
minimize(CycleTime)! Solution printing
writeln("Minimum cycle time: ", getobjval)
writeln("Sequence of batches:\nBatch Duration Cleaning")
first:=1
repeatsecond:= integer(sum(j in JOBS | first<>j) j*getsol(succ(first,j)) )writeln(" ",first, strfmt(DUR(first),8), strfmt(CLEAN(first,second),9))first:=second
until (second=1)
end-model

The listing above again includes the solution printing and it makes use of some new features: we use a
repeat-until loop for enumerating the batches in the order that they have been scheduled. For every

Scheduling problems

98

Applications of optimization with Xpress-MP

batch (first) we calculate its successor (second) that then becomes on its turn the one for which wesearch a successor. The loop stops when we are back at the starting point (batch 1). The calculation of thesuccessor is based on the solution values of the successor variables. This sum is of type real, but has aninteger value, and we need to transform it to the type integer. The function integer that is employed
here truncates the value that it is applied to (in many cases it is preferable to use the function roundinstead, which rounds to the nearest integer).

7.5.3

Results
The minimum cycle time for this problem is 243 minutes which is achieved with the following sequenceof batches: 1 → 4 → 3 → 5 → 2 → 1. This time includes 202 minutes of (incompressible) processing
time and 41 minutes of cleaning. This problem is highly combinatorial: for NJ batches, there are (NJ − 1)!
possible cycles. In this simple example, it would be possible to enumerate by hand the 24 possibilities but
already for 10 batches there are 9! = 362,880 different sequences and for 20 batches, 19! ≈ 1. 2 · 1017
sequences.

7.6

Assembly line balancing
An electronics factory produces an amplifier on an assembly line with four workstations. An amplifier isassembled in twelve operations between which there are certain precedence constraints. The following
table indicates the duration of every task (in minutes) and the list of its immediate predecessors (theabbreviation PCB used in this table stands for ‘printed circuit board’).
The production manager would like to distribute the tasks among the four workstations, subject to theprecedence constraints, in order to balance the line to obtain the shortest possible cycle time, that is, thetotal time required for assembling an amplifier. Every task needs to be assigned to a single workstationthat has to process it without interruption. Every workstation deals with a single operation at a time. We
talk about cycle time because the operations on every workstation will be repeated for every amplifier.
When an amplifier is finished, the amplifiers at stations 1 to 3 advance by one station, and the assemblyof a new amplifier is started at the first workstation.
Table 7.8: List of tasks and predecessors

7.6.1

Task

Description

1
2
3
4
5
6
7
8
9
10
11
12

Preparing the box
PCB with power module
PCB with pre-amplifierFilter of the amplifier
Push-pull circuit
Connecting the PCBs
Integrated circuit of the pre-amplifier
Adjusting the connections
Heat sink of the push-pullProtective gridElectrostatic protection
Putting on the coverDuration

Predecessors

3
6
7
6
4
8
9
11
2
13
4
3

–
1
1
2
2
2,3
3
6
4,5,8
8,11
7
9,10

Model formulation
Let TASKS be the set of tasks, DURi the duration of task i, MACH the set of workstations (numberedin the order of the production flow). The precedence relations can be represented as a directed graph
G = (TASKS, ARCS), where ARCS denotes the set of arcs. An arc (i, j) from task i to j symbolizes that i is theimmediate predecessor of j.
To assign the tasks, we define binary variables processim with processim = 1 if and only if task i is assigned to workstation m. The constraints (7.6.1) are required to assure that every task goes to a single
workstation.
X∀i ∈ TASKS :processim = 1(7.6.1)m∈MACHAn assignment is valid only if it fulfills the precedence constraints, which means that for every arc (i, j)
the workstation processing i must have a number that is smaller or equal to the workstation number for

Scheduling problems

99

Applications of optimization with Xpress-MP

6
4
6

2

2

9

4

3

5

1

3

8

11

6

8

12

3

10
13

7
7

11

9

4

Figure 7.8: Graph of tasks with durations

j. This relation is established with the constraints (7.6.2). Note how the numbers of the workstations fori and j are calculated with sums of the assignment variables: due to constraints (7.6.1) only the index mfor which processim = 1 will be counted.

X∀(i, j) ∈ ARCS :k · processim ≤m∈MACHX

k · processjm(7.6.2)m∈MACH

We now introduce a real variable cycle ≥ 0 that represents the cycle time. The constraints (7.6.3) indicatethat this variable is an upper bound on the workload assigned to every workstation.

X∀m ∈ MACH :DURi · processim ≤ cycle(7.6.3)i∈TASKSWe thus obtain the following mathematical model, where all variables with the exception of cycle arebinaries. The objective function (7.6.4) consists of minimizing cycle.
minimize cycle(7.6.4)X∀i ∈ TASKS :processim = 1(7.6.5)m∈MACH∀(i, j) ∈ ARCS :X

k · processim ≤m∈MACH∀m ∈ MACH :X

X

k · processjm(7.6.6)m∈MACH

DURi · processim ≤ cycle(7.6.7)i∈TASKS∀i ∈ TASKS, m ∈ MACH : processim ∈ {0, 1}(7.6.8)cycle ≥ 0

(7.6.9)In this problem formulation, we assume the the production line already exists with its NM workstations
and that we wish to balance the workload. For (re-)designing production lines, it would possible to workwith a given market demand (number of amplifiers per day) and a given cycle time to minimize thenumber of workstations that form the new production line. To be able to generate solutions for thisproblem, the maximum duration of tasks obviously must not exceed the given cycle time.
To solve this new version of the problem, we may take the preceding model, delete the objective function
and replace the variable cycle by a constant (the desired cycle time). We then try to solve the system (7.6.5)
to (7.6.9), initializing the number of machines NM with some lower bound, such as the total duration of
all tasks divided by the cycle time and rounding the resulting value to the next larger integer. If thissystem has no solution, we increment the value of NM and restart the solution algorithm. This process
converges since a trivial solution exists with one task per workstation.

7.6.2

Implementation
The following Mosel program implements the mathematical model defined by lines (7.6.4) – (7.6.9) inthe previous section. It poses a frequently recurring problem, namely the encoding of the graph G. One

Scheduling problems

100

Applications of optimization with Xpress-MP

possibility consists of defining a binary matrix B (called an adjacency matrix), with Bij = 1 if and only if
the arc (i, j) is in G. The second method that is used here lists only the existing arcs and therefore has alower memory consumption for sparse graphs like G: the graph is represented by a table ARC with a line
per arc and two columns. The arc represented by the line a has the source ARCa1 and the sink ARCa2 . The
first data line contains the number 1 and 2 for the arc (1,2), the second 1 and 3 for (1,3) and so on. Since
we initialize the arc numbers dynamically with the data read from file (the range of arc numbers RA is
not fixed in the program), the index-tuples for the entries of the array ARC need to be given in the data
file b6linebal.dat.
model "B-6 Assembly line balancing"uses "mmxprs"
declarations
MACH=1..4
TASKS=1..12! Set of workstations! Set of tasks

DUR: array(TASKS) of integer! Durations of tasks
ARC: array(RA:range, 1..2) of integer ! Precedence relations between tasks
process: array(TASKS,MACH) of mpvar
cycle: mpvar
end-declarations! 1 if task on machine, 0 otherwise! Duration of a production cycle

initializations from ’b6linebal.dat’
DUR ARC
end-initializations! One workstation per task
forall(i in TASKS) sum(m in MACH) process(i,m) = 1! Sequence of tasks
forall(a in RA) sum(m in MACH) m*process(ARC(a,1),m) <=sum(m in MACH) m*process(ARC(a,2),m)! Cycle time
forall(m in MACH) sum(i in TASKS) DUR(i)*process(i,m) <= cycle
forall(i in TASKS, m in MACH) process(i,m) is_binary! Minimize the duration of a production cycle
minimize(cycle)
end-model

7.6.3

Results
The mixed integer solver finds a minimum cycle time of 20 minutes. The assignment of tasks to workstations and the resulting workload per workstation are given in the following table. It is easy to verify thatthe precedence constraints hold.
Table 7.9: Assignment with minimum cycle timeWorkstation number
Assigned tasks
Workload

7.7

1

2

3

4

1,2,4,5
19

3,7,11
20

6,8
19

9,10,1218

References and further materialWith the exception of the stadium construction all problems in this chapter are very difficult combinatorial problems, so-called NP-hard problems (the exact definition of this notion is given in [GJ79]). There isa large variety of scheduling. The works of Carlier and Chretienne [CC88], Lopez and Roubellat [LR99],
Błazewicz et al. [BESW93], and French [Fre82] are good entry points for this very active domain.
The stadium construction problem (Section 7.1) is a project scheduling problem that is now easily solved
and for which a number of software tools are available, such as Project by Microsoft and PSN by ScitorLe Bihan. To answer question 1, there are simple solution methods based on graphs. The graph of a
project may be coded in two different ways. The first consists of representing tasks by arcs and the nodes

Scheduling problems

101

Applications of optimization with Xpress-MP

correspond to the stages of the project. This graph, called AOA (activities on arcs) is used by the PERT
method [Eva64]. In the representation that we have used the nodes correspond to the tasks, this is the
AON (activities on nodes) encoding that is used by the Method of Potentials [Pri94a] [GM90].
Historically, the PERT method stems from the USA and the Method of Potentials from France. Besides
the encoding of the graph, the two methods are largely equivalent, to the point that certain software
provides both methods. Nevertheless, the AOA representation becomes quite impractical outside thecontext of project scheduling, for instance for workshop planning. In this case, the AON notation is usedsystematically.
The question 2 concerns the PERT-cost problem, or scheduling with project crashing, in which the objective is to minimize the total cost if the tasks have flexible durations and the costs depend on thesedurations. Depending on the assumptions, several model formulations are possible: for example, an additional cost that is proportional to the number of days saved per task, or a cost inversely proportional tothe duration of a task. The reader may find other practical cases in Wasil [WA88].
The flow shop problem (Section 7.2) is a classical scheduling problem. The mathematical formulation we
have used is due to Wagner [Wag59]. Other solution methods are described by French [Fre82], pages
132-135, and Pinedo [Pin95a], pages 99-101. Curiously, these works forget to mention the constraints(7.2.6) – (7.2.7) that are absolutely necessary for obtaining correct results.
The flow-shop problem is NP-hard in the general case, and Mathematical Programming only allows solving problems limited to about thirty operations. For large cases, simple heuristics like NEH (from thename of the authors Nawaz, Enscore and Ham [NEH83]) or metaheuristics like tabu search [WH89] areused. The two-machine case is easy and may be solved by an algorithm by Johnson that is described in
the two books just quoted.
The problem in Section 7.3 is a classical scheduling problem called job shop. This problem is NP-hardin the general case. Modeling as a MIP is often inefficient for large problems. Nevertheless, the genericmodel formulation given here is used by Applegate and Cook with cutting plane techniques [AC91] where
the authors obtain interesting results for difficult instances. But the problem is usually solved by listbased heuristics or with exact methods based on a formulation with disjunctive graphs [CCE+ 93], [Fre82],[Pin95b]. Note that the generic model formulation is also valid for the general flow shop problem (whilst
the model of Section 7.2 only applies to the permutation flow shop).
Different solution algorithms are available for one-machine problems (Section 7.4). The aim of this exercise was to give a common model formulation for the three objectives. Other models using variables
relative to the position of jobs may be found in [LQ92] [Sev98].
The paint production problem (Section 7.5) is a representative of a family of scheduling problems where
the preparation before every task depends on the chosen order (sequence-dependent setup times). We
have in fact used one of the models for the famous traveling salesman problem (TSP). In this problem,
the jobs (batches) become cities and the cleaning times the cost of traveling from one city to the next.
The objective is to find the minimum cost for the journey of a sales representative that starts from hishome in city 1, visits every city once, and finally returns to his starting point.
Here we have considered the asymmetric case, that is, the cleaning time between batch i and batch j is in
general different from the one between j and i. In Chapter 11, we study a symmetric TSP, with distances
between cities as the crow flies. It is solved with a different technique, by progressively adding constraintsof type (5). Since the TSP is NP-hard, the models presented often take too long to be solved for more thantwenty cities. Beyond this limit, specialized tree search methods [HK70] need to be used. Demonstration
versions of such methods are available [Ros85] [EM92], and a Pascal code is given by Syslo [SDK83]. For
cases with hundreds of cities one needs to give up on optimality and use heuristic techniques, some ofwhich like tabu search (e.g. [GTdW93]) or simulated annealing ([KGV83]) are in practice quite efficient.
Such heuristics are explained with some detail in [Pri94a].
The problem of assembly line balancing (Section 7.6) is NP-hard for the two given versions (minimizing thecycle time or the number of workstations). Without precedence constraints, the first version is equivalent
to a scheduling problem with parallel machines (see the wagon loading problem in Chapter 9) whilst thesecond is reduced to a bin-packing problem (the problem of saving files to disks in Chapter 9). Problems of
about twenty tasks can be dealt with in Mathematical Programming, beyond this limit heuristics may be
used like the one by Dar-El [DE73], described together with another heuristic in the book on production
planning by Buffa and Sarin, pages 660-671 [BS87]. Scholl has recently published a book dedicated toassembly line balancing problems [Sch99].

Scheduling problems

102

Applications of optimization with Xpress-MP

Chapter 8

Planning problems
The term planning is often used with three different meanings in the context of production management:
long-term planning that is used for taking strategic decisions, mid-term planning that corresponds tothe tactical level, and short-term planning for operational decisions. Long-term planning concerns the
direction a company is taking over several years, like the depot location problem in Chapter 10. Shortterm planning takes detailed production decisions for the shop floor, with a time horizon of a few days.
Most importantly it comprises scheduling problems such as those described in Chapter 7. The present
chapter is dedicated to tactical planning, currently referred to as production planning.
This decision level often covers a time horizon of one or several months, working for instance with timeslices of the length of a week. It takes a macroscopic view of the production system: the detailed constraints for workshops and machines are ignored, the operations are dealt with in an aggregated fashion.
The objective is to determine the best quantities to produce in every time period whilst, in a majority ofcases, minimizing the cost incurred through production and storage.
Section 8.1 presents a simple case of single product planning, namely bicycles. The following section(8.2) deals with the production of several products (different types of glasses). In Section 8.3 we plan the
production of toy lorries with all their components. This is a typical case of Material Requirement Planning(MRP). The two following problems have more complex cost functions: the production of electronic
components in Section 8.4 includes costs for changing the level of production, and for the production offiberglass in Section 8.5 the cost depends on the time period. The problem of Section 8.6 stands apart: it
concerns the assignment of product batches to machines of different capacities and speeds.

8.1

Planning the production of bicycles
A company produces bicycles for children. The sales forecast in thousand of units for the coming year are
given in the following table. The company has a capacity of 30,000 bicycles per month. It is possible toaugment the production by up to 50% through overtime working, but this increases the production cost
for a bicycle from the usual B
C 32 to B
C 40.
Table 8.1: Sales forecasts for the coming year in thousand units
Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

30

15

15

25

33

40

45

45

26

14

25

30

Currently there are 2,000 bicycles in stock. The storage costs have been calculated as B
C 5 per unit held instock at the end of a month. We assume that the storage capacity at the company is virtually unlimited(in practice this means that the real capacity, that is quite obviously limited, does not impose any limits in
our case). We are at the first of January. Which quantities need to be produced and stored in the courseof the next twelve months in order to satisfy the forecast demand and minimize the total cost?

8.1.1

Model formulation
The variables that we need to determine are the numbers of bicycles pnormt and povert to be produced
respectively in normal working hours and in overtime hours during month t, and the number of bicycles
storet held in stock at the end of the month. Let MONTHS = {1, . . . , 12} be the set of time periods. The
objective is to minimize the total cost, that is, the sum of cost of production (in normal and additional
hours) and the storage cost. We write CNORM and COVER for the production cost of a bicycle in normal

103

Applications of optimization with Xpress-MP

and overtime hours, and CSTOCK the monthly storage cost per bicycle. With these conventions we obtain
the objective function (8.1.1).
minimize

X(CNORM · pnormt + COVER · povert + CSTOCK · storet )(8.1.1)t∈MONTHS

In every month t, the available quantity of bicycles is given as the sum of bicycles in stock at the end ofmonth t − 1 and the bicycles produced in normal and overtime hours during this month. This sum must
equal the number of bicycles sold in the course of month t, plus the number of bicycles held in stock at
the end of month t. This key relation is called the inventory balance equation, and forms an importantcomponent of mathematical models for production planning problems.
Let us call the initial stock level ISTOCK and DEMt the forecasted demand for month t. The balance
constraint for the first month is then given by relation (8.1.2)
pnorm1 + pover1 + ISTOCK = DEM1 + store1(8.1.2)The constraints (8.1.3) establish this relation for all following months.
∀t ∈ MONTHS, t 6= 1 : pnormt + povert + storet−1 = DEMt + storet(8.1.3)The production capacity in normal working hours is limited by the given capacity CAP, and the additional
production in overtime hours is limited to 50% of this capacity; hence the constraints (8.1.4) and (8.1.5).
∀t ∈ MONTHS : pnormt ≤ CAP(8.1.4)∀t ∈ MONTHS : povert ≤ 0. 5 ∗ CAP(8.1.5)

And finally, we note that all variables are non-negative (constraints (8.1.6)).
∀t ∈ MONTHS : pnormt ≥ 0, povert ≥ 0, storet ≥ 0

8.1.2(8.1.6)

Implementation
The mathematical model may be translated into the following Mosel program.
model "C-1 Bicycle production"
uses "mmxprs"
declarations
TIMES = 1..12! Range of time periods

DEM: array(TIMES) of integer
CNORM,COVER: integer
CSTOCK: integer
CAP: integer
ISTOCK: integer! Demand per months! Prod. cost in normal / overtime hours! Storage cost per bicycle! Monthly capacity in normal working hours! Initial stock

pnorm:array(TIMES) of mpvar
pover:array(TIMES) of mpvar
store:array(TIMES) of mpvar
end-declarations! No. of bicycles produced in normal hours! No. of bicycles produced in overtime hours! No. of bicycles stored per monthinitializations from ’c1bike.dat’
DEM CNORM COVER CSTOCK CAP ISTOCK
end-initializations! Objective: minimize production cost
Cost:= sum(t in TIMES) (CNORM*pnorm(t) + COVER*pover(t) + CSTOCK*store(t))! Satisfy the demand for every period
forall(t in TIMES)
pnorm(t) + pover(t) + if(t>1, store(t-1), ISTOCK) = DEM(t) + store(t)! Capacity limits on normal and overtime working hours per monthforall(t in TIMES) do
pnorm(t) <= CAP
pover(t) <= 0.5*CAP
end-doPlanning problems

104

Applications of optimization with Xpress-MP

! Solve the problem
minimize(Cost)
end-model

In this implementation we use the inline if function to condense the balance constraints for the first andall following periods into a single constraint. For t = 1 the expression evaluates topnorm(1) + pover(1) + ISTOCK = DEM(1) + store(1)while for the rest of the periods we havepnorm(t) + pover(t) + store(t-1) = DEM(t) + store(t)8.1.3

Results
The minimum cost for production and storage is B
C 11,247,000. The following table list the production
plan for the coming year (in thousands) — its first line recalls the forecast monthly demand. In January
to April and September to December the demands can be satisfied with production in normal hours. In
April, 3000 bicycles need to be stored to satisfy the higher demand in May. In June to August overtime
work is required to produce the required quantities.
Table 8.2: Quantities to produce and store in thousand units
DemandNormalAdditional
Store

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

30
28
–
–

15
15
–
–

15
15
–
–

25
28
–
3

33
30
–
–

40
30
10
–

45
30
15
–

45
30
15
–

26
26
–
–

14
14
–
–

25
25
–
–

30
30
–
–

Note that in practice one might not like a solution like this: the inventory at the month end is almost
always 0, so if there were a surge in demand (e.g. unusually good weather) the company would be introuble.

8.2

Production of drinking glasses
The main activity of a company in northern France is the production of drinking glasses. It currently
sells six different types (V1 to V6), that are produced in batches of 1000 glasses, and wishes to plan its
production for the next 12 weeks. The batches may be incomplete (fewer than 1000 glasses). The demandin thousands for the 12 coming weeks and for every glass type is given in the following table.
Table 8.3: Demands for the planning period (batches of 1000 glasses)
Week

1

2

3

4

5

6

7

8

9

10

11

12

V1
V2
V3
V4
V5
V6

20
17
18
31
23
22

22
19
35
45
20
18

18
23
17
24
23
20

35
20
10
38
15
19

17
11
9
41
10
18

19
10
21
20
22
35

23
12
23
19
18
0

20
34
15
37
30
28

29
21
10
28
28
12

30
23
0
12
7
30

28
30
13
30
15
21

32
12
17
37
10
23

For every glass type the initial stock is known, as well as the required final stock level (in thousands). Per
batch of every glass type, the production and storage costs in B
C are given, together with the required
working time for workers and machines (in hours), and the required storage space (measured in numbersof trays).
The number of working hours of the personnel is limited to 390 hours per week, and the machines havea weekly capacity of 850 hours. Storage space for up to 1000 trays is available. Which quantities of thedifferent glass types need to be produced in every period to minimize the total cost of production andstorage?

Planning problems

105

Applications of optimization with Xpress-MP

Table 8.4: Data for the six glass types

V1
V2
V3
V4
V5
V6

8.2.1

Production
cost

Storage
cost

Initial
stock

Final
stock

Timeworker

Timemachine

Storage
space

100
80
110
90
200
140

25
28
25
27
10
20

50
20
0
15
0
10

10
10
10
10
10
10

3
3
3
2
4
4

2
1
4
8
11
9

4
5
5
6
4
9

Model formulation
This problem is a generalization of the bicycle problem in Section 8.1: we now have several products,
each using the same resources.
To express the model in a simple way, we need to consider the time periods one after the other. Let
PRODS be the set of products (glass types) and WEEKS = {1, . . . , NT} the set of time periods. We write
DEMpt for the demand for product p in time period t. These demands are given above in Table 8.3. We
also have CPRODp and CSTOCKp the production and storage cost for glass type p. This cost is identical for
all time periods, but it would be easy to model a different cost per time period by adding an index forthe time period.
TIMEWp and TIMEMp denote the worker and machine times respectively required per unit of product p,
and correspondingly, SPACEp the storage area. The initial stock ISTOCKp is given, as is the desired final
stock level FSTOCKp per product. All these data are listed in Table 8.4 above. We write CAPW and CAPM
for the capacities of workers and machines respectively, and CAPS for the capacity of the storage area.
To solve this problem, we need variables producept to represent the production of glass type p in time
period t. The variables corresponding to the stock level of every product p at the end of period t are
called storept . By convention, the initial stock level ISTOCKp may be considered as the stock level at theend of time period 0 and we use the notation storep0 to simplify the formulation of the stock balance
constraints (8.2.1):∀p ∈ PRODS, t ∈ WEEKS : storept = storep,t−1 + producept − DEMpt(8.2.1)

These stock balance constraints state that the quantity storept of product that is held in stock at the endof a time period t equals the stock level storep,t−1 at the end of the preceding period plus the production
producept of the time period t minus the demand DEMpt of this time period.
The company wishes to have a certain amount of product in stock at the end of the planning period.
These constraints on the final stock levels are expressed by the constraints (8.2.2):(8.2.2)∀p ∈ PRODS : storep,NT ≥ FSTOCKp

We now formulate the various capacity constraints for every time period. The following constraints
guarantee that the capacity limits on manpower (8.2.3), machine time (8.2.4), and storage space (8.2.5)
are kept:∀t ∈ WEEKS :X

TIMEWp · producept ≤ CAPW

(8.2.3)

TIMEMp · producept ≤ CAPM

(8.2.4)SPACEp · producept ≤ CAPS

(8.2.5)p∈PRODS∀t ∈ WEEKS :X
p∈PRODS∀t ∈ WEEKS :X
p∈PRODSWe may now formulate the cost function that is to be minimized (8.2.6). This function is the sum of
production and storage costs for all products and time periods.
minimize

X

X

CPRODp · producept + CSTOREp · storept

(8.2.6)p∈PRODS t∈WEEKSWe obtain the complete mathematical model by combining lines (8.2.1) to (8.2.6), to which we needto add the non-negativity constraints for the production variables (8.2.13) and for the stored quantities

Planning problems

106

Applications of optimization with Xpress-MP

(8.2.14).

X

X



(8.2.7)∀p ∈ PRODS, t ∈ WEEKS : storept = storep,t−1 + producept − DEMpt(8.2.8)∀p ∈ PRODS : storep,NT ≥ FSTOCKp(8.2.9)X

TIMEWp · producept ≤ CAPW

(8.2.10)

TIMEMp · producept ≤ CAPM

(8.2.11)SPACEp · producept ≤ CAPS

(8.2.12)minimize

CPRODp · producept + CSTOREp · storeptp∈PRODS t∈WEEKS∀t ∈ WEEKS :p∈PRODS∀t ∈ WEEKS :X
p∈PRODS∀t ∈ WEEKS :X
p∈PRODS

8.2.2∀p ∈ PRODS, t ∈ WEEKS : producep,t ≥ 0(8.2.13)∀p ∈ PRODS, t ∈ WEEKS : storep,t ≥ 0(8.2.14)

Implementation
The mathematical model may be implemented with Mosel as follows.
model "C-2 Glass production"
uses "mmxprs"
declarations
NT = 12WEEKS = 1..NT
PRODS = 1.. 6! Number of weeks in planning period! Set of products

CAPW,CAPM: integer! Capacity of workers and machines
CAPS: integer! Storage capacity
DEM: array(PRODS,WEEKS) of integer! Demand per product and per week
CPROD: array(PRODS) of integer! Production cost per product
CSTOCK: array(PRODS) of integer! Storage cost per product
ISTOCK: array(PRODS) of integer! Initial stock levels
FSTOCK: array(PRODS) of integer! Min. final stock levels
TIMEW,TIMEM: array(PRODS) of integer ! Worker and machine time per unit
SPACE: array(PRODS) of integer! Storage space required by products
produce: array(PRODS,WEEKS) of mpvar ! Production of products per week
store: array(PRODS,WEEKS) of mpvar! Amount stored at end of week
end-declarations
initializations from ’c2glass.dat’
CAPW CAPM CAPS DEM CSTOCK CPROD ISTOCK FSTOCK TIMEW TIMEM SPACE
end-initializations! Objective: sum of production and storage costs
Cost:=
sum(p in PRODS, t in WEEKS) (CPROD(p)*produce(p,t) + CSTOCK(p)*store(p,t))! Stock balances
forall(p in PRODS, t in WEEKS)
store(p,t) = if(t>1, store(p,t-1), ISTOCK(p)) + produce(p,t) - DEM(p,t)! Final stock levels
forall(p in PRODS) store(p,NT) >= FSTOCK(p)! Capacity constraints
forall(t in WEEKS) dosum(p in PRODS) TIMEW(p)*produce(p,t) <= CAPW
sum(p in PRODS) TIMEM(p)*produce(p,t) <= CAPM
sum(p in PRODS) SPACE(p)*store(p,t)<= CAPS
end-do! Workers! Machines! Storage! Solve the problem
minimize(Cost)
end-model

Planning problems

107

Applications of optimization with Xpress-MP

In the model formulation above as before we use the if function in the formulation of the stock balance
constraints. It is also possible to employ the notation storep0 for the initial stock levels as in the mathematical model. Since store is an array of variables, we still need to use the array ISTOCK for reading in
the data from file and then fix the storage variables for time period 0 to these values (additional set ofbound constraints):
declarations
store: array(PRODS,0..NT) of mpvar
end-declarations! Amount stored at end of week

! Stock balances
forall(p in PRODS, t in WEEKS)
store(p,t) = store(p,t-1) + produce(p,t) - DEM(p,t)! Fix the initial stock levels
forall(p in PRODS) store(p,0) = ISTOCK(p)In this model, it is easy to change the objective function, for instance to take into account only the
production costs (first part of the objective) or the storage costs (second part of the objective)
Remark: in the implementation of this model, it may be tempting to define names like prod for the
production variables or PROD for the set of products, but both these names cannot be used because
prod is a keyword of the Mosel language (see the complete list of reserved words in Section 5.2.3).

8.2.3

Results
The solution of this problem gives us a total cost of B
C 185,899.
Table 8.5 displays the quantities of the different glass types to produce in every week of the planning
period and the quantities held in stock at the end of every week. Some production and storage quantities
are fractional but every unit corresponds to 1000 glasses. We may therefore round the results to three
digits after the decimal point to obtain a result with integral numbers of glasses without modifying thetotal cost too much.
Taking for example the quantities to be produced in the first period: 8,760 glasses of type 1, 18,000
glasses of type 3, 16,000 glasses of type 4, 47,680 glasses of type 5, 12,000 glasses of type 6 and noneof type 2 (the demand for this type is covered by the initial stock). When looking at the constraints inmore detail, we find that the available manpower is fully used most of the time (in the first week, 351
hours are worked, in all other weeks the limit of 390 hours is reached) and the machines are used to theirfull capacity in certain time periods (weeks 1-3 and 5), but the available storage space is in excess of theactual needs.
Table 8.5: Quantities to produce and store of every glass type (in thousands)
Week
1 Prod.
Store
2 Prod.
Store
3 Prod.
Store
4 Prod.
Store
5 Prod.
Store
6 Prod.
Store

1

2

8.76
5.48
38.76 22.24
0
16
3
–
18
35
–
–
16
45
–
–
47.68 14.64
24.68 19.32
12
18
–
–

3

4

5

6

7

8

0.56
30.2 27.36 8.64
23
20
4.8
10.36
–
–
–
23
20
11
10
12
34
–
–
–
–
–
–
17
10
9
21
23
15
–
–
–
–
–
–
24
38
41
20
19
37
–
–
–
–
–
–
35.08 14.35 23.48 22.77 43.75
031.4 30.75 44.23
4570.75 40.75
20
19
18
35
0.75 27.25
–
–
–
–
0.75
–

9
29
–
21
–
10
–
28
–
26.5
39.25
12
–

10

11

12

30
28
42
–
–
10
23
30
22
–
–
10
0
13
27
–
–
10
12
30
47
–
–
10
2.75
0
0
35
20
10
49 29.25 5.75
19 27.25 10

As opposed to the bicycle problem in Section 8.1, in this problem final stock levels have been given. The
specification of final stock levels is quite a tricky issue in production planning models. If one does not
specify levels, then models will typically run stocks down to zero at the end of the planning horizon, asituation which will usually be unacceptable in practice. If the final stock levels are different from theinitial stock levels, as in this example, then care must be taken in interpreting average costs, as somebenefit has accrued if stock levels are being run down, and vice versa if stock is being increased.

Planning problems

108

Applications of optimization with Xpress-MP

8.3

Material Requirement Planning
The company Minorette produces two types of large toy lorries for children: blue removal vans and redtank lorries. Each of these lorries is assembled from thirteen items. See Figure 8.1 for the breakdown ofcomponents (also called a Gozinto graph or Parts explosion) and the following Table 8.6 for the prices of
the components.
Table 8.6: Prices of components
Wheel
B
C 0.30

Steel bar
B
C1

BumperB
C 0.20ChassisB
C 0.80Cabin
B
C 2.75

Door window
B
C 0.10Windscreen
B
C 0.29Blue containerB
C 2.60

Red tank
B
C3Blue motor
B
C 1.65

Red motor
B
C 1.65

Headlight
B
C 0.15

Blue lorry(or red)1

1
Blue container(or red tank)

Assembled
chassis2

2

BumperAxle

1

1
Chassis

2

1

Wheel

Steel bar

1

Assembledcabin

2Blue motor(or red)

1

2

Cabin

Door
Windscreen
window

Headlight

1

Figure 8.1: Breakdown of components (Gozinto graph)

The subsets (axles, chassis, blue or red cabin) may be assembled by the company using the components,
or subcontracted. The following table lists the costs for assembling and for subcontracting these subsets,
together with the capacities of the company. The assembly costs do not take into account the buying
prices of their components.
Table 8.7: Subcontracting and assembly costs, assembly capacities
Subcontracting
Assembly
Capacity

Axle

Assembled chassis

Assembled cabinBlue lorry

Red lorryB
C 12.75
B
C 6.80
600

B
C 30
B
C 3.554000

B
C3B
C 3.203000–
B
C 2.204000–
B
C 2.605000

For the next month, Minorette has the same demand forecast of 3000 pieces for the two types of lorries.
At present, the company has no stock. Which quantities of the different items should Minorette buy or
subcontract to satisfy the demand while minimizing the production cost?

8.3.1

Model formulation
Let ITEMS be the complete set of all products with the subsets FINAL of final products, ASMBL of products
resulting from assembly, and PREPROD of products used in the assembly of others. We write REQqp forthe requirement of component p in the assembly of q. For instance, for an axle we need two wheels andone steel bar, that is REQaxle,wheel = 2 and REQaxle,steelbar = 1.
Let CBUYp be the price paid when buying item p. Since subcontracting a product means buying it fromits producer, we consider that this cost is the price for buying a preproduct that is not assembled and is
the cost of subcontracting for the assembled pieces.
We write CPRODp for the cost of assembling a product p. The variables that we need to determine are
the quantities to produce producep of the assembled products p and those to buy (or subcontract) buyp

Planning problems

109

Applications of optimization with Xpress-MP

of every preproduct p. This gives the following mathematical model:Xminimize

CBUYp · buyp +p∈PREPRODX

CPRODp · producep(8.3.1)p∈ASMBL∀p ∈ FINAL : producep ≥ DEMp

(8.3.2)∀p ∈ PREPROD :(8.3.3)p ∈ ASMBL : buyp + producep ≥X

REQqp · produceqq∈ASMBLX

p 6∈ ASMBL : buyp ≥REQqp · produceqq∈ASMBL∀p ∈ ASMBL : producep ≤ CAPp

(8.3.4)∀p ∈ ASMBL : producep ∈ IN

(8.3.5)∀p ∈ PREPROD : buyp ∈ IN

(8.3.6)The objective function (8.3.1) minimizes the production cost, that is, the buying price or cost for subcontracting and the assembly cost. The constraints (8.3.2) indicate that the produced quantities of the
final products (blue and red lorries) must be greater than or equal to the demand DEMp . The constraints(8.3.3) guarantee that the total quantity of item p that is bought or produced is sufficient to produce allarticles q that contain p.
The constraints (8.3.4) establish the limits on the assembly capacities CAPp for every assembled product p.
The constraints (8.3.5) and (8.3.6) are the integrality constraints for the variables.
Note that an alternative formulation to the model displayed above consists of defining buying and assembly costs for all items: preproducts that cannot be assembled receive the value ‘infinity’ as their assembly
cost, and this value is also assigned as the buying price to the final products. With these costs, we may
define variables producep and buyp for all products and in the objective function (1) and the constraints(3) simply sum over all variables — the ‘infinity’ cost values force the corresponding variables to take thevalue 0. Whilst easier to write down, this second formulation introduces large coefficient values into themodel that typically lead to numerical instabilities and should therefore be avoided.

8.3.2

Implementation
The following Mosel program implements the mathematical model of lines (8.3.1) to (8.3.6). All data andindex sets are defined in the data file and initialized dynamically when executing the program.
model "C-3 Toy production"
uses "mmxprs"
declarations
ITEMS: set of string
FINAL: set of string
ASMBL: set of string
PREPROD: set of string! Set of all products! Set of final products! Set of assembled products! Set of preproducts

CAP: array(ASMBL) of integer! Capacity of assembly lines
DEM: array(FINAL) of integer! Demand of lorries
CPROD: array(ASMBL) of real
! Assembly costs
CBUY: array(ITEMS) of real
! Purchase costs
REQ: array(ASMBL,PREPROD) of integer ! Items req. for assembling a productend-declarations
initializations from ’c3toy.dat’
DEM CBUY REQ[CPROD, CAP] as ’ASSEMBLY’
end-initializations
finalize(ASMBL); finalize(PREPROD); finalize(FINAL); finalize(ITEMS)
declarations
produce: array(ASMBL) of mpvar
buy: array(PREPROD) of mpvar
end-declarations! Quantity of items produced! Quantity of items bought! Objective: total costs
Cost:=sum(p in PREPROD) CBUY(p)*buy(p) + sum(p in ASMBL) CPROD(p)*produce(p)Planning problems

110

Applications of optimization with Xpress-MP

! Satisfy demands
forall(p in FINAL) produce(p) >= DEM(p)! Assembly balance
forall(p in PREPROD) buy(p) + if(p in ASMBL, produce(p), 0) >=sum(q in ASMBL) REQ(q,p)*produce(q)! Limits on assembly capacity
forall(p in ASMBL) produce(p) <= CAP(p)forall(p in PREPROD) buy(p) is_integer
forall(p in ASMBL) produce(p) is_integer! Solve the problem
minimize(Cost)
end-model

As opposed to most other models presented in this book, in this implementation we do not fix theindex sets directly in the Mosel program but initialize all data dynamically from file. We only define thevariables once the data has been read in and the index sets have been finalized — that is, the arrays ofvariables are created with fixed sizes. If we defined the variables in the same declarations block as
the data arrays, this array would be created as a dynamic array, just like the data arrays. But unlike thedata arrays, in a dynamic array of variables every entry needs to be created explicitly, using the function
create. It is therefore also possible to implement the Mosel program as shown below, though it should
be noted that the above version with fixed-size arrays is slightly more efficient:declarations
ITEMS: set of string
FINAL: set of string
ASMBL: set of string
PREPROD: set of string! Set of all products! Set of final products! Set of assembled products! Set of pre-products

CAP: array(ASMBL) of integer! Capacity of assembly lines
DEM: array(FINAL) of integer! Demand of lorries
CPROD: array(ASMBL) of real
! Assembly costs
CBUY: array(ITEMS) of real
! Purchase costs
REQ: array(ASMBL,PREPROD) of integer ! Items req. for assembling a productproduce: array(ASMBL) of mpvar
buy: array(PREPROD) of mpvar
end-declarations! Quantity of items produced! Quantity of items bought

initializations from ’c3toy.dat’
DEM CBUY REQ[CPROD, CAP] as ’ASSEMBLY’
end-initializations
forall(p in ASMBL) create(produce(p))
forall(p in PREPROD) create(buy(p))

In this example, the data of the arrays CPROD and CAP are contained in a single record ASSEMBLY andthese two arrays are therefore read jointly using the keyword as to indicate the label of the record.

8.3.3

Results
The minimum total cost for satisfying the demand of toy lorries is B
C 238,365. The next two tables summarize the quantities of products that are bought, subcontracted, or assembled at the factory.
Table 8.8: Quantities of preproducts bought

Planning problems

Wheel
1200

Steel bar
600

Bumper600

Chassis
300

Cabin
0

Door window0

Windscreen
0Blue container
3000

Red tank
3000

Blue motor
3000

Red motor
3000

Headlight
12000

111

Applications of optimization with Xpress-MP

Table 8.9: Production and subcontracting of assembled products
Produce
Subcontract

8.4

Axle

Assembled chassis

Assembled cabinBlue lorry

Red lorry

600
0

300
5700

0
6000

3000
–

3000
–

Planning the production of electronic components
To augment its competitiveness a small business wishes to improve the production of its best selling
products. One of its main activities is the production of cards with microchips and electronic badges. The
company also produces the components for these cards and badges. Good planning of the productionof these components therefore constitutes a decisive success factor for the company. The demand forcomponents is internal in this case and hence easy to anticipate.
For the next six months the production of four products with references X43-M1, X43-M2, Y54-N1, Y54-N2
is to be planned. The production of these components is sensitive to variations of the level of production, and every change leads to a non-negligible cost through controls and readjustments. The company
therefore wishes to minimize the cost associated with these changes whilst also taking into account the
production and storage costs.
The demand data per time period, the production and storage costs, and the initial and desired final
stock levels for every product are listed in the following table. When the production level changes,
readjustments of the machines and controls have to be carried out for the current month. The cost
incurred is proportional to the increase or reduction of the production compared to the preceding month.
The cost for an increase of the production is B
C 1 per unit but only B
C 0.50 for a decrease of the productionlevel.
Table 8.10: Data for the four products
Month

1

2

X43-M1
X43-M2
Y54-N1
Y54-N2

1500
1300
2200
1400

3000
800
1500
1600

Demands
3
4
2000
800
2900
1500

4000
1000
1800
1000

5

6

2000
1100
1200
1100

2500
900
2100
1200

Cost
Production Storage
20
25
10
15

0.4
0.5
0.3
0.3

Stock levels
Initial Final
10
0
0
0

50
10
10
10

What is the production plan that minimizes the sum of costs incurred through changes of the productionlevel, production and storage costs?

8.4.1

Model formulation
The mathematical model developed for the problem in Section 8.2 has many similarities to this model.
We denote the variables and constants in the same way. Let PRODS be the set of components, MONTHS ={1, . . . , NT} the set of planning periods, DEMpt the demand for product p in time period t, CPRODp and
CSTOCKp respectively the cost of producing and storing one unit of product p, and ISTOCKp and FSTOCKp
the initial and final stock levels of product p.
The variables producept and storept represent respectively the quantity produced and the amount of
product p in stock at the end of period t. With the convention storep0 = ISTOCKp the stock balance
constraints are given by (8.4.1) and the final stock levels are guaranteed by constraints (8.4.2).
∀p ∈ PRODS, t ∈ MONTHS : storept = storep,t−1 + producept − DEMpt(8.4.1)∀p ∈ PRODS : storep,NT ≥ FSTOCKp(8.4.2)

To measure the changes to the level of production, we need to additional sets of variables, reducet andaddt that represent the reduction or increase of the production in time period t. A change the the levelof production is simply the difference between the total quantity produced in period t and the total
quantity produced in period t − 1. If this difference is positive, we have an increase in the productionlevel, and a reduction otherwise. The change is expressed through the value of addt − reducet .
Since the level of production cannot increase and reduce at the same time, one of the two variables is
automatically at 0; whilst the other represents either the increase or the reduction of the production.
This leads to the equation (8.4.3).
∀t ∈ {2, . . . , NT} :X

producept −p∈PRODS

Planning problems

X

producep,t−1 = addt − reducet(8.4.3)p∈PRODS

112

Applications of optimization with Xpress-MP

The objective function is like the one of problem 8.2 with, in addition, the cost of changes to the production level. We write CADD and CRED for the cost of increasing and reducing the production respectively.
These costs are proportional to the changes. They form the second part of the objective function (8.4.4).

X

minimizeX

CPRODp · producept + CSTOREp · storept

p∈PRODS t∈MONTHS

+

NT
X

CRED · reducet + CADD · addt

(8.4.4)t=2Note that there are no variables reduce1 or add1 because there is no change at the beginning of theplanning period.
The complete linear program contains the lines (8.4.4) to (8.4.3), to which we need to add the nonnegativity constraints (8.4.9) and (8.4.10) for all variables:XminimizeX

CPRODp · producept + CSTOREp · storept

p∈PRODS t∈MONTHS

+

NT
X

CRED · reducet + CADD · addt

(8.4.5)t=2∀p ∈ PRODS, t ∈ MONTHS : storept = storep,t−1 + producept − DEMpt(8.4.6)∀p ∈ PRODS : storep,NT ≥ FSTOCKp(8.4.7)∀t ∈ {2, . . . , NT} :X

producept −p∈PRODS

8.4.2

X

producep,t−1 = addt − reducet(8.4.8)p∈PRODS∀t ∈ {2, . . . , NT} : reducet ≥ 0, addt ≥ 0(8.4.9)∀p ∈ PRODS, t ∈ MONTHS : producep,t ≥ 0, storep,t ≥ 0(8.4.10)

Implementation
As we have noted, the model of this problem is close to the one of the problem in Section 8.2. A similar
likeness can be observed between the two implementations with Mosel.
model "C-4 Electronic components"
uses "mmxprs"
declarations
NT = 6MONTHS = 1..NT
PRODS = 1..4

! Number of time periods (months)! Set of components

DEM: array(PRODS,MONTHS) of integer! Demand of components per month
CPROD: array(PRODS) of integer! Production costs
CSTOCK: array(PRODS) of real
! Storage costs
CADD,CRED: real
! Cost of additional/reduced prod.
ISTOCK,FSTOCK: array(PRODS) of integer ! Initial and final stock levels
produce: array(PRODS,MONTHS) of mpvar
store: array(PRODS,MONTHS) of mpvar
reduce: array(MONTHS) of mpvar
add: array(MONTHS) of mpvar
end-declarations! Quantities produced per month! Stock levels at end of every month! Reduction of production per month! Increase of production per monthinitializations from ’c4compo.dat’
DEM CPROD CSTOCK CADD CRED ISTOCK FSTOCK
end-initializations! Objective: total cost of production, storage, and change of prod. level
Cost:= sum(p in PRODS, t in MONTHS) (CPROD(p)*produce(p,t) +CSTOCK(p)*store(p,t)) +sum(t in 2..NT) (CRED*reduce(t) + CADD*add(t))! Stock balance constraints (satisfy demands)
forall(p in PRODS, t in MONTHS)
store(p,t) = if(t>1, store(p,t-1), ISTOCK(p)) + produce(p,t) - DEM(p,t)! Changes to the level of production

Planning problems

113

Applications of optimization with Xpress-MP

forall(t in 2..NT)sum(p in PRODS) (produce(p,t) - produce(p,t-1)) = add(t) - reduce(t)! Guarantee final stock levels
forall(p in PRODS) store(p,NT) >= FSTOCK(p)! Solve the problem
minimize(Cost)
end-model

8.4.3

Results
When solving the problem we obtain a total cost of B
C 683,929. The production level for the first four
periods is 7060 units in total, in period 5 it decreases to 6100 units and remains at this level for the lastperiod. The following table lists the corresponding production plan (there are several possible plans).
Table 8.11: Optimal production plan
1

2

3

4

5

6

X43-M1
X43-M2
Y54-N1
Y54-N2

1490
1300
2150
2120

3000
800
1500
1760

2000
800
3640
620

4000
1000
1060
1000

2000
1100
1200
1800

2550
910
2130
510

Total

7060

7060

7060

7060

6100

6100

The stock levels are relatively low. With the exception of the last period where a final stock is required,products X43-M1 and X43-M2 are never held in stock. The product Y54-N1 is held in stock in periods 3
and 6 (740 and 30 units), and the product Y54-N2 is held in stock in the periods 1, 2, 5, and 6 with 720,
880, 700, and 10 units respectively.

8.5

Planning the production of fiberglass
A company produces fiberglass by the cubic meter and wishes to plan its production for the next six
weeks. The production capacity is limited, and this limit takes a different value in every time period.
The weekly demand is known for the entire planning period. The production and storage costs also takedifferent values depending on the time period. All data are listed in the following table.
Table 8.12: Data per week
Week

Production
capacity (m3 )Demand(m3 )Production
cost (BC/m3 )

Storagecost (BC/m3 )

1
2
3
4
5
6

140
100
110
100
120
100

100
120
100
90
120
110

5
8
6
6
7
6

0.2
0.3
0.2
0.25
0.3
0.4

Which is the production plan that minimizes the total cost of production and storage?

8.5.1

Model formulation
This problem is not much different from the previous ones and it would be easy to solve it using oneof the formulations presented earlier in this chapter. To introduce some variation, we are going touse a transshipment flow formulation. Other models of this family will be discussed in the context oftransport problems in Chapter 10. To start, we draw a network that represents the problem (Figure 8.2).
This network consists of six nodes for the production in every time period, and six more nodes for thedemand in every time period. In the general case, we obtain a network of 2·NT nodes if NT is the numberof time periods.
The production nodes form the upper half of the graph. The have been assigned the odd indices 1 to 11
corresponding to time periods 1 to 6. Every node n has a weight CAPn that corresponds to the production

Planning problems

114

Applications of optimization with Xpress-MP

140

100

110

100

120

100

1

3

5

7

9

11

5

8

6

6

7

6

2

4

6

8

10

12

100

120

100

90

120

110

Figure 8.2: Representation of the network

capacity of the period. The demand nodes are in the lower half of the graph, with the even indices 2 to 12
equally corresponding to time periods 1 to 6. Every demand node n has a weight DEMn that corresponds
to the demand of the time period. An arc with the cost of the production in a period goes from every
production node to the demand node of the same period. A second set of arcs weighted with the storage
costs connects every demand node to the demand node of the following time period.
A production plan corresponds to a flow in this network. Let flowmn denote the flow on the arc (m, n). The
flow on a vertical arc represents the quantity (in cubic meters) of fiberglass produced in the correspondingtime period. The flow on a horizontal arc represents the quantity of product carried over to the next time
period (stock level). The aim is to compute the minimum cost flow that satisfies the demands and does
not exceed the production capacities. The resulting model is a transshipment problem because there
are no capacities on the arcs. With COSTmn the cost of an arc (m, n), we obtain the following objectivefunction (total cost of the flow):X

minimize

COSTmn · flowmn(8.5.1)(m,n)∈ARCS

For every time period, the amount carried over to the next period (stock) equals the stock at the beginning of the period plus the production of the period minus the demand of the period. This equation
results in the constraint (8.5.2) for the first period since there is no initial stock. In other words, the quantity stored at the end of the period (flow on arc (2,4)) equals the amount produced in period 1 (flow onarc (1,2)) minus the demand DEM2 .
(8.5.2)flow24 = flow12 − DEM2
The general stock balance equation (8.5.3) applies to all other periods but the last. For every even index n,
the flow between n − 2 and n represents the production in period n / 2, whilst the flow between n − 1 andn represents the stock carried over from the preceding period. The flow between n and n + 2 represents
the stock at the end of the current period. LAST denotes the last node in the graph (= node with thelargest sequence number).
∀n = 4, . . . , LAST − 2, n even : flown,n+2 = flown−2,n + flown−1,n − DEMn

(8.5.3)For the last period, we do not wish to create any final stock, so we obtain the constraint (8.5.4).
flowLAST−2,LAST + flowLAST−1,LAST − DEMLAST = 0(8.5.4)We also need to fulfill the constraints on the production capacity (8.5.5).
∀n = 1, . . . , LAST − 1, n odd : flown,n+1 ≤ CAPn

(8.5.5)

We obtain the following mathematical model. It is not required to state that the flow variables flowmn
are integer since this property is automatically given in the optimal solution to the linear problem (thisresults from LP theory that is not covered in this book).
minimizeX(8.5.6)

COSTmn · flowmn(m,n)∈ARCS

Planning problems

flow24 = flow12 − DEM2

(8.5.7)∀n = 4, . . . , LAST − 2, n even : flown,n+2 = flown−2,n + flown−1,n − DEMn

(8.5.8)flowLAST−2,LAST + flowLAST−1,LAST − DEMLAST = 0(8.5.9)∀n = 1, . . . , LAST − 1, n odd : flown,n+1 ≤ CAPn

(8.5.10)∀(m, n) ∈ ARCS : flowmn ≥ 0(8.5.11)115

Applications of optimization with Xpress-MP

8.5.2

Implementation
The mathematical model leads to the following Mosel program. The demand and capacity data (= node
weights) have been grouped into a single array WEIGHT, indexed by the node number. The graph withthe cost of the arcs is encoded as a two-dimensional array ARC, an entry ARCmn of which is defined if thearc (m, n) is in the network. The value of the entry corresponds to the cost of the arc. The array ARC isgiven in sparse format in the data file c5fiber.dat.
model "C-5 Fiberglass"uses "mmxprs"
declarations
NODES: range! Production and demand nodes! odd numbers: production capacities! even numbers: demands

ARC: array(NODES,NODES) of real
WEIGHT: array(NODES) of integer! Cost of flow on arcs
! Node weights (capacities/demand)

flow: array(NODES,NODES) of mpvar
end-declarations! Flow on arcs

initializations from ’c5fiber.dat’
ARC WEIGHT
end-initializations
forall(m,n in NODES | exists(ARC(m,n))) create(flow(m,n))! Objective: total cost of production and storage
Cost:= sum(m,n in NODES | exists(ARC(m,n))) ARC(m,n)*flow(m,n)! Satisfy demands (flow balance constraints)
forall(n in NODES | isodd(n)=FALSE)
if(n<getlast(NODES), flow(n,n+2), 0) =if(n>2, flow(n-2,n), 0) + flow(n-1,n) - WEIGHT(n)! Production capacities
forall(n in NODES | isodd(n)) flow(n,n+1) <= WEIGHT(n)! Solve the problem
minimize(Cost)
end-model

In this Mosel program, the variables are defined as a dynamic array and therefore need to be createdonce their index range is known, that is, after the data defining the arcs of the network has been readfrom file.
We remind the reader that the use of the function exists allows Mosel to enumerate only the defined
entries of a sparse array, provided that the index sets are named and are the same as those used for thedeclaration of the array.
The implementation above introduces two new functions: isodd indicates whether the integer value it
is applied to is even or odd. getlast returns the last entry of an array (that is, the entry with the highest
index value).

8.5.3

Results
The optimal solution calculated by the Mosel program is a total cost of B
C 3,988 with the quantities offiberglass to be produced and stored in every period shown in Table 8.13.
Table 8.13: Production and storage per week

Planning problems

Week

1

2

3

4

5

6

Production
Stock

140
40

80
0

110
10

100
20

110
10

100
0

116

Applications of optimization with Xpress-MP

8.6

Assignment of production batches to machines
Having determined a set of ten batches to be produced in the next period, a production manager islooking for the best assignment of these batches to the different machines in his workshop. The fiveavailable machines in the workshop may each process any of these batches, but since they are all models
from different years of manufacture the speed with which they process the batches changes from one
machine to another. Furthermore, due to maintenance periods and readjustments, every machine may
only work a certain number of hours in the planning period. Table 8.14 lists the processing times of the
production batches on all machines and the capacities of the machines.
Table 8.14: Processing durations and capacities (in hours)
Machine1

2

3

Production batches
4
5
6
7

8

9

10

Capacity

1
2
3
4
5

8
15
21
20
8

15
7
20
11
13

14
23
6
8
13

23
22
22
14
13

25
10
9
19
16

9
17
21
19
16

17
16
14
7
17

18
19
25
19
20

8
11
24
9
10

16
11
10
5
20

8
12
24
6
25

The production cost of a batch depends on the machine that processes it. The hourly usage cost forevery machine depends on its technology, its age, its consumption of consumables (such as electricity,machine oil) and the personnel required to operate it. These differences are amplified by the variationin the processing durations of a batch depending on the machine. Table 8.15 lists the production costsin thousand B
C. On which machine should each batch be executed if the production manager wishes tominimize the total cost of production?
Table 8.15: Production cost depending on the assignment (in kB
C)

8.6.1

Machine1

2

3

Production batches
4
5
6
7

8

9

10

1
2
3
4
5

17
23
16
19
18

21
16
20
19
19

22
21
16
22
15

18
16
25
22
15

18
25
19
17
16

19
18
19
21
23

18
21
18
19
15

24
17
24
20
21

15
16
16
16
25

20
19
17
19
16

Model formulation
In its generic form, this problem is a generalized assignment problem. Let MACH be the set of machines
and PRODS the set of batches to produce. We write DURmp for the processing duration of batch p on
machine m and COSTmp for the cost incurred if batch p is assigned to machine m. CAPm denotes the
maximum capacity (in hours) of the machine m. We introduce binary variables usemp that take the value
1 if and only if the batch p is assigned to the machine m. The objective function (8.6.1) that is to beminimized is the total production cost.

X

minimizeX

COSTmp · usemp(8.6.1)m∈MACH p∈PRODS

As a first set of constraints, we need to ensure that every batch is assigned to a single machine. This isdone with constraints (8.6.2). This type of constraint is typical for classical assignment problems, just likethe problem of flight connections in Chapter 11.

X∀p ∈ PRODS :usemp = 1

(8.6.2)m∈MACH

Every machine may only work within its capacity limits, and this constraint is established by relations(8.6.3). In these constraints we sum up the durations of the batches assigned to the same machine andspecify that this sum does not exceed the maximum total capacity of the machine. This type of constraint
is a knapsack constraint (see Chapter 9).
∀m ∈ MACH :X

DURmp · usemp ≤ CAPm

(8.6.3)p∈PRODS

Planning problems

117

Applications of optimization with Xpress-MP

This gives the following complete model:
minimize

X

X

COSTmp · usemp(8.6.4)m∈MACH p∈PRODS∀p ∈ PRODS :X

usemp = 1

(8.6.5)

DURmp · usemp ≤ CAPm

(8.6.6)m∈MACH∀m ∈ MACH :Xp∈PRODS∀m ∈ MACH, ∀p ∈ PRODS : usemp ∈ {0, 1}

8.6.2(8.6.7)

Implementation
The conversion of such a compact mathematical model into a Mosel program is straightforward.
model "C-6 Machine assignment"
uses "mmxprs"
declarations
MACH = 1..5
PRODS = 1..10! Set of machines! Set of production batches

CAP: array(MACH) of integer! Machine capacities
DUR: array(MACH,PRODS) of integer ! Production durations
COST: array(MACH,PRODS) of integer ! Production costs
use: array(MACH,PRODS) of mpvar! 1 if machine assigned to batch,! 0 otherwise

end-declarations
initializations from ’c6assign.dat’
DUR CAP COST
end-initializations! Objective: total production cost
Cost:= sum(m in MACH, p in PRODS) COST(m,p)*use(m,p)! Assign a single machine to every batch
forall(p in PRODS) sum(m in MACH) use(m,p) = 1! Limits on machine capacities
forall(m in MACH) sum(p in PRODS) DUR(m,p)*use(m,p) <= CAP(m)forall(m in MACH, p in PRODS) use(m,p) is_binary! Solve the problem
minimize(Cost)
end-model

8.6.3

Results
After solving this problem, we obtain a total cost of B
C 173k. The following table (there are several
solutions with the same objective value) shows the assignment:
Table 8.16: Optimal assignment

Planning problems

MachineAssigned batches

Total duration

1
2
3
4
5

1, 9
2, 5
3, 6, 8
7, 10
4

17
18
25
13
13

118

Applications of optimization with Xpress-MP8.7

References and further material
There is plenty of literature about production planning, for instance research articles and monographs,
such as Hax and Candea [HC84] and the one by Buffa and Taubert [BT79]. The recent book by Voß andWoodruff [VW02] on Supply Chain Management provides implementations of several typical problems
with Mosel and other modeling languages.
Historically, planning problems have first been treated by taking into account the stock management
almost exclusively. They were to answer the two questions: how much to order and when to order, tomaintain the chosen stock level. The first model was the Economic Order Quantity model. It determines
the quantity to produce according to a simple mathematical formula due to Wilson, that takes into
account the cost of the order, the storage cost, and the demand of the period under consideration. The
answer to the second question is based on the economic order quantity, the demand and the duration of
the period. But this type of model is valid, among others, only if the demand is deterministic (known forthe period) and does not change over time, if the parameters used for the calculations are also constantand do not depend on the quantity ordered, and if the resupply takes place instantaneously. These
drawbacks have lead to multi-period models in which the time horizon is discretized over a sequenceof time periods. By indexing the quantities with the time periods, these models are suitable for Linear
Programming.
The simplest multi-period models have cost that are proportional to the quantities (this are usually production and storage costs). In other words, there are no fix costs like adjustments before starting the
production of a batch. Constraints on limited resources (personnel, machine time etc.) are frequent. The
problem of planning the production of bicycles in Section 8.1 is a one-product case that falls under thiscategory. The production of fiberglass in Section 8.5 has costs that depend on the time period and is wellmodeled by flows. The glass production problem of Section 8.2 is a typical multi-product case. All thesebasic models work with a fixed personnel level (fixed workforce models). The first model of this typethat appears in the literature is due to Bowman [Bow56].
Let us cite some extensions of multi-product models with proportional costs. Researchers early on studied
variable workforce models, see for instance [HH60]. Later appear costs for additional hours, recruitment,
licensing and training. The problem of electronic components in Section 8.4 also forms an interesting
extension with cost caused by changes of the production level. A different level of generalization isreached by multi-stage production planning systems [JM74], [Can77].
In production planning, lot sizing problems form a class of their own that has not been mentioned in thischapter. These very hard problems are characterized by setup costs that apply every time the productionof a product is restarted. These costs that are due to the division of a lot (batch) into several sub-lots areadded to the usual production costs. The only easy case considers a single product and an unlimited production capacity. It is solved approximately with the heuristic of Silver and Meal [SPP98], and to optimality
by a dynamic programming method of Wagner and Whitin [WW58]. Among the difficult extensions, let
us cite the uncapacitated multi-product case [Kao79], and the cases with limited production capacities
with the heuristics by Walker [Wal76] and the column generation method of Lasdon and Terjung [LT71].
The 70s have seen the birth of two major concepts that are ‘en vogue’ in production planning: the
MRP method and Just-in-Time. MRP (Material Requirement Planning) was invented by Orlicky at IBM[Orl75]. This computerized stock management system has the advantage of taking into account thesubsets, components and raw material required by a product. It produces a master production schedulethat answers the two questions (how much and when to order). The use of Linear Programming for MRP
is more recent; the problem of the production of toy lorries in Section 8.3 gives an example for this.
Just-in-Time (JIT) is more of a philosophy or a state of mind that aims at eliminating any waste of time orresource use. JIT reduces the stock levels and increases the frequency of renewal of the latter by trying
to produce the desired product with the required quantity at the right moment. Establishing this wayof functioning is not easy but often leads to major improvements. The ‘philosophical’ aspect is attractive
and fairly common sense.
A problem that is rarely mentioned are the conflicts between different levels of production planningand scheduling. The aggregation of operations is required in planning to reduce the complexity but,
most of all, because it is useless to go into too much detail with a planning horizon of several months
for which the orders are not known precisely. Due to this aggregation, planning may unfortunately
result in a production plan that proves to be infeasible for scheduling when the detailed constraints of
the workshops are integrated into the model. Some recent methods allow the planner to avoid these
problems [DPL94].
The generalized assignment problem of Section 8.6 is not really a production planning problem. We haveplaced it into this chapter because it appears when the batches calculated by the planning have to be

Planning problems

119

Applications of optimization with Xpress-MP

assigned to machines or workshops with different capacities and processing times. Instances of large size
have to be treated with specialized techniques, like the tree-based method by Fisher et al. that uses the
Lagrangian relaxation of the linear program [FJVW86].

Planning problems

120

Applications of optimization with Xpress-MP

Chapter 9

Loading and cutting problems
Loading or packing problems consist of placing objects of different sizes into containers of known capacities. Depending on the context, the objects and containers may be of a different nature: programs
to store in memory, files to save onto a disk, boxes to be put into containers or onto shelves, etc. The
term packing usually refers to non-fragmentable objects that may cohabitate in a single container (suchas boxes in a lorry), whilst loading applies to fluid products (liquids and gas in tanks, powder in silos),that are fragmentable but may not be mixed in a single container.
In the closely related cutting stock problem the aim is to cut a set of objects from larger aggregates:
cable segments from a bobbin, orders of glass, sheet metal or photographic film cut from large sheets,pieces cut from metal blocks, etc. A collective name for loading and cutting stock problems is placement
problems.
In the simplest versions of these problems, the sizes and capacities only have a single dimension (weightof a box, size of a file in bytes, length of a cable, etc.) and the containers all have the same capacity.
Obviously, problems exist with several types of containers and objects with two, three, or more dimensions. Frequently, these are geometric dimensions (sheets, boxes), or various physical properties (weight,
volume).
Common objectives are to fill as much as possible of a container knowing that it is impossible to takeeverything, or to place objects into a minimum number of containers, or to balance the load. When
cutting material, one aims to minimize the trim loss or to produce large offcuts that may be used for
cutting further objects.
The first problem, presented in Section 9.1, is a loading problem in which the load of wagons has to bebalanced. The problem of Section 9.2 concerns loading a barge. The problem of loading tanks in Section
9.3 consists of assigning petrol products to a set of tanks (some of which are partially filled) so as tomaximize the number of tanks that is left empty. The subject of Section 9.4 is the problem of saving files
onto disks. The two last problems deal with cutting rectangular sheets from large metal sheets (Section
9.5) and metal bars for desks (Section 9.6). These two are typical examples of pattern selection problems,that is, problems where the containers may only be cut according to a limited number of patterns.

9.1

Wagon load balancing
Three railway wagons with a carrying capacity of 100 quintals (1 quintal = 100 kg) have been reserved totransport sixteen boxes. The weight of the boxes in quintals is given in the following table. How shall theboxes be assigned to the wagons in order to keep to the limits on the maximum carrying capacity and tominimize the heaviest wagon load?
Table 9.1: Weight of boxes
Box
Weight

1
34

2
6

3
8

4
17

5
16

6
5

7
13

8
21

Box
Weight

9
25

10
31

11
14

12
13

13
33

14
9

15
25

16
25

Before implementing a Mathematical Programming solution, one may wish to try to see whether it ispossible to solve this problem instance with the following simple heuristic: until all boxes are distributed
to the wagons we choose in turn the heaviest unassigned box and put it onto the wagon with the least
load.

121

Applications of optimization with Xpress-MP

9.1.1Model formulation
This packing problem is similar to bin packing problems (see also the tank loading problem in Section 9.3
and the backup of files in Section 9.4). Here the number of containers (wagons) is fixed, and the objective
is to minimize the heaviest load. In bin packing problems, the capacity of the containers (bins) is fixed,
but the objective is to minimize the number of bins used.
Let BOXES be the set of boxes, WAGONS the set of wagons, WEIGHTb the weight of box b and WMAX
the maximum carrying load of a wagon. The assignment of the boxes to the wagons can be defined
through binary variables loadbw that take the value 1 if and only if box b is assigned to wagon w.
In this problem we wish to minimize the maximum load of the wagons. Such an objective is sometimes
referred to as minimax objective. Similarly, there are maximin problems in which the objective is tomaximize a minimum, like the assignment of personnel to workposts in Chapter 11. Minimax or maximin
optimization problems are also called bottleneck problems. The procedure to represent a bottleneck
criterion in a linear model is always the same:• We define a non-negative variable maxweight to represent the maximum weight over all thewagon loads.
• Constraints are established to set maxweight as the upper bound on every wagon load.
• The objective function consists of minimizing maxweight.
By proceeding this way, in the optimal solution the minimization will force maxweight to take the valuethat corresponds to the weight of the heaviest wagon load. We derive the following 0-1 problem. The
constraints (9.1.2) ensure that every box is assigned to a single wagon. The constraints (9.1.3) establish
maxweight as the upper bound on the wagon loads. The objective (9.1.1) is to minimize maxweight. Allvariables are binary (9.1.4) with the exception of maxweight which is simply a non-negative real.
(9.1.1)minimize maxweightX∀b ∈ BOXES :loadbw = 1(9.1.2)w∈WAGONSX∀b ∈ BOXES :WEIGHTb · loadbw ≤ maxweight(9.1.3)w∈WAGONS∀b ∈ BOXES, ∀w ∈ WAGONS : loadbw ∈ {0, 1}(9.1.4)

maxweight ≥ 0(9.1.5)For two containers (wagons), it would be possible to define a much simpler mathematical model. The
problem then turns into the task of partitioning the boxes into two subsets with weights
P as close as possible. In the best case, we obtain subsets of the weight TOTALW / 2 (with TOTALW = b∈BOXES WEIGHTb
the total weight of all boxes). We may therefore reduce the problem to choosing boxes to place onto
the first wagon in such a way as to obtain a load as close as possible (from below) to TOTALW / 2. For
this purpose, we need to define binary variables loadb with loadb = 1 if box b goes onto wagon 1, andloadb = 0 (or 1 − loadb = 1) if the box is loaded onto wagon 2. The boxes that are not loaded onto wagon
1 go onto wagon 2. Thus, if the first wagon receives a load of weight TOTALW / 2 − ∆, the second wagon
has the load TOTALW / 2 + ∆.
maximize

X

WEIGHTb · loadb(9.1.6)b∈BOXESX

WEIGHTb · loadb ≤ TOTALW / 2(9.1.7)b∈BOXES∀b ∈ BOXES : loadb ∈ {0, 1}(9.1.8)This slightly simpler problem where we try to fill as much as possible of a single container (here the firstwagon with a capacity limited by TOTALW / 2) is called a knapsack problem (see also the barge loading
problem in Section 9.2).

9.1.2

Implementation
The mathematical model of lines (9.1.1) – (9.1.4) is easily translated into a Mosel program. The role ofthe lower bound on the maximum weight variable that has been added to the model formulation is
explained in the discussion of the results in the next section.

Loading and cutting problems

122

Applications of optimization with Xpress-MP

Before defining and solving the MP problem, we test whether we can find heuristically a distribution ofloads to the wagons that fits the given capacity limit WMAX. The solution heuristic requires the boxes
to be given in decreasing order of their weight. We therefore implement a sorting heuristic using a Shellsort method:• First sort, by straight insertion, small groups of numbers.
• Next, combine several small groups and sort them (possibly repeat this step).
• Finally, sort the whole list of numbers.
The spacings between the numbers of groups sorted during each pass through the data are called theincrements. A good choice is the sequence that can be generated by the recurrence i1 = 1, ik+1 = 3·ik +1, k =1, 2, . . .
Our implementation of the sorting algorithm assumes that the entries of the array to sort are numbered
1, . . . , N.
model "D-1 Wagon load balancing"uses "mmxprs"
forward function solve_heur:real
forward procedure shell_sort(A:array(range) of integer)
declarations
BOXES = 1..16
WAGONS = 1..3

! Set of boxes! Set of wagons

WEIGHT: array(BOXES) of integer
WMAX: integer! Box weights! Weight limit per wagon

load: array(BOXES,WAGONS) of mpvar ! 1 if box loaded on wagon, 0 otherwise
maxweight: mpvar! Weight of the heaviest wagon load
end-declarations
initializations from ’d1wagon.dat’
WEIGHT WMAX
end-initializations! Solve the problem heuristically and terminate the program if the! heuristic solution is good enoughif solveheur<=WMAX thenwriteln("Heuristic solution fits capacity limits")
exit(0)end-if! Every box into one wagon
forall(b in BOXES) sum(w in WAGONS) load(b,w) = 1! Limit the weight loaded into every wagon
forall(w in WAGONS) sum(b in BOXES) WEIGHT(b)*load(b,w) <= maxweight! Bounds on maximum weightmaxweight <= WMAX
maxweight >= ceil((sum(b in BOXES) WEIGHT(b))/3)
forall(b in BOXES,w in WAGONS) load(b,w) is_binary! Minimize the heaviest load
minimize(maxweight)!----------------------------------------------------------------! Heuristic solution: one at a time place the heaviest unassigned box! onto the wagon with the least loadfunction solve_heur:realdeclarations
ORDERW: array(BOXES) of integer! Box weights in decreasing order
Load: array(WAGONS,range) of integer ! Boxes loaded onto the wagons
CurWeight: array(WAGONS) of integer ! Current weight of wagon loads
CurNum: array(WAGONS) of integer! Current number of boxes per wagon
end-declarations! Copy the box weights into array ORDERW and sort it in decreasing order

Loading and cutting problems

123

Applications of optimization with Xpress-MP

forall(b in BOXES) ORDERW(b):=WEIGHT(b)shellsort(ORDERW)! Distribute the loads to the wagons using the LPT heuristic
forall(b in BOXES) do
v:=1! Find wagon with the smallest load
forall(w in WAGONS) v:=if(CurWeight(v)<CurWeight(w), v, w)
CurNum(v)+=1! Increase the counter of boxes on v
Load(v,CurNum(v)):=b! Add box to the wagon
CurWeight(v)+=ORDERW(b)! Update current weight of the wagon
end-doreturned:= max(w in WAGONS) CurWeight(w)end-function! Return the solution value!----------------------------------------------------------------! Sort an array in decreasing order using a Shell sort method
procedure shell_sort(A:array(range) of integer)
N:=getsize(A)inc:=1! Determine the starting increment
repeatinc:=3*inc+1
until (inc>N)
repeatinc:=inc div 3
forall(i in inc+1..N) do
v:=A(i)j:=iwhile (A(j-inc)<v) do
A(j):=A(j-inc)
j -= incif j<=inc then break; end-if
end-doA(j):= v
end-dountil (inc<=1)
end-procedure! Loop over the partial sorts! Outer loop of straight insertion! Inner loop of straight insertion

end-model

In this Mosel program for the first time we use a function (all subroutines defined so far were procedures,that is, they do not have a return value). What changes besides the keyword function instead ofprocedure is that the type of the return value (here: real) is declared in the subroutine prototype
and the actual value is assigned to returned in the body of the function.
In the implementation of the Shell sort algorithm we also encounter some new features:• The function getsize returns the size (= number of defined elements) of an array or set.
• The main sorting loop uses the three different types of loops available in Mosel: repeat-until,
forall, and while. Like the forall loop, the while loop has an inline version that applies to asingle statement and the version while-do to loop over a block of statements.
• The break statement can be used to interrupt one or several loops. In our case it stops the inner
while loop. Since we are jumping out of a single loop, we could just as well write break 1. If wewrote break 3, the break would make the algorithm jump 3 loop levels higher, that is outside of
the repeat-until loop.
Another feature introduced by this implementation is the Mosel function exit to terminate the execution of a model: if the heuristic solves the problem, there is no need to define the mathematical modeland start the optimization algorithm.

9.1.3

Results
The heuristic returns a maximum load of 101 quitals (the two other wagons are loaded with 97 quintalseach). This means that the heuristic solution does not satisfy the capacity limit of the wagons (100 quintals) and we need to use an optimizin algorithm to find out whether it is possible to transport all theboxes.

Loading and cutting problems

124

Applications of optimization with Xpress-MP

If we try running the mathematical model above without the lower bound constraint on the variablemaxweight, the MIP search is not finished after several hundred thousand nodes (or several minutes ofrunning time on a Pentium III PC) because the formulation is very weak (there are, for instance, manyequivalent solutions, simply obtained through permuting the numbering of boxes): an optimal MIP solution is found quickly1 but it takes a long time to prove its optimality.
A lowerPbound on the maximum weight is given by the ideal case that all wagons take the same load,that isWEIGHTb / 3. Since all box weights are integers and boxes cannot be fragmented, web∈BOXEScan improve
thisbound by rounding it to the next larger integer (expressed through ceil). We haveP
WEIGHTb / 3 = ceil 295 / 3 = ceil(98. 33333) = 99. With this bound, the optimal LP
ceilb∈BOXESsolution has the value 99. Due to this lower bound, the MIP search stops as soon as an integer feasible
solution with the value 99 is found. Since the boxes cannot be divided, it was not obvious that we would
be able to find an integer solution within the carrying capacity of 100 quintals per wagon.
Table 9.2: Wagon loads
Wagon

Total weight

1
2
3

99
98
98

Boxes
2, 3, 7, 9, 12, 14, 16
4, 5, 6, 8, 11, 15
1, 10, 13

An alternative to the additional lower bound constraint on the maximum weight is to adapt the cutoff
value used by the optimizer: whenever an integer feasible solution is found during the Branch andBound search, the optimizer uses this value as the new, upper bound on the objective function. Since inthis problem we can only have integer-valued solutions, we can deduce a value (close to) 1 better than
this bound, to force the optimizer to look for the next integer solution. We add the following line to ourMosel program before calling the optimization algorithm:
setparam("XPRS_MIPADDCUTOFF",-0.99999)

Once the MIP search has found a solution with value 99, the optimizer sets the new upper bound to
99 − 0. 99999 = 98. 00001. The optimal solution of the root LP (without the lower bound constraint on
maxweight) is 98.33333 which is larger than this bound, and the search therefore terminates.
Table 9.2 displays one of many possible assignments of boxes to wagons with the objective value 99.

9.2

Barge loading
A shipper on the river Rhine owns a barge of carrying capacity 1500 m3 . Over time he has specialized
in the transport of wheat. He has seven regular customers who load and unload practically at the sameplaces. The shipper knows his costs of transport from long experience and according to his personal
preferences has concluded agreements with his clients for the price charged to them for the transport oftheir wheat. The following table summarizes the information about the seven clients. Every client wishes
to tranport a certain number of lots, deciding himself the size of his lots in m3 . The table lists the pricecharged by the shipper for every transported lot. The last column of the table contains the cost incurredby the shipper per transported m3 . This cost differs depending on the distance covered.
Table 9.3: Lots to transport
ClientAvailable quantity(no. of lots)

Lot size(in m3 )Price perlot (in B
C)

Transport
cost (in B
C/m3 )

1
2
3
4
5
6
7

12
31
20
25
50
40
60

10
8
6
9
15
10
12

1000
600
600
800
1200
800
1100

80
70
85
80
73
70
80

The objective of the shipper is to maximize his profit from transporting the wheat with lots that may be
divided.
1 The reader is reminded that it is possible to visualize the output of the optimizer by adding the line
setparam("XPRS_VERBOSE",true) before the optimization statement. In Xpress-IVE, the tree search can be visualized directly.

Loading and cutting problems

125

Applications of optimization with Xpress-MP

Question 1: As a first step, assuming that every client has an unlimited quantity of wheat, which clients’
wheat should be transported?
Question 2: If in addition the actual availability of wheat lots from the customers is taken into account,which strategy should the shipper adopt?
Question 3: What happens if the lots cannot be divided?

9.2.1

Model formulation
The models for all three questions are given in this subsection. In each case, the objective is to maximize
the shipper’s profit. Let CAP be the total carrying capacity of the barge. We write CLIENTS for the set ofclients, AVAILc for the number of lots available from a client c, SIZEc the corresponding lot size, PRICEc
the price charged to this customer, and COSTc the transport cost per m3 incurred by the shipper. With thegiven data, we calculate the shipper’s profit PROFc per transported lot of client c. This profit is obtained
easily via equation (9.2.1): from the price charged to a customer we need to deduct the transport cost fora lot stemming from this customer.
(9.2.1)∀c ∈ CLIENTS : PROFc = PRICEc − COSTc · SIZEc

The shipper’s profit per transported lot of every client calculated through this equation is summarized in
the following table.
Table 9.4: Profit per lot
Client

1

2

3

4

5

6

7

Profit/lot (in B
C)
Profit/m3 (in B
C)

200
20

40
5

90
15

80
8.8889

105
7

100
10

140
11.6667

We introduce the decision variables loadc whose value is the number of lots transported for client c.
Since the lots may be divided, these variables may take fractional values. The model only has a single
constraint, namely to limit the volume of wheat transported by the barge (constraints (9.2.3)). Using the
profit data calculated above, we may now easily express the objective function (9.2.2).
maximizeX

PROFc · loadc(9.2.2)c∈CLIENTS

X

SIZEc · loadc ≤ CAP(9.2.3)c∈CLIENTS∀c ∈ CLIENTS : loadc ≥ 0(9.2.4)

If we have to take into account the number of lots available from each client, we simply have to add
the bound constraints (9.2.5) that state that the transported number of lots loadc is within the limit ofavailability.
∀c ∈ CLIENTS : loadc ≤ AVAILc(9.2.5)In the last case, the lots may not be fragmented, that is, the variables loadc may only take integer values.
We therefore have to add the following constraints (9.2.6) to the model.
∀c ∈ CLIENTS : loadc ∈ IN

9.2.2

(9.2.6)

Implementation
To simplify the statement of the objective function, in the following Mosel implementation of the mathematical model the profit per client is calculated separately, as is done in the previous section. The resulting
model is very simple because it contains only a single constraint. The additional constraints are progressively added to the problem definition. After every optimization run we call the procedure print_sol
to display the results.
model "D-2 Ship loading"uses "mmxprs"
forward procedure print_sol(num:integer)
declarations

Loading and cutting problems

126

Applications of optimization with Xpress-MP

CLIENTS = 1..7

! Set of clients

AVAIL: array(CLIENTS) of integer
SIZE: array(CLIENTS) of integer
PRICE: array(CLIENTS) of integer
COST: array(CLIENTS) of integer
PROF: array(CLIENTS) of integer
CAP: integer! Number of lots per client! Lot sizes! Prices charged to clients! Cost per client! Profit per client! Capacity of the ship

load: array(CLIENTS) of mpvar
end-declarations! Lots taken from clients

initializations from ’d2ship.dat’
AVAIL SIZE PRICE COST CAP
end-initializations
forall(c in CLIENTS) PROF(c):= PRICE(c) - COST(c)*SIZE(c)Profit:= sum(c in CLIENTS) PROF(c)*load(c)! Limit on the capacity of the shipsum(c in CLIENTS) SIZE(c)*load(c) <= CAP! Problem 1: unlimited availability of lots at clientsmaximize(Profit)
print_sol(1)! Problem 2: limits on availability of lots at clients
forall(c in CLIENTS) load(c) <= AVAIL(c)maximize(Profit)
print_sol(2)! Problem 3: lots must be integer
forall(c in CLIENTS) load(c) is_integer
maximize(Profit)
print_sol(3)!----------------------------------------------------------------! Solution printing
procedure print_sol(num:integer)
writeln("Problem ", num, ": profit: ", getobjval)
forall(c in CLIENTS)
write( if(getsol(load(c))>0 , " " + c + ":" + getsol(load(c)), ""))
writeln
end-procedure
end-model

9.2.3

Results
We examine the results obtained for the three questions in turn. In the first case, the optimizer returns a
total profit of B
C 30,000 and the only load taken are 150 lots from client 1. The other clients have to look
for a different transport opportunity. This answer is quite obvious: the barge should be filled with thewheat from the client that provides the highest profit for the space it occupies (ratio PROFc / SIZEc , values
displayed in the second line of Table 9.4).
But in reality, the available quantities of wheat are limited. When we solve this second, more constrained
problem, the value of the objective function decreases to B
C 17,844.44. The transported quantities belong
to five clients as shown in the table below. In this case, we are also able to deduce the solution throughlogic. It suffices to load the hold starting with the client who provides the best profit-per-volume ratio(PROFc / SIZEc ). We continue to load the hold with the wheat from the second best client (in the order ofdecreasing profit-per-volume ratio) and so on, until the hold is full. The process stops when the hold isfull, so the last lot is likely to be incomplete.
In our example, the order of clients according to the decreasing profit-per-volume ratio is 1, 3, 7, 6, 4, 5,
and 2. All wheat of clients 1,3,7, and 6 is taken, filling 1360 m3 of the hold. The remaining 140 m3 arefilled with wheat from client 4, that is 15.5556 of his lots.
If the lots are forced to take integer values only (question 3), the solution is similar to the previous one,

Loading and cutting problems

127

Applications of optimization with Xpress-MP

Table 9.5: Transported lots
Client
Question 1
Question 2
Question 3

1

2

3

4

5

6

7

150
12
12

0
0
0

0
20
20

0
15.5556
15

0
0
1

0
40
39

0
60
60

with a few small changes. The transported quantities for clients 1, 3, and 7 remain the same. Only 15 lotsof client 4 and 39 of client 6 are taken and in addition 1 lot of client 5. The value of the objective function
correspondingly decreases to B
C 17,805. Note that in the general case, there may be large differences
between the lots chosen in the cases of fragmentable or integer lots. This type of problem is called aknapsack problem.

9.3

Tank loading
Five tanker ships have arrived at a chemical factory. They are carrying loads of liquid products that mustnot be mixed: 1200 tonnes of Benzol, 700 tonnes of Butanol, 1000 tonnes of Propanol, 450 tonnes ofStyrene, and 1200 tonnes of THF. Nine tanks of different capacities are available on site. Some of them
are already partially filled with a liquid. The following table lists the characteristics of the tanks (intonnes). Into which tanks should the ships be unloaded (question 1) to maximize the capacity of thetanks that remain unused, or (question 2) to maximize the number of tanks that remain free?
Table 9.6: Characteristics of tanks

9.3.1

Tank

1

2

3

4

5

6

7

8

9

Capacity
Current product
Quantity

500
–
0

400
Benzol
100

400
–
0

600
–
0

600
–
0

900
–
0

800
THF
300

800
–
0

800
–
0

Model formulation
Let TANKS be the set of tanks and LIQ the set of liquid products. We write ARRl for the quantity of liquidl that is about to arrive at the factory. CAPt is the capacity of tank t, which initially contains a quantity
QINITt of liquid type TINITt (this type is only defined if QINITt > 0). All quantities are in tonnes. The model
formulation relies on the following property: it is possible to fill the tanks optimally by giving priority to
the tanks that already contain a liquid l for unloading the corresponding liquid before it is filled into any
empty tanks.
To understand why this correct, assume that a quantity x of liquid l is filled into an empty tank a, andanother tank b already contains liquid l with a remaining capacity of y > 0. If x > y, we do not makeany changes to the empty tanks (concerning their number or total capacity) if we first fill b before fillingx − y into a. If x ≤ y, we save tank a by filling x into tank b. We therefore obtain a solution that is atleast as good by giving priority to b until this tank is full or all of liquid l has been unloaded.
After the prioritized filling of the partially filled tanks, our problem remains to fill the empty tanks withquantities RESTl defined by the relations (9.3.1): these are the quantities of product that remain oncethe capacities of the partially filled tanks are exhausted. As for the rest of the data, we suppose that thevalues RESTl are larger than zero, otherwise it suffices to remove from the problem the tanks that arecompletely filled and the liquids that have been entirely unloaded.
∀l ∈ LIQ : RESTl = ARRl −X(CAPt − QINITt )(9.3.1)t∈TANKSTINITt =l

The model can then be formulated in a straightforward manner using binary variables loadlt that are
defined only for tanks t that are empty after the prioritized filling (9.3.5). A variable loadlt is 1 if theliquid l is unloaded into the tank t. The constraints (9.3.4) guarantee that every empty tank is filled withat most one liquid. The constraints (9.3.3) ensure that the set of tanks that are filled with the liquid lhave a sufficiently large total capacity. The objective function (9.3.2) minimizes the total capacity of thetanks that are used and so answers the first question, namely maximize the total capacity of the tanks
remaining empty. To maximize the number of unused tanks, it suffices to remove the coefficients CAPt

Loading and cutting problems

128

Applications of optimization with Xpress-MP

from the sum in (9.3.2).
minimize∀l ∈ LIQ :

X

X

l∈LIQ

t∈TANKSQINITt =0

X

CAPt · loadlt(9.3.2)

CAPt · loadlt ≥ RESTl(9.3.3)X(9.3.4)t∈TANKSQINITt =0∀t ∈ TANKS, QINITt = 0 :loadlt ≤ 1l∈LIQ∀l ∈ LIQ, t ∈ TANKS, QINITt = 0 : loadlt ∈ {0, 1}

9.3.2(9.3.5)

Implementation
The following Mosel program implements the mathematical model developed in the previous section.
Since the loadlt variables are only created for the tanks that are entirely empty at the beginning, there isno need to test the condition QINITt = 0 in the sums over these variables.
model "D-3 Tank loading"uses "mmxprs"
forward procedure print_sol
declarations
TANKS: range
LIQ: set of string! Set of tanks! Set of liquids

CAP: array(TANKS) of integer
TINIT: array(TANKS) of string
QINIT: array(TANKS) of integer
ARR: array(LIQ) of integer
REST: array(LIQ) of integer! Tank capacities! Initial tank contents type! Quantity of initial contents! Arriving quantities of chemicals! Rest after filling part. filled tanks

load: array(LIQ,TANKS) of mpvar! 1 if liquid loaded into tank,! 0 otherwise

end-declarations
initializations from ’d3tanks.dat’
CAP ARR[TINIT, QINIT] as ’FILLED’
end-initializations
finalize(LIQ)
forall(t in TANKS | QINIT(t)=0, l in LIQ) do
create(load(l,t))load(l,t) is_binary
end-do! Complete the initially partially filled tanks and calculate the remaining! quantities of liquids
forall(l in LIQ)
REST(l):= ARR(l) - sum(t in TANKS | TINIT(t)=l) (CAP(t)-QINIT(t))! Objective 1: total tank capacity used
TankUse:= sum(l in LIQ, t in TANKS) CAP(t)*load(l,t)! Objective 2: number of tanks used
TankNum:= sum(l in LIQ, t in TANKS) load(l,t)! Do not mix different liquids
forall(t in TANKS) sum(l in LIQ) load(l,t) <= 1
! Load the empty tanks within their capacity limits
forall(l in LIQ) sum(t in TANKS) CAP(t)*load(l,t) >= REST(l)! Solve the problem with objective 1
minimize(TankUse)
print_sol

Loading and cutting problems

129

Applications of optimization with Xpress-MP

! Solve the problem with objective 2
minimize(TankNum)
print_sol!----------------------------------------------------------------! Solution printing
procedure print_sol
writeln("Used capacity: ", getsol(TankUse) +sum(t in TANKS | QINIT(t)>0) CAP(t)," Capacity of empty tanks: ", sum(t in TANKS) CAP(t) getsol(TankUse) sum(t in TANKS | QINIT(t)>0) CAP(t))
writeln("Number of tanks used: ", getsol(TankNum) +sum(t in TANKS | QINIT(t)>0) 1)
forall(t in TANKS)
if(QINIT(t)=0) thenwrite(t, ": ")forall(l in LIQ) write( if(getsol(load(l,t))>0 , l, ""))
writelnelsewriteln(t, ": ", TINIT(t))
end-ifend-procedure
end-model

9.3.3

Results
When evaluating the results, we need to take into account that our model only works with the tanks
that are completely empty initially: the capacities (or for objective 2 the number) of the tanks that are
partially filled in the beginning need to be added to the objective values calculated by the program.
For the first objective, we obtain a total capacity used of 5200 tonnes, tank 4 with a capacity of 600 tonnes
remains empty (there are several equivalent solutions). Table 9.7 shows the distribution of chemicals tothe other tanks and the unused capacity possibly remaining if these tanks are not completely filled.
Table 9.7: Optimal tank filling
Product

Tanks

Remaining capacity

BenzolButanol
PropanolStyrene
THF

2, 6
9
3, 5
1
7, 8

0
100
0
50
100

In the graphical representation of the solution in Figure 9.1 the unused tank capacities are represented
by grey-shaded areas and the initial fill heights are indicated with dashed lines.

Styr.

Benzol

Prop.

1

2

3

4

Prop.

Benzol

THF

THF

Butan.

5

6

7

8

9

Figure 9.1: Optimal tank filling

In total, 8 tanks are used in the optimal solution for the first objective. After solving the problem with thesecond objective, we find that 8 is indeed the smallest number of tanks that may be used for unloading
all liquids.

Loading and cutting problems

130

Applications of optimization with Xpress-MP

9.4

Backing up files
Before leaving on holiday, you wish to backup your most important files onto floppy disks. You have got
empty disks of 1.44Mb capacity. The sixteen files you would like to save have the following sizes: 46kb,
55kb, 62kb, 87kb, 108kb, 114kb, 137kb, 164kb, 253kb, 364kb, 372kb, 388kb, 406kb, 432kb, 461kb, and851kb.
Assuming that you do not have any program at hand to compress the files and that you have got asufficient number of floppy disks to save everything, how should the files be distributed in order tominimize the number of floppy disks used?

9.4.1

Model formulation
Let FILES be the set of files to backup and DISKS = {1, . . . , ND} the set of floppy disks. Let CAP be thecapacity of a disk and SIZEf the size of file f in kbyte. We use variables savefd that take the value 1 if file
f is saved to disk d, and 0 in all other cases. We also define a variable used for every disk d that takes thevalue 1 if any files are saved to this disk and 0 otherwise.
The objective is to minimize the number of disks that are used, which corresponds to minimizing the sumof variables used , subject to the constraints:a) A file must be saved onto a single disk. This constraint corresponds to the relation (9.4.1).

X∀f ∈ FILES :savefd = 1(9.4.1)d∈DISKSb) The capacity of disks is limited, which translates into the constraints (9.4.2).

X∀d ∈ DISKS :SIZEf · savefd ≤ CAP · used

(9.4.2)f∈FILESThe variable used on the right side of this constraint links the variables savefd and used : if savefd has thevalue 1, then the file f is saved onto the disk d and hence, the disk d is used and variable used needs
to take the value 1. The relation (9.4.2) forces used to take the value 1 if at least one variable savefd(f ∈ FILES) has the value 1. We obtain the following MIP model:
minimizeX(9.4.3)usedd∈DISKSX∀f ∈ FILES :savefd = 1(9.4.4)

SIZEf · savefd ≤ CAP · used

(9.4.5)d∈DISKS∀d ∈ DISKS :X
f∈FILES∀d ∈ DISKS, f ∈ FILES : savefd ∈ {0, 1}(9.4.6)∀d ∈ DISKS : used ∈ {0, 1}(9.4.7)

It is possible to model this problem with fewer variables: we keep the Boolean variables savefd from thepreceding model (but not the variables used ) and introduce an additional variable diskuse that corresponds to the number of disks that are used. We obtain a second model formulation as follows:
minimize diskuse(9.4.8)∀f ∈ FILES : diskuse ≥X

d · savefd(9.4.9)d∈DISKSX∀f ∈ FILES :savefd = 1(9.4.10)

SIZEf · savefd ≤ CAP(9.4.11)∀d ∈ DISKS, f ∈ FILES : savefd ∈ {0, 1}(9.4.12)

diskuse ≥ 0(9.4.13)d∈DISKSX∀d ∈ DISKS :f∈FILESThe objective function is very simple in this case (9.4.8). We suppose that the disks are filled starting withthe disk numbered d = 1, then disk number 2, and so on. Due to the constraints (9.4.1) or (9.4.10), theLoading and cutting problems

131

Applications of optimization with Xpress-MP

disk that contains the file f has the index k that may be calculated through the relation (9.4.14).

Xk=(9.4.14)d · savefdd∈DISKSThe value of diskuse must be at least as large as the highest index number of the disks that are used, hence
the constraints (9.4.9). The constraints (9.4.10), that indicate that a file must be stored onto a single disk,
are identical to the constraints (9.4.1) of the previous model. The capacity limits are established throughthe constraints (9.4.11).
Finally, all variables savefd must be binaries and diskuse non-negative (it is not necessary to define thisvariable explicitly as integer because this constraints will automatically be satisfied in the optimal solution). In this model, the minimization will reduce the value of diskuse as much as possible and backup
the files in such a way that the index value of the largest disk becomes as small as possible. This model
has ND Boolean variables fewer than the previous one.

9.4.2

Implementation
The following Mosel program implements the second mathematical model (lines (9.4.8) to (9.4.13)). Notethat based on the given file sizes we calculate heuristically an upper bound on the number of disks that
we may need for backing up all the files: we divide the sum of all file sizes by the capacity of a floppy
disk, round this value to the next larger integer. The resulting value may be used as an upper boundin cases like the given data set where most of the files that need to be saved are very small comparedto the capacity of a disk. The only bound value for the number of disks that is save in the general caseis the number of files that need to be saved. (Consider, for instance, the case of 5 files of size 0.75Mb:
5 · 0. 75 = 3. 75, dividing this by 1.44 and rounding the result to the next larger integer we obtain 3, but
we actually need 5 disks since only a single file fits onto every disk.)model "D-4 Bin packing"uses "mmxprs"
declarations
ND: integer
FILES = 1..16
DISKS: range
CAP: integer
SIZE: array(FILES) of integer
end-declarations! Number of floppy disks! Set of files! Set of disks! Floppy disk size
! Size of files to be saved

initializations from ’d4backup.dat’
CAP SIZE
end-initializations! Provide a sufficiently large number of disks
ND:= ceil((sum(f in FILES) SIZE(f))/CAP)DISKS:= 1..ND
declarations
save: array(FILES,DISKS) of mpvar
diskuse: mpvar
end-declarations! 1 if file saved on disk, 0 otherwise! Number of disks used

! Limit the number of disks used
forall(f in FILES) diskuse >= sum(d in DISKS) d*save(f,d)! Every file onto a single disk
forall(f in FILES) sum(d in DISKS) save(f,d) = 1! Capacity limit of disks
forall(d in DISKS) sum(f in FILES) SIZE(f)*save(f,d) <= CAP
forall(d in DISKS,f in FILES) save(f,d) is_binary! Minimize the total number of disks used
minimize(diskuse)end-model

Loading and cutting problems

132

Applications of optimization with Xpress-MP

9.4.3Results
In the optimal solution, three disks are used. The files may be distributed to the disks as shown in thefollowing table (there are several possible solutions).
Table 9.8: Distribution of files to disks
Disk

File sizes (in kb)

Used space (in Mb)

1
2
3

46 87 137 164 253 364 388
55 62 108 372 406 432
114 461 851

1.439
1.435
1.426

In problems like this, where the objective function is very weak and we have essentially a feasibility
problem, it may speed the Branch and Bound search to break the symmetry of the problem. The files
have been numbered in an arbitrary way, as have the disks, so there is no loss of generality if we fix
one assignment, for example assigning the biggest file to the first disk. Such symmetry breaking devices
can be quite useful in combinatorial problems, as they sometimes prevent the LP relaxation from taking
fractional combinations of solutions that would otherwise be integer feasible.

9.5

Cutting sheet metal
A sheet metal workshop cuts pieces of sheet metal from large rectangular sheets of 48 decimeters × 96
decimeters (dm). It has received an order for 8 rectangular pieces of 36 dm × 50 dm, 13 sheets of 24 dm
× 36 dm, 5 sheets of 20 dm × 60 dm, and 15 sheets of 18 dm × 30 dm. Theses pieces of sheet metal need
to be cut from the available large pieces. How can this order by satisfied by using the least number oflarge sheets?

9.5.1

Model formulation
In this type of highly combinatorial problem, one exploits the fact that only a relatively small numberof combinations of rectangles can be cut from the large sheets. These cutting patterns can easily be
enumerated; Figure 9.2 represents the sixteen subsets that can be found. The sets shown are maximal
subsets, that is, no further small rectangle of the order may be added to them. The Figure only displays
one of several possible arrangements of every pattern.
Let SIZES be the set of rectangles of different sizes of the order, and PATTERNS the set of cutting patterns
represented in the graphic. For every size, the demand DEMs is given. Every pattern has an associatedcost COSTp (COSTp = 1 if we simply wish to minimize the number of large sheets used). The compositionof the cutting patterns is given by the matrix CUTsp defined by Table 9.9.
Table 9.9: Summary of cutting patterns
Pattern

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

36 x 50
24 x 36
20 x 60
18 x 30

1
2
0
0

1
1
0
1

1
0
0
3

0
2
2
0

0
1
2
1

0
0
2
3

0
3
1
0

0
2
1
2

0
1
1
3

0
0
1
5

0
5
0
0

0
4
0
1

0
3
0
3

0
2
0
5

0
1
0
6

0
0
0
8

Once having finished the tedious work of enumerating the cutting patterns, writing the mathematicalmodel is easy. We need to calculate the number of sheets to cut with every pattern in order to produce
all ordered rectangular pieces whilst minimizing the total number of sheets used.
We introduce integer variables usep (9.5.3) that denote the number of times a cutting pattern p is used.
The constraints (9.5.2) indicate that the number of rectangles of every type needs to satisfy the demand.
The objective function (9.5.1) is the total cost (or simply the number) of large sheets used in cutting.

X

minimize(9.5.1)

COSTp · usepp∈PATTERNSX∀s ∈ SIZES :CUTsp · usep ≥ DEMs

(9.5.2)p∈PATTERNS∀p ∈ PATTERNS : usep ∈ IN

Loading and cutting problems

133(9.5.3)

Applications of optimization with Xpress-MP

24x36
24x36

24x36

18x30

20x60

24x36

24x36

24x36

P7

18x30

24x36

18x30

18x30

18x30

P15

18x30
18x30

24x36

P14

24x36

P12

18x30

18x30

24x36

18x30

18x30

24x36

18x30

24x36

24x36

24x36

18x30

24x36

24x36

20x60

24x36

24x36

18x30

24x36

18x30

18x30

18x30

24x36

18x30

18x30

24x36

P13

24x36

P11

18x30

24x36

24x36

18x30

20x60

18x30

P10

18x30

18x30

18x30

P9
20x60

18x30
18x30

20x60

24x36

18x30

P8
20x60

18x30
20x60

P6

P5
24x36

20x60

18x30
18x30

18x30

18x30

18x30

18x30

P16

20x60

18x30

P4

36x50

24x36

18x30

20x60

P3

36x50

18x30

P2

36x50

18x30

P1

18x30

18x30

Figure 9.2: Graphical representation of the 16 cutting patterns

Loading and cutting problems

134

Applications of optimization with Xpress-MP

9.5.2

Implementation
The following Mosel program implements the mathematical model described in the previous section.
model "D-5 Sheet metal cutting"uses "mmxprs"
declarations
PATTERNS = 1..16
SIZES = 1..4

! Set of cutting patterns! Set of sheet sizes

DEM: array(SIZES) of integer
CUT: array(SIZES,PATTERNS) of integer! Demands for the different sizes! Cutting patterns

use: array(PATTERNS) of mpvar
end-declarations! Use of cutting patterns

initializations from ’d5cutsh.dat’
DEM CUT
end-initializations! Objective: total number of sheets used
Sheets:= sum(p in PATTERNS) use(p)! Satisfy demands
forall(s in SIZES) sum(p in PATTERNS) CUT(s,p)*use(p) >= DEM(s)forall(p in PATTERNS) use(p) is_integer! Solve the problem
minimize(Sheets)
end-model

9.5.3

Results
The solver finds an optimal integer solution that uses 11 large sheets out of which the pattern 1 is cut 6times, patterns 2 and 3 once, pattern 6 twice and pattern 10 once (note that there are several equivalentsolutions). From this distribution come exactly the numbers of small sheets required to satisfy the order.
We have thus minimized the total number of large sheets and produced exactly the required quantity,
which is not necessarily the case in larger examples.
The mathematical model is very compact in its generic form after the cutting patterns have been enumerated. In practical applications it may be too difficult to carry out this task by hand and some specialized
program will have to be employed to generate these patterns in order to prevent the omission of admissible combinations.

9.6

Cutting steel bars for desk legs
The company SchoolDesk produces desks of different sizes for kindergartens, primary and secondary
schools, and colleges. The legs of the desks all have the same diameter, with different lengths: 40 cm forthe smallest ones, 60 cm for medium height, and 70 cm for the largest ones. These legs are cut from steel
bars of 1.5 or 2 meters. The company has received an order for 108 small, 125 medium and 100 large
desks. How should this order be produced if the company wishes to minimize the trim loss?

9.6.1

Model formulation
This problem is similar to the previous problem of cutting sheet metal. The model formulation exploits
the fact that only a small number of cutting patterns exist for the steel bars. The shorter bars of 1.5 m
may, for instance, be cut into two pieces of 70 cm which leaves an offcut of 10 cm that cannot be used
for any other type of leg. The same bar could alternatively be cut into one piece of 60cm and one of 70
cm, a loss of 20 cm. The following table summarizes the different possibilities.
Let SIZES be the set of leg/desk heights and DEMs the demand for desks of a size s. Since every desk has
four legs, we need to multiply the demand by 4 to obtain the required number of desk legs. The sets

Loading and cutting problems

135

Applications of optimization with Xpress-MP

Table 9.10: Possible cutting patterns for every bar type
Patternnumber40cm

Leg types
60cm 70cm

Loss(in cm)Bar type 1(1.5m)

1
2
3
4
5
6

0
0
2
0
2
3

0
1
0
2
1
0

2
1
1
0
0
0

10
20
0
30
10
30

Bar type 2
(2m)

7
8
9
10
11
12

0
0
1
3
0
5

1
2
0
0
3
0

2
1
2
1
0
0

0
10
20
10
20
0

PAT1 and PAT2 of cutting patterns for bars of type 1 and 2 respectively are combined to form the set of
all patterns PATTERNS = PAT1 ∪ PAT2. We write LENb for the length of bar type b.
The objective is to minimize the trim loss, that is, the difference between the total length of the original
steel bars used for cutting and the total length of the desk legs ordered. With variables usep denoting thenumber of times a cutting pattern p is used, the first two terms of the objective function (9.6.1) represent
the total length of the bars that are used and the last term the total length of the ordered desk legs.
minimize

X

LEN1 · usep +p∈PAT1X∀s ∈ SIZES :X

LEN2 · usep −p∈PAT2X

4 · DEMs · s(9.6.1)s∈SIZES

CUTps · usep ≥ 4 · DEMs

(9.6.2)p∈PATTERNS∀p ∈ PATTERNS : usep ∈ IN

(9.6.3)In the objective function, the term corresponding to the length of the ordered desk legs may be omitted
because it is a constant value that has no influence on the minimization process. The constraints (9.6.2), in
which CUTps indicates the number of legs of length s contained in the cutting pattern p, guarantee thatthe demands are satisfied. The constraints (9.6.3) force the variables usep to take non-negative integer
values only.

9.6.2

Implementation
A translation of the mathematical model to a Mosel is given by the following program. Note that the set
operator + is used to build the union of two sets (PAT1 and PAT2).
model "D-6 Cutting steel bars"
uses "mmxprs"
declarations
PAT1 = 1..6; PAT2 = 7..12
PATTERNS = PAT1 + PAT2
SIZES: set of integer! Sets of cutting patterns! Set of all cutting patterns! Desk heights

DEM: array(SIZES) of integer
CUT: array(PATTERNS,SIZES) of integer
LEN: array(range) of integer! Demands for the different heights! Cutting patterns! Lengths of original steel bars

use: array(PATTERNS) of mpvar
end-declarations! Use of cutting patterns

initializations from ’d6cutbar.dat’
DEM CUT LEN
end-initializations! Objective: total loss
Loss:= sum(p in PAT1) LEN(1)*use(p) + sum(p in PAT2) LEN(2)*use(p) sum(s in SIZES) 4*DEM(s)*s! Satisfy demands

Loading and cutting problems

136

Applications of optimization with Xpress-MP

forall(s in SIZES) sum(p in PATTERNS) CUT(p,s)*use(p) >= 4*DEM(s)forall(p in PATTERNS) use(p) is_integer! Solve the problem
minimize(Loss)
end-model

In this model we not only use ranges of integers as index sets (PAT1, PAT2, PATTERNS, and the unnamed
index set of LEN) as in most other examples but also a set of integers, SIZES. The form set of integer
is more general than a range set that always contains consecutive integer numbers. In the present case,we want SIZES to contain only the three different leg heights (and not in addition all the integer values
between the smallest and the largest height). This is a nice way of using data directly as index values.

9.6.3

Results
The minimal loss calculated by the optimization is 2020cm. The ordered desk legs are cut from 2 bars of
1.5m (using cutting patterns 1 and 3 once) and 385 bars of 2m (using 195 times pattern 7, 7 times pattern
8, 97 times pattern 11, and 86 times pattern 12). With this combination of cutting patterns, exactly thequantity of legs is produced that is needed to satisfy the order of desks.

9.7

References and further material
All problems in this chapter are NP-hard and Mathematical Programming is only able to deal with problems of moderate size. The wagon loading problem in Section 9.1 is also known as the problem ofscheduling n nonpreemptive tasks on m identical machines, also called the m processor problem. Minimizing the maximum load of the wagons corresponds to minimizing the total duration of the schedule.
Using a formulation as a linear program it is hardly possible to deal with more than three machines andthirty tasks. For two machines, a dynamic programming method (a kind of recursive optimization) ofcomplexity O(nB) (with B the total duration of all items) solves problems with up to one hundred objects[MT90].
Certain specialized tree search methods are able to cope with up to a hundred tasks [HW95]. When the
optimal methods take too long, it is possible to find good solutions with the LPT (Longest ProcessingTime) heuristic that at every iteration places the unassigned object with the longest processing time onto
the machine with the least load. The ratio of the LPT solution compared to the optimal solution is only
1
4
− 3m
[Gra69]. As we have seen in Section 9.1 the LPT heuristic can easily be implemented with Mosel:
3
the heuristic solution is close to the optimum, but not sufficiently good for the given capacity limits(which demonstrates the usefulness of optimization!). Sorting algorithms like the Shell sort used in our
implementation of the LPT heuristic are described with some detail in [PFTVed].
The barge loading problems in Section 9.2 are knapsack problems; they are recognizable by their single
capacity constraint. The knapsack problem with fractional variables is easy: the container must simply befilled in the order of decreasing cost per size unit. Integer knapsack problems are NP-hard, but cases ofconsiderable size can be solved with dynamic programming or tree search methods [SDK83], [MT79].
The tank loading problem is described in a book by Christofides et al. [CMT79a]. There a tree searchmethod is described for large instances (35 liquids, 70 tanks), and also algorithms for the dynamic casethat take into account sequences of loading and unloading operations.
The problem of backing up files onto disks in Section 9.4 is a so-called bin-packing problem in which
one searches to distribute n objects i of weight Wi into a minimum number of boxes among m available,
every one with the same capacity. This problem is NP-hard. It is usually solved heuristically. The interested
reader may be recommended the work by Coffman et al. [CGJ96] that contains a summary of over ahundred references, and the one by Martello and Toth [MT90].
Cutting stock problems like those in Sections 9.5 and 9.6 are highly combinatorial and usually containlarge numbers of variables. The constraints are quite varied, like the one that cuts must go from border
to border (guillotine cuts).
Among the optimal methods, there are tree-based methods, see for instance [HZ96]. Another exact
method, called column generation, was introduced by Gilmore and Gomory ([GG61] and [GG63]) to solve
one-dimensional problems. This method consists of firstly solving a problem that contains a very reduced
subset of the columns of the complete mathematical program, the latter possibly being too large to be

Loading and cutting problems

137

Applications of optimization with Xpress-MP

generated in its entirety. The promising columns are then added progressively. When the method works
well, the optimum is found after generating only a small fraction of the columns of the complete model.
For large-sized instances one needs to use metaheuristics like tabu search [LMV99] or genetic algorithms[Jak96]. Patterns that are equivalent in terms of trim loss may have different cutting costs. See Chu
for methods that minimize the cost [CA99]. The formulation in 9.5 is a set covering problem like theplacement of mobile phone transmitters in Chapter 12. This formulation is relatively efficient and works
with up to a hundred patterns, but this limit is small compared to the enormous number of patterns
that arise if the rectangles that need to be cut are small compared to the original sheets. Sweeney etal. provide a comprehensive bibliography with over 400 references for loading and cutting problems[SRP92].

Loading and cutting problems

138

Applications of optimization with Xpress-MP

Chapter 10

Ground transport
As with air transport in Chapter 11, transport by road and rail is rich in optimization problems. The maindifference between the two types of networks is the high density of the network and the multiplicityof players in ground transport. The opening of frontiers and strong competition between transport
providers has made the use of optimization methods a vital means of reducing transport costs and thus
being able to stand out from competitors.
Section 10.1 presents a vehicle rental problem in which the cars have to be returned to the agencies at theleast cost in order to establish the ideal fleet strengths. In Section 10.2 a problem of distributing among
different modes of transport is described: a given quantity of goods has to be transported between
two points in a network which provides different modes of transport with a known cost and limited
capacities. Section 10.3 deals with a classical problem at the strategic level, the choice of depot locations
that minimizes the cost of opening depots and of delivering to clients. In Section 10.4 we solve a problem
optimizing the routes for the delivery of heating oil. Section 10.5 describes a combined (intermodal)transport problem that differs from the problem in 10.2 through costs for changing the mode. The
problem of planning a fleet of vans terminates this chapter.

10.1

Car rental
A small car rental company has a fleet of 94 vehicles distributed among its 10 agencies. The location of
every agency is given by its geographical coordinates X and Y in a grid based on kilometers. We assume
that the road distance between agencies is approximately 1.3 times the Euclidean distance (as the crowflies). The following table indicates the coordinates of all agencies, the number of cars required the nextmorning, and the stock of cars in the evening preceding this day.
Table 10.1: Description of the vehical rental agencies
Agency

1

2

3

4

5

6

7

8

9

10

X coordinate
Y coordinate

0
0

20
20

18
10

30
12

35
0

33
25

5
27

5
10

11
0

2
15

Required cars
Cars present

10
8

6
13

8
4

11
8

9
12

7
2

15
14

7
11

9
15

12
7

Supposing the cost for transporting a car is B
C 0.50 per km, determine the movements of cars that allow thecompany to re-establish the required numbers of cars at all agencies, minimizing the total cost incurred
for transport.

10.1.1Model formulation
For every agency a in the given set AGENTS, we write Xa and Ya for its geographic coordinates. REQa isthe number of cars required at an agency a and STOCKa the present number of cars for each agency. The
difference between these two values indicates whether there is an excess number (positive value) or aneed for additional vehicles (negative value). The problem consists of finding the minimum cost flow ofvehicles from the set EXCESS of agencies with an excess of cars to the set NEED of agencies that have a
deficit of cars. A flow that re-establishes the required numbers of cars necessarily exists because the sumof excesses is equal to the sum of deficits. As a first step, we define variables moveab to represent theflow between two agencies:∀a ∈ EXCESS, b ∈ NEED : moveab ∈ IN
(10.1.1)

139

Applications of optimization with Xpress-MP

Every agency with an excess number of cars needs to get rid of these (10.1.2), and every agency in needof cars has to complete its fleet by receiving the missing number (10.1.3).
∀a ∈ EXCESS :X

moveab = STOCKa − REQa

(10.1.2)

moveab = STOCKb − REQb

(10.1.3)b∈NEEDX∀b ∈ NEED :a∈EXCESSThe objective function to be minimized is the total cost of transporting cars (10.1.4), where COST denotes
the transport cost per car per kilometer and DISTab is the distance between two agencies a and b.

XminimizeX

COST · DISTab · moveab(10.1.4)a∈EXCESS b∈NEEDThis minimum cost flow problem is a transportation problem recognizable by a set of sources with availabilities and a set of sinks (destinations) with demands. For minimum cost flow problems, the simplex
algorithm always finds an integer solution when solving the LP. The constraints (10.1.1) may therefore be
replaced by simple non-negativity constraints.

10.1.2

Implementation
The following Mosel program implements the mathematical model above. After reading in the data, wetest whether the number of cars in stock and the required number of cars are the same and stop the
program if this is not the case. Otherwise, the two subsets of agencies with an excess or in need of cars
are calculated. These sets are used subsequently in the declaration of the array of decision variables andthe distance matrix.
model "E-1 Car rental"
uses "mmxprs"
declarations
AGENTS = 1..10
REQ: array(AGENTS) of integer
STOCK: array(AGENTS) of integer
X,Y: array(AGENTS) of integer
COST: real
NEED: set of integer
EXCESS: set of integer
end-declarations! Car rental agencies! Required number of cars
! Number of cars in stock! Coordinates of rental agencies! Cost per km of moving a car! Agencies needing more cars
! Agencies with too many cars

initializations from ’e1carrent.dat’
REQ STOCK X Y COST
end-initializations
if sum(a in AGENTS) (STOCK(a)-REQ(a)) <> 0 thenwriteln("Problem is infeasible")exit(0)end-if! Calculate sets of agencies with excess or deficit of cars
forall(a in AGENTS)
if STOCK(a) - REQ(a) < 0 then
NEED += {a}elif STOCK(a) - REQ(a) > 0 then
EXCESS += {a}end-if
finalize(NEED); finalize(EXCESS)
declarations
DIST: array(EXCESS,NEED) of real
move: array(EXCESS,NEED) of mpvar
end-declarations! Distance between agencies! Cars exchanged between agencies! Calculate distances between agencies
forall(a in EXCESS,b in NEED)
DIST(a,b):= 1.3*sqrt((X(a)-X(b))^2 + (Y(a)-Y(b))^2)

Ground transport

140

Applications of optimization with Xpress-MP

! Objective: total transport cost
Cost:= sum(a in EXCESS,b in NEED) COST*DIST(a,b)*move(a,b)! Agencies with excess availability
forall(a in EXCESS) sum(b in NEED) move(a,b) = STOCK(a) - REQ(a)! Agencies in need of cars
forall(b in NEED) sum(a in EXCESS) move(a,b) = REQ(b) - STOCK(b)forall(a in EXCESS,b in NEED) move(a,b) is_integer! Solve the problem
minimize(Cost)
end-modelThis implementation introduces the exponential operator ˆ. Note that √instead of using the predefined
Mosel function sqrt(r) we could also write rˆ0.5 for the square root r.

10.1.3Results
The optimizer calculates a total transport cost of B
C 152.64. The following table displays the required
movements of cars between agencies to obtain the desired distribution of the fleet with this minimum
cost.
Table 10.2: Optimal plan for transporting vehicles

10.2

→

1

3

4

6

7

10

Excess

2
5
8
9

0
0
0
2

1
0
0
3

0
3
0
0

5
0
0
0

1
0
0
0

0
0
4
1

7
3
4
6

Need

2

4

3

5

1

5

Choosing the mode of transport
A company in the south-west of France needs to transport 180 tonnes of chemical products stored indepots D1 to D4 to the three recycling centers C1, C2, and C3. The depots D1 to D4 contain respectively
50, 40, 35, and 65 tonnes, that is 190 tonnes in total. Two modes of transport are available: road and rail.
Depot D1 only delivers to centers C1 and C2 and that by road at a cost of B
C 12k/t and B
C 11k/t. Depot D2
only delivers to C2, by rail or road at B
C 12k/t and B
C 14k/t respectively. Depot D3 delivers to center C2 byroad (BC 9k/t) and to C3 by rail or road for B
C 4k/t and B
C 5k/t respectively. The depot D4 delivers to center C2
by rail or road at a cost of B
C 11k/t and B
C 14k/t, and to C3 by rail or road at B
C 10k/t and B
C 14k/t respectively.
Its contract with the railway company for the transport of chemical products requires the company totransport at least 10 tonnes and at most 50 tonnes for any single delivery. Besides the standard security
regulations, there are no specific limitations that apply to road transport. How should the company
transport the 180 tonnes of chemicals to minimize the total cost of transport?

10.2.1Model formulation
We are going to model this problem as a minimum cost flow problem with a fixed total throughput.
We first construct a graph G = (NODES, ARCS). To start, we put into the set of nodes NODES a layer ofnodes for the depots and a second one for the recycling centers (follow the construction with the helpof Figure 10.1). The set of arcs ARCS contains the possible connections between depots and recycling
centers. A transport plan corresponds to a flow in G, that is a flow flowij on every arc (i, j). An arc (i, j)
is characterized by a minimum flow MINCAPij (0 except for rail transport), a capacity or maximum flow
MAXCAPij (infinity except for rail transport), and a transport cost COSTij per tonne.
The two modes of transport from a depot to a center require two different arcs. Such a graph, with at
most p arcs in the same sense between two nodes is called a p-graph. Such a graph cannot be coded as a
(two-dimensional) matrix: for instance the element COSTij of a cost matrix can only define a single cost.
To obtain a graph with at most one arc between any pair of nodes, it is sufficient to create a fictitious

Ground transport

141

Applications of optimization with Xpress-MP

node per mode of transport, for every connection between a depot i and a center j. For instance, there
is one connection by road and one by rail between depot D2 (node 3) and the center C1 (node 12). To
avoid generating a 2-graph with two arcs (3,12), we create a node 6 for the rail transport and node 7
for road transport. The railway connection becomes the path (3,6,12), the road connection (3,7,12). The
capacities and costs are only established for the arcs (3,6) and (3,7).

DepotsRail/Road
6)

D1(0,-,12

2

12
0,
,5

(0

7

,50

)

11

,0)

,-,

(0

)

0,0

C1
12
(0,-,0)(0,-,0)

road

,14)

(0,-

D2

(0,4

(0,-,0)

rail

0

(1

)

Centers3(0,-,0)

8

C2

rail
1

(0,3

(0,-,0)

15

13
5,0

)

D3

(0,-,9)(10,5

4(0,-,0)

9
0,4)

road

50

,1

,0)

1)

,65

(0
(0,-,0)

10

)

0,
(1

(0,-,0)

4
-,1

(0,

)

0,10

5

(10,5

D4

(0,-,14)

(0
,-,

5)

rail

C3
14

(0,-,0)11
road

Figure 10.1: Graph of the network

The graph does not take into account the stock levels at the depots. To include these, we create a source
node (fictitious node 1) that is connected to every depot d by an arc (1, d) with a capacity MAXCAP1d that
corresponds to the stock level at d. Therefore, the flow leaving the depot d cannot exceed this value.
To facilitate the formulation of the mathematical model we also create a sink node (fictitious node 15)
to which every center is connected. The resulting graph is represented in Figure 10.1, with the triple(MINCAPij , MAXCAPij , COSTij ) for every arc (i, j). A ‘–’ stands for infinite capacity.
The mathematical model contains the flow conservation constraints (10.2.2), also called Kirchhoff’s law:
the sum of the incoming flows at every node (except for source and sink) equals the sum of outgoingflows. The flow on every arc has at least the minimum value MINCAPij (constraints (10.2.3)), without
exceeding the maximum capacity MAXCAPij (constraints (10.2.4)). The constraint (10.2.5) imposes a total
flow quantity MINQ = 180 tonnes. It stipulates that the total flow leaving the source (node 1) equals
MINQ. It would equally be possible to establish the equivalent constraint that the total flow into the sinkis equal to MINQ, since the flow is conserved through the network. In this constraint we could replace
the equality sign by ≥, since the total quantity transported will be forced to its lower bound when thecost, and hence the sum of flows, is minimized.
We are left with the explanation of the objective function in line (10.2.1). Since COSTij is the cost per
tonne, the cost of a quantity flowij transported via the arc (i, j) is COSTij · flowij . And hence the total costof transport to be minimized is the sum of the flow quantities in the entire set of arcs.
Finally, due to the way we have defined the graph, we obtain a fairly compact mathematical model. Notethat the non-negativity constraints are implicitly given through the constraints (10.2.3).
minimizeX(10.2.1)

COSTij · flowij(i,j)∈ARCS

X∀i ∈ NODES, i 6= SOURCE, SINK :(j,i)∈ARCS∀(i, j) ∈ ARCS : flowij ≥ MINCAPij

Ground transport

142

flowji =X

flowij(10.2.2)(i,j)∈ARCS(10.2.3)

Applications of optimization with Xpress-MP∀(i, j) ∈ ARCS : flowij ≤ MAXCAPij

X(10.2.4)(10.2.5)

flowSOURCE,i = MINQ

(SOURCE,i)∈ARCS

An alternative to this graph-based formulation of the problem would be to represent the quantity of
product transported from a depot d to a client c using mode m as a decision variable transportdcm for every admissible triple (d, c, m). The total transported quantity is then given by the sum of all defined variables transportdcm , and the objective function is the sum of COSTdcm · transportdcm for all admissible triples(d, c, m). The minimum and maximum capacities of certain modes of transport are given as bound constraints on the corresponding variables transportdcm , similarly to constraints (10.2.3) and (10.2.4) above.
For our specific problem, this formulation may be easier than the graph-based formulation. However, for
the implementation and further discussion of this problem we use the generic minimum cost flow model
given by lines (10.2.1) and (10.2.5).

10.2.2

Implementation
This problem raises a classical problem, namely the coding of a graph. It is possible to represent a graphthrough an N × N matrix (where N is the total number of nodes), and define the flow variables for thepairs of nodes (i, j) that are connected by an arc. However, for sparse graphs like the present one, it isoften preferable (and more efficient) to represent the graph as a list of arcs. This representation is usedin the following Mosel implementation, as opposed to the representation in the mathematical model
above. An arc a = (i, j) is referred to by its counter a and not via the node pair (i, j). The list of arcs is givenin the form of a two-dimensional array A with Aa1 = i and Aa2 = j. The flow variables are defined oncethe data has been read (and hence, the set of arcs is known). Note that the nodes in this implementation
are not numbered but labeled ‘SOURCE’, ‘D2’, ‘C4’ etc. to ease the interpretation of the results.
model "E-2 Minimum cost flow"uses "mmxprs"
declarations
NODES: set of string! Set of nodes
MINQ : integer! Total quantity to transport
A: array(ARCS:range,1..2) of string! Arcs
COST: array(ARCS) of integer! Transport cost on arcs
MINCAP,MAXCAP: array(ARCS) of integer ! Minimum and maximum arc capacities
end-declarations
initializations from ’e2minflow.dat’
A MINQ MINCAP MAXCAP COST
end-initializations
finalize(ARCS)! Calculate the set of nodes
NODES:= union(a in ARCS) {A(a,1),A(a,2)}
declarations
flow: array(ARCS) of mpvar
end-declarations! Flow on arcs

! Objective: total transport cost
Cost:= sum(a in ARCS) COST(a)*flow(a)! Flow balance: inflow equals outflow
forall(n in NODES | n<>"SOURCE" and n<>"SINK")
sum(a in ARCS | A(a,2)=n) flow(a) = sum(a in ARCS | A(a,1)=n) flow(a)! Min and max flow capacities
forall(a in ARCS | MAXCAP(a) > 0) doflow(a) >= MINCAP(a)flow(a) <= MAXCAP(a)end-do! Minimum quantity to transportsum(a in ARCS | A(a,1)="SOURCE" ) flow(a) >= MINQ
! Solve the problem
minimize(Cost)

Ground transport

143

Applications of optimization with Xpress-MP

end-modelThis model introduces another aggregate operator of Mosel: the set of nodes NODES is constructed as theunion of all nodes connected by arcs in the set ARCS.

10.2.3Results
The minimum cost is B
C 1,715k. Figure 10.2 displays the solution: arcs that are used for transporting goods
are labeled with the transported quantities, unused arcs and nodes are printed with dotted lines. For
example, the entire stock of 50 tonnes at depot D1 is transported by road to the recycling center C2.

DepotsRail/Road
6

D1
30

Centers
30

C1

rail

2

12
7
30

50

D2
3

50

8

30

15

C2

rail
1

65

15

35

D3

9

4
35

15

13
50

road

65

85

10

85

rail
5

14

50

D4

C3

11
Figure 10.2: Optimal transport plan

It may be noted that such a problem with minimum flows on its arcs may well have no solution. This is
the case, for instance, if MINQ = 10 because the four arcs of rail transport leaving the depots each requireat least a minimum flow of 10 tonnes.

10.2.4

Extension
This model may be used to solve any type of minimum cost flow problem. For example, it would bepossible to add demands for the centers. For a center c, it is sufficient to place the demand as a minimum
value for the flow on the arc (c, sink) to guarantee the satisfaction of the demand. However, a necessary
condition for the existence of a solution needs to be fulfilled: the sum of product availabilities at the
depots must be greater than or equal to the sum of the demands at the centers.

10.3

Depot location
A large company wishes to open new depots to deliver to its sales centers. Every new set-up of a depot
has a fixed cost. Goods are delivered from a depot to the sales centers close to the site. Every delivery has
a cost that depends on the distance covered. The two sorts of cost are quite different: set-up costs arecapital costs which may usually be written off over several years, and transport costs are operating costs.
A detailed discussion of how to combine these two costs is beyond the scope of this book — we assume
here that they have been put on some comparable basis, perhaps by taking the costs over a year.

Ground transport

144

Applications of optimization with Xpress-MP

There are 12 sites available for the construction of new depots and 12 sales centers need to receive
deliveries from these depots.
The following Table 10.3 gives the costs (in thousand B
C) of satisfying the entire demand of each customer(sales center) from a depot (not the unit costs). So, for instance, the cost per unit of supplying customer 9(who has a total demand of 30 tonnes according to Table 10.5) from depot 1 is B
C 60000/30t, i.e. B
C 2000/t.
Certain deliveries that are impossible are marked with the infinity symbol ∞.
Table 10.3: Delivery costs for satisfying entire demand of customers
Depot

1

2

3

4

5

Customer
6
7

8

9

10

11

12

1
2
3
4
5
6
7
8
9
10
11
12

100
120
140
160
190
200
100
120
140
160
190
200

80
90
110
125
150
180
80
90
110
125
150
180

50
60
80
100
130
150
50
60
80
100
130
150

50
70
80
100
∞
∞
50
70
80
100
∞
∞

60
65
75
80
∞
∞
60
65
75
80
∞
∞

100
110
130
150
∞
∞
100
110
130
150
∞
∞

90
110
125
150
180
80
90
110
125
150
180
80

60
80
100
130
150
50
60
80
100
130
150
50

70
80
100
∞
∞
50
70
80
100
∞
∞
50

65
75
80
∞
∞
60
65
75
80
∞
∞
60

110
130
150
∞
∞
100
110
130
150
∞
∞
100

120
140
160
190
200
100
120
140
160
190
200
100

In addition, for every depot we have the following information: the fixed cost for constructing the depot
that needs to be included into the objective function and its capacity limit, all listed in Table 10.4.
Table 10.4: Fix costs and capacity limits of the depot locations
Depot
Cost (kBC)
Capacity (t)

1

2

3

4

5

6

7

8

9

10

11

12

3500
300

9000
250

10000
100

4000
180

3000
275

9000
300

9000
200

3000
220

4000
270

10000
250

9000
230

3500
180

The quantities demanded by the sales centers (customers), are summarized in the following table.
Table 10.5: Demand data
Customer

1

2

3

4

5

6

7

8

9

10

11

12

Demand (t)

120

80

75

100

110

100

90

60

30

150

95

120

In every case, the demand of a customer needs to be satisfied but a sales center may be delivered tofrom several depots. Which depots should be opened to minimize the total cost of construction and ofdelivery, whilst satisfying all demands?

10.3.1Model formulation
To formulate the mathematical model, we write DEMc for the demand by customer (sales center) c, and
CAPd for the maximum capacity of depot d. The fixed cost of constructing depot d is given by CFIXd , thecost of delivery from depot d to customer c as COSTcd . Furthermore, let DEPOTS be the set of all possible
depot locations and CUST the set of customers to be delivered to from these depots.
To solve the problem, we need to know which depots will be opened. So we define binary variablesbuildd that take the value 1 if site d is chosen, and otherwise 0. In addition, we also need to know which
depot(s) deliver(s) goods to a customer. We introduce variables fflowdc for the fraction of the demandof customer c that is satisfied from depot d. These variables take their values in the interval [0, 1]; wetherefore have the constraints (10.3.1).
∀d ∈ DEPOTS, c ∈ CUST : fflowdc ≤ 1(10.3.1)

The demand of every customer needs to be entirely satisfied:∀c ∈ CUST :X

fflowdc = 1(10.3.2)d∈DEPOTS

Ground transport

145

Applications of optimization with Xpress-MP

We now have to model the fact that the total amount leaving depot d must be no more than its capacity
CAPd , but that the flow is zero if the depot is not built. The constraints (10.3.3) capture this.

X∀d ∈ DEPOTS :DEMc · fflowdc ≤ CAPd · buildd(10.3.3)c∈CUST

Let us now see why. Since fflowdc is the fraction of customer c’s satisfied demand, fflowdc · DEMc is theamount going from d to c. The total outflow from d must be no more than CAPd if d is built (buildd = 1),
and must be 0 if buildd = 0.
The total cost to be minimized consists of the costs for constructing the depots and the delivery costs.
These are the two sums comprising the objective function (10.3.4). The complete mathematical model isgiven by the following.

X

minimizeX

CFIXd · buildd +X

COSTdc · fflowdc∀d ∈ DEPOTS, c ∈ CUST : fflowdc ≤ 1∀c ∈ CUST :X(10.3.4)d∈DEPOTS c∈CUSTd∈DEPOTS(10.3.5)

fflowdc = 1(10.3.6)

DEMc · fflowdc ≤ CAPd · buildd(10.3.7)d∈DEPOTS∀d ∈ DEPOTS :Xc∈CUST

10.3.2∀d ∈ DEPOTS, c ∈ CUST : fflowdc ≥ 0(10.3.8)∀d ∈ DEPOTS : buildd ∈ IN

(10.3.9)

Implementation
The following Mosel program is a straightforward translation of the mathematical model.
model "E-3 Depot location"
uses "mmxprs"
declarations
DEPOTS = 1..12
CUST = 1..12! Set of depots! Set of customers

COST: array(DEPOTS,CUST) of integer
CFIX: array(DEPOTS) of integer
CAP: array(DEPOTS) of integer
DEM: array(CUST) of integer! Delivery cost
! Fix cost of depot construction! Depot capacity! Demand by customers

fflow: array(DEPOTS,CUST) of mpvar
build: array(DEPOTS) of mpvar
end-declarations! Perc. of demand supplied from depot! 1 if depot built, 0 otherwise

initializations from ’e3depot.dat’
COST CFIX CAP DEM
end-initializations! Objective: total cost
TotCost:= sum(d in DEPOTS, c in CUST) COST(d,c)*fflow(d,c) +sum(d in DEPOTS) CFIX(d)*build(d)! Satisfy demands
forall(c in CUST) sum(d in DEPOTS) fflow(d,c) = 1! Capacity limits at depots
forall(d in DEPOTS) sum(c in CUST) DEM(c)*fflow(d,c) <= CAP(d)*build(d)forall(d in DEPOTS) build(d) is_binary
forall(d in DEPOTS, c in CUST) fflow(d,c) <= 1
! Solve the problem
minimize(TotCost)
end-model

We could add an additional set of constraints stating the relation ‘if there is any delivery from depot d,then this depot must be built’ (and its inverse ‘if a depot is not built, then there is no delivery from thisGround transport

146

Applications of optimization with Xpress-MP

depot’). This relation is implied by the constraints (10.3.3) but the additional (disaggregated) constraints
provide a tighter formulation. That is, if these constraints are added to the model they draw the solution
value of the LP relaxation closer to the MIP solution. The additional constraints may be added directly
to the model, but since the model is fully stated through the version printed above, we could turn theminto model cuts as shown below. By defining these constraints as model cuts we leave the choice to the
optimizer whether to use these additional constraints or not. Note that any constraint that is to becomea model cut needs to be named and declared globally in the Mosel program as shown in the following
program extract (linear constraints in Mosel have the type linctr).
declarations
modcut: array(DEPOTS,CUST) of linctr
end-declarations
forall(d in DEPOTS, c in CUST) do
modcut(d,c):= fflow(d,c) <= build(d)setmodcut(modcut(d,c))
end-do10.3.3Results
The optimization algorithm calculates a minimum total cost of B
C 18,103k. The five depots 1, 5, 8, 9, and
12 are built. The following table details how the customers are delivered to from these depots. All depots
built, except depot 5 are used to their maximum capacity.
Table 10.6: Delivery plan

10.4

Depot

1

2

3

4

5

Customer
6
7

8

9

10

11

12

1
5
8
9
12

–
120
–
–
–

5
40
35
–
–

75
–
–
–
–

100
–
–
–
–

–
–
–
110
–

–
–
100
–
–

–
–
–
–
60

–
–
–
–
30

–
–
85
65
–

–
–
–
95
–

120
–
–
–
–

–
–
–
–
90

Heating oil delivery
A transporter has to deliver heating oil from the refinery at Donges to a certain number of clients in thewest of France. His clients are located at Brain-sur-l’Authion, Craquefou, Guérande, la Haie Fouassière,
Mésanger and les Ponts-de-Cé. The following table lists the demands in liters for the different sites.
Table 10.7: Demands by clients (in liters)
Brain-sur-l’Authion

Craquefou

Guérande

Haie Fouassière

Mésanger

Ponts-de-Cé

14000

3000

6000

16000

15000

5000

The next table contains the distance matrix between the clients and the refinery.
Table 10.8: Distance matrix (in km)
Brain-surDonges l’Authion
DongesBrain-s.-l’Authion
Craquefou
Guérande
Haie Fouassière
Mésanger
Ponts-de-Cé

0
148
55
32
70
140
73

148
0
93
180
99
12
72

Craquefou

Guérande

Haie
Fouassière

Mésanger

Pontsde-Cé

55
93
0
85
20
83
28

32
180
85
0
100
174
99

70
99
20
100
0
85
49

140
12
83
174
85
0
73

73
72
28
99
49
73
0

The transport company uses tankers with a capacity of 39000 liters for the deliveries. Determine the tours
for delivering to all clients that minimize the total number of kilometers driven.

Ground transport

147

Applications of optimization with Xpress-MP

10.4.1

Model formulation
This problem may be seen as a generalization of the famous Traveling Salesman Problem (TSP), an example of which we see in Section 11.5. In this case there are several ‘salesmen’, who each have to have atour starting and finishing at Donges. We have to determine not only the cities in the tours, but also how
many tours there are to be, and the order the cities in each tour are visited.
We introduce variables precij that take the value 1 if town i immediately precedes town j in a tour, and0 otherwise. Let SITES = {1, . . . , NS} be the number of sites. Site 1 is the refinery, so that we havethe subset CLIENTS = {2, . . . , NS} of sites to which we deliver. Let DISTij be the distance between two
towns i and j, DEMi the quantity ordered by client i, and CAP the maximum capacity of the tankers.
We also use variables quanti for the total amount of oil delivered on the route that includes client i up
to and including client i. For example, if the route including 10 is 1,3,11,10,6,1, then quant10 would be
DEM3 + DEM11 + DEM10 . With these notations, we may formulate the following mathematical model:
minimizeX

X(10.4.1)

DISTij · preciji∈SITES j∈SITES,i6=j∀j ∈ CLIENTS :X

precij = 1(10.4.2)

i∈SITES,i6=j∀i ∈ CLIENTS :X

precij = 1(10.4.3)j∈SITES,j6=i∀i ∈ CLIENTS : DEMi ≤ quanti ≤ CAP(10.4.4)∀i ∈ CLIENTS : quanti ≤ CAP + (DEMi − CAP) · prec1i(10.4.5)∀i, j ∈ CLIENTS, i 6= j :quantj ≥ quanti + DEMj − CAP + CAP · precij + (CAP − DEMj − DEMi ) · precji(10.4.6)∀i ∈ CLIENTS : quanti ≥ 0(10.4.7)∀i, j ∈ SITES, i 6= j : precij ∈ {0, 1}(10.4.8)

The objective (10.4.1) of this problem is to minimize the total number of kilometers driven. Every customer site has to be delivered to once. This is expressed through the two sets of constraints (10.4.2) and(10.4.3) that make the delivery enter and leave every town (except the depot) exactly once.
The quantity quanti must be at least as large as the quantity ordered by client i and within the capacity
limit CAP of the tankers (10.4.4).
Furthermore, if client i is the first of a tour, then quanti is equal to the quantity ordered by this client.
This constraint is expressed through the two sets of constraints (10.4.4) and (10.4.5). Indeed, if i is thefirst client of a tour, then prec1i is 1 and, after simplification, the constraint (10.4.5) is equivalent to theconstraint (10.4.9).
quanti ≤ DEMi
(10.4.9)
From (10.4.9) and (10.4.4) it follows that quanti is equal to the demand of client i. If i is not the first of
a tour, prec1i is 0 and the constraint (10.4.5) is equivalent to the constraint (10.4.10), which is redundantsince it is already expressed in constraint (10.4.4).
quanti ≤ CAP(10.4.10)

Let us now consider the case where i is not the first customer of the tour. Then quanti must equal thesum of quantities delivered between the refinery and i inclusively. This means that if client j comes afterclient i in a tour, we can write that quantj must be equal to the quantity delivered on the tour from therefinery to i, plus the quantity ordered by j. This relation is stated by the constraint (10.4.6). If indeed
j is the immediate successor of i in a tour, then precij is 1 and precji is 0, and the constraint (10.4.6) isequivalent to the constraint (10.4.11).
quantj ≥ quanti + DEMj

(10.4.11)

When j does not come immediately after i, constraint (10.4.6) still remains valid. If j is the immediate
predecessor of i, the constraint (10.4.6) becomes (10.4.12).
quantj ≥ quanti − DEMi

(10.4.12)

This constraint means that the quantity delivered from the refinery up to j is no less than the quantity
delivered between the depot and the successor i of j on the tour, a quantity that needs to be reduced by

Ground transport

148

Applications of optimization with Xpress-MP

the delivery at i. If j is the immediate predecessor of i, then i is the immediate successor of j. We therefore
obtain in addition to (10.4.12) the constraint (10.4.13) by swapping the indices in (10.4.11).
quanti ≥ quantj + DEMi

(10.4.13)

The combination of constraints (10.4.12) and (10.4.13) is equivalent to the equation (10.4.14).
(10.4.14)

quanti = quantj + DEMi

If i and j are not next to each other on a tour, we obtain the constraint (10.4.15). Since the terms on theright hand side of the inequality sign are less than or equal to DEMj , this constraint is redundant since it
is subsumed by the constraint (10.4.4).
quantj ≥ quanti + DEMj − CAP(10.4.15)

And finally, constraints (10.4.7) and (10.4.8) indicate that the variables quanti are non-negative and thatthe precij are binary variables.

2
4

3

1
RefineryFigure 10.3: Example of an infeasible solution

To close we would like to remark that the assignment of variables quanti to every node i guarantees
that the capacity limits of the tankers are not exceeded, whilst making any tour impossible that does
not include the depot. Without these variables, it would be possible to obtain solutions like the one
represented in Figure 10.3.
This solution satisfies the constraints (10.4.2) and (10.4.3) because every node is entered and left exactly
once, but it is infeasible because the tour does not pass through the refinery. Assigning the strictlyincreasing values quanti all along a tour excludes this type of solution.

10.4.2

Implementation
The following Mosel program implements the mathematical model of lines (10.4.1) – (10.4.8).
model "E-4 Oil delivery"
uses "mmxprs"
declarations
NS = 7
SITES = 1..NS
CLIENTS = 2..NS! Set of locations, 1=refinery

DEM: array(SITES) of integer! Demands
DIST: array(SITES,SITES) of integer ! Distances between locations
CAP: integer! Lorry capacity
prec: array(SITES,SITES) of mpvar
quant: array(CLIENTS) of mpvar
end-declarations! 1 if i immediately precedes j,! 0 otherwise! Quantity delivered up to i

initializations from ’e4deliver.dat’
DEM DIST CAP
end-initializations! Objective: total distance driven

Ground transport

149

Applications of optimization with Xpress-MP

Length:= sum(i,j in SITES | i<>j) DIST(i,j)*prec(i,j)! Enter and leave every city only once (except the depot)
forall(j in CLIENTS) sum(i in SITES| i<>j) prec(i,j) = 1
forall(i in CLIENTS) sum(j in SITES| i<>j) prec(i,j) = 1! If i is the first client of a tour, then quant(i)=DEM(i)forall(i in CLIENTS) quant(i) <= CAP + (DEM(i)-CAP)*prec(1,i)! If j comes just after i in a tour, then quant(j) is greater than the! quantity delivered during the tour up to i plus the quantity to be! delivered at j (to avoid loops and keep capacity limit of the tanker)
forall(i,j in CLIENTS| i<>j) quant(j) >= quant(i) + DEM(j) - CAP +CAP*prec(i,j) + (CAP-DEM(j)-DEM(i))*prec(j,i)
forall(i in CLIENTS) do
quant(i) <= CAP
quant(i) >= DEM(i)end-doforall(i,j in SITES | i<>j) prec(i,j) is_binary! Solve the problem
minimize(Length)
end-model

It is possible to define an additional set of constraints stating that quanti is greater than or equal to thesum of the quantities to deliver to client i and to his predecessor on the tour. These constraints may help
improve the lower bound on the variables quanti for sites i that are not visited first in a tour. Since themodel is fully stated through the version printed above we turn them into model cuts, leaving the choice
to the optimizer whether to use these additional constraints or not. The reader is reminded that any
constraint that is to become a model cut needs to be named and declared globally in the Mosel program.
declarations
modcut: array(CLIENTS) of linctr
end-declarations
forall(i in CLIENTS) do
modcut(i):= quant(i) >= DEM(i) + sum(j in SITES| i<>j) DEM(j)*prec(j,i)
setmodcut(modcut(i))
end-do10.4.3Results
The optimal solution involves making two delivery tours. One tour delivers a total of 22000 liters ofheating oil, visiting first Guérande and then Haie Fouassière. The other tour goes from the refinery to
Mésanger, then Brain-sur-l’Authion, Les Ponts-de-Cé, and Craquefou, delivering a total of 37000 liters. A
total distance of 497 km needs to be driven for these tours.

10.5

Combining different modes of transport
A load of 20 tonnes needs to be transported on a route passing through five cities, with a choice of threedifferent modes of transport: rail, road, and air. In any of the three intermediate cities it is possible tochange the mode of transport but the load uses a single mode of transport between two consecutive
cities. Table 10.9 lists the cost of transport in $ per tonne between the pairs of cities.
Table 10.9: Transport costs with different modes
1–2
Rail
Road
Air

30
25
40

Pairs of cities
2–3 3–4 4–5
25
40
20

40
45
50

60
50
45

The next table (10.10) summarizes the costs for changing the mode of transport in $ per tonne. The costis independent of location.

Ground transport

150

Applications of optimization with Xpress-MP

Table 10.10: Cost for changing the mode of transport
from \ to

Rail

Road

Air

Rail
Road
Air

0
8
15

5
0
10

12
10
0

How should we organize the transport of the load at the least cost?

10.5.1Model formulation
Let MODES be the set of modes of transport. The connection between a pair of consecutive cities isreferred to as a leg of the route, the complete transport trajectory being given by the set LEGS of all legs.
Let CTRANSml be the transport cost using mode m on leg l, and CCHGmn the cost for changing from modem to mode n which is independent of the site where it takes place in our example. We need two typesof binary variables to handle the two types of costs: a first group (10.5.1) of variables useml that havethe value 1 if mode m is used for leg l of the total trajectory, and a second group (10.5.2) of variables
changemnl with value 1 if there is a change from mode m to mode n between legs l and l + 1.
∀m ∈ MODES, l ∈ LEGS : useml ∈ {0, 1}(10.5.1)∀m, n ∈ MODES, l ∈ {1, . . . , NL − 1} : changemnl ∈ {0, 1}(10.5.2)

A single mode of transport has to be used on each leg (10.5.3).

X∀l ∈ LEGS :useml = 1

(10.5.3)m∈MODESA single change of the mode of transport may take place at every intermediate city (10.5.4). In this case,the change from a mode to the same mode does not cost anything since the load stays on board the samevehicle. It would of course be possible to have a non-zero cost if the load changed to a different vehicleof the same type.
X(10.5.4)∀l ∈ {1, . . . , NL − 1} :changemnl = 1m,n∈MODESIn the city between the legs l and l + 1, we have a change of the mode from m to n (proposition A) if
and only if the mode m is used for leg l and mode n is used on leg l + 1 (proposition B). The constraints(10.5.5) provide a linear formulation of the implication A ⇒ B: if changemnl = 1, then useml = 1 andusen,l+1 = 1. Theoretically, taking these constraints isolated from the rest, it would be possible to haveuseml = usen,l+1 = 1 with changemnl = 0, but this case is excluded through the constraints (10.5.4).
∀m, n ∈ MODES, l ∈ {1, . . . , NL − 1} : useml + usen,l+1 ≥ 2 · changemnl(10.5.5)

An alternative formulation of these constraints are the two sets of constraints (10.5.6) and (10.5.7).
∀m, n ∈ MODES, l ∈ {1, . . . , NL − 1} : useml ≥ changemnl(10.5.6)∀m, n ∈ MODES, l ∈ {1, . . . , NL − 1} : usen,l+1 ≥ changemnl(10.5.7)

These alternative constraints are stronger: they exclude more fractional solutions than the constraints(10.5.5), at the cost of defining twice as many constraints.
The objective function (10.5.8), in $ per tonne, comprises the transport cost for every leg of the trajectory(depending on the mode of transport used) and the sum of the costs for changing the mode in theintermediate cities.
NL−1

minimize

X

X

CTRANSml · useml +m∈MODES l∈LEGS

10.5.2

X

X

X

CCHGmn · changemnl(10.5.8)m∈MODES n∈MODES l=1

Implementation
The mathematical model translates into the following Mosel program. The implementation uses the‘weak’ constraints (10.5.5).

Ground transport

151

Applications of optimization with Xpress-MP

model "E-5 Combined transport"
uses "mmxprs"
declarations
NL = 4LEGS = 1..NL
MODES: set of string! Legs of the transport! Modes of transport

CTRANS: array(MODES,LEGS) of integer
CCHG: array(MODES,MODES) of integer
end-declarations! Transport cost
! Cost of changing mode of transport

initializations from ’e5combine.dat’
CTRANS CCHG
end-initializations
finalize(MODES)
declarations
use: array(MODES,LEGS) of mpvar! 1 if a mode is used, 0 otherwisechange: array(MODES,MODES,1..NL-1) of mpvar ! 1 if change from mode m to n
! at end of leg, 0 otherwise
end-declarations! Objective: total cost
Cost:= sum(m in MODES, l in LEGS) CTRANS(m,l)*use(m,l) +sum(m,n in MODES,l in 1..NL-1) CCHG(m,n)*change(m,n,l)! One mode of transport per leg
forall(l in LEGS) sum(m in MODES) use(m,l) = 1! Change or maintain mode of transport between every pair of legs
forall(l in 1..NL-1) sum(m,n in MODES) change(m,n,l) = 1! Relation between modes used and changes
forall(m,n in MODES,l in 1..NL-1) use(m,l) + use(n,l+1) >= 2*change(m,n,l)
forall(m in MODES, l in LEGS) use(m,l) is_binary
forall(m,n in MODES,l in 1..NL-1) change(m,n,l) is_binary! Solve the problem
minimize(Cost)
end-model

10.5.3Results
The minimum total cost calculated by the optimizer is $104/t. This value is obtained from $100/t oftransport costs and $4/t for changing the mode of transport. The following table displays the results indetail.
Table 10.11: Modes of transport used and respective costs
1–2
Mode of transport
Cost (in $/t)

Change at 2

2–3

0

rail
25

rail
30

Change at 3

3–4

0

rail
40

Change at 4

4–5

5

road
50

10.6 Fleet planning for vans
A chain of department stores uses a fleet of vans rented from different rental agencies. For the next six
months period it has forecast the following needs for vans (Table 10.12):
Table 10.12: Requirements for vans for six months
January

February

March

April

May

June

430

410

440

390

425

450

At the 1st January, the chain has 200 vans, for which the rental period terminates at the end of February.

Ground transport

152

Applications of optimization with Xpress-MP

To satisfy its needs, the chain has a choice among three types of contracts that may start the first dayof every month: 3-months contracts for a total cost of $1700 per van, 4-months contracts at $2200 pervan, and 5-months contracts at $2600 per van. How many contracts of the different types need to be
started every month in order to satisfy the company’s needs at the least cost and to have no remaining
vans rented after the end of June?
rent34
rent33
rent43
rent42
rent52
rent51

Month

1

2

3

4

5

6

Figure 10.4: Running contracts in month 5 (May)

10.6.1Model formulation
Let NINIT be the number of vans already rented at the beginning of the planning period. We define
integer variables rentcm to denote the number of contracts of type c (c ∈ CONTR = {3, 4, 5}) to start at the
beginning of month m (m ∈ MONTHS = {1, . . . , 6}). To clarify the constraint formulation we are using, let
us first look at a specific formulation. Take for example the constraint for month 5 (May) represented in
Figure 10.4: the running contracts may be 5-months contracts signed in January, 4 or 5-months contracts
signed in February, 3 or 4-months contracts signed in March, or 3-months contracts signed in April. These
contracts must satisfy the requirement of 425 vans for the month of May. Note that for instance a 5month contract cannot be signed in March because it does not terminate at the end of June, nor can any
contract be signed in May.
m=1:

NINT + rent31 + rent41 + rent51 ≥ 430

m=2:

NINT + rent31 + rent41 + rent51 + rent32 + rent42 + rent52 ≥ 410

m=3:

rent31 + rent41 + rent51 + rent32 + rent42 + rent52 + rent33 + rent43 ≥ 440

m=4:

rent41 + rent51 + rent32 + rent42 + rent52 + rent33 + rent43 + rent34 ≥ 390

m=5:

rent51 + rent42 + rent52 + rent33 + rent43 + rent34 ≥ 425

m=6:

rent52 + rent43 + rent34 ≥ 450

It is possible to write these constraints in a generic way. Let NM denote the number of months, COSTc
the cost per van of contracts of duration c, and REQm the requirement of vans in month m.
minimize

X

X

COSTc · rentcm(10.6.1)

c∈CONTR m∈MONTHS
min(m,NM−c+1)X∀m = 1, 2 : NINIT +X

rentcn ≥ REQm

(10.6.2)c∈CONTR n=max(1,m−c+1)min(m,NM−c+1)∀m = 3, . . . , NM :X

X

rentcn ≥ REQm

(10.6.3)c∈CONTR n=max(1,m−c+1)∀c ∈ CONTR, m ∈ MONTHS : rentcm ∈ IN

(10.6.4)

The objective function is to minimize the total cost for all contracts that are signed (10.6.1). A contract
lasts c months, therefore the requirement of vans in any month m is covered by the contracts of type csigned in the months max(1, m − c + 1) and min(m, NM − c + 1). The constraints (10.6.3) specify that startingfrom March the signed contracts have to cover the need for vans of the month. For January and February,
the constraints (10.6.2) also include the number of vans initially available NINIT.
The constraints (10.6.4) establish the integrality constraints for the variables. From LP theory it is possible
to show that this mathematical program is equivalent to a minimum cost flow problem that is known tohave integer values at the (linear) optimum. It is therefore possible to replace the constraints (10.6.4) bysimple non-negativity conditions.

Ground transport

153

Applications of optimization with Xpress-MP

10.6.2

Implementation
The constraints (10.6.2) and (10.6.3) of the mathematical model are combined into a single statement in
the following Mosel program.
model "E-6 van rental"
uses "mmxprs"
declarations
NM = 6MONTHS = 1..NM
CONTR = 3..5

! Months! Contract types

REQ: array(MONTHS) of integer
COST: array(CONTR) of integer
NINIT: integer! Monthly requirements! Cost of contract types! Vans rented at beginning of plan

rent: array(CONTR,MONTHS) of mpvar
end-declarations! New rentals every monthinitializations from ’e6vanrent.dat’
REQ COST NINIT
end-initializations! Objective: total cost
Cost:= sum(c in CONTR, m in MONTHS) COST(c)*rent(c,m)! Fulfill the monthly requirements
forall(m in MONTHS)
if(m<=2, NINIT, 0) +sum(c in CONTR, n in maxlist(1,m-c+1)..minlist(m,NM-c+1)) rent(c,n) >=REQ(m)! Solve the problem
minimize(Cost)
end-model

In this implementation, we use the Mosel functions maxlist and minlist to calculate respectively the
maximum and minimum value of a list of numbers. In this example, each time only two numbers are
given as arguments to these functions, but in the general case the lists may have any finite number of
entries, such as maxlist(2,-10,A(3),B(7,-5)) (where A and B are assumed to be arrays of integers
or reals). These two functions should not be confused with the aggregate operators max and min thatprovide a similar functionality, but are used with set expressions, such as max(i in ISET) A(i) (where
A is an array of integers or reals indexed by ISET).

10.6.3Results
The optimization algorithm finds a minimum total cost of $1,261,000. The table shows the new contracts
signed every month and the resulting totals.
Table 10.13: Plan of van rentals
New contracts

January

February

March

April

May

June

3 months
4 months
5 months

230
0
0

0
0
0

0
210
0

240
0
0

0
0
0

0
0
0

Total

430

430

440

450

450

450

Note that the requirement to have no vehicles on contract at the start of July is very distorting.

10.7

References and further material
In our transport example of Section 10.1 (car rental), the availability is equal to the total demand. Whendealing with an unbalanced case like an offer that is larger than the demand, the constraints (10.1.2)
must be changed into inequalities ensuring that no origin delivers more than it has available. Before

Ground transport

154

Applications of optimization with Xpress-MP

Linear Programming was invented, transport problems had already been studied in the 1930s and 40s, inthe USA by Hitchcock [Hit41], in the USSR by Kantorovitch. There are efficient specialized algorithms likethe stepping stone algorithm [BJ90].
The choice of modes of transport in Section 10.2 is a minimum cost flow problem in a graph. Fast algorithms for this problem that work directly on the graph are available. An algorithm with a good
simplicity/performance ratio is the one by Busacker and Gowen; a Pascal source is provided in [Pri94a].
Other algorithms are given by Ahuja et al [AMO93].
The problem of depot location (facility location problem) in Section 10.3 is a mixed-integer problem with
continuous variables for the transported quantities and binary variables for the construction decisions.
This problem is NP-hard, even with depots of infinite capacity. Tree-based methods like the one by
Erlenkotter [Erl78] are able to solve problems of a certain size (100 customers). A book by Daskin describes
a large number of other location problems [Das95].
The heating oil delivery problem in Section 10.4 is a typical case of a vehicle routing problem (VRP). The
formulation given here is only suitable for small instances (20-30 customers). Some specialized tree searchmethods are able to solve to optimality instances with up to 100 customers [LDN84] [BMR94], even for thecase that the delivery to clients needs to take place in certain time windows [DDS92]. Beyond this limitof 100 customers, it is recommended to use heuristic methods like the one by Clarke and Wright [CW64].
An overview on classical heuristics is given by Christofides [CMT79b]. Metaheuristics like tabu search findgood solutions for large sized instances [GHL94].
The mathematical program for the intermodal transport problem in Section 10.5 uses an excessively large
number of variables when the number of cities and especially the number of modes of transport growlarger. Luckily, it may be solved efficiently with a Dynamic Programming method (a kind of recursive
optimization) [Kas98]. The case studied here is relatively simple because the cities form a unique path.
The problem gets very hard for an arbitrary graph.
Like the problem of personnel planning for a construction site in Chapter 14, the van fleet planningproblem in Section 10.6 belongs to a category of optimization problems in which the requirements fora period are satisfied by resources that are used during more than one period. Another problem ofthis type consists of assigning shifts or other tasks to persons that are available for a certain number ofhours. The associated mathematical programs have a series of consecutive coefficients of value 1 in theircolumns. It is possible to show that they are equivalent to minimum cost flow problems [AMO93].

Ground transport

155

Applications of optimization with Xpress-MP

Chapter 11

Air transport
The domain of air transport is a fertile ground for original and difficult optimization problems. The
fierce competition between airlines has led to the development of Operations Research departments at
the largest ones (such as American Airlines, Delta Airlines, British Airways, and more recently, the renaissance of such a department at Air France). This chapter illustrates the diversity of applications, involving
passenger flows, flight crews, aircraft movements, location of hubs (flight connection platforms), andflight trajectories.
In the first problem (Section 11.1), the incoming planes at an airport need to be assigned to the flights
leaving from the airport in order to minimize the number of passengers (and hence the luggage) who
have to change planes. The problem in Section 11.2 consists of forming flight crews according to various compatibility and performance criteria. Section 11.3 describes an interesting and original problem,
namely scheduling the landing sequence of planes on a runway. The choice of the architecture of a network (location of hubs) is dealt with in Section 11.4. The subject of Section 11.5 is a case of emergency
logistics (the provision of a disaster-stricken country with fresh supplies).

11.1

Flight connections at a hub
The airline SafeFlight uses the airport Roissy-Charles-de-Gaulle as a hub to minimize the number of flightconnections to European destinations. Six Fokker 100 airplanes of this airline from Bordeaux, ClermontFerrand, Marseille, Nantes, Nice, and Toulouse are landing between 11am and 12:30pm. These aircraft
leave for Berlin, Bern, Brussels, London, Rome, and Vienna between 12:30 pm and 13:30 pm. The numbersof passengers transferring from the incoming flights to one of the outgoing flights are listed in Table 11.1.
Table 11.1: Numbers of passengers transferring between the different flights

Origins

Bordeaux
Clermont-Ferrand
Marseille
NantesNice
Toulouse

Berlin

Bern

35
25
12
38
–
–

12
8
8
15
9
–

Destinations
Brussels
London
16
9
11
14
8
–

38
24
27
30
25
14

Rome

Vienna

5
6
3
2
10
6

2
8
2
9
5
7

For example, if the flight incoming from Bordeaux continues on to Berlin, 35 passengers and their luggage may stay on board during the stop at Paris. The flight from Nice arrives too late to be re-usedon the connection to Berlin, the same is true for the flight from Toulouse that cannot be used for thedestinations Berlin, Bern and Brussels (the corresponding entries in the table are marked with ‘–’).
How should the arriving planes be re-used for the departing flights to minimize the number of passengers
who have to change planes at Roissy?

11.1.1Model formulation
Let PLANES be the set of aircraft (the number of which also corresponds the numbers of flight origins
and flight destinations) and PASSij the number of passengers transferring at the hub from origin i to theflight with destination j. We introduce binary variables contij that take the value 1 if and only if the plane

156

Applications of optimization with Xpress-MP

coming from i continues its journey to destination j. The following LP represents the problem:
maximize

X

X

PASSij · contij(11.1.1)

i∈PLANES j∈PLANES∀j ∈ PLANES :X

contij = 1(11.1.2)

contij = 1(11.1.3)∀i, j ∈ PLANES : contij ∈ {0, 1}(11.1.4)i∈PLANES∀i ∈ PLANES :X
j∈PLANES

The initial objective was to minimize the number of passengers that change planes, but this objective
is equivalent to the objective (11.1.1) that maximizes the number of passengers staying on board theirplane at the hub. The constraints (11.1.2) indicate that every destination is served by exactly one flight,
and the constraints (11.1.3) that one and only one flight leaves every origin. The constraints (11.1.4)
specify that the variables are binaries. Note that since this problem is an instance of the well known
assignment problem the optimal LP solution calculated by the simplex algorithm always takes integer
values. It is therefore sufficient simply to define non-negativity constraints for these variables. The upperbound of 1 on the variables results from the constraints (11.1.2) and (11.1.3).

11.1.2

Implementation
In the data array PASS, that is read in from the file f1connect.dat by the following Mosel implementation of the mathematical model, the ‘–’ of the table are replaced by large negative coefficients (–1000).
This negative cost prevents the choice of inadmissible flight connections. Another possibility would be toleave the corresponding entries of the array PASS undefined and to test in the definition of the variables
whether a connection may be used. This formulation has the advantage of using fewer variables if not
all flight connections are feasible, but it cannot be used if there are admissible flight connections thatare not taken by any passengers.
model "F-1 Flight connections"uses "mmxprs"
declarations
PLANES = 1..6

! Set of airplanes

PASS: array(PLANES,PLANES) of integer ! Passengers with flight connections
cont: array(PLANES,PLANES) of mpvar
end-declarations! 1 if flight i continues to jinitializations from ’f1connect.dat’PASS
end-initializations! Objective: number of passengers on connecting flights
Transfer:= sum(i,j in PLANES) PASS(i,j)*cont(i,j)! One incoming and one outgoing flight per plane
forall(i in PLANES) sum(j in PLANES) cont(i,j) = 1
forall(j in PLANES) sum(i in PLANES) cont(i,j) = 1! Solve the problem: maximize the number of passengers staying on board
maximize(Transfer)
end-model

11.1.3Results
In the optimal solution, 112 passengers stay on board their original plane. The following table lists thecorresponding flight connections

Air transport

157

Applications of optimization with Xpress-MP

Table 11.2: Optimal flight connections

11.2

Plane arriving
fromcontinues todestination

Bordeaux
Clermont-Ferrand
Marseille
NantesNice
Toulouse

LondonBern
Brussels
BerlinRome
ViennaNumber ofpassengers
38
8
11
38
10
7

Composing flight crews
During the Second World War the Royal Air Force (RAF) had many foreign pilots speaking different
languages and more or less trained on the different types of aircraft. The RAF had to form pilot/co-pilot
pairs (‘crews’) for every plane with a compatible language and a sufficiently good knowledge of theaircraft type. In our example there are eight pilots. In the following table every pilot is characterized bya mark between 0 (worst) and 20 (best) for his language skills (in English, French, Dutch, and Norwegian),and for his experience with different two-seater aircraft (reconnaissance, transport, bomber, fighterbomber, and supply planes).
Table 11.3: Ratings of pilots
Pilot

1

2

3

4

5

6

7

8

EnglishFrenchDutch
NorwegianPlane type ReconnaissanceTransport
BomberFighter-bomber
Supply plane

20
12
0
0
18
10
0
0
0

14
0
20
0
12
0
17
0
0

0
0
12
0
15
9
0
14
0

13
10
0
0
0
14
11
0
0

0
15
8
17
0
15
13
0
12

0
20
11
0
0
8
10
12
18

8
8
14
0
8
12
0
16
0

8
9
12
16
0
13
0
0
18

Language

A valid flight crew consists of two pilots that both have each at least 10/20 for the same language and10/20 on the same aircraft type.
Question 1: Is it possible to have all pilots fly?
Subsequently, we calculate for every valid flight crew the sum of their scores for every aircraft type forwhich both pilots are rated at least 10/20. This allows us to define for every crew the maximum scoreamong these marks. For example, pilots 5 and 6 have marks 13 and 10 on bombers and 12 and 18 on
supply planes. The score for this crew is therefore max(13 + 10, 12 + 18) = 30.
Question 2: Which is the set of crews with maximum total score?

11.2.1Model formulation
Let PILOTS be the set of pilots. This type of problem is easily modeled through an undirected compatibility graph G = (PILOTS, ARCS). Every node represents a pilot, two nodes p and q are connected by anundirected arc (or edge) a = [p, q] if and only if pilots p and q are compatible, that is, they have a language and a plane type in common for which both are rated at least 10/20. The arcs are assigned weights
corresponding to the maximum score SCOREa of the flight crew. Figure 11.1 shows the resulting graphwith the scores for question 2.
A valid set of crews corresponds in G to a subset of arcs such that any two among them have no nodein common. In Graph Theory, such a set is called a matching. For question 1 we are looking for the
maximum cardinality matching, for question 2 for the matching with maximum total weight. The graphsuggests we formulate the model as follows.
maximize

X

CREWa · flya(11.2.1)

a∈ARCS

Air transport

158

Applications of optimization with Xpress-MP

30

1

27

2

3

28
24

25

4

26

27

30

21

29
30

5

28

6

7

36
25

30
8

Figure 11.1: Compatibility graph for pilots∀r ∈ PILOTS :X

flya ≤ 1(11.2.2)

a=[p,q]∈ARCS
p=r∨q=r∀a ∈ ARCS : flya ∈ {0, 1}(11.2.3)

For every edge a = [p, q] in the graph, a binary variable flya indicates whether this edge is used or not(11.2.3). Through constraints (11.2.2) every node r is contained in at most one edge. The objective function (11.2.1) accumulates the weight of the chosen edges for question 2. For question 1, we maximize thenumber of flight crews to see whether all pilots are taken: the coefficients SCOREa need to be removed
from the objective function. Note that the matching of maximum cardinality is a special case of matchingwith maximum weight, with all weights equal to 1.

11.2.2

Implementation
The Mosel program below first calculates the set of admissible crews based on the language skills andflight experience of the pilots: for every pair of pilots p and q the algorithm first checks whether they
have compatible language skills, and if this is the case then their flight experience is compared. The twopilots are retained as an admissible crew if they both have sufficient experience on the same aircraft type.
The crews are saved in a two-dimensional array CREW indexed by the set of arcs, that is, the two pilots p
and q of an arc a = [p, q] are represented by CREW(a,1) and CREW(a,2). With the convention p < q,every pair of pilots is listed only once. The following program solves the problem twice, once with theobjective function for question 1 and once for question 2.
model "F-2 Flight crews"
uses "mmxprs"
forward procedure print_sol
declarations
PILOTS = 1..8
ARCS: range
RL, RT: set of string! Set of pilots! Set of arcs representing crews! Sets of languages and plane types

LANG: array(RL,PILOTS) of integer ! Language skills of pilots
PTYPE: array(RT,PILOTS) of integer ! Flying skills of pilots
CREW: array(ARCS,1..2) of integer ! Possible crews
end-declarations
initializations from ’f2crew.dat’
LANG PTYPE
end-initializations! Calculate the possible crews
ct:=1
forall(p,q in PILOTS| p<q and(or(l in RL) (LANG(l,p)>=10 and LANG(l,q)>=10)) andAir transport

159

Applications of optimization with Xpress-MP

(or(t in RT) (PTYPE(t,p)>=10 and PTYPE(t,q)>=10)) ) do
CREW(ct,1):=p
CREW(ct,2):=q
ct+=1
end-dofinalize(ARCS)
declarations
fly: array(ARCS) of mpvar
end-declarations! 1 if crew is flying, 0 otherwise! First objective: number of pilots flying
NFlying:= sum(a in ARCS) fly(a)! Every pilot is member of at most a single crew
forall(r in PILOTS) sum(a in ARCS | CREW(a,1)=r or CREW(a,2)=r) fly(a) <= 1
forall(a in ARCS) fly(a) is_binary! Solve the problem
maximize(NFlying)! Solution printing
writeln("Number of crews: ", getobjval)
print_sol! **** Extend the problem ****
declarations
SCORE: array(ARCS) of integer
end-declarations! Maximum scores of crews

forall(a in ARCS)
SCORE(a):= max(t in RT | PTYPE(t,CREW(a,1))>=10 and PTYPE(t,CREW(a,2))>=10)(PTYPE(t,CREW(a,1)) + PTYPE(t,CREW(a,2)))! Second objective: sum of scores
TotalScore:= sum(a in ARCS) SCORE(a)*fly(a)! Solve the problem
maximize(TotalScore)
writeln("Maximum total score: ", getobjval)
print_sol!----------------------------------------------------------------! Solution printing
procedure print_sol
forall(a in ARCS)
if(getsol(fly(a))>0) thenwriteln(CREW(a,1), " - ", CREW(a,2))
end-ifend-procedure
end-model

A feature of Mosel shown by the above implementation is the incremental definition of the array CREW
that stores the list of crews (and hence of its index set ARCS). After the calculation of the crews is completed the (dynamic) index set ARCS is finalized so that the array of variables fly is defined as a static
array on this index set.
Another feature that is new in this model is the cumulative or used for testing the pilots’ compatibility.
The expressionor(l in RL) (LANG(l,p)>=10 and LANG(l,q)>=10)

evaluates to true if for at least one l the values of LANG(l,p) and LANG(l,q) are both greater than orequal to 10.

Air transport

160

Applications of optimization with Xpress-MP

11.2.3

Results
In answer to the first question, the program finds that four crews are flying, that is, all eight pilots. For
the second question, the optimization calculates a maximum total score of 125 for the four crews [1,2],[3,7], [4,5], and [6,8].

11.3

Scheduling flight landings
The plane movements in large airports are subject to numerous security constraints. The problem presented in this section consist of calculating a schedule for flight landings on a single runway. More general
problems have been studied but they are fairly complicated (dynamic cases, for instance through planes
arriving late, instances with several runways, etc.).
Ten planes are due to arrive. Every plane has an earliest arrival time (time when the plane arrives above
the zone if traveling at maximum speed) and a latest arrival time (influenced among other things by its
fuel supplies). Within this time window the airlines choose a target time, communicated to the public
as the flight arrival time. The early or late arrival of an aircraft with respect to its target time leads to
disruption of the airport and causes costs. To take into account these cost and to compare them moreeasily, a penalty per minute of early arrival and a second penalty per minute of late arrival are associatedwith every plane. The time windows (in minutes from the start of the day) and the penalties per plane
are given in the following table.
Table 11.4: Characteristics of flight time windows
Plane

1

2

3

4

5

6

7

8

9

10

Earliest arrival
Target time
Latest Arrival
Earliness penalty
Lateness penalty

129
155
559
10
10

195
258
744
10
10

89
98
510
30
30

96
106
521
30
30

110
123
555
30
30

120
135
576
30
30

124
138
577
30
30

126
140
573
30
30

135
150
591
30
30

160
180
657
30
30

Due to turbulence and the duration of the time during which a plane is on the runway, a security interval
has to separate any two landings. An entry in line p of column q in the following Table 11.5 denotes theminimum time interval (in minutes) that has to lie between the landings of planes p and q, even if they
are not consecutive. Which landing schedule minimizes the total penalty subject to arrivals within thegiven time windows and the required intervals separating any two landings?
Table 11.5: Matrix of minimum intervals separating landings
1
2
3
4
5
6
7
8
9
10

11.3.1

1

2

3

4

5

6

7

8

9

10

–
3
15
15
15
15
15
15
15
15

3
–
15
15
15
15
15
15
15
15

15
15
–
8
8
8
8
8
8
8

15
15
8
–
8
8
8
8
8
8

15
15
8
8
–
8
8
8
8
8

15
15
8
8
8
–
8
8
8
8

15
15
8
8
8
8
–
8
8
8

15
15
8
8
8
8
8
–
8
8

15
15
8
8
8
8
8
8
–
8

15
15
8
8
8
8
8
8
8
–

Model formulation
Let PLANES be the set of planes due to arrive at the airport. A plane p has an arrival time window[STARTp , STOPp ] with the target arrival time TARGETp . The penalty CEARLYp applies per minute of earlyarrival, and CLATEp per minute of late arrival. The minimum interval between the landings of two planes
p and q is denoted by DISTpq . In the given data instance any two time windows are overlapping. We
therefore formulate a model for this overlapping situation and describe later how to generalize it forany type of time windows.
We need to define variables landp for the landing time of every plane p. They are bounded by the earliest

Air transport

161

Applications of optimization with Xpress-MP

and latest arrival times (11.3.1).
∀p ∈ PLANES : STARTp ≤ landp ≤ STOPp(11.3.1)

To account for the interval separating the landing times of any two planes p and q, binary variables precpq
are required with precpq = 1 if the landing of aircraft p precedes the landing of q.
∀p, q ∈ PLANES, p 6= q : precpq ∈ {0, 1}(11.3.2)

The constraints (11.3.3) specify that p arrives before q or q arrives before p. The constraints (11.3.4)
guarantee the separation of the landings; these are classical disjunctive (exclusion) constraints that areused for instance in scheduling to prevent the overlapping of two tasks on the same machine. M denotes
some large positive constant.
∀p, q ∈ PLANES, p 6= q : precpq + precqp = 1(11.3.3)∀p, q ∈ PLANES, p 6= q : landp + DISTpq − Mpq · precqp ≤ landq(11.3.4)

If the plane p arrives before q, we have precpq = 1, and hence precqp = 0 through (11.3.3). The constraint(11.3.4) for p and q results in landp + DISTpq ≤ landq ensuring the required separation of the two flights.
If precpq = 0, then precqp = 1 and the constraint (11.3.4) is trivially satisfied because the left side of theinequality is a large negative value. To avoid problems with numerical stability, the value of Mpq should
not be chosen too large, Mpq = STOPp + DISTpq − STARTq is sufficient for every constraint (11.3.4).
To halve the number of variables, we define the binary variables precpq with p < q to take the value 1
if p lands before q and 0 otherwise. The constraints (11.3.2) are replaced by (11.3.5) and the constraints(11.3.3) become redundant.
(11.3.5)∀p, q ∈ PLANES, p < q : precpq ∈ {0, 1}
The constraints (11.3.4) are rewritten to (11.3.6) and (11.3.7). (11.3.6) are the disjunctive constraints
for the pairs of planes (p, q) with p > q and (11.3.7) the disjunctions for p < q. If for instance, p < q(constraints (11.3.7)) and p arrives before q, then precpq = 1 and hence 1 − precpq = 0, and landp + DISTpq ≤landq . If p < q but q arrives before p, then precpq = 0 and (11.3.7) is trivially satisfied.
∀p, q ∈ PLANES, q < p : landp + DISTpq ≤ landq + M · precqp(11.3.6)∀p, q ∈ PLANES, p < q : landp + DISTpq ≤ landq + M · (1 − precpq )(11.3.7)

Taking into account the early or late arrival of a plane p with respect to its targeted arrival time TARGETp
is a delicate matter. We introduce a variable earlyp for the earliness and a variable latep for the lateness.
We may then write the objective function (11.3.8) with the penalties per minute of early or late arrival,
CEARLYp and CLATEp .
X
minimize(CEARLYp · earlyp + CLATEp · latep )(11.3.8)p∈PLANES

The variables earlyp and latep are bounded from above (constraints (11.3.9) and (11.3.10)) so that thearrival is scheduled within the time window [STARTp , STOPp ].
∀p ∈ PLANES : 0 ≤ earlyp ≤ TARGETp − STARTp(11.3.9)∀p ∈ PLANES : 0 ≤ latep ≤ STOPp − TARGETp(11.3.10)

The landing time is linked to the earliness or lateness by the constraints (11.3.11).
∀p ∈ PLANES : landp = TARGETp − earlyp + latep(11.3.11)

The fact that we are minimizing prevents earlyp and latep from both being non-zero simultaneously in(11.3.11).
We finally obtain a MIP model consisting of the lines (11.3.1), and (11.3.5) – (11.3.11).

11.3.2

Generalization to arbitrary types of time windows
In the general case, there are three sets OVERLAP, SEP, and NONSEP of plane pairs (p, q) with p < q.
OVERLAP contains all plane pairs with overlapping time windows. The variables precpq are used to decide
whether p arrives before q. As a simplification, the preceding model deals with the case that all pairs oftime windows are in the set OVERLAP.

Air transport

162

Applications of optimization with Xpress-MP

There may also be a set SEP of pairs (p, q) with disjoint time windows and guaranteed separation: for
instance the time windows [10,50], [70,110] and a separation time of 15 that will obviously be satisfied.
More formally, the relation defining such a pair of time windows is (STOPp + DISTpq < STARTq ) ∨ (STOPq +DISTqp < STARTp ). The third set NONSEP of plane pairs that may be defined contains all pairs with disjoint
time windows without guaranteed separation, such as [10,50], [70,110] with a separation interval of 30.
To obtain a general model, we need to add the following sets of constraints (11.3.12) to (11.3.15) to theconstraints (11.3.6) – (11.3.7) that only concern the plane pairs in OVERLAP. The constraints (11.3.12) and(11.3.13) force the variables to be 0 or 1 in the case of disjoint time windows. The constraints (11.3.14)
and (11.3.15) are simple precedence constraints as in the stadium construction problem of Chapter 7.

11.3.3∀(p, q) ∈ SEP ∪ NONSEP, STOPp < STARTq : precpq = 1(11.3.12)∀(p, q) ∈ SEP ∪ NONSEP, STOPq < STARTp : precpq = 0(11.3.13)∀(p, q) ∈ NONSEP, STOPp < STARTq : landp + DISTpq ≤ landq(11.3.14)∀(p, q) ∈ NONSEP, STOPq < STARTp : landq + DISTqp ≤ landp(11.3.15)

Implementation
The following Mosel program implements the mathematical model of Section 11.3.1. The variables precpq
are defined for all pairs p and q, but only those precpq with p < q are used in the constraints so that theremainder of these variables (all pairs with p ≥ q) do not appear in the problem.
model "F-3 Landing schedule"
uses "mmxprs"
declarations
PLANES = 1..10! Set of airplanes

START, STOP: array(PLANES) of integer ! Start, end of arrival time windows
TARGET: array(PLANES) of integer! Planned arrival times
CEARLY, CLATE: array(PLANES) of integer ! Cost of earliness/lateness
DIST: array(PLANES,PLANES) of integer ! Minimum interval between planes
M: array(PLANES,PLANES) of integer! Sufficiently large positive values
prec: array(PLANES,PLANES) of mpvar
land: array(PLANES) of mpvar
early,late: array(PLANES) of mpvar
end-declarations! 1 if plane i precedes j! Arrival time
! Earliness/latenessinitializations from ’f3landing.dat’
START STOP TARGET CEARLY CLATE DIST
end-initializations
forall(p,q in PLANES) M(p,q):= STOP(p) + DIST(p,q) - START(q)! Objective: total penalty for deviations from planned arrival times
Cost:= sum(p in PLANES) (CEARLY(p)*early(p) + CLATE(p)*late(p))! Keep required intervals between plan arrivals
forall(p,q in PLANES | p>q)land(p) + DIST(p,q) <= land(q) + M(p,q)*prec(q,p)
forall(p,q in PLANES | p<q)land(p) + DIST(p,q) <= land(q) + M(p,q)*(1-prec(p,q))! Relations between earliness, lateness, and effective arrival time
forall(p in PLANES) do
early(p) >= TARGET(p) - land(p)late(p) >= land(p) - TARGET(p)land(p) = TARGET(p) - early(p) + late(p)end-doforall(p in PLANES) do
START(p) <= land(p); land(p) <= STOP(p)early(p) <= TARGET(p)-START(p)late(p) <= STOP(p)-TARGET(p)end-doforall(p,q in PLANES | p<q) prec(p,q) is_binary! Solve the problem

Air transport

163

Applications of optimization with Xpress-MP

minimize(Cost)
end-model

As mentioned earlier in this book, the obvious pair START – END for naming the beginning and end of thetime windows cannot be used because END is a reserved word in Mosel (see Section 5.2.3 for the complete
list of reserved words).

11.3.4

Results
The program calculates a total deviation cost of 700. The following table lists the scheduled arrivals
together with the targeted arrival times and the resulting deviations (planes arriving earlier or later than
the announced target time). It may be worth mentioning that the LP solution to this problem has an
objective value of 0, with many fractional variables precpq .
Table 11.6: Arrival times and deviations
Plane

1

2

3

4

5

6

7

8

9

10

Scheduled arrival
Target time
Deviation

165
155
10

258
258
0

89
98
0

106
106
0

118
123
-5

126
135
-9

134
138
-4

142
140
2

150
150
0

180
180
0

11.4 Airline hub location
The airline FAL (French Air Lines) specializes in freight transport. The company links the major French
cities with cities in the United States, namely: Atlanta, Boston, Chicago, Marseille, Nice, and Paris. The
average quantities in tonnes transported every day by this company between these cities are given in thefollowing table.
Table 11.7: Average quantity of freight transported between every pair of cities
Atlanta
BostonChicagoMarseille
Nice
Paris

Atlanta

BostonChicagoMarseille

Nice

Paris

0
1500
400
300
400
350

500
0
510
600
100
1020

1000
250
0
810
420
260

300
630
460
0
730
580

400
360
320
820
0
380

1500
1140
490
310
970
0

We shall assume that the transport cost between two cities i and j is proportional to the distance that
separates them. The distances in miles are given in the next table.
Table 11.8: Distances between pairs of cities
Atlanta
BostonChicagoMarseille
Nice

BostonChicagoMarseille

Nice

Paris

945

605
866

4667
3726
4471

4749
3806
4541
109

4394
3448
4152
415
431

The airline is planning to use two cities as connection platforms (hubs) to reduce the transport costs. Everycity is then assigned to a single hub. The traffic between cities assigned to a given hub H1 to the cities
assigned to the other hub H2 is all routed through the single connection from H1 to H2 which allows theairline to reduce the transport cost. We consider that the transport cost between the two hubs decreases
by 20%. Determine the two cities to be chosen as hubs in order to minimize the transport cost.

11.4.1Model formulation
We write CITIES for the set of cities and NHUBS for the number of hubs. Let DISTij be the distance between
two cities i and j and QUANTij the quantity to be transported from i to j. The transport cost per tonne of

Air transport

164

Applications of optimization with Xpress-MPfreight depends on the cities that are chosen as hubs. The freight to be transported from any city i to any
city j transits through two (not necessarily distinct) hubs k and l. Let COSTijkl be the transport cost from ito j through the hubs k and l. This cost is equal to the transport cost from i to k, plus the cost from k to l,plus the transport cost from l to j. The cost from k to l corresponds to 80% of the normal cost from k to l
displayed in Table 11.8 since this is an inter-hub transport.
Let the binary variable flowijkl be 1 if the freight from i to j is transported via the hubs k and l in thisorder, and 0 otherwise. We also introduce variables hubi that are 1 if city i is a hub and 0 otherwise.
The objective function (11.4.1) minimizes the total transport cost. The constraint (11.4.2) indicates that
we wish to create exactly NHUBS hubs. Through the constraints (11.4.3) every pair of cities (i, j) is assigned
to a single pair of hubs. The constraints (11.4.4) and (11.4.5) imply that if a variable flowijkl is at 1, thenthe variables hubk and hubl are 1. In other words, any freight to be transported from i to j may onlytransit through k and l if both, k and l, are hubs. To complete the model, the constraints (11.4.6) and(11.4.7) define the variables as binaries.
Remark: in the constraints (11.4.3) k may be equal to l which this means that freight may transit via asingle hub. This is the case if two origin/destination cities are assigned to the same hub. The inter-hub
transport cost is 0 in this case.
minimize

X

X

X

X

COSTijkl · QUANTij · flowijkl(11.4.1)

i∈CITIES j∈CITIES k∈CITIES l∈CITIESX

hubi = NHUBS(11.4.2)

i∈CITIES∀i, j ∈ CITIES :X

X(11.4.3)

flowijkl = 1

k∈CITIES l∈CITIES

11.4.2∀i, j, k, l ∈ CITIES : flowijkl ≤ hubk

(11.4.4)∀i, j, k, l ∈ CITIES : flowijkl ≤ hubl

(11.4.5)∀i ∈ CITIES : hubi ∈ {0, 1}(11.4.6)∀i, j, k, l ∈ CITIES : flowijkl ∈ {0, 1}(11.4.7)

Implementation
In the following Mosel program implementation of the mathematical model, first the cost COSTijkl for
freight transport from i to j via the hubs k and l is calculated, using the inter-hub transport reduction
factor FACTOR that is defined in the data file.
model "F-4 Hubs"
uses "mmxprs"
declarations
CITIES = 1..6
NHUBS = 2

! Cities! Number of hubs

COST: array(CITIES,CITIES,CITIES,CITIES) of real ! (i,j,k,l) Transport cost
! from i to j via hubs k and l
QUANT: array(CITIES,CITIES) of integer ! Quantity to transport
DIST: array(CITIES,CITIES) of integer ! Distance between cities
FACTOR: real
! Reduction of costs between hubs
flow: array(CITIES,CITIES,CITIES,CITIES) of mpvar ! flow(i,j,k,l)=1 if! freight from i to j goes via k & lhub: array(CITIES) of mpvar! 1 if city is a hub, 0 otherwise
end-declarations
initializations from ’f4hub.dat’
QUANT DIST FACTOR
end-initializations! Calculate costs
forall(i,j,k,l in CITIES)COST(i,j,k,l):= DIST(i,k)+FACTOR*DIST(k,l)+DIST(l,j)! Objective: total transport cost
Cost:= sum(i,j,k,l in CITIES) QUANT(i,j)*COST(i,j,k,l)*flow(i,j,k,l)! Number of hubs

Air transport

165

Applications of optimization with Xpress-MP

sum(i in CITIES) hub(i) = NHUBS! One hub-to-hub connection per freight transport
forall(i,j in CITIES) sum(k,l in CITIES) flow(i,j,k,l) = 1! Relation between flows and hubs
forall(i,j,k,l in CITIES) doflow(i,j,k,l) <= hub(k)flow(i,j,k,l) <= hub(l)end-doforall(i in CITIES) hub(i) is_binary
forall(i,j,k,l in CITIES) flow(i,j,k,l) is_binary! Solve the problem
minimize(Cost)
end-model

11.4.3

Revised formulation
The formulation of the mathematical model in lines (11.4.1) to 11.4.7), and as a consequence its implementation with Mosel, uses a large number of variables for this relatively small problem (more than 1300
binaries). Looking at the distance data more carefully, one can see that the six airports are clusteredgeographically: the American airports are relatively close to each other and far from the European ones,
that are again close to each other. We may therefore reasonably assume that one hub will be located in
the US and one in Europe. The American airports will be connected via the US hub and the European
airports will use the European hub. This implies that we are able to exclude a large number of indexcombinations (e.g. EU airports connected to a US hub). We define the sets EU of European airports andUS of US airports, the union of which is the set CITIES of our previous model formulation.
In addition to the geographical considerations, we may reduce even further the number of variables byaccumulating the quantities to be transported between any pair of destinations to a single value andonly working with destination pairs i and j where i < j.
The index tuples for variables flow that remain after these operations are: ∀i, j, k ∈ US, i < j : (i, j, k, k)(intra-American flights) ∀i, j, k ∈ EU, i < j : (i, j, k, k) (intra-European flights), and ∀i, k ∈ US, j, l ∈ EU :(i, j, k, l) (inter-continental flights). For our problem there are just more than 100 variables.
All constraints of the previous problem remain the same except that we only sum over the variables thathave been defined. We may even add two additional sets of constraints based on the observation thatthe intra-American (11.4.8) and intra-European (11.4.9) flights will use only a single hub:∀i, j ∈ US, i < j :Xflowijkk = 1(11.4.8)

flowijkk = 1(11.4.9)k∈US∀i, j ∈ EU, i < j :Xk∈EUThe Mosel implementation of the revised formulation is given below. Note that the array of decisionvariables flow is defined as a dynamic array so that we only create the entries that are required.
Another change from the previous model is the replacement of the COST array by a function to reduce
the storage space required by the data (for larger problem sizes than ours this may be a significant
advantage).
model "F-4 Hubs (2)"uses "mmxprs"
forward function calc_cost(i,j,k,l:integer):realdeclarations
US = 1..3; EU = 4..6
CITIES = US + EU
NHUBS = 2

! Cities! Number of hubs

QUANT: array(CITIES,CITIES) of integer ! Quantity to transport
DIST: array(CITIES,CITIES) of integer ! Distance between cities
FACTOR: real
! Reduction of costs between hubs
flow: dynamic array(CITIES,CITIES,CITIES,CITIES) of mpvar

Air transport

166

Applications of optimization with Xpress-MP

hub: array(CITIES) of mpvar
end-declarations! flow(i,j,k,l)=1 if freight! from i to j goes via k and l! 1 if city is a hub, 0 otherwise

initializations from ’f4hub.dat’
QUANT DIST FACTOR
end-initializations
forall(i,j in CITIES | i<j) QUANT(i,j):=QUANT(i,j)+QUANT(j,i)
forall(i,j,k in US | i<j) create(flow(i,j,k,k))
forall(i,j,k in EU | i<j) create(flow(i,j,k,k))
forall(i,k in US, j,l in EU) create(flow(i,j,k,l))! Objective: total transport cost
Cost:= sum(i,j,k,l in CITIES | exists(flow(i,j,k,l)))
QUANT(i,j)*calc_cost(i,j,k,l)*flow(i,j,k,l)! Number of hubssum(i in CITIES) hub(i) = NHUBS! One hub-to-hub connection per freight transport
forall(i,j in CITIES | i<j) sum(k,l in CITIES) flow(i,j,k,l) = 1
forall(i,j in US | i<j) sum(k in US) flow(i,j,k,k) = 1
forall(i,j in EU | i<j) sum(k in EU) flow(i,j,k,k) = 1! Relation between flows and hubs
forall(i,j,k,l in CITIES | exists(flow(i,j,k,l))) doflow(i,j,k,l) <= hub(k)flow(i,j,k,l) <= hub(l)end-doforall(i in CITIES) hub(i) is_binary
forall(i,j,k,l in CITIES | exists(flow(i,j,k,l))) flow(i,j,k,l) is_binary! Solve the problem
minimize(Cost)!----------------------------------------------------------------! Transport cost from i to j via hubs k and lfunction calc_cost(i,j,k,l:integer):real
returned:=DIST(i,k)+FACTOR*DIST(k,l)+DIST(l,j)
end-function
end-model

11.4.4

Results
ChicagoParis
BostonMarseille

Nice

AtlantaFigure 11.2: Choice of hubs

Both models have the same solution. To minimize the total cost of transport the cities Boston and Paris
have to be used as hubs. Not surprisingly, Atlanta and Chicago are assigned to the hub Boston, Niceand Marseille to Paris. This means, the freight from Atlanta, Boston, and Chicago is gathered at the hub
Boston. There, any freight destined for any of the American cities is directly sent to its destination. All
merchandise for European destinations is first shipped to the hub at Paris and from there distributed to
the European cities. The freight originating from Europe and with destination USA first transits throughthe hub at Paris and then through Boston before arriving at its destination. The total cost of transport isAir transport

167

Applications of optimization with Xpress-MP

B
C 42,153,794.

11.5

Planning a flight tour
A country in south-east Asia is experiencing widespread flooding. The government, with international
help, decides to establish a system of supply by air. Unfortunately, only seven runways are still in a usable
state, among which is the one in the capital.
The government decides to make the planes leave from the capital, have them visit all the other six
airports and then come back to the capital. The following table lists the distances between the airports.
Airport A1 is the one in the capital. In which order should the airports be visited to minimize the total
distance covered?
Table 11.9: Distance matrix between airports (in km)
A1
A2
A3
A4
A5
A6

11.5.1

A2

A3

A4

A5

A6

A7

786

549
668

657
979
316

331
593
607
890

559
224
472
769
386

250
905
467
400
559
681

Model formulation
To model this problem, we denote by CITIES = {1, . . . , 7} the set of airports to be served and DISTij the
distance between the airports Ai and Aj . Since the journey is carried out by plane, the distance matrix issymmetric. We use binary variables flyij that take the value 1 if the plane flies from i to j. The tour weare looking for is a cycle that visits every airport once and once only. At every airport i, the cycle comesfrom a single predecessor and continues to a single successor airport. The constraints (11.5.1) and (11.5.2)
express these conditions.
∀j ∈ CITIES :X

flyij = 1

(11.5.1)

i∈CITIES,i6=j∀i ∈ CITIES :X

flyij = 1

(11.5.2)

j∈CITIES,i6=j

If we only use the constraints (11.5.1) and (11.5.2), the cycle may be fragmented into several sub-cycles,
for instance going from airport 1 to airport 2 and from there back to 1. This sub-cycle satisfies thesetwo constraints but does not lead to a valid solution. It is therefore necessary to prevent the formingof sub-cycles from occurring. This is the aim of the constraints (11.5.3): every sub-cycle that appears ina subset S of airports is ‘broken’ by requiring that the number of arcs in S must be strictly less than the
cardinality of S.
If N is the number of cities, for even a relatively small N it is simply impossible to test all sub-cycles that
may be formed (their number is in the order of 2N ). To start with, we are therefore going to solve theproblem without the constraints (11.5.3), which usually leads to a solution comprising sub-cycles, andthen one by one add constraints that render the forming of the sub-cycles impossible. This procedure will
be described in more detail in the next section.
∀S :X

flyij ≤ |S| − 1(11.5.3)(i,j)∈S

The objective function (11.5.4) is the total length of the cycle passing through all cities, that is the lengthof all arcs that are used.

X

minimizeX

DISTij · flyij(11.5.4)

i∈CITIES j∈CITIES∀j ∈ CITIES :X

flyij = 1

(11.5.5)i∈CITIES,i6=j∀i ∈ CITIES :X

flyij = 1

(11.5.6)

j∈CITIES,i6=j

Air transport

168

Applications of optimization with Xpress-MP

∀S :

X

flyij ≤ |S| − 1(11.5.7)(i,j)∈S∀i, j ∈ CITIES : flyij ∈ {0, 1}

11.5.2(11.5.8)

Implementation and results
The following Mosel program corresponds to the mathematical model without the constraints (11.5.3).
Since the distance matrix is symmetric, the data file only contains the upper triangle (distance from i to j
for which i < j); the other half is completed after reading in the data.
model "F-5 Tour planning"uses "mmxprs"
declarations
NCITIES = 7
CITIES = 1..NCITIES! Cities

DIST: array(CITIES,CITIES) of integer
NEXTC: array(CITIES) of integer! Distance between cities! Next city after i in the solution

fly: array(CITIES,CITIES) of mpvar
end-declarations! 1 if flight from i to jinitializations from ’f5tour.dat’DIST
end-initializations
forall(i,j in CITIES | i<j) DIST(j,i):=DIST(i,j)! Objective: total distance
TotalDist:= sum(i,j in CITIES | i<>j) DIST(i,j)*fly(i,j)! Visit every city once
forall(i in CITIES) sum(j in CITIES | i<>j) fly(i,j) = 1
forall(j in CITIES) sum(i in CITIES | i<>j) fly(i,j) = 1
forall(i,j in CITIES | i<>j) fly(i,j) is_binary! Solve the problem
minimize(TotalDist)
end-model

The solution to this problem is represented in Figure 11.3. It contains three subcycles that we need to
exclude. The strategy to adopt suppresses one sub-cycle at a time, starting with the smallest one.
We have implemented the following procedure break_subtour that is added to the Mosel program
above and called after the solution of the initial problem. The procedure first saves the successor of every
city into the array NEXTC. It then calculates the set of cities that are on the tour starting from city 1: if thesize of the set corresponds to the total number of cities NCITIES, we have found the optimal solution
and the algorithm stops, otherwise there are subtours.
The algorithm then enumerates all subtours to find the smallest one: if a subtour with two cities is foundwe can stop (there are no subtours of length 1), otherwise the tour starting from the next city that has
not yet been enumerated as part of a tour (set ALLCITIES) is calculated. When the smallest subtour has
been found, we add a constraint of type (11.5.3) for this set and re-solve the problem.
procedure break_subtour
declarations
TOUR,SMALLEST,ALLCITIES: set of integer
end-declarations
forall(i in CITIES)NEXTC(i):= integer(round(getsol(sum(j in CITIES) j*fly(i,j) )))! Get (sub)tour containing city 1
TOUR:={}
first:=1
repeat

Air transport

169

Applications of optimization with Xpress-MP

A2

A6
A5
A3

A1
A4
A7

Figure 11.3: Initial solution with three sub-cycles

TOUR+={first}
first:=NEXTC(first)
until first=1
size:=getsize(TOUR)! Find smallest subtour
if size < NCITIES then
SMALLEST:=TOURif size>2 then
ALLCITIES:=TOUR
forall(i in CITIES) do
if(i not in ALLCITIES) then
TOUR:={}
first:=irepeatTOUR+={first}
first:=NEXTC(first)
until first=i
ALLCITIES+=TOURif getsize(TOUR)<size then
SMALLEST:=TOUR
size:=getsize(SMALLEST)
end-ifif size=2 thenbreak
end-ifend-ifend-doend-if! Add a subtour breaking constraintsum(i in SMALLEST) fly(i,NEXTC(i)) <= getsize(SMALLEST) - 1! Re-solve the problem
minimize(TotalDist)! New round of subtour elimination
break_subtour
end-ifend-procedure

During its first execution the subtour elimination procedure adds the following constraint to the model
to eliminate the subtour (2,6,2):
fly(2,6) + fly(6,2) <= 1

Air transport

170

Applications of optimization with Xpress-MP

With this additional constraint we obtain the solution represented in the following figure: it still consistsof three (different) subtours.

A2

A6
A5
A3

A1
A4
A7

Figure 11.4: Second solution

When the subtour elimination procedure is called again, it adds a constraint to exclude the subcycle(1,7,1):fly(1,7) + fly(7,1) <= 1

After this second execution, we finally obtain a single tour with a total length of 2575 km. The arcs thatare directed in the model may be replaced by undirected arcs (edges) without any consequence for thesolution since the distance matrix is symmetric.

A2

A6
A5
A3

A1
A4
A7

Figure 11.5: Optimal solution

The Mosel implementation of the subtour elimination procedure uses a certain number of set operatorsand other functionality related to sets: the function getsize returns the size (= number of elements) ofa set or array. {} denotes the empty set. A set may be assigned using the usual assignment operator :=,e.g. TOUR:={}, and a set is added to another set using the operator +=.

Air transport

171

Applications of optimization with Xpress-MP11.6

References and further material
The problem of flight connections in Section 11.1 is a classical assignment problem. This problem is wellsolved by the Hungarian algorithm described, for instance, by Papadimitriou [PS98] or by techniques
searching for the maximum flow in a transport network [Pri94a].
The composition of flight crews in Section 11.2 is a problem of matching in an arbitrary graph whilst theassignment of personnel to workposts in Chapter 14 is a matching problem in a bipartite graph: the edges
connect nodes of two types, the persons and the workposts. The matrix of the linear program is totally
unimodular in the bipartite case, which means that the variables automatically take the values 0 or 1 at
the optimum. In the non-bipartite case of the flight crew, the simplex algorithm may find solutions withvariables at 0.5 in the optimum. The problems of maximum cardinality or maximum weight matchingin an arbitrary graph (bipartite or not) are all polynomial. An algorithm in O(n3 ) (with n the number ofnodes) with a Fortran code is given in Minoux et al. [MB86]. A simpler algorithm for maximum cardinality
matching, but still of complexity O(n3 ) is given by Syslo [SDK83] with a Pascal source code.
The problem of scheduling flight landings in Section 11.3 represents a class of scheduling problems withtime windows and distance constraints between tasks. It has been studied by Beasley et al. [BKSA00]
for the case of multiple runways. The authors describe a specialized tree search method for large size
instances. They are able to solve realistic instances with 50 planes and four runways to optimality. Beasleyhas also published an interesting article [BC96] concerning crew scheduling. Based on a set of flights withgiven dates and durations this problem consists of defining the sequence of flights for the crews in order
to cover all the flights but without exceeding a limited duration of work.
With the formulation presented in Section 11.4 for the hub location problem, the number of variables
used may be very large. If it is possible to transform any city into a hub, the number of the variables
flowijkl is in the order of O(n4 ) (with n the number of cities). And hence, a problem with just over thirty
cities would require a million variables. In Daskin [Das95] and Campbell [Cam96] several heuristics for thistype of problem are given.
The problem in Section 11.5 is the famous Traveling Salesman Problem (TSP). It consists of organizing atour that visits every node of a graph once and only once, whilst minimizing the total cost of the arcs
used on the tour. The flight tour planning problem presented here is symmetrical because the cost matrix
is a symmetrical distance matrix. The paint production problem in Chapter 7 is an asymmetrical TSP. It issolved via a direct formulation, that is, without progressively adding constraints. The TSP is such a wellknown NP-hard problem problem that it is used for testing new optimization techniques. The CD thatcomes with the book contains a data file with a distance matrix for several French cities and a slightly
modified version of the Mosel program of Section 11.5, adapted to the structure of this file.
The TSP is so hard that the MIP models of this book take a very long time to solve for more than twenty
cities. Beyond this number, it becomes necessary to use specialized tree search methods [HK70]. Pedagogical examples of such methods are given by Roseaux [Ros85] and Evans [EM92], and a Pascal codeis given by Syslo [SDK83]. Certain Euclidean problems with hundreds of cities can be dealt with by themethod called Branch and Cut [GH91]. Otherwise, one has to give up trying to find optimal solutions
and use heuristic techniques, some of which, like tabu search or simulated annealing, are very efficientin practice. Such heuristics are described in detail with listings of Pascal code in the book about graphs
by Prins [Pri94a].

Air transport

172

Applications of optimization with Xpress-MP

Chapter 12

Telecommunication problems
In recent years, the domain of telecommunications has experienced an explosive growth. It is extraordinarily rich in original optimization problems, which explains the fact that this book devotes an entire
chapter to this topic. The short term exploitation of telecom networks are problems of on-line control
and problems tackled by the theory of telecommunication protocols. But design problems, the sizing
or location of resources, and planning are well suited to optimization approaches. These are therefore
described here.
The design of a network needs to satisfy the traffic forecasts between nodes, minimize the construction
costs and fulfill conditions on its reliability. It starts with the choice of the location of nodes. Section 12.6
deals with a mobile phone network in which transmitters need to be placed to cover the largest possible
part of the population within given budget limits. Once the nodes are chosen, one needs to select thepairs of nodes that will be connected with direct lines for the transmission of data. In Section 12.4, acable network with a tree structure is constructed with the objective of minimizing the total cost. The
problem in Section 12.2 studies the connection of cells to a main ring in a mobile phone network withcapacity constraints. Reliability considerations demand multiple connections between every cell and thering.
Other optimization problems arise in the analysis and the exploitation of existing networks. A natural
question dealt with in the problem of Section 12.3 is to know the maximum traffic that a network withknown capacity limits may support. The problem in Section 12.1 concerns the reliability of the data exchange between two nodes of a given network. This reliability is measured by the number k of disjoint
paths (that is, paths without any intermediate node in common) between two nodes. If k > 1, then communication is still possible even if k − 1 nodes break down. The problem in Section 12.5 is an example oftraffic planning: the data packets in the repeater of a telecommunications satellite need to be scheduled.

12.1

Network reliability
We consider the military telecommunications network represented in Figure 12.1. This network consistsof eleven sites connected by bidirectional lines for data transmission.
For reliability reasons in the case of a conflict, the specifications require that the two sites (nodes) 10
and 11 of the network remain able to communicate even if any three other sites of the network aredestroyed. Does the network in Figure 12.1 satisfy this requirement?

12.1.1Model formulation
This problem can be converted into a maximum flow problem. The latter is dealt with in detail in
Chapter 15 (Problem of water supply), but we recap here briefly. Let a directed graph be given as
G = (NODES, ARCS), where NODES is a set of nodes, ARCS a set of arcs. An arc connecting node n tonode m is written (n, m) and has an integer capacity of CAPnm .
Imagine that a liquid enters the network at a given node SOURCE and leaves it at another node SINK. By
spreading over the network, this liquid creates a flow that can be defined as a flow flownm in every arc(n, m). This flow is valid if it stays within the capacity limits of the arcs and if it does not incur any loss
when passing through intermediate nodes (that is, nodes other than SOURCE and SINK). The maximum
flow problem consists of finding a flow that maximizes the total throughput injected at SOURCE (orarriving at SINK since there are no losses). If this problem is formulated as an LP, the simplex algorithm
automatically finds an optimal solution with integer flows.

173

Applications of optimization with Xpress-MP

1

2

8

3

11

10

5

9

4

6

7

Figure 12.1: Telecommunications network

At first sight, our reliability problem does not quite resemble the maximum flow problem. Let us imagine
a modified network H with capacities equal to 1 on all connections: the optimal flows on arcs therefore
will take the values 0 or 1. We also impose a capacity of 1 at every node: through this, two units of flow
leaving S = 10 through two different arcs will follow two paths to T = 11, without passing through any
common node. Such paths are called node-disjunctive.
Since a flow on every arc leaving SOURCE is either 0 or 1, the maximum number of disjunctive paths from
SOURCE to SINK in G is equal to the throughput of a maximum flow in the modified graph H. If wefind k paths, the network is resilient to k − 1 broken nodes: in the worst case, these nodes will be on
k − 1 disjunctive paths, but communication will remain possible via the one remaining path. In the end,
the answer to our reliability problem consists of comparing k − 1 with the maximum number of node
breakdowns allowed by the specifications.
The only minor difficulty is that the flow problem is usually defined in a directed graph. But in our
telecommunications network the bidirectional connections are not directed. To distinguish the flow inboth senses on a single connection between the nodes n and m, we replace the connection by two arcs
(n, m) and (m, n). We therefore obtain a directed graph G = (NODES, ARCS) that allows us to formulate
the problem in a simple manner.

X

maximize(12.1.1)

flowSOURCE,mm succ.of SOURCE∀n 6= SOURCE, SINK :X

m succ.of n∀n 6= SOURCE, SINK :X

Xflownm =flowmn(12.1.2)

m pred.of nflownm ≤ 1(12.1.3)

m succ.of nXflowm,SOURCE = 0(12.1.4)

m pred.of SOURCE∀(n, m) ∈ ARCS : flownm ∈ {0, 1}(12.1.5)(12.1.6)

The objective function (12.1.1) is to maximize the total throughput, the sum of all flows on the arcs
leaving SOURCE (we could also sum the flows arriving at SINK). The constraints (12.1.2) model the flow
conservation law at every intermediate node n (also called Kirchhoff’s law): the total flow arriving fromthe predecessors must be equal to the total flow leaving to the successors. The constraints (12.1.3) limit
the flow through every intermediate node to 1; we consider here the flow leaving every node.
The constraint (12.1.4) is necessary to avoid the flow injected at SOURCE returning to this node. It is not
required in the water supply problem in Chapter 15 because in this problem the source has no predecessors. And finally, the constraints (12.1.5) specify that all flow variables flownm are binary. Since this is a
maximum flow problem, the LP solution will automatically have integer values. Furthermore, the constraints (12.1.3) imply that the variables are upper bounded by 1. The constraints (12.1.5) could therefore
be replaced by the usual non-negativity constraints.

Telecommunication problems

174

Applications of optimization with Xpress-MP

12.1.2

Implementation
The following Mosel program implements the mathematical model. A graph with N nodes may be encoded in two different ways: in the form of a list of arcs or as an adjacency matrix . The first method
is employed in this book, for example, in the line balancing problem in Section 7.6 and the problem
of forming flight crews in Section 11.2. We use here the matrix representation that is also used in the
problem of gritting roads in Section 15.4.
The adjacency matrix is a binary matrix ARC of size N × N, with ARCnm = 1 if and only if the arc (n, m)
exists. This array is read from the data file in sparse format, that is, only the non-zero elements or ARC
are given, in the form (n m) ARC(n,m). The entries of ARC that are not defined have the default value
0 if they are addressed in the program, but we simply test for existing entries of ARC using the Mosel
function exists that enumerates the array in a very efficient way, especially if only relatively few of theN2 possible entries are indeed defined.
To obtain a more compact model and hence a smaller LP that solves faster, only the variables flownm that
correspond to existing arcs are defined. Note that it is not necessary to repeat this condition in every sum
over these variables.
The data file contains the representation of the graph in an undirected form: the bidirectional connection
between n and m only appears in one sense in the file (with the convention n < m). We therefore
construct the oriented version of the graph that is required for the maximum flow formulation: for every
defined arc (n, m) we define its opposite (m, n). The rest of the model is a straight translation from themathematical formulation. This model is valid for any choice of SOURCE and SINK among the nodes of
the network.
model "G-1 Network reliability"
uses "mmxprs"
declarations
NODES: range
SOURCE = 10; SINK = 11

! Set of nodes! Source and sink nodes

ARC: array(NODES,NODES) of integer! 1 if arc defined, 0 otherwise

flow: array(NODES,NODES) of mpvar
end-declarations! 1 if flow on arc, 0 otherwise

initializations from ’g1rely.dat’ARC
end-initializations
forall(n,m in NODES | exists(ARC(n,m)) and n<m ) ARC(m,n):= ARC(n,m)
forall(n,m in NODES | exists(ARC(n,m)) ) create(flow(n,m))! Objective: number of disjunctive paths
Paths:= sum(n in NODES) flow(SOURCE,n)! Flow conservation and capacities
forall(n in NODES | n<>SOURCE and n<>SINK) dosum(m in NODES) flow(m,n) = sum(m in NODES) flow(n,m)
sum(m in NODES) flow(n,m) <= 1
end-do! No return to SOURCE nodesum(n in NODES) flow(n,SOURCE) = 0
forall(n,m in NODES | exists(ARC(n,m)) )flow(n,m) is_binary! Solve the problem
maximize(Paths)! Solution printing
writeln("Total number of paths: ", getobjval)
forall(n in NODES | n<>SOURCE and n<>SINK)
if(getsol(flow(SOURCE,n))>0) thenwrite(SOURCE, " - ",n)nnext:=nwhile (nnext<>SINK) do
nnext:=round(getsol(sum(m in NODES) m*flow(nnext,m)))Telecommunication problems

175

Applications of optimization with Xpress-MP

write(" - ", nnext)
end-dowriteln
end-ifend-model

The solution display for this example uses a small algorithm to print all paths between the SOURCE and
SINK nodes: for every node to which there is a flow from the SOURCE node, we calculate the chain of its
successors until we arrive at the SINK node. To obtain the successor of a node n, we exploit the fact thatevery intermediate node only appears in a single path and hence only has a single successor. We calculate
the index number of this successor as the sum of flows leaving node n multiplied by the node index. The
result of getsol is type real and needs to be transformed to an integer using the function round.

1

2

8

3

11

10

5

9

4

6

7

Figure 12.2: Disjunctive paths

12.1.3Results
The solution algorithm calculates a maximum throughput of 4, which means there are 4 disjunctive paths
between nodes 10 and 11 and hence, the two nodes may continue to communicate in spite of the destruction of three intermediate sites. The specifications are satisfied. The four disjunctive paths between
10 and 11 are represented in Figure 12.2.

12.2

Dimensioning of a mobile phone network
Figure 12.3 represents the typical architecture of a mobile phone network. Every elementary geographical zone or cell is served by a transmitter-receiver called a relay. The calls originating from a mobile
phone first pass through these relays. Every relay is connected by cable or electro-magnetic waves to atransit node (hub). One of the hubs controls the network, this is the MTSO (Mobile Telephone Switching
Office). A very reliable ring of fiber optic cable connects the hubs and the MTSO with high capacity links.
It is capable of re-establishing itself in the case of a breakdown (self-healing ring) and there is no needto replicate it.
At the present state of technology, there are no dynamic connections between the relays and the MTSO.
The connections are fixed during the design phase, so it is necessary to choose the nodes of the ring thata relay should be connected to. The number of links between a cell c and the ring is called the diversityof the cell c, denoted by CNCTc . A diversity larger than 1 is recommended for making the system morereliable.
The traffic in this kind of system is entirely digitized, expressed in values equivalent to bidirectional
circuits at 64kbps (kilobit per second). This capacity corresponds to the number of simultaneous calls
during peak periods. The ring has edges of known capacity CAP. The traffic TRAFc from a cell c is divided
into equal parts (TRAFc / CNCTc ) among the connections to the ring. This traffic is transmitted via the ring

Telecommunication problems

176

Applications of optimization with Xpress-MP

Cell 1
2 connections

Cell 2
1 connection

Relay

Relay

Hub 4

Hub 3
Ring
Hub 2

Hub 1

MTSO

Figure 12.3: Structure of a mobile phone network

to the MTSO, that then routes it to another cell or to a hub that serves as the interface to the fixed-line
telephone network. A relay may be connected directly to the MTSO that also has the functions of an
ordinary hub.
We consider a network of 10 cells and a ring of 5 nodes with a capacity of CAP = 48 circuits. The MTSO is
at node 5. The following table indicates the traffic, the required number of connections and the cost perconnection in thousand $ per cell. For example, cell 1 is connectable with node 1 for a cost of $15,000.
Its diversity is 2, which means it must be connected to at least two nodes of the ring. Its traffic capacity
is of 22 simultaneous circuits. The objective is to define the connections of the cells to the ring thatminimize the connection costs whilst remaining within the capacity limits and satisfying the constraints
on the number of connections.
Table 12.1: Connection costs, traffic and number of connections per cell12.2.1

Cell

1

2

3

4

5

6

7

8

9

10

Hub 1
Hub 2
Hub 3
Hub 4
Hub 5 (MTSO)
Traffic
Connections

15
8
7
11
10
22
2

9
11
8
5
14
12
2

12
6
7
15
15
20
2

17
5
9
18
24
12
2

8
22
21
19
6
15
3

7
25
15
9
17
25
1

19
25
21
20
22
15
3

20
9
15
18
25
14
2

21
22
14
16
20
8
2

25
24
13
4
11
22
2

Model formulation
We write CELLS for the set of cells to be connected, NODES = HUBS ∪ {MTSO} the set of nodes, composed of the set of simple hubs and an MTSO, and COSTcn the cost of connecting cell c to node n. The
connections can be defined through binary variables connectcn , that are 1 if and only if cell c is connected
to node n (12.2.4). The objective function (12.2.1) measures the total connection cost. The constraints(12.2.2) ensure that every cell is connected to the required number of nodes.
minimize

X

X

COSTcn · connectcn(12.2.1)c∈CELLS n∈NODES∀c ∈ CELLS :X

connectcn = CNCTc(12.2.2)· connectcn ≤ 2 · CAP(12.2.3)n∈NODESXX TRAFcc∈CELLS n∈HUBS

CNCTc∀c ∈ CELLS, n ∈ NODES : connectcn ∈ {0, 1}(12.2.4)

Constraint (12.2.3) is a necessary condition for keeping within the capacity limits of the ring. It is basedon the fact that all demands of any origin are routed through the MTSO, following the ring in one sense

Telecommunication problems

177

Applications of optimization with Xpress-MP

or the other. Since every edge of the ring has a capacity CAP, the total traffic on the ring may not exceed
2 · CAP. Note that the traffic of a cell directly connected to the MTSO does not enter the ring.
For the problem to have any solution the ring must have a certain minimum capacity to route all the
traffic. The least possible traffic in the ring is obtained if every cell c has a direct connection with the MTSO
among its CNCTc required connections with the ring. The ring must be able to handle the remaining
traffic, which results in the minimum capacity constraint (12.2.5): for example, a cell with diversity 2routes at least 1 − 21 = 12 of its traffic through the ring; the traffic of a cell with diversity 1 in the bestcase (i.e. if it is connected to the MTSO) does not enter the ring at all. This relation does not contain any
variable, it is therefore not included in the mathematical model. It is however necessary to check whether
the given data fulfill this condition, which is the case here.

XTRAFc · 1 −c∈CELLS

1
CNCTc



≤ 2 · CAP

(12.2.5)

The non-specialist in telecommunications might raise the question of what has happened to the return
traffic. The traffic from a cell c is expressed in bidirectional circuits. If someone in cell c calls someone
in cell d, then the call traces its path from c to the MTSO, reserving a path for the answer. The MTSO
then transmits the call to cell d, still reserving a return path. When the called person in d picks up thephone, an access path in both directions is established, consuming one circuit from the capacity of c andone circuit from the capacity of d.

12.2.2

Implementation
The mathematical model is easily translated into a Mosel program. The cost data array from table 12.1 isgiven in transposed form in the data file to match the format of the array COSTcn .
Before defining the model we check whether the ring has a sufficiently large capacity to satisfy all traffic
demands (inequality (12.2.5)). If this condition does not hold then the program is stopped using thefunction exit.
model "G-2 Mobile network dimensioning"uses "mmxprs"
declarations
HUBS = 1..4
MTSO = 5
NODES = HUBS+{MTSO}CELLS = 1..10! Set of hubs
! Node number of MTSO
! Set of nodes (simple hubs + MTSO)! Cells to connect

CAP: integer
COST: array(CELLS,NODES) of integer
TRAF: array(CELLS) of integer
CNCT: array(CELLS) of integer! Capacity of ring segments! Connection cost
! Traffic from every cell
! Connections of a cell to the ringconnect: array(CELLS,NODES) of mpvar! 1 if cell connected to node,! 0 otherwise

end-declarations
initializations from ’g2dimens.dat’
CAP COST TRAF CNCT
end-initializations! Check ring capacity
if not (sum(c in CELLS) TRAF(c)*(1-1/CNCT(c)) <= 2*CAP) thenwriteln("Ring capacity not sufficient")exit(0)end-if! Objective: total cost
TotCost:= sum(c in CELLS, n in NODES) COST(c,n)*connect(c,n)! Number of connections per cell
forall(c in CELLS) sum(n in NODES) connect(c,n) = CNCT(c)! Ring capacity
sum(c in CELLS, n in HUBS) (TRAF(c)/CNCT(c))*connect(c,n) <= 2*CAP
forall(c in CELLS, n in NODES) connect(c,n) is_binary

Telecommunication problems

178

Applications of optimization with Xpress-MP

! Solve the problem
minimize(TotCost)
end-model

12.2.3Results
The optimal MIP solution has a total cost of $249,000. The traffic on the ring is 94, compared to a capacityof 96. The following table lists the details of the connections to be established between cells and nodesof the ring.
Table 12.2: Connection costs, traffic and number of connections per cell

12.3

Cell

1

2

3

4

5

6

7

8

9

10

Host nodes

3,5

3,4

2,5

2,3

1,4,5

5

1,4,5

2,3

3,5

4,5

Routing telephone calls
A private telephone company exploits a network, represented in Figure 12.4, between five cities: Paris,
Nantes, Nice, Troyes, and Valenciennes.

200
300

Valenciennes
70

Paris

NantesTroyes
300
120

80

Nice
Figure 12.4: Structure of the network of the company

The number beside each edge (connection) is the capacity of the link in terms of circuits. This issue
is worth some further explanation. Suppose that a person A at Nantes calls a person B at Nice. The
company needs to find a path formed by non-saturated edges from Nantes to Nice to route the binary
flow corresponding to the digitized voice of A. But the conversation is only possible if a flow from Niceto Nantes is established so that B may answer A. In digital telephone networks, the binary flows all
have the same throughput standard (often 64 kbps). The associated capacity is called a channel. The
return channel uses the same edges as the first channel, but in the opposite direction. This linked pair ofchannels necessary for a telephone conversation is a circuit.
The routing of the return channel through the same edges is due to the fact that the call could fail if
one waited until the callee picked up the phone before searching for a non-saturated return path. This
is why, at the moment of the call, the routing system constructs the path edge by edge and immediately
reserves the edges of the return channel. As a consequence, as the capacity of an edge is consumed inincrements of a bidirectional circuit, we do not consider any directed flows. For example, there may be
10 persons calling from Nantes to Nice and 20 from Nice to Nantes, or the opposite: in both cases, 30
circuits are used.
At a given moment, the company is facing demands for circuits given in the following table. Is it possible
to satisfy the demands entirely? If this is not possible, try to transmit as much as possible. In every case,indicate the corresponding routing, that is, the access paths used.

12.3.1Model formulation
In the maximum flow problem in Section 15.1 (water supply management) the task is to maximize thetotal throughput of a single product flow that traverses a limited capacity network between an origin

Telecommunication problems

179

Applications of optimization with Xpress-MP

Table 12.3: Demand of circuits
Names of the cities

Circuits

Nantes-Nice
Nantes-Troyes
Nantes-Valenciennes
Nice-Valenciennes
Paris-Troyes

100
80
75
100
70

(source) and a destination (sink). Here, we are dealing with a generalization called the multi-commodity
network flow (MCNF) problem: several flows that may not be mixed and that correspond to different
products use the same network, every flow having its own source and sink. The objective is still tomaximize the total throughput. Here, a ‘product’ corresponds to the totality of calls exchanged between
two given cities. The calls between two different pairs of cities may not be mixed: calls between Nantes
and Nice must not end in Troyes!
The MCNF problem is a difficult combinatorial problem. In the classical maximum flow problem, Linear
Programming uses flow variables flownm for every arc (n, m). The simplex algorithm automatically finds
integer-valued optimal flows if the arc capacities are integers. It is possible to generalize this model tosolve the MCNF problem in a directed graph with variables flowknm for the flow of product k on the arc(n, m) but the values at the end of the LP are not usually integers. Optimally rounding the fractionalflows is very difficult because there may be several flows that are routed through the same saturated arc.
Furthermore, the model with three-index variables requires certain tricks for dealing with an undirected
network as is the case in the present example. We are therefore going to use a simpler formulation called
arc-paths.
This formulation is based on different elementary paths (without intermediate nodes in common) that
may be used between pairs of cities that are communicating. Instead of using variables flowknm for every
pair of cities k and every arc (n, m), this formulation uses a single variable flowp for the flow all along apath p, where this path corresponds to a known pair of cities. The flow conservation laws per node are
therefore implicitly fulfilled. This formulation cannot be employed in dense networks with an enormous
number of possible paths. But for telecommunication networks that have a skeleton-like structure it isbetter than modeling with three-index flow variables.
For the data, we use ARCS for the set of (undirected) arcs, CALLS for the set of city pairs between which
there are demands for calls, and PATHS the set of paths. Let CAPa denote the capacity of an arc and DEMc
the demand for a city pair c. We also need to know which pair of cities is at the ends of every path,given by the call index CINDEXp . The arc-paths formulation is then remarkably easy. It is called ‘arc-paths’
because it works with arcs and paths, but not with nodes.
maximizeX(12.3.1)

flowp

p∈PATHS∀a ∈ ARCS :X

flowp ≤ CAPa

(12.3.2)

flowc ≤ DEMc

(12.3.3)

p∈PATHSa∈p∀c ∈ CALLS :Xp∈PATHSCINDEXp =c∀p ∈ PATHS : flowp ∈ IN

(12.3.4)

The objective function is to maximize (12.3.1), the sum of the flows on all paths that may be used. The
constraints (12.3.2) consider every arc a: the sum of flows on the paths passing through a must not exceed
the capacity of a. The constraints (12.3.3) concern every pair of cities c: the sum of flows exchanged
between them over the various paths must not exceed the demand for circuits. The integrality constraints(12.3.4) are necessary because the MCNF problem is not automatically integer at its linear optimum.

12.3.2

Implementation
The translation of the mathematical model for Mosel is given in the following program. This is a typicalcase where after the formulation of the mathematical model some work still is left to do for the formulation of the model with Mosel since no direct representation of the paths is immediately available. The
pairs of cities are named as represented in Table 12.3. Similarly, the arcs are named with the cities they
connect: Nantes-Paris, Nantes-Nice, Paris-Nice, Paris-Valenciennes, Troyes-Nice and Troyes-Valenciennes.
The paths are coded as a two-dimensional array ROUTEpa , every line of which corresponds to a path andTelecommunication problems

180

Applications of optimization with Xpress-MP

contains the list of edges it uses. In the worst case, a path may contain all the arcs since the paths are elementary and do not visit any city twice. The size of the second dimension of the array ROUTE is therefore
given by the number NARC of arcs in the problem. The following table lists the complete set of paths
and the corresponding city pairs at the ends of every path (array CINDEXp ). With this representation, themodel can now be implemented easily.
Table 12.4: Definition of paths
PathCity pair

List of arcs

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

Nantes-Nice
Nantes-Nice
Nantes-Nice
Nantes-Troyes
Nantes-Troyes
Nantes-Troyes
Nantes-Troyes
Nantes-Valenciennes
Nantes-Valenciennes
Nantes-Valenciennes
Nantes-Valenciennes
Nice-Valenciennes
Nice-Valenciennes
Nice-Valenciennes
Paris-Troyes
Paris-Troyes
Paris-Troyes

Nantes-Nice
Nantes-Paris, Paris-Nice
Nantes-Paris, Paris-Valenciennes, Valenciennes-Troyes, Troyes-Nice
Nantes-Paris, Paris-Valenciennes, Valenciennes-Troyes
Nantes-Paris, Paris-Nice, Nice-Troyes
Nantes-Nice, Nice-Troyes
Nantes-Nice, Nice-Paris, Paris-Valenciennes, Valenciennes-Troyes
Nantes-Paris, Paris-Valenciennes
Nantes-Nice, Nice-Paris, Paris-Valenciennes
Nantes-Paris, Paris-Nice, Nice-Troyes, Troyes-Valenciennes
Nantes-Nice, Nice-Troyes, Troyes-Valenciennes
Nice-Nantes, Nantes-Paris, Paris-Valenciennes
Nice-Paris, Paris-Valenciennes
Nice-Troyes, Troyes-Valenciennes
Paris-Valenciennes, Valenciennes-Troyes
Paris-Nantes, Nantes-Nice, Nice-Troyes
Paris-Nice, Nice-Troyes

Making the path inventory is quite tedious, but the collection of data is usually a major problem inMathematical Programming. In industrial applications, the data files must be constructed automaticallyby some special purpose software (in the examples of Sections 11.2 or 12.5 we use Mosel for doingsome data preprocessing) or be extracted from databases or spreadsheets. The full distribution of Mosel
includes modules for accessing databases or spreadsheets directly.
model "G-3 Routing telephone calls"
uses "mmxprs"
declarations
CALLS: set of string
ARCS: set of string
PATHS: range
CAP: array(ARCS) of integer
DEM: array(CALLS) of integer
CINDEX: array(PATHS) of string
end-declarations! Set of demands! Set of arcs
! Set of paths (routes) for demands! Capacity of arcs
! Demands between pairs of cities! Call (demand) index per path index

initializations from ’g3routing.dat’
CAP DEM INDEX
end-initializations
finalize(CALLS); finalize(ARCS); finalize(PATHS)
NARC:=getsize(ARCS)
declarations
ROUTE: array(PATHS,1..NARC) of string ! List of arcs composing the routes
flow: array(PATHS) of mpvar! Flow on paths
end-declarations
initializations from ’g3routing.dat’
ROUTE
end-initializations! Objective: total flow on the arcs
TotFlow:= sum(p in PATHS) flow(p)! Flow within demand limits
forall(d in CALLS) sum(p in PATHS | CINDEX(p) = d) flow(p) <= DEM(d)Telecommunication problems

181

Applications of optimization with Xpress-MP

! Arc capacities
forall(a in ARCS)
sum(p in PATHS, b in 1..NARC | ROUTE(p,b)=a) flow(p) <= CAP(a)forall(p in PATHS) flow(p) is_integer! Solve the problem
maximize(TotFlow)
end-model

12.3.3Results
The LP solution to this problem has integer values. 380 out of the required 425 calls are routed, thatis all but 45 Nantes-Troyes calls. The following table lists the corresponding routing of telephone calls(there are several equivalent solutions). On the connections Nantes-Paris, Nantes-Nice, Paris-Nice, andTroyes-Valenciennes there is unused capacity; the other arcs are saturated.
Table 12.5: Optimal routing of telephone calls
City pair

12.4

DemandRoutedNantes-Nice
Nantes-Troyes

100
80

100
35

Nantes-Valenciennes
Nice-Valenciennes

75
100

Paris-Troyes

70

75
20
80
70

Path
Nantes-Paris, Paris-Nice
Nantes-Nice, Nice-Paris, Paris-Valenciennes,
Valenciennes-Troyes
Nantes-Paris, Paris-Valenciennes
Nice-Paris, Paris-Valenciennes
Nice-Troyes, Troyes-Valenciennes
Paris-Valenciennes, Valenciennes-Troyes

Construction of a cabled network
A university wishes to connect six terminals located in different buildings of its campus. The distances, inmeters, between the different terminals are given in Table 12.6.
Table 12.6: Distances between the different terminals (in meters)
Terminal 1
Terminal 2
Terminal 3
Terminal 4
Terminal 5
Terminal 6

Terminal 1

Terminal 2

Terminal 3

Terminal 4

Terminal 5

Terminal 6

0
120
92
265
149
194

120
0
141
170
93
164

92
141
0
218
103
116

265
170
218
0
110
126

149
93
103
110
0
72

194
164
116
126
72
0

These terminals are to be connected via underground cables. We suppose the cost of connecting two
terminals is proportional to the distance between them. Determine the connections to install to minimize
the total cost.

12.4.1Model formulation
We define the undirected labeled graph G = (TERMINALS, DIST, CONNECTIONS) where the set of nodes
TERMINALS corresponds to the set of terminals and the set of edges CONNECTIONS contains the possible
connections [s, t] between these terminals which are labeled with the distance DISTst that separates them.
We define binary variables connectst that are 1 if and only if terminals s and t are directly connected. Since
the connections are undirected, it is sufficient to define the variables for s < t. The objective is to connect
all the terminals at the least cost. The objective function is therefore given by (12.4.1).
minimize

X

Xs∈TERMINALSt∈TERMINALSs<t

DISTst · connectst(12.4.1)

To connect NTERM terminals, the most economical network is a tree of NTERM − 1 connections. A tree
connecting NTERM nodes is a connected graph without any cycles or, equivalently a connected graph

Telecommunication problems

182

Applications of optimization with Xpress-MP

with NTERM − 1 connections. The constraint (12.4.2) imposes these NTERM − 1 connections.

X

Xs∈TERMINALSt∈TERMINALS
s<tconnectst = NTERM − 1(12.4.2)

Furthermore, every terminal must be connected to at least one other terminal in the tree. A first idea isto add the constraints (12.4.3).

X∀s ∈ TERMINALS :connectst ≥ 1(12.4.3)t∈TERMINALS
s<t

However, these constraints are not sufficient. The constraints (12.4.2) and (12.4.3) may lead to an infeasible solution as shown in Figure 12.5 in which a cycle is created.

T4
T2

T5

T6
T1

T3

Figure 12.5: An infeasible unconnected solution

To avoid this type of solution, we need to impose the constraint (12.4.4) for any subset S of the set ofterminals. The translation of this constraint into a program is unfortunately a difficult task: on one hand
there is no possibility of enumerating all subsets, and on the other hand the number of constraints of thistype is enormous (2NTERM , that is, about one million for twenty nodes).
∀S ⊆ TERMINALS :X X

connectst ≤ |S| − 1(12.4.4)s∈S t∈S, s<t

There follows therefore, a different way of preventing cycles. Consider a tree with its edges directed
departing from a root r, that is, a node that is connected to a single other node, like T1 in Figure 12.6
(any root will do). We may now assign every node t a level value levelt that may be interpreted as thelength (in terms of number of edges) of the path connecting t with r (in a tree this path exists and is
unique). For example, we have levelr = 0 and levelt = 1 for any node t directly connected to r. The anticycle constraints (12.4.5) are based on positive level value variables. For this formulation, the connections
must be directed: we have either connectst = 1 or connectts = 1 for two directly connected nodes s and t.
This has no impact on the solution.
∀s, t ∈ TERMINALS, s 6= t : levelt ≥ levels + 1 − NTERM + NTERM · connectst(12.4.5)

To understand these constraints, let us suppose that a solution of the mathematical model contains acycle, for instance the cycle 1 → 2 → 5 → 1 of Figure 12.5, oriented in this sense. The constraints (12.4.5)
for which connectst = 1 in this cycle result in:level2 ≥ level1 + 1(12.4.6)

level5 ≥ level2 + 1(12.4.7)

level1 ≥ level5 + 1(12.4.8)

By summing up term by term, we obtain a contradiction: 0 ≥ 3 ! A solution containing a cycle therefore
violates the constraints (12.4.5). On the contrary, if we have a tree, values of levelt exist that satisfy theconstraints, for instance the level numbers obtained by traversing the tree starting from an arbitrary rootr. If s is directly before t on the path from r to t, we have levelt = levels + 1 and connectst = 1. The

Telecommunication problems

183

Applications of optimization with Xpress-MP

constraint (12.4.5) for s and t is satisfied since it reduces to levelt = levels + 1. If s is not directly connected
to t, we have connectst = 0 and the constraint (12.4.5) reduces to levelt −levels ≥ 1−NTERM, an inequalitythat is trivially satisfied since the level numbers range between 0 and NTERM − 1.
The constraints (12.4.5) only detect cycles with connections directed around the cycle. For instance, they
do not exclude the cycle 1 → 2 ← 5 → 1. Such a cycle cannot occur when we add the following constraints
that direct all connections to the (arbitrarily chosen) root node.

X∀s = 2, . . . , NTERM :connectst = 1(12.4.9)t∈TERMINALS
s6=t

In the formulation of the constraints (12.4.9) we have chosen node 1 as the root node. Every node must
be connected to at least one other node. Since a tree does not contain cycles there must be a single pathfrom every node in the tree to the root node. That means, every node s other than the root node has
exactly one outgoing connection. In other words, for every s 6= 1 exactly one variable connectst must beat 1.
We thus obtain the following mathematical model.

XminimizeX(12.4.10)

DISTst · connectsts∈TERMINALS t∈TERMINALS

X

X(12.4.11)

connectst ≤ NTERM − 1s∈TERMINALS t∈TERMINALS∀s, t ∈ TERMINALS, s 6= t : levelt ≥ levels + 1 − NTERM + NTERM · connectst∀s = 2, . . . , NTERM :X(12.4.12)(12.4.13)

connectst = 1

t∈TERMINALS
s6=t

12.4.2∀s, t ∈ TERMINALS : connectst ∈ {0, 1}(12.4.14)∀t ∈ TERMINALS : levelt ≥ 0(12.4.15)

Implementation
The translation of the mathematical model into a Mosel program is direct.
model "G-4 Cable connections between terminals"
uses "mmxprs"
declarations
NTERM = 6
TERMINALS = 1..NTERM! Set of terminals to connect

DIST: array(TERMINALS,TERMINALS) of integer! Distance between terminals

connect: array(TERMINALS,TERMINALS) of mpvar ! 1 if direct connection! between terminals, 0 otherwise
level: array(TERMINALS) of mpvar! level value of nodes
end-declarations
initializations from ’g4cable.dat’DIST
end-initializations! Objective: length of cable used
Length:= sum(s,t in TERMINALS) DIST(s,t)*connect(s,t)! Number of connectionssum(s,t in TERMINALS | s<>t) connect(s,t) = NTERM - 1! Avoid subcycle
forall(s,t in TERMINALS | s<>t)level(t) >= level(s) + 1 - NTERM + NTERM*connect(s,t)! Direct all connections towards the root (node 1)
forall(s in 2..NTERM) sum(t in TERMINALS | s<>t) connect(s,t) = 1
forall(s,t in TERMINALS | s<>t) connect(s,t) is_binary

Telecommunication problems

184

Applications of optimization with Xpress-MP

! Solve the problem
minimize(Length)
end-model

12.4.3Results
The optimal solution connects the terminals as shown in Figure 12.6. The total length of cable required
is 470 meters.

T4
110

T2

93
T5

72

103

T6

92

T1

T3

Figure 12.6: Optimal network of connections

12.5

Scheduling of telecommunications via satellite
A digital telecommunications system via satellite consists of a satellite and a set of stations on earth which
serve as interfaces with the terrestrial network. With SS-TDMA (Satellite-Switch, Time Division Multiple
Access) access technology, the satellite divides its time among the stations. Consider for example thetransmissions from four transmitter stations in the US to four receiver stations in Europe. The following
table gives a possible 4 × 4 traffic matrix. TRAFtr is the quantity of data transmitted from station t tostation r. It is expressed in seconds of transmission duration, because all lines have the same constanttransmission rate.
Table 12.7: Matrix TRAF with its lower bound LB
TRAF
1
2
3
4
colr

1
0
15
17
6
38

2
7
8
12
13
40

3
11
13
6
15
45

4
15
9
10
4
38

rowt
33
45
45
38
LB = 45

The satellite has a switch that allows any permutation between the four transmitters and the four receivers. Figure 12.8 gives an example of a permutation connecting the transmitters 1 to 4 to the receivers
3, 4, 1, and 2 respectively. These connections allow routing a part of the traffic matrix, called a mode. A
part of a matrix entry transmitted during a mode is a packet of data. A mode is thus a 4 × 4 matrix M
with at most one non-zero packet per row and column and such that Mtr ≤ TRAFtr for all t, r. To every
mode corresponds a slice of a schedule as shown in Figure 12.8.
Table 12.8: Example of a mode and associated schedule
1
2
3
4
colr

Telecommunication problems

1
0
0
15
0
38

2
0
0
0
13
40

3
11
0
0
0
45

4
0
9
0
0
38

Stations
1 to 32 to 43 to 14 to 2

Packets
11
9
15
13

LB = 45

185

Applications of optimization with Xpress-MP

A valid schedule of transmissions defines a sequence of permutations for the on-board switch that routes
all the traffic of the matrix TRAF. In other words, this boils down to decomposing TRAF into a sum ofmode matrices. An element of TRAF may be fragmented, like TRAF31 that is only partially transmitted in
the mode represented in Figure 12.8. A fragmented element will appear in several packets and several
modes of the final decomposition. The duration of a mode is the length of its longest packet. The
objective is to find a schedule with minimal total duration.

12.5.1Model formulation
Let TRANSM = {1, . . . , NT} be the set of transmitter stations and RECV = {1, . . . , NT} the set of receiver stations. We shall first formulate a model for this problem when fragmentations are not allowed. With a square matrix as in the present case, NT modes are sufficient for decomposing TRAF.
We define binary variables flowtrm that are 1 if the element TRAFtr is transmitted in mode numberm ∈ MODES = {1, . . . , NT} (12.5.7). The constraints (12.5.2) indicate that every TRAFtr must be transmitted in a single mode (no preemption). The constraints (12.5.3) and (12.5.4) ensure that every mode isvalid, with a single entry per row and per column.
The continuous variables durm represent the duration of every mode, that is, the duration of its largest
element. The constraints (12.5.5) bound every TRAFtr transmitted in mode m by durm . The objectivefunction (12.5.1) denotes the total duration of the decomposition to modes, that is, the sum of thedurations of the modes. The mathematical model we obtain is a bottleneck assignment problem withthree indices. It requires NT 3 + NT variables and NT 3 + 3NT 2 constraints.

X

minimize(12.5.1)

durmm∈MODES∀t ∈ TRANSM, r ∈ RECV :X

flowtrm = 1(12.5.2)m∈MODES∀t ∈ TRANSM, m ∈ MODES :X

flowtrm = 1(12.5.3)

r∈RECV∀r ∈ RECV, m ∈ MODES :X

flowtrm = 1(12.5.4)t∈TRANSM∀t ∈ TRANSM, r ∈ RECV, m ∈ MODES : TRAFtr · flowtrm ≤ durm

(12.5.5)∀m ∈ MODES : durm ≥ 0(12.5.6)∀t ∈ TRANSM, r ∈ RECV, m ∈ MODES : flowtrm ∈ {0, 1}(12.5.7)

The problem described initially in fact allows preemptions. A formulation in the form of a single mathematical model is still possible in this case, but with an even larger number of variables, since theory showsthat of the order of NT 2 modes may be required for decomposing TRAF with a minimal total duration.
This is why we are going to use an iterative algorithm due to Gonzalez and Sahni [GS76]. Every iteration
requires us to solve an LP to extract a mode from TRAF.
Given that the on-board switch connects a transmitter with a receiver station, the elements from thesame row or column of TRAF cannot be transmitted in parallel. The sum rowt of the elements of a row
t is therefore a lower bound on the duration of any decomposition. The same remark holds for the sumcolr of every column r of TRAF. The maximum of these sums defines an even better lower bound LB. The
table 12.7 displays these sums: we find LB = 45.
The algorithm is to find a schedule with duration LB. This duration is optimal since it is not possible to findanything shorter than the lower bound LB. For the algorithm to work correctly, the traffic matrix TRAF
must be converted into a matrix for which the sum of every row and every column is LB. To obtain thisso-called quasi bistochastic (QBS) matrix TQBS, we need to add fictitious traffic to the matrix elements sothat the row and columns add up to LB. This may be done systematically with the following algorithm.
• Calculate the row and column sums rowt and colr of matrix TRAF
• Calculate LB, the maximum of all row and column sums rowt and colr
• ∀t ∈ TRANS, r ∈ RECV:– Calculate q, the minimum of LB − rowt and LB − colr
– Set TQBStr = TRAFtr + q– Add q to rowt and to colr

Telecommunication problems

186

Applications of optimization with Xpress-MP

Table 12.9: Matrix TQBS
1
7
15
17
6

TQBS
1
2
3
4

2
12
8
12
13

3
11
13
6
15

4
15
9
10
11

The following table contains the matrix TQBS for our example. The elements of all rows and columns
sum to 45.
The algorithm of Gonzalez and Sahni has the following general structure:• Calculate LB, the maximum of the row and column sums of TRAF
• Convert TRAF into a QBS matrix TQBS by adding fictitious traffic• While TQBS is non-zero do– Calculate a mode m with NT non-zero elements using Mathematical Programming– Calculate pmin, the minimum packet length among the elements of m
– Create a slice of the schedule with the packets of m cut at pmin
– Deduce pmin from every element of TQBS contained in mode m
end-do• Remove the fictitious traffic from the modes that have been obtainedThis algorithm converges because at every extraction of a mode at least one element of TQBS becomes 0:
the one that sets the value of pmin. The algorithm is optimal because after the subtraction of pmin, TQBS
remains a QBS matrix with the sums of rows and columns equal to LB − pmin. The decomposition createsa slice of the schedule of duration pmin and leaves a matrix with the bound LB − pmin. Continuing thisway, we obtain a total duration equal to LB.
Any mode may be extracted at every iteration, but to have the least number of modes in the decomposition it is desirable to construct modes that are well filled with traffic. A possibility is to calculate maximin
modes in which the size of the smallest element is maximized. This may be done using Mathematical
Programming. The following model formulates a maximin assignment problem analogous to the one in
Chapter 14 for the assignment of personnel to machines.

maximize pmin∀t ∈ TRANSM :(12.5.8)

X

flowtr = 1(12.5.9)r∈RECV
TQBStr >0X∀r ∈ RECV :flowtr = 1(12.5.10)

r∈RECV
TQBStr >0∀t ∈ TRANSM :X

TQBStr · flowtr ≥ pmin

(12.5.11)t∈TRANSM∀t ∈ TRANSM, r ∈ RECV, TQBStr > 0 : flowtr ∈ {0, 1}(12.5.12)

pmin ≥ 0(12.5.13)

A binary variable flowtr (12.5.12) has the value 1 if and only if the element TQBStr is taken in the mode.
Right from the beginning the matrix TQBS may contain zero-valued elements and in any case, zeros willappear during the iterations of the algorithm of Gonzalez and Sahni. This is why the variables flowtr
are only defined for the non-zero TQBStr . The constraints (12.5.9) make it impossible to take more than
one element per row, the constraints (12.5.10) play the same role for the columns. A continuous variable
pmin bounds the chosen entries (12.5.11) from below. This minimum packet length variable is maximizedin the objective function (12.5.8).
To apply the algorithm of Gonzalez and Sahni, the preceding mathematical model must be solved for thequasi bistochastic matrix TQBS. The flowtr at 1 indicate the elements TQBStr chosen for the mode. The
solution value of pmin is then subtracted from the chosen TQBStr and the MIP problem is solved again
with the remaining traffic of TQBS. The process stops when TQBS is completely 0.

Telecommunication problems

187

Applications of optimization with Xpress-MP

12.5.2

Implementation
The following Mosel program implements the algorithm of Gonzalez and Sahni described in the previous
section. After every solution of the MIP problem, the corresponding elements of TQBS are updated, andat the next loop the problem is redefined correspondingly. Since the entire MIP problem is redefinedin every iteration of the while loop, all constraints are named in this problem so that their previous
definition may be replaced by the new one. Proceeding in such a way is necessary because the problem
definition in Mosel is incremental: constraints may only be removed by explicitly resetting them. Oncedefined, the variables flow(t,r) cannot be removed, but through the tests TQBS(t,r)>0 in the definition of the constraints only the variables actually required are used in the definition of the current
problem.
In this program, we save all the solutions found (elements and durations of the modes) for printing out
the complete solution in the form of a diagram at the end of the program. During the solution printout,
the scheduled packets are compared with the original traffic matrix TRAF so that the additional fictitious
traffic in TQBS is not shown, but only the original traffic demands.
model "G-5 Satellite scheduling"uses "mmxprs"
declarations
TRANSM = 1..4
RECV = 1..4
TRAF: array(TRANSM,RECV) of integer
TQBS: array(TRANSM,RECV) of integer
row: array(TRANSM) of integer
col: array(RECV) of integer
LB: integer
end-declarations! Set of transmitters! Set of receivers! Traffic betw. terrestrial stations! Quasi bistochastic traffic matrix! Row sums
! Column sums
! Maximum of row and column sums

initializations from ’g5satell.dat’TRAF
end-initializations! Row and column sums
forall(t in TRANSM) row(t):= sum(r in RECV) TRAF(t,r)
forall(r in RECV) col(r):= sum(t in TRANSM) TRAF(t,r)
LB:=maxlist(max(r in RECV) col(r), max(t in TRANSM) row(t))! Calculate TQBS
forall(t in TRANSM,r in RECV) do
q:= minlist(LB-row(t),LB-col(r))
TQBS(t,r):= TRAF(t,r)+q
row(t)+=q
col(r)+=q
end-dodeclarations
MODES: range
flow: array(TRANSM,RECV) of mpvar! 1 if transmission from t to r,! 0 otherwise
pmin: mpvar! Minimum exchange
onerec, minexchg: array(TRANSM) of linctr ! Constraints on transmitters! and min exchange
onerec: array(RECV) of linctr! Constraints on receivers

solflowt: array(TRANSM,MODES) of integer
solflowr: array(RECV,MODES) of integer
solpmin: array(MODES) of integer
end-declarations! Solutions of every iteration! Solutions of every iteration! Objective value per iteration

forall(t in TRANSM,r in RECV) flow(t,r) is_binary
ct:= 0while(sum(t in TRANSM,r in RECV) TQBS(t,r) > 0) do
ct+=1! One receiver per transmitter
forall(t in TRANSM) onerec(t):= sum(r in RECV | TQBS(t,r)>0) flow(t,r) =1

Telecommunication problems

188

Applications of optimization with Xpress-MP

! One transmitter per receiver
forall(r in RECV) onetrans(r):= sum(t in TRANSM | TQBS(t,r)>0) flow(t,r) =1! Minimum exchange
forall(t in TRANSM)
minexchg(t):= sum(r in RECV | TQBS(t,r)>0) TQBS(t,r)*flow(t,r) >= pmin
! Solve the problem: maximize the minimum exchange
maximize(pmin)! Solution printing
writeln("Round ", ct, " objective: ", getobjval)! Save the solution
solpmin(ct):= round(getobjval)
forall(t in TRANSM,r in RECV | TQBS(t,r)>0)
if(getsol(flow(t,r))>0) thensolflowt(t,ct):= tsolflowr(t,ct):= r
end-if! Update TQBS
forall(t in TRANSM)
TQBS(solflowt(t,ct),solflowr(t,ct)) -= solpmin(ct)end-do! Solution printing
writeln("\nTotal duration: ", sum(m in MODES) solpmin(m))
write("")
forall(i in 0..ceil(LB/5)) write(strfmt(i*5,5))
writelnforall(t in TRANSM) do
write("From ", t, " to: ")forall(m in MODES)
forall(i in 1..solpmin(m)) dowrite(if(TRAF(solflowt(t,m),solflowr(t,m))>0, string(solflowr(t,m)),"-"))
TRAF(solflowt(t,m),solflowr(t,m))-=1
end-dowriteln
end-doend-model

12.5.3Results
The following tables show the successive contents of TQBS during the repeated solution of the MIP. The
boxed elements are the elements chosen for the mode. The duration of every mode pmin is given above
the corresponding table.
The decomposition terminates after eight iterations and the resulting schedule has the expected length
LB = 45. The upper half of Figure 12.7 shows the schedule obtained with the fictitious traffic of thematrix TQBS, the lower half displays the result for the original traffic data without any fictitious traffic(matrix TRAF).

12.6

Location of GSM transmitters
A mobile phone operator decides to equip a currently uncovered geographical zone. The management
allocates a budget of B
C 10 million to equip this region. A study shows that only 7 locations are possible
for the construction of the transmitters and it is also known that every transmitter only covers a certain
number of communities. Figure 12.8 represents a schematic map of the region with the division into
communities and the possible locations for transmitters. Every potential site is indicated by a black dot
with a number, every community is represented by a polygon. The number in the center of a polygon is
the number of the community.
Certain geographical and topological constraints add to the construction cost and reduce the reach of
the GSM transmitters. Table 12.11 lists the communities covered and the cost for every site.

Telecommunication problems

189

Applications of optimization with Xpress-MP

Table 12.10: Decomposition of TQBS into modes
Mode 1, pmin = 13
7

12

11

Mode 2, pmin = 11
15

7

11

12

Mode 3, pmin = 7
7

2

12

0

2

15

8

13

9

15

8

0

9

4

8

0

9

17

12

6

10

4

12

6

10

4

1

6

10

6

13

15

11

6

0

15

11

6

0

15

0

Mode 4, pmin = 6

Mode 5, pmin = 3

Mode 6, pmin = 3

0

12

0

2

0

6

0

2

0

3

0

4

1

0

9

4

1

0

3

4

1

0

0

4

1

6

3

4

1

0

3

1

1

0

3

6

0

8

0

0

0

8

0

0

0

5

0

Mode 7, pmin = 12Mode 8, pmin = 1

0

0

0

2

0

0

0

1

1

1

0

0

0

1

0

0

1

1

0

0

1

0

0

0

0

0

2

0

0

0

1

0

From 1 to:

4

3

1

2

2

2 44

From 2 to:

3

1

2

4

4

1 12

From 3 to:

1

2

4

3

1

4 21

From 4 to:

2

4

3

1

3

3 33

0

13

24

31

37 40 43 45

From 1 to:

4

3

2

From 2 to:

3

1

2

From 3 to:

1

2

4

3

From 4 to:

2

3

1

4

0

13

24

4
4

31

1 2
1

4 21
3

37

45

Figure 12.7: Schedule with and without fictitious traffic

Table 12.11: Cost and communities covered for every site
Site
Cost (in million B
C)
Communities covered

Telecommunication problems

1

2

3

4

5

6

7

1.8
1,2,4

1.3
2,3,5

4.0
4,7,8,10

3.5
5,6,8,9

3.8
8,9,12

2.6
7,10,11,12,15

2.1
12,13,14,15

190

Applications of optimization with Xpress-MP

1

11

7

4

3

1

6
10
15
12

8

2

5

5

7

2

4

14
9

3

13

6

Figure 12.8: Map of the region to coverFor every community the number of inhabitants is known (see Table 12.12). Where should the transmitters be built to cover the largest population with the given budget limit of B
C 10M?
Table 12.12: Inhabitants of the communities

12.6.1

Community

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Population (in 1000)

2

4

13

6

9

4

8

12

10

11

6

14

9

3

6

Model formulation
Let COMMS be the set of communities, PLACES the potential sites for constructing transmitters and
BUDGET the budget allocated by the management. We further write COSTp for the cost of building
a transmitter a site p, POPc the number of inhabitants of the community c and COVERpc a binary constantthat indicates whether a transmitter placed at p covers community c (COVERpc = 1) or not (COVERpc = 0).
Two types of binary variables are required: variables coveredc that are 1 if and only if a community c iscovered by a transmitter, and variables buildp that are 1 if and only if a transmitter is built at site p.
We formulate the constraints for the coverage of the communities first . We need to translate the equivalence ‘community c receives a GSM signal ⇔ at least at one site covering this community is a transmitteris built’, or ‘coveredc = 1 ⇔ there is at least one p with COVERpc · buildp = 1’. This type of equivalencecannot be translated directly into a linear form. Since a community may be covered by more than one
transmitter, the constraints (12.6.1) express one direction of the equivalence by specifying that the sum
COVERpc · buildp is greater than or equal to coveredc .

X∀c ∈ COMMS :COVERpc · buildp ≥ coveredc(12.6.1)

p∈PLACES

The other direction of the equivalence is ensured through the maximization of the population covered(12.6.2). The optimization algorithm will not leave any coveredc at 0 if any transmitter that covers thiscommunity is built.
It is also necessary to remain within the budgetary limits. The constraint (12.6.4) bounds the total construction cost for transmitters with the given maximum budget. And finally, the constraints (12.6.5) and(12.6.6) define all variables to be binaries.
maximize

X

POPc · coveredc(12.6.2)

X(12.6.3)c∈COMMS∀c ∈ COMMS :COVERpc · buildp ≥ coveredcp∈PLACESXCOSTp · buildp ≤ BUDGET(12.6.4)∀c ∈ COMMS : coveredc ∈ {0, 1}(12.6.5)∀p ∈ PLACES : buildp ∈ {0, 1}(12.6.6)p∈PLACES

Telecommunication problems

191

Applications of optimization with Xpress-MP

12.6.2

Implementation
The translation of the mathematical model into a Mosel program is straightforward. The matrix COVER
is stored in sparse format in the data file (that is, only the entries with value 1 are given).
model "G-6 Transmitter placement"
uses "mmxprs"
declarations
COMMS = 1..15
PLACES = 1..7

! Set of communities! Set of possible transm. locations

COST: array(PLACES) of real
! Cost of constructing transmitters
COVER: array(PLACES,COMMS) of integer ! Coverage by transmitter locations
POP: array(COMMS) of integer! Number of inhabitants (in 1000)
BUDGET: integer! Budget limit
build: array(PLACES) of mpvar
covered: array(COMMS) of mpvar
end-declarations! 1 if transmitter built, 0 otherwise! 1 if community covered, 0 otherwise

initializations from ’g6transmit.dat’
COST COVER POP BUDGET
end-initializations! Objective: total population covered
Coverage:= sum(c in COMMS) POP(c)*covered(c)! Towns covered
forall(c in COMMS) sum(p in PLACES) COVER(p,c)*build(p) >= covered(c)! Budget limitsum(p in PLACES) COST(p)*build(p) <= BUDGET
forall(p in PLACES) build(p) is_binary
forall(c in COMMS) covered(c) is_binary! Solve the problem
maximize(Coverage)
end-model

12.6.3Results
When solving this problem, we find that 109,000 persons can be covered by the transmitters. Four transmitters are built, at the sites 2, 4, 6, and 7. Communities 1 and 4 are not covered by these transmitter
locations. This operation requires a total budget of B
C 9.5 million which remains under the limit set by the
management of the GSM operator.
In practice, for the choice of transmitter locations other factors also have to be taken into account. The
areas covered by neighboring transmitters typically overlap each other: these overlappings should neither
be too large (cost inefficient) nor too small (some overlapping is required to pass on calls). Another typicalissue is that the transmitters must be placed and dimensioned in such a way that they may be allocatedfrequencies (out of a set attributed to the operator) that do not cause any interference problems.

12.7

References and further material
The recent work by Sansò et al. [SS98] is dedicated to optimization problems in the design of telecommunication networks. The problem of independent routings in Section 12.1 is a problem of connectivity in agraph. An undirected graph is k-connected regarding its nodes if at least k disjoint chains exist between
any pair of nodes. Efficient graph algorithms for simple connectivity and 2-connectivity are given in thebooks by Baase [Baa88] and Prins [Pri94a]. The general case is dealt with for instance by Papadimitriou etal. [PS98] and Ahuja et al. [AMO93].
The problem from Section 12.2 of connecting mobile phone relays to a central network in ring form isdealt with in an article by Dutta et al. [DK99]. The maximization of the throughput of a network in Section 12.3 is a multi-commodity network flow (MCNF) problem. The main MCNF problems and solution

Telecommunication problems

192

Applications of optimization with Xpress-MP

methods are given by Gondran [GM90] and Ahuja et al. [AMO93]. For large telecommunications networks, a compatible MCNF algorithm is described by Minoux [Min75], and an algorithm for the minimum
cost MCNF problem by Gersht et al. [GS87]. A column generation method for MCNF in telecommunications is described in [BHJS95].
The problem of the cable network in Section 12.4 is a classical minimum weight spanning tree problem.
Its model formulation by mathematical LP/MIP is heavy and only valid for small problem (n ≤ 20 nodes). It
is in fact easily solved by graph algorithms like the one by Prim in O(n2 ) or the one by Kruskal in O(m log n)
where m denotes the number of edges. These two algorithms are described by Ahuja et al. [AMO93].
The problem of telecommunications via satellite in Section 12.5 is also known as a workshop schedulingproblem (open shop). The transmitter stations become the machines, the receiver stations the products
to produce. TRAFtr indicates the processing duration of a product r on a machine t. A machine may only
produce a single product at a time, and a product may only be processed on a single machine at any time.
As opposed to the flow shop and job shop problems in Chapter 7, the sequence of processing a producton the different machines is free. The objective is to find the schedule that minimizes the total duration.
The open shop problem is NP-hard if preemptions are not allowed. Heuristics have to be used beyond thelimit of ten machines/ten products. See for instance Guéret and Prins [GP98]. The major variants of open
shop to be found in telecommunications by satellite are presented in an overview article by Prins [Pri94b].
An example of planning traffic for a longer term, solved by Mathematical Programming, is given by Scottet al. [SSR00]. The bottleneck assignment model with three indices presented at the beginning of Section
12.5.1 for the nonpreemptive case stems from Balas and Landweer [BL83].
The choice of locations for GSM transmitters in Section 12.6 is a covering problem. The closely related
problems of covering and partitioning are classics in 0-1 (M)IP, summarized in the section ‘References andfurther material’ to Chapter 15. We introduce a partitioning problem (electoral districts) in Chapter 15
and another type of covering problem (cutting sheet metal) in Chapter 9.
The covering problem, although NP-hard, is relatively well solved since the simplex algorithms finds an
optimal LP solution where the majority of the binary variables have integer values, which renders thetree search on the remaining fractional variables easier. See for instance Beasley [Bea87]. It is possible
to solve problems with several hundred variables to optimality. Very efficient heuristics based on LP/MIP
have recently been described for covering problems of very large size [CFT99]. Syslo describes heuristics
and exact tree search methods for two types of problems, with source code in Pascal [SDK83].

Telecommunication problems

193

Applications of optimization with Xpress-MP

Chapter 13

Economics and finance
Economists often work with uncertain data and naturally turn towards statistical tools but frequently
a simple mathematical model can solve a problem with many variables very quickly to optimality. It isthen possible to restart the optimization by varying certain parameters (those that are uncertain). In
this case Mathematical Programming becomes an efficient simulation tool. The problems in this chapter
concern applications in finance and marketing, aimed at financial decision makers in companies but also
for individuals.
Section 13.1 describes a problem of choosing loans to finance expansion projects. Sections 13.2 and 13.5
deal with the optimal use of a given budget, in the first case for organizing an efficient publicity campaign, in the second case to maximize the amount of money a family may spend on leisure. The selectionof the most profitable expansion projects over several time periods subject to limited annual budgets isthe topic of Section 13.6. The problems in Sections 13.3 and 13.7 concern the optimal composition of
investment portfolios according to different criteria (maximum total return and minimum variance). The
problem in Section 13.7 differs from all other examples in this book in that it has a quadratic objectivefunction. For both portfolio problems it is shown how the models may be used for repeated (simulation)runs by varying certain parameters. Section 13.4 also deals with the choice of investments by looking at
a multi-period time horizon and with a different objective: every time period a certain amount of money
needs to be available to finance an early retirement scheme.

13.1

Choice of loans
Mr Chic, director of a chain of shops selling clothes, wishes to open three new shops: one in London, one
in Munich, and one in Rome. To open a new shop costs respectively B
C 2.5 million, B
C 1 million and B
C 1.7
million. To finance his project, he calls at three different banks.
Table 13.1: Rates offered by the banks for the different projects
Shop in London

Shop in Munich

Shop in Rome

5%
5.2%
5.5%

6.5%
6.2%
5.8%

6.1%
6.2%
6.5%

Bank 1
Bank 2
Bank 3

Depending on the location of the shops and the evaluation of the related risks, each bank decides tofinance at most B
C 3 million over 8 years with different interest rates for the shops (Table 13.1). Determine
the amount to borrow from each bank for financing each shop in order to minimize the total expensesof Mr Chic.

13.1.1Model formulation
Let borrowbs be the amount borrowed from bank b to finance shop s. Let BANKS and SHOPS respectively
be the sets of banks willing to finance and of new shops. Let DUR be the number of years over which
the repayment of the credits stretches, VMAX the maximum amount that every bank is prepared tofinance, RATEbs the interest rate offered by bank b for shop s, and PRICEs the cost of opening shop s. The
mathematical model is the following:
minimize

X

X

b∈BANKS s∈SHOPS

194

borrowbs ·RATEbs1 − (1 + RATEbs )−DUR(13.1.1)

Applications of optimization with Xpress-MP

X∀s ∈ SHOPS :borrowbs = PRICEs(13.1.2)

borrowbs ≤ VMAX

(13.1.3)∀b ∈ BANKS, s ∈ SHOPS : borrowbs ≥ 0(13.1.4)b∈BANKSX∀b ∈ BANKS :s∈SHOPSThe objective is to minimize Mr. Chic’s expenses, that is, to minimize the sum of annual payments he has
to make. If the amount borrowbs is borrowed from a bank b to finance the shop s at the rate RATEbs over
DUR years, then the annual payment paybs to be made by Mr Chic for this shop to the bank b is calculated
as follows: the net present value of all DUR annual payments paybs for shop s to the bank b must equal
the value of the loan borrowbs :borrowbs =

DUR
X
t=1

paybs
1 − (1 + RATEb )−DUR
=
pay
·
bs
(1 + RATEb )t
RATEb(13.1.5)

By dividing the equation by the fraction on the right hand side we obtain the following equation for theannuities paybs :RATEbsborrowbs ·= paybs(13.1.6)
1 − (1 + RATEbs )−DUR
The annual payments paybs have to be made for all shops s to all banks b during DUR years. But since theannual payments are the same every year, it is sufficient to minimize the sum of the payments for one
year. We thus obtain the objective function as the sum of paybs over all b and s. By replacing paybs withthe left hand side of equation (13.1.6) we obtain the objective function in the form (13.1.1).
Every shop has to be completely financed, which means that the amounts borrowed for financing every
shop s must correspond to its cost PRICEs . This constraint is expressed through (13.1.2). The constraint(13.1.3) indicates that every bank must not finance more than VMAX and the constraints (13.1.4) state
the non-negativity of the variables.

13.1.2

Implementation
The following Mosel program gives the translation of the mathematical model above.
model "H-1 Loan choice"uses "mmxprs"
declarations
BANKS = 1..3
! Set of banks
SHOPS = {"London", "Munich", "Rome"} ! Set of shops
DUR: integer! Duration of credits
PRICE: array(SHOPS) of integer
RATE: array(BANKS,SHOPS) of real
VMAX: integer! Price of shops! Interest rates offered by banks! Maximum loan volume per bankborrow: array(BANKS,SHOPS) of mpvar
end-declarations! Loan taken from banks per project

initializations from ’h1loan.dat’
PRICE RATE VMAX DUR
end-initializations! Objective: interest payments
Interest:=sum(b in BANKS, s in SHOPS) borrow(b,s)*RATE(b,s)/(1-(1+RATE(b,s))^(-DUR))! Finance all projects
forall(s in SHOPS) sum(b in BANKS) borrow(b,s) = PRICE(s)! Keep within maximum credit volume per bankforall(b in BANKS) sum(s in SHOPS) borrow(b,s) <= VMAX! Solve the problem
minimize(Interest)
end-model

Economics and finance

195

Applications of optimization with Xpress-MP

13.1.3

Results
To minimize his expenses, Mr. Chic finances the shop in London (shop 1) entirely through a credit frombank 1 and the shop in Munich (shop 2) through a credit from bank 3. For the shop in Rome (shop 3) heborrows B
C 0.5 million from bank 1 and B
C 1.2 million from bank 2. The sum of the annual payments to bemade by Mr. Chic is then B
C 822,181.

13.2

Publicity campaign
The small company, Pronuevo, launches a new product into a regional market and wishes to have apublicity campaign using different media. It therefore contacts a regional publicity agency, PRCo, thatspecializes in this type of regional campaign and completely hands over this task for a total budget ofB
C 250,000. The agency knows the market well, that is, the impact of publicity in a local magazine or over
the radio, or as a TV spot on the regional channel. It suggests addressing the market for two months
through six different media. For each medium, it knows the cost and the number of people on which thismedium has an impact. An index of the quality of perception of the campaign is also known for every
medium.
The publicity agency establishes a maximum number of uses of every medium (for instance, not morethan eight transmissions of a TV spot). The following Table 13.2 lists all this information. Pronuevo wants
the impact of the publicity campaign to reach at least 100,000 people. Which media should be chosen
and in which proportions to obtain a maximum index of perception quality?
Table 13.2: Data for the publicity campaign

13.2.1

NumberMedia type

Peoplepotentially reached

Unit
cost

Maximum
use

Perception
quality

1
2
3
4
5
6

Free weekly newspaper
Monthly magazine
Weekly magazine
Radio spot
Billboard 4x3 m
TV spot

12,000
1,500
2,000
6,000
3,000
9,000

1,500
8,000
12,0009,000
24,00051,0004 weeks
2 months
8 weeks
60 broadcasts
4 boards
8 broadcasts

3
7
8
2
6
9

Model formulation
We write MEDIA for the set of media that may be used in this campaign. The constants REACHm , COSTm ,MAXUSEm , and SCOREm denote respectively the number of people potentially reached, the unit cost ofusing a medium, the maximum number of uses and the quality index for every medium used. The decisionvariables usem that we choose are the number of uses made of every medium m during this campaign.
To start with, the budgetary constraint needs to be fulfilled: the constraint (13.2.2) indicates that the sumof the costs for every medium must not exceed the total allocated budget BUDGET. Every medium may
be used only up to its maximum authorized number of uses (constraints (13.2.3)). The targeted numberof people TARGET needs to be reached, that means, that the sum of the numbers of people potentiallyreached through every medium must be at least as large as the targeted 100,000 persons (constraint(13.2.4)). The objective function (13.2.1) has to maximize the sum of the quality indices of every media
used. The decision variables usem must be integer, constraints (13.2.5).

X

maximize

SCOREm · usem

(13.2.1)

COSTm · usem ≤ BUDGET(13.2.2)∀m ∈ MEDIA : usem ≤ MAXUSEm(13.2.3)m∈MEDIAXm∈MEDIAX

REACHm · usem ≥ TARGET(13.2.4)m∈MEDIA∀m ∈ MEDIA : usem ∈ IN

Economics and finance196(13.2.5)

Applications of optimization with Xpress-MP

13.2.2

Implementation
This model could be implemented quite simply in a non-generic way but we have given preference to a
generic form (using data arrays read from file) that is easier to modify for future extensions.
model "H-2 Publicity"uses "mmxprs"
declarations
MEDIA = 1..6

! Set of media types

REACH: array(MEDIA) of integer
COST: array(MEDIA) of integer
MAXUSE: array(MEDIA) of integer
SCORE: array(MEDIA) of integer
BUDGET: integer
TARGET: integer! Number of people reached! Unitary cost
! Maximum use! Quality rating (best=highest value)! Available publicity budget! Number of people to be reacheduse: array(MEDIA) of mpvar
end-declarations! Use made of different media

initializations from ’h2publ.dat’
REACH COST MAXUSE SCORE BUDGET TARGET
end-initializations! Objective: quality of perception of the campaign
Perceive:= sum(m in MEDIA) SCORE(m)*use(m)! Budgetary limitsum(m in MEDIA) COST(m)*use(m) <= BUDGET! Outreach of campaignsum(m in MEDIA) REACH(m)*use(m) >= TARGET
forall(m in MEDIA) douse(m) is_integer
use(m) <= MAXUSE(m)end-do! Solve the problem
maximize(Perceive)end-model

13.2.3Results
With the program above we find a quality index of 122 units. The budget is fully used and 103,000 people
are reached. As detailed in the following table, all media types except the TV spot are used. Given thatthe problem is of very moderate size, the LP directly calculates an integer solution.
Table 13.3: Use made of different media types
Media type

Number of uses

Free weekly newspaper
Monthly magazine
Weekly magazine
Radio spot
Billboard 4x3 m
TV spot

4
2
8
4
4
0

We should note that the problem described in this section is a much simplified version of reality. For
instance, it is unlikely that the effectiveness of radio advertising is proportional to the number of broadcasts of a particular advert. Nor are adverts in different media independent of each other. In practice,much more sophisticated models will be used.

Economics and finance

197

Applications of optimization with Xpress-MP13.3

Portfolio selection
A consultant in finance has to choose for one of his wealthy female clients a certain number of shares inwhich to invest. She wishes to invest B
C 100,000 in 6 different shares. The consultant estimates for her the
return on investment that she may expect for a period of six months. The following table gives for eachshare its country of origin, the category (T: technology, N: non-technology) and the expected return on
investment (ROI). The client specifies certain constraints. She wishes to invest at least B
C 5,000 and at mostB
C 40,000 into any share. She further wishes to invest half of her capital in European shares and at most
30% in technology. How should the capital be divided among the shares to obtain the highest expected
return on investment?
Table 13.4: List of shares

13.3.1

NumberOriginCategory

Expected ROI

1
2
3
4
5
6

Japan
UK
FranceUSA
Germany
France

T
T
T
N
N
N

5.3%
6.2%
5.1%
4.9%
6.5%
3.4%

Model formulation
In the mathematical formulation, we write SHARES for the set of shares chosen for the investment,
CAPITAL the total capital that is to be invested, RETs the expected ROI. EU denotes the subset of theshares that are of European origin and TECHNOLOGY the set of technology values. The decision variables
buys denote the amount of money invested in share s. Every variable is bounded, at least VMIN = 5000
and at most VMAX = 40000 have to be invested into every share. The constraint (13.3.1) establishes thesebounds on the variables buys .
(13.3.1)∀s ∈ SHARES : VMIN ≤ buys ≤ VMAXAt most 30% of the values may be technology values, i.e. the sum invested into this category must
not exceed B
C 30,000, constraint (13.3.2). The investor also insists on spending at least 50% on European
shares, that is, at least B
C 50,000, constraint (13.3.3).

Xbuys ≤ 0. 3 · CAPITAL(13.3.2)s∈TECHNOLOGY

X

buys ≥ 0. 5 · CAPITAL(13.3.3)s∈EUThe constraint (13.3.8) specifies the the total invested sum must correspond to the initial capital CAPITAL.
The constraints (13.3.9) are the usual non-negativity constraints. The objective function (13.3.4) maximizes the return on investment of all shares.

X

maximizes∈SHARES

RETs· buys100(13.3.4)∀s ∈ SHARES : VMIN ≤ buys ≤ VMAX

X

buys ≤ 0. 3 · CAPITAL(13.3.5)(13.3.6)s∈TECHNOLOGY

X

buys ≥ 0. 5 · CAPITAL(13.3.7)s∈EUXbuys = CAPITAL(13.3.8)∀s ∈ SHARES : buys ≥ 0(13.3.9)s∈SHARES

After some further discussion, it turns out that the client actually does not wish to invest VMIN into every
share the consultant has chosen for her. She is unwilling to have small share holdings. If she buys any
shares in a company she wants to invest at least B
C 5,000. The model therefore needs to be modified:instead of constraining every variable buys to take a value between VMIN and VMAX, it must either lie in

Economics and finance198

Applications of optimization with Xpress-MP

this interval or take the value 0. This type of variable is known as semi-continuous variable. In this newmodel, we replace (13.3.5) by the following constraint (13.3.10):∀s ∈ SHARES : buys = 0 ∨ VMIN ≤ buys ≤ VMAX

13.3.2(13.3.10)

Implementation
The translation of the mathematical model into the following Mosel program is straightforward. In theimplementation, we have turned the limits on technology values and European shares into parametersof the model (MAXTECH and MINEU). Similarly, the bounds VMIN and VMAX are defined as parameters.
model "H-3 Portfolio"
uses "mmxprs"
parameters
MAXTECH = 0.3
MINEU = 0.5
VMIN = 5000
VMAX = 40000
end-parameters! Maximum investment into tech. values! Minimum investment into European shares! Minimum amount for a single value! Maximum amount for a single value

declarations
SHARES = 1..6

! Set of shares

RET: array(SHARES) of real
CAPITAL: integer
EU: set of integer
TECHNOLOGY: set of integer! Estimated return in investment! Capital to invest! European values among the shares! Technology values among sharesbuy: array(SHARES) of mpvar
end-declarations! Amount of values taken into portfolio

initializations from ’h3portf.dat’
RET CAPITAL EU TECHNOLOGY
end-initializations! Objective: total return
Return:= sum(s in SHARES) RET(s)/100*buy(s)! Requirements concerning portfolio composition
sum(s in TECHNOLOGY) buy(s) <= MAXTECH*CAPITAL
sum(s in EU) buy(s) >= MINEU*CAPITAL! Total capital to investsum(s in SHARES) buy(s) = CAPITAL
forall(s in SHARES) do
VMIN <= buy(s)buy(s) <= VMAX
end-do! Solve the problem
maximize(Return)
end-model

To define the variables buy(s) as semi-continuous variables, we need to replace the lower bound constraints
VMIN <= buy(s)by the linebuy(s) is_semcont(VMIN)

13.3.3Results
The solution to the problem indicates the sums that should be invested into the different shares. The
total expected ROI with the initial problem formulation is B
C 5,755 for six months, for the problem withsemi-continuous variables it is B
C 5,930. The funds are divided among the shares as shown in the following
table.

Economics and finance199

Applications of optimization with Xpress-MP

Table 13.5: Optimal portfolio
SharenumberInvested amount (in BC)buys ≥ VMIN
buys semi-cont.
5,000
20,0005,000
25,00040,000
5,000

1
2
3
4
5
6

0
30,000030,00040,0000

As mentioned in the introduction to this chapter, Mathematical Programming may be used as a simulation
tool in financial applications. In our case, what would happen if we allowed investing up to 40% into
technology values? Since the corresponding limit MAXTECH has been defined as a parameter, there is noneed to modify the Mosel program to test the result for this value. It is sufficient to set the new value ofthe parameter when calling Mosel. With the Mosel Command Line Interpreter this may be done with thefollowing command to execute the model:
mosel -c "exec h3portf ’MAXTECH=0.4’"

With this increased limit on technology values, the expected return becomes B
C 5,885 for the first model(and BC 6,060 for the second). Compared to the previous plan, B
C 10,000 are transfered from share 4 to share
2, and the amounts invested in the other shares remain the same. By varying the different parameters(it is possible to set new values for several parameters at the same time), the consultant may be able tosatisfy his client.

13.4

Financing an early retirement scheme
The National Agricultural Bank (NAB) decides to establish an early retirement scheme for fifteen employees who are taking early retirement. These employees will retire over a period of seven years startingfrom the next year. To finance this early retirement scheme, the bank decides to invest in bonds for thisseven-year period. The necessary amounts to cover the pre-retirement leavers are given in the following
Table 13.6; they have to be paid at the beginning of every year.
Table 13.6: Amounts required every year
Year
Amount (in 1000 B
C)

1

2

3

4

5

6

7

1000

600

640

480

760

1020

950

For the investments, the bank decides to take three different types of bonds, SNCF bonds, Fujitsu bonds,
and Treasury bonds. The money that is not invested in these bonds is placed as savings with a guaranteedyield of 3.2%. Table 13.7 lists all information concerning the yield and the durations of the bonds andalso the value of a bond. In this type of bond, it is only possible to buy an integer number of bonds, andthe invested capital remains locked in for the total duration of the bond. Every year, only the intereston the capital is distributed. The person in charge of the retirement plan decides to buy bonds at the
beginning of the first year, but not in the following years. How should he organize the investments inorder to spend the least amount of money to cover the projected retirement plan?
Table 13.7: Information about loans
Loan
SNCF
Fujitsu
Treasury

13.4.1

Value of bonds (in 1000 B
C)Interest

Duration

1.0
0.8
0.5

7.0%
7.0%
6.5%

5 years
4 years
6 years

Model formulation
Let YEARS = {1, . . . , NT} be the time period over which the early retirement scheme stretches, and BONDS
the set of loans. capital denotes the total capital that needs to be invested in the first year in order tobe able to finance the scheme over all seven years. Let VALUEb be the unit price for a bond b, DURb its
duration, and RATEb the annual interest on bond b. The variable buyb represents the number of bondsof type b bought in the first year, the variable investt the sum invested (into forms of savings other thanthe bonds) in year t. The funds required by the early retirement plan per year t are denoted by DEMt .

Economics and finance

200

Applications of optimization with Xpress-MP

The general principle of the constraints is the following: every year, the set of interest payments has
to cover the investments of this year and the amount required by the retirement plan. The constraint(13.4.1) represents the investment in the first year, the year during which the bonds are bought. The total
invested capital minus the price paid for the bonds and minus the investment of the first year must be
equal to the funds required in the first year.

X

capital −VALUEb · buyl − invest1 = DEM1

(13.4.1)

b∈BONDS

While no bond has come to term, that is during the three years following the first, no more money isinvested into bonds, but we receive the interest (first sum of constraint (13.4.2)). We also receive theinterest and the amount invested into other forms of savings than bonds in the preceding year. Fromthese benefits we have to subtract the investments (other than loans) and the funds required for thecurrent year to obtain an equilibrium, constraints (13.4.2).
∀t = 2, . . . , 4 :X

VALUEb · buyb ·b∈BONDS

RATEb
+
100


1+

3. 2
100


· investt−1 − investt = DEMt

(13.4.2)

At the beginning of year 5, the Fujitsu bond falls due and similarly at the beginning of year 6, the EDF
bond falls due. The capital invested into the corresponding bond is recovered and available for a newinvestment. We also receive the interest for the preceding year for all bonds. The rest of constraint(13.4.3) is identical to the constraints (13.4.2).

X∀t = 5, 6 :VALUEb · buyl · 1 +b∈BONDS
DURb =t−1


+ 1+

3. 2
100

RATEb
100



X

+

VALUEb · buyb ·b∈BONDS
DURb ≥t

RATEb
100


· investt−1 − investt = DEMt

(13.4.3)

In the last year, only the Treasury bond that ran over 6 years remains. We recover the capital initiallyinvested into this bond and the investments from the preceding year. In this year, there is no new investment (constraint 13.4.4)

X



VALUEb · buyb · 1 +b∈BONDS
DURb =6

RATEb
100

 
+

1+

3. 2
100


· invest6 = DEM7

(13.4.4)

To the constraints (13.4.1) – (13.4.4) we still need to add the objective function (13.4.5) that minimizes
the initially invested capital, and to declare the definition spaces of the variables (constraints (13.4.6) –(13.4.8))

13.4.2minimize capital(13.4.5)cap ≥ 0(13.4.6)∀t ∈ YEARS : investt ≥ 0(13.4.7)∀b ∈ BONDS : buyb ∈ IN

(13.4.8)

Implementation
In the following Mosel program, the constraints for the annual balances of years 2-7 are combined into asingle constraint expression selecting the corresponding constraint terms through the inline if function.
To simplify the notation of the constraints, we define an auxiliary array RET representing the annualinterest per bond.
model "H-4 Retirement"
uses "mmxprs"
declarations
BONDS = {"SNCF","Fujitsu","Treasury"} ! Set of bonds
NT = 7! Length of planning period
YEARS = 1..NT
DEM: array(YEARS) of integer
VALUE: array(BONDS) of real

Economics and finance! Annual payments for retirement! Unit price of bonds

201

Applications of optimization with Xpress-MP

RATE: array(BONDS) of real
RET: array(BONDS) of real
DUR: array(BONDS) of real
INTEREST: real

! Remuneration rates paid by bonds! Unit annual interest of bonds! Duration of loans! Interest for other secure investment

buy: array(BONDS) of mpvar
invest: array(YEARS) of mpvar
capital: mpvar
end-declarations! Number of bonds acquired! Other annual investment! Total capital required

initializations from ’h4retire.dat’
DEM VALUE RATE DUR INTEREST
end-initializations
forall(b in BONDS) RET(b):= VALUE(b)*RATE(b)/100! Annual balances
capital - sum(b in BONDS) VALUE(b)*buy(b) - invest(1) = DEM(1)forall(t in 2..NT)sum(b in BONDS | DUR(b)+1>=t) (RET(b)*buy(b) +if(DUR(b)+1=t, VALUE(b)*buy(b), 0)) +(1+INTEREST/100)*invest(t-1) - if(t<NT, invest(t), 0) = DEM(t)forall(b in BONDS) buy(b) is_integer! Solve the problem: minimize invested capital
minimize(capital)
end-model

13.4.3Results
The initially required capital is B
C 4548.92k. The optimal solution consists of investing B
C 911k in SNCF bonds(911 bonds), B
C 597.6k in Fujitsu bonds (747 bonds), and B
C 880k in Treasury coupons (1760 coupons). The
following table lists the investments into forms of savings other than loans for the first six years.
Table 13.8: Investments (other than loans)Year
Amount (in 1000 B
C)

13.5

1

2

3

4

5

6

1160.317

760.249

307.379

0.017

0.420

12.403

Family budget
The mother of a family wishes to use her son’s computer. On the internet she has found optimization
software and would now like to formulate a mathematical model to help plan their annual budget. She
has prepared a list of the monthly expenses and receipts of money. Every month the expenses for living
are B
C 550. The monthly rent for the apartment is B
C 630. She also budgets B
C 135 for telephone every twomonths, B
C 850 for gas and electricity bills every six months, B
C 340 per month for the car and B
C 100 of tax
every four months. Her receipts of money are a monthly payment of B
C 150 of state allowance for families
with dependent children and a net salary of B
C 1900 per month. She knows that she pays at least B
C 165
for leisure every month (subscription to the swimming pool for the older children, football club for the
youngest, gym for herself) but she would like to spend more (restaurant, cinema, holidays). How should
the budget be balanced all through the year to maximize the money available for leisure?

13.5.1Model formulation
Let MONTHS be the set of twelve month representing the year to plan and ITEMS the set of expense
types. We write EXPENSEi for the amount of an expense and FREQi for its periodicity (e.g. FREQi = 6means that i is a 6-monthly expense). The salary is denoted by INCOME, and the allowance by ALLOW.
We use two types of variables: hobbym , the amount that may be spent for leisure in month m, and savem ,the savings of month m. Every month, the family wishes to spend at least HMIN = B
C 165 for leisure. Thisis expressed by the constraints (13.5.1).
∀m ∈ MONTHS : hobbym ≥ HMIN

Economics and finance202(13.5.1)

Applications of optimization with Xpress-MP

For the system to work, the budget must be balanced. The expenses in every month must not exceed thereceipts of money of this month, but it is possible to put aside some money to save it for a month withhigher expenses. The constraints (13.5.2) express the fact that the expenses of the current month plus thesavings of this month and the money spent for leisure must be less than or equal to the perceived salary,plus the allowance, plus the savings from the preceding month. Since no information about savings at
the beginning of the planning period (variable save0 ) is given, we assume them to be 0. The operator
mod denotes the remainder of integer division (e.g. 7 mod 6 = 1, that is, 6-monthly payments are only
taken into account in months 6 and 12).

X∀m ∈ MONTHS :EXPENSEi + savem + hobbym ≤ INCOME + ALLOW + savem−1(13.5.2)i∈ITEMSm mod FREQi =0

The objective function is to maximize the total amount available for leisure in the entire year (13.5.3).
The non-negativity constraints for the variables hobbym are superseded by the constraints (13.5.1).

X

maximize(13.5.3)

hobbymm∈MONTHS∀m ∈ MONTHS : hobbym ≥ HMIN

(13.5.4)∀m ∈ MONTHS :X

EXPENSEi + savem + hobbym ≤ INCOME + ALLOW + savem−1(13.5.5)i∈ITEMSm mod FREQi =0∀m ∈ MONTHS : savem ≥ 0

13.5.2(13.5.6)

Implementation
This model is very similar to a production planning model with transfer of stocks from one time period tothe next. To deal with the different periodicities of the expenses, we use the Mosel operator mod.
For the savings at the beginning of the first time period, we may either define a variable save0 and fix its
value to 0 or, as shown below, directly employ the value 0 in the corresponding constraint.
model "H-5 Family budget"uses "mmxprs"
declarations
MONTHS = 1..12
ITEMS: set of string! Time period! Set of shops

INCOME, ALLOW: integer
HMIN: integer
EXPENSE: array(ITEMS) of integer
FREQ: array(ITEMS) of integer! Monthly income and allowance! Min. amount required for hobbies! Expenses! Frequency (periodicity) of expenses

hobby: array(MONTHS) of mpvar
save: array(MONTHS) of mpvar
end-declarations! Money to spend for leisure/hobbies! Savings

initializations from ’h5budget.dat’
INCOME ALLOW HMIN
[EXPENSE, FREQ] as ’PAYMENT’
end-initializations! Objective: money for hobby
Leisure:= sum(m in MONTHS) hobby(m)! Monthly balances
forall(m in MONTHS)
sum(i in ITEMS | m mod FREQ(i) = 0) EXPENSE(i) + hobby(m) +save(m) <= INCOME + ALLOW + if(m>1, save(m-1), 0)
forall(m in MONTHS) hobby(m) >= HMIN
! Solve the problem
maximize(Leisure)
end-model

Economics and finance203

Applications of optimization with Xpress-MP

13.5.3

Results
In the course of the year, the family may spend up to B
C 3,550 for leisure. The following table shows apossible plan of their monthly leisure budget (there are other equivalent solutions).
Table 13.9: Monthly leisure budget and savings

13.6

Month

1

2

3

4

5

6

7

8

9

10

11

12

Hobby
Savings

530
0

395
0

405
125

165
255

165
620

165
0

530
0

295
0

405
125

165
355

165
720

165
0

Choice of expansion projects
The large company Tatayo in the north of Italy has specialized in the construction of cars for more than ten
years. The company wishes to expand and has issued internally a call for proposals for expansion projects
for a planning period of five years. Among the many, often cranky, propositions the management hasretained five projects. Every project has an annual cost and is designed to produce a benefit after five
years. Table 13.10 gives a list of the projects with short descriptions and the expected benefit after
five years. The forecast annual costs of the projects for the next five years are detailed in Table 13.11,
together with the funds available. Which project(s) should the management choose now to maximize
the total benefit after five years?
Table 13.10: Estimated benefits of the projects (in million BC)
Project

Description

Expected benefit

1
2
3
4
5

Expand assembly line
Reorganize the main shop
New painting facilities
Research for a new concept car
Reorganize the logistics chain

10.8
4.8
3.2
4.44
12.25

Table 13.11: Annual costs of projects and available funds (in million B
C)

13.6.1

Project

Year 1

Year 2

Year 3

Year 4

Year 5

1
2
3
4
5

1.8
1.2
1.2
1.4
1.6

2.4
1.8
1.0
1.4
2.1

2.4
2.4
0
1.2
2.5

1.8
0.6
0.48
1.2
2.0

1.5
0.5
0
1.2
1.8

Funds

4.8

6.0

4.8

4.2

3.5

Model formulation
We write PROJECTS for the set of projects retained by the management, and TIME for the set of time
periods (years) within the planning horizon. The benefit from every project p is RETp and COSTpt thecost of project p in year t. CAPt denotes the capital available for funding projects in year t. The binary
variable choosep has the value 1 if project p is chosen and 0 otherwise. The only constraints given inthis problem are the limits on the capital that the management is prepared to invest every year. The
constraints (13.6.2) ensure that the sum of the costs COSTpt of the chosen projects (choosep = 1) does notexceed the available capital CAPt in year t. The objective function (13.6.1) maximizes the total profit, thatis, the sum of the benefits RETp from the chosen projects. The variables choosep are binary (13.6.3).

X

maximize

RETp · choosep(13.6.1)

p∈PROJECTS∀t ∈ TIME :X

COSTpt · choosep ≤ CAPt

(13.6.2)p∈PROJECTS∀p ∈ PROJECTS : choosep ∈ {0, 1}

Economics and finance204(13.6.3)

Applications of optimization with Xpress-MP

13.6.2

Implementation
The following Mosel program implements the mathematical model above.
model "H-6 Expansion"uses "mmxprs"
declarations
PROJECTS = 1..5
TIME = 1..5

! Set of possible projects! Planning period

COST: array(PROJECTS,TIME) of real
CAP: array(TIME) of real
RET: array(PROJECTS) of real
DESCR: array(PROJECTS) of string! Annual costs of projects! Annually available capital! Estimated profits! Description of projects

choose: array(PROJECTS) of mpvar
end-declarations! 1 if project is chosen, 0 otherwise

initializations from ’h6expand.dat’
COST CAP RET DESCR
end-initializations! Objective: Total profit
Profit:= sum(p in PROJECTS) RET(p)*choose(p)! Limit on capital availability
forall(t in TIME) sum(p in PROJECTS) COST(p,t)*choose(p) <= CAP(t)forall(p in PROJECTS) choose(p) is_binary! Solve the problem
maximize(Profit)
end-model

13.6.3Results
When solving the above program as is, we obtain a solution with a total profit of B
C 19.89 million: theprojects ‘New painting facility’, ‘Concept car’, and ‘Reorganize logistics’ are chosen.
We may try to experiment a little with this problem. If we solve the model again after modifying the callto the optimization as follows:
maximize(XPRS_LIN, Profit)

the linear relaxation of the problem is solved (that is, neglecting the integrality condition for the variables
choosep ). Looking at the solution of this relaxed problem, we find that the variables for choosing projects‘New paint facility’ and ‘Reorganize logistics’ are at 1, the variable for ‘Expand assembly line’ is almost
1, more precisely 0.9542, and the value of ‘Reorganize shop’ is almost 0 (0.0042). The total profit of thelinear solution is B
C 25.775M. One might wonder what would happen if we ‘rounded’ the variable for‘Expand assembly line’ to 1, thus forcing the acceptance of project 1. This may be done by adding thefollowing line to the Mosel program before calling the optimization algorithm:
choose(1)=1

With this additional constraint, the solution consists of choosing the projects ‘Expand assembly line’ (theproject that was forced to be accepted) and ‘New painting facility’, with a total profit of B
C 14M. Thissolution demonstrates that rounding fractional variables to the closest integer value is not always a goodheuristic for creating integer feasible solutions from the solution to the LP relaxation.

13.7

Mean variance portfolio selection
An investor wishes to invest a certain amount of money. He is evaluating four different securities (assets) for his investment. The securities are US Treasury Bills (‘T-bills’), a computer hardware company, acomputer software company, and a high-risk investment in a theater production. He estimates the mean
yield on each dollar he invests in each of the securities, and also adopts the Markowitz idea of getting estimates of the variance/covariance matrix of estimated returns on the securities. (For example, hardware

Economics and finance

205

Applications of optimization with Xpress-MP

and software company worths tend to move together, but are oppositely correlated with the success oftheatrical production, as people go to the theater more when they have become bored with playing with
their new computers and computer games.) The return on theatrical productions are highly variable,
whereas the T-bill yield is certain. The estimated returns and the variance/covariance matrix are given in
Table 13.12
Table 13.12: Estimated returns and variance/covariance matrix
Hardware

SoftwareShow-biz

T-bills

Estimated return

8

9

12

7

Hardware
SoftwareShow-biz
T-bills

4
3
-1
0

3
6
1
0

-1
1
10
0

0
0
0
0

Question 1: Which investment strategy should the investor adopt to minimize the variance subject togetting some specified minimum target yield?
Question 2: Which is the least variance investment strategy if the investor wants to choose at most two
different securities (again subject to getting some specified minimum target yield)?

13.7.1Model formulation for question 1
The objective of this problem is to minimize the variance subject to getting some specified minimum
target yield, say TARGET. Let RETs be the expected return of a security s and VARst the variance/covariancematrix of estimated returns on the securities. If we introduce decision variables fracs to be the fraction of
the total capital that is invested in security s, then we obtain the following Quadratic Program:
minimize

X X

VARst · fracs · fract(13.7.1)

s∈SECS t∈SECS

X

fracs = 1

(13.7.2)

RETs · fracs ≥ TARGET(13.7.3)s∈SECS

Xs∈SECS∀s ∈ SECS : fracs ≥ 0(13.7.4)

The objective function (13.7.1) is the variance of the estimated returns. Constraint (13.7.2) says that theinvestor wishes to spend all the money. The desired target yield is enforced through constraint (13.7.3).
The constraints (13.7.4) are the usual non-negativity constraints.
Note that this is not a Linear Programming problem as we have product terms in the objective function.

13.7.2

Implementation for question 1
The Mosel implementation of this problem is straightforward. Notice that in addition to the XpressOptimizer (module mmxprs) we are using the Mosel module mmquad which contains all that is required
to handle QP problems.
model "H-7 QP Portfolio"
uses "mmxprs", "mmquad"parameters
TARGET = 7.0
end-parameters
declarations
SECS = 1..4

! Minimum target yield! Set of securities

RET: array(SECS) of real
VAR: array(SECS,SECS) of real

! Expected yield of securities! Variance/covariance matrix of! estimated returns

frac: array(SECS) of mpvar
buy: array(SECS) of mpvar
end-declarations! Fraction of capital used per security! 1 if asset is in portfolio, 0 otherwise

Economics and finance206

Applications of optimization with Xpress-MP

initializations from ’h7qportf.dat’
RET VAR
end-initializations! Objective: mean variance
Variance:= sum(s,t in SECS) VAR(s,t)*frac(s)*frac(t)! Spend all the capitalsum(s in SECS) frac(s) = 1! Target yieldsum(s in SECS) RET(s)*frac(s) >=TARGET! Solve the problem
minimize(Variance)
end-model

13.7.3Results for question 1
Running the model thus, so as to specify a target yield of 8.5%
mosel -s -c "exec h7qportf ’TARGET=8.5’"

we get a minimum variance of 0.72476. We should invest 15.14% into hardware, 4.32% into software,
25.24% into show-biz, and 55.29% into T-bills.

13.7.4Model formulation for question 2
For the second problem we need to express the fact that no more than a certain number MAXASSETS ofdifferent securities may be bought. The previous model may be extended by adding binary variables buys
to represent whether we do (buys = 1) or do not (buys = 0) buy any of the asset for our portfolio. If we
add to the model above the constraints

Xbuys ≤ MAXASSETS(13.7.5)s∈SECS∀s ∈ SECS : fracs ≤ buys

(13.7.6)∀s ∈ SECS : buys ∈ {0, 1}(13.7.7)

then we can see that for a fracs to be bigger than 0, buys must be bigger than 0 (constraints (13.7.6)) andso must be 1 as it is binary (13.7.7). The constraint (13.7.5) establishes the limit on the number of different
securities bought.
This problem is actually a Mixed Integer Quadratic Program (MIQP) as it has elements of Quadratic Programming and also Integer Programming.

13.7.5

Implementation for question 2
The Mosel implementation of the second problem is similar to the first, but since there are changes atseveral places in the model, we show the entire program again.
model "H-7 QP Portfolio (2)"uses "mmxprs", "mmquad"parameters
TARGET = 7.0
MAXASSETS = 4
end-parameters
declarations
SECS = 1..4

! Minimum target yield! Maximum number of assets in portfolio! Set of securities

RET: array(SECS) of real
VAR: array(SECS,SECS) of real

! Expected yield of securities! Variance/covariance matrix of! estimated returns

frac: array(SECS) of mpvar
buy: array(SECS) of mpvar
end-declarations! Fraction of capital used per security! 1 if asset is in portfolio, 0 otherwise

Economics and finance207

Applications of optimization with Xpress-MP

initializations from ’h7qportf.dat’
RET VAR
end-initializations! Objective: mean variance
Variance:= sum(s,t in SECS) VAR(s,t)*frac(s)*frac(t)! Spend all the capitalsum(s in SECS) frac(s) = 1! Target yieldsum(s in SECS) RET(s)*frac(s) >=TARGET! Limit the total number of assetssum(s in SECS) buy(s) <= MAXASSETS
forall(s in SECS) dobuy(s) is_binary
frac(s) <= buy(s)end-do! Solve the problem
minimize(Variance)
end-model

13.7.6

Results for question 2
Running the model with the following parameter settings,
mosel -s -c "exec h7qportf ’TARGET=8.5, MAXASSETS=2’"that is, a target yield of 8.5% and at most two assets in the portfolio, we get a minimum variance of
0.9. We need to invest 30% into show-biz and 70% into T-bills. The results for questions 1 and 2 aresummarized in Table 13.13.
Table 13.13: Results with TARGET = 8. 5
Percentage of total investment
Question 1
Question 2
Hardware
SoftwareShow-biz
T-bills

15.14%
4.33%
25.24%
55.29%

0%
0%
30%
70%

Variance

0.72

0.9

Notice that because we are restricting ourselves to holding at most two securities, the variance for a given
yield has gone up compared with the case where we can hold any number of securities.
The second model is probably the more realistic one: in practical applications, there are many thousandsof possible securities, and constraints that specify that no more than a certain number of different securities may be bought stop us having the managerial problem of being forced to look after too many
securities.

13.7.7

Extension
Finally, we should note that in practice deciding on a reasonable value for an acceptable TARGET / Variancepair is not easy, and the decision maker might want to study a large range of possible combinations. In
the implementation we may simply add a loop around the optimization (and solution output). TARGET is
no longer a parameter and is therefore now declared in the declarations block. The possible values of
TARGET range between the smallest and the largest value of the mean returns of all securities, enumerated in steps of 0.1. Note that in this case the constraint on the target yield must be named (Rmean), so
it gets replaced each time through the loop.
declarations
TARGET: real
end-declarations
TARGET:=min(s in SECS) RET(s)Economics and finance208

Applications of optimization with Xpress-MP

repeatRmean:= sum(s in SECS) RET(s)*frac(s) >= TARGET
minimize(Variance)
writeln("Target: ", TARGET, " minimum variance: ", getobjval)
TARGET+=0.1
until(TARGET>max(s in SECS) RET(s) )The results of such a study may be plotted as a graph, for instance using the graphing capabilities of
Xpress-IVE.

13.8

References and further material
The problems presented here are relatively simple, but they bring into play uncertainity in the data. Being
able to solve a model quickly to optimality allows us to use the model in simulations.
Often the players in the world of finance (mainly bankers) do not worry much about optimization problems. They use the software and procedures established by their management. The programs developed
internally usually do not perform any optimization, but more likely a simple simulation based on priority
rules. These rules are based on the principle that one needs to invest into the product(s) with the best
ratio of benefit to cost. The problem hence resides in the reliability of the data and in the interpretationof the results that is made afterwards. The events on the stock markets over the last few years, especiallyfor technology values, are an additional proof of the need to consider data subject to changes.
The Quadratic Programming (QP) problem in Section 13.7 stems from Manne [Man86]. In general, QP
problems occur much less frequently than LP or MIP problems, but many of these occurrences are problems related to finance.
Operations Research techniques have become widespread in the financial sector. For an overview, see theexcellent book by Winston [Win98] and the one by Zenios [Zen96].

Economics and finance209

Applications of optimization with Xpress-MP

Chapter 14

Timetabling and personnel planning
Personnel management is a very sensitive subject in the life of a company. It is determined by a large
number of qualitative and psychological factors that, a priori, do not make this field well suited for quantitative modeling through Mathematical Programming. One may even put forward two major reasonsthat put Operations Research on hold in the 70s, until its renewed flowering in the 90s: serious mistakes
were committed by treating humans as a resource just like others in the models, and furthermore, theresearchers at that time did not have the computational means to satisfy their ambitions.
Optimization is currently considered as providing useful help for timetabling of personnel, which can bea real headache, and also for staffing, that is, the planning of the requirements for human resources.
The key to preventing negative reactions/rejections is never to impose a solution found by the computer
by force. Indeed, many subtle sociological and psychological constraints are very difficult to take into
account. This makes interactive decision making tools a preferred choice, where the mathematical model
makes suggestions and the human being plans by interactively adjusting the solution to his exact needs.
The first problem consists of assigning persons to machines to maximize the productivity of a workshop.
We shall then see three timetabling problems: for nurses in a hospital (Section 14.2), for college courses(Section 14.3), and for exams in a technical university (Section 14.4). The problem in Section 14.5 shows
how the benefit from a production unit may be optimized by moving personnel from one line to another.
The chapter ends with the planning of recruitment and leaving personnel at a construction site.

14.1

Assigning personnel to machines
An operator needs to be assigned to each of the six machines in a workshop. Six workers have been
pre-selected. Everyone has undergone a test of her productivity on every machine. Table 14.1 lists the
productivities in pieces per hour. The machines run in parallel, that is, the total productivity of theworkshop is the sum of the productivities of the people assigned to the machines.
Table 14.1: Productivity in pieces per hour
Workers

1

2

Machines
3
4

5

6

1
2
3
4
5
6

13
18
20
23
28
19

24
25
20
26
33
36

31
30
27
28
34
25

40
43
34
37
38
45

29
22
33
30
20
24

19
15
25
18
17
27

The objective is to determine an assignment of workers to machines that maximizes the total productivity.
We may start by calculating a (non-optimal) heuristic solution using the following fairly natural method:choose the assignment p → m with the highest productivity, cross out the line p and the column m (since
the person has been placed and the machine has an operator), and restart this process until we haveassigned all persons. The problem should then be solved to optimality using Mathematical Programming.
And finally, solve the same problem to optimality, but for machines working in series.

210

Applications of optimization with Xpress-MP

14.1.1

Model formulation
14.1.1.1 Parallel machines
This problem type is well known under the name of the assignment problem. It also occurs in Chapter 11(flight connections). Let PERS be the set of workers, MACH the set of machines (both of the same size N),
and OUTPpm the productivity of worker p on machine m. A first idea for encoding an assignment would
be to define N integer variables assignp taking values in the set of machines, where assignp denotes thenumber of the machine to which the person p is assigned. With these variables it is unfortunately notpossible to formulate in a linear form the constraints that a person has to be on a single machine and amachine only takes one operator. These constraints become easy to express if we instead use N2 binary
variables assignpm that are 1 if and only if person p is assigned to machine m (14.1.1).
∀p ∈ PERS, m ∈ MACH : assignpm ∈ {0, 1}(14.1.1)

The fact that a person p is assigned to a single machine m is by (14.1.2). Similarly, to express that amachine m is operated by a single person p, we have constraints (14.1.3).

X∀p ∈ PERS :assignpm = 1(14.1.2)

assignpm = 1(14.1.3)m∈MACHX∀m ∈ MACH :p∈PERS

The objective function to be maximized sums the OUTPpm for the variables assignpm that are at 1 (14.1.4).
The resulting mathematical model is formed by the lines (14.1.1) to (14.1.4).
maximize

X

X

OUTPpm · assignpm(14.1.4)

p∈PERS m∈MACH

The assignment problem may be looked at as a flow problem. It is sufficient to define a bipartite labeledgraph G = (PERS, MACH, ARCS, OUTP), where PERS is a set of nodes representing the personnel, MACH a
set of nodes for the machines, and ARCS the set of arcs describing the possible assignments of workers tomachines. In our case, there are N2 arcs, but it would be possible to prohibit certain assignments in thegeneral case. Every arc (p, m) has infinite capacity and is labeled with the cost (productivity) OUTPpm . We
create a source node SOURCE that is connected to every person-node p by an arc (SOURCE, p) of capacity
1. We then also create a sink node SINK to which are connected all machine-nodes m by arcs (m, SINK),
also of capacity 1. An optimal assignment corresponds to a flow in G of throughput N with maximum
total cost. Due to the unit capacities, the flow will trace N disjunctive paths from SOURCE to SINK thatindicate the assignments.
The linear solution to the assignment problem is automatically integer. The constraints (14.1.1) may
therefore be replaced by simple non-negativity conditions. It is not necessary to specify that the variables assignpm must not be larger than 1 because this is guaranteed through the constraints (14.1.2) and(14.1.3). The model can be modified to deal with sets of personnel and machines of different size. If, forinstance, we have more workers than machines, we keep the constraints (14.1.3) so that every machine
receives an operator, but we replace the constraints (14.1.2) by (14.1.5).

X∀p ∈ PERS :assignpm ≤ 1(14.1.5)m∈MACH

Certain assignments may be infeasible. In such a case the value of the corresponding variable assignpm
will be forced to 0, or a highly negative number (that is, with a very large absolute value) is used for
OUTPpm . If the graph of possible assignments is given instead of the productivity matrix, it is also possible
to generate the variables assignpm merely for the arcs (p, m) that are part of the graph.

14.1.1.2 Machines working in series
If the machines work in series, the least productive worker on the machine she has been assigned todetermines the total productivity of the workshop. An assignment will still be described by N2 variables
assignpm (14.1.1) and the constraints (14.1.2) and (14.1.3). To this we add a non-negative variable pmin forthe minimum productivity. The objective is to maximize pmin. This type of optimization problem where
one wants to maximize a minimum is called maximin, or bottleneck. To obtain a linear formulation ofsuch problem, the procedure is always the same:

Timetabling and personnel planning

211

Applications of optimization with Xpress-MP

• A variable bounding the productivities on every machine from below is defined (here pmin).
• Constraints are added that bound every productivity by pmin.
• We maximize pmin which implies that in the optimum at least one productivity is equal to pmin
and hence defines the bottleneck.
Two groups of constraints are possible to bound the productivities by pmin: the N2 constraints (14.1.6) or
the N constraints (14.1.7). In the latter, a single variable assignpm has the value 1 due to the constraints(14.1.2). The resulting mathematical model is given by the lines (14.1.1) to (14.1.3), the constraints (14.1.6)or (14.1.7), the non-negativity condition for pmin and the new objective function: maximize pmin. Dueto the constraints (14.1.6) or (14.1.7), the mathematical model is no longer a classical assignment problem and integrality of the LP solution values is no longer guaranteed. We therefore have to keep theconstraints (14.1.1).
∀p ∈ PERS, m ∈ MACH : OUTPpm · assignpm ≥ pmin
∀p ∈ PERS :X

OUTPpm · assignpm ≥ pmin

(14.1.6)(14.1.7)m∈MACH

14.1.2

Implementation
The following Mosel program first implements and solves the model for the case of parallel machines.
Afterwards, we define the variable pmin that is required for solving the problem for the case that themachines work in series. We also add the necessary bounding constraints (choosing constraints 14.1.7)
for the variables assignpm and turn these variables into binaries.
model "I-1 Personnel assignment"
uses "mmxprs"
declarations
PERS = 1..6
MACH = 1..6
OUTP: array(PERS,MACH) of integer
end-declarations! Personnel! Machines! Productivity

initializations from ’i1assign.dat’
OUTP
end-initializations! **** Exact solution for parallel assignment ****
declarations
assign: array(PERS,MACH) of mpvar! 1 if person assigned to machine,! 0 otherwise

end-declarations! Objective: total productivity
TotalProd:= sum(p in PERS, m in MACH) OUTP(p,m)*assign(p,m)! One machine per person
forall(p in PERS) sum(m in MACH) assign(p,m) = 1! One person per machine
forall(m in MACH) sum(p in PERS) assign(p,m) = 1! Solve the problem
maximize(TotalProd)
writeln("Parallel machines: ", getobjval)! **** Exact solution for serial machines ****
declarations
pmin: mpvar
end-declarations! Minimum productivity! Calculate minimum productivity
forall(p in PERS) sum(m in MACH) OUTP(p,m)*assign(p,m) >= pmin
forall(p in PERS, m in MACH) assign(p,m) is_binary

Timetabling and personnel planning

212

Applications of optimization with Xpress-MP

! Solve the problem
maximize(pmin)
writeln("Serial machines: ", getobjval)
end-model

The following procedure parallel_heur may be added to the above program. If called after the data
has been initialized, it heuristically calculates a (non-optimal) solution to the parallel assignment problem
using the intuitive procedure described in the introduction to Section 14.1.
procedure parallel_heurdeclarations
ALLP, ALLM: set of integer
HProd: integer
pmax,omax,mmax: integer
end-declarations! Copies of sets PERS and MACH
! Total productivity value! Copy the sets of workers and machines
forall(p in PERS) ALLP+={p}forall(m in MACH) ALLM+={m}! Assign workers to machines as long as there are unassigned persons
while (ALLP<>{}) do
pmax:=0; mmax:=0; omax:=0! Find the highest productivity among the remaining workers and machines
forall(p in ALLP, m in ALLM)
if OUTP(p,m) > omax thenomax:=OUTP(p,m)
pmax:=p; mmax:=m
end-ifHProd+=omax
ALLP-={pmax}; ALLM-={mmax}
writeln("
end-do! Add to total productivity! Remove person and machine from sets",pmax, " operates machine ", mmax, " (", omax, ")")

writeln(" Total productivity: ", HProd)
end-procedure

14.1.3Results
The following table summarizes the results found with the different solution methods for the two problems of parallel and serial machines. The very intuitive heuristic method is likely to be employed by people
who do not know of Mathematical Programming. However, there is a notable difference between theheuristic and the exact solution to the problem with parallel machines.
Table 14.2: Optimal assignment found

14.2

Person4Algorithm

1

2

3

5

6

Productivity

Parallel Machines

HeuristicExact

4 (19)
3 (31)

1 (18)
5 (43)

6 (33)
4 (25)

2 (26)
6 (30)

3 (34)
1 (28)

5 (45)
2 (36)

175
193

Serial Machines

Exact

6 (29)

3 (30)

5 (34)

2 (26)

1 (28)

4 (27)

26

Scheduling nurses
Mr. Schedule has been asked to organize the schedule of nurses for the Cardiology service at St. Joseph’s
hospital. A working day in this service is subdivided into twelve periods of two hours. The personnel
requirement changes from period to period: for instance, only a few nurses are required during thenight, but the total number must be relatively high during the morning to provide specific care to thepatients. The following table lists the personnel requirement for every time period.
Question 1: Determine the minimum number of nurses required to cover all the requirements, knowingthat a nurse works eight hours per day and that she is entitled to a break of two hours after she has
worked for four hours.

Timetabling and personnel planning

213

Applications of optimization with Xpress-MP

Table 14.3: Personnel requirement per time period
NumberTime interval

Minimum number of nurses

0
1
2
3
4
5
6
7
8
9
10
11

00am – 02am
02am – 04am
04am – 06am
06am – 08am
08am – 10am
10am – 12pm
12pm – 02pm
02pm – 04pm
04pm – 06pm
06pm – 08pm
08pm – 10pm
10pm – 12am

15
15
15
35
40
40
40
30
31
35
30
20

Question 2: The service only has 80 nurses, which is not sufficient with the given requirements. Mr.
Schedule therefore proposes that part of the personnel works two additional hours per day. These two
additional hours follow immediately after the last four hours, without any break. Determine the scheduleof the nurses in this service that minimizes the number of nurses working overtime.

14.2.1Model formulation for question 1
Let startt be the number of nurses starting work in time period t (first period = time period from 0am–
2am, second period = time period from 2am–4am etc.), NT be the number of time periods, and TIME ={0, . . . , NT − 1} the set of time periods. The objective function is given by (14.2.1).
minimizeX(14.2.1)

starttt∈TIMEWe need to make sure that a sufficiently large number of nurses is available during every time period.
For instance, in the period t = 6 (12pm–2pm) of the day 40 nurses are required. The number of nurses
working during this time period is equal to the total of the number of nurses starting work in this period,
the immediately preceding time period, and in the two periods 3 and 4 intervals earlier. That is, during
time period 6, the nurses starting in the intervals 2, 3, 5, and 6 are working. For instance, a nurse starting
in interval 3 will work in the intervals 3 and 4, have her break in interval 5, and work again in the intervals
6 to 7. If we write REQt for the personnel requirement in interval t, we obtain the following constraint
for the time period 6:(14.2.2)
start2 + start3 + start5 + start6 ≥ REQ6
Similarly, we can formulate the personnel requirement constraints for all other time periods. If WORK
denotes the set of intervals that have an impact on the time period under consideration (that is WORK ={0, −1, −3, −4}), we may write these constraints in the generic form (14.2.3).
∀t ∈ TIME :X

start(t+i+NT) mod NT ≥ REQt

(14.2.3)i∈WORKFor the first four time periods every day, we need to take into account the nurses who started working in
the last four time intervals of the previous day. The expression (t + 1 + NT) mod NT transposes the indicescorrectly, so that for instance for the first interval of a day, the periods 8, 9, 11, and 0 are taken into
account by the corresponding constraint (t = 0).
The constraints (14.2.4) indicate that all variables startt must be integer.
∀t ∈ TIME : startt ∈ IN

14.2.2(14.2.4)

Implementation of question 1
The following Mosel program is a straightforward implementation of the mathematical model. WORK isan example of sets in Mosel that contain negative values and also the value 0.
model "I-2 Scheduling nurses 1"
uses "mmxprs"

Timetabling and personnel planning

214

Applications of optimization with Xpress-MP

declarations
NT = 12
TIME = 0..NT-1
WORK: set of integer! Number of time periods

REQ: array(TIME) of integer
start: array(TIME) of mpvar
end-declarations! Nurses started in other time periods! that are working during a period! Required number of nurses per time period! Nurses starting work in a period

initializations from ’i2nurse.dat’REQ
end-initializations
WORK:= {0, -1, -3, -4}! Objective: total personnel required
Total:= sum(t in TIME) start(t)! Nurses working per time period
forall(t in TIME) sum(i in WORK) start((t+i+NT) mod NT) >= REQ(t)forall(t in TIME) start(t) is_integer! Solve the problem
minimize(Total)
end-model

14.2.3Results for question 1
100 nurses are required in total for the Cardiology service. Their working periods may be scheduled asshown in Table 14.4 (there are several equivalent solutions).
Table 14.4: Nurse schedule covering all requirements

14.2.4

Time intervalStarting work

Total

00am – 02am
02am – 04am
04am – 06am
06am – 08am
08am – 10am
10am – 12pm
12pm – 02pm
02pm – 04pm
04pm – 06pm
06pm – 08pm
08pm – 10pm
10pm – 12am

6
17
12
17
0
11
0
13
7
17
0
0

30
40
29
35
40
40
40
30
31
35
30
20

Model formulation for question 2
The optimal solution involves employing 100 nurses. Unfortunately, the service currently only has 80
nurses. They will therefore have to work extra hours.
We retain the variables startt introduced for the answer to Question 1. In addition, we introduce variables
overtt corresponding to the number of nurses starting to work in interval t and working two overtime
hours at the end of their normal period of duty.
Our objective now is to minimize the number of nurses that have to work overtime. We thus obtain theobjective function (14.2.5).
X
minimizeovertt(14.2.5)t∈TIMEThe number of nurses may not exceed the available total staff of NUM = 80. This constraint is expressed

Timetabling and personnel planning

215

Applications of optimization with Xpress-MP

by the relation (14.2.6).

X(14.2.6)

startt ≤ NUM

t∈TIMEFor every time interval t, the number of nurses overtt starting in this period who work overtime may atmost be as large as the total number of nurses startt starting their work during this time period. We
therefore obtain the constraints (14.2.7)∀t ∈ TIME : overtt ≤ startt(14.2.7)

As before, the personnel requirement of every time slice needs to be covered. We obtain the constraints(14.2.8) that are similar to the constraints (14.2.3) of Question 1, but in which we also take into accountthe nurses working extra hours. For example, the nurses on duty in interval 6 started work either in theintervals 2, 3, 5, or 6 as in the first question, or in the interval 1 if they are working overtime.
∀t ∈ TIME : overt(t−5+NT) mod NT +X

start(t+i+NT)modNT ≥ REQt

(14.2.8)i∈WORK

To complete the model for Question 2, the constraints (14.2.9) and (14.2.10) specify that all variables startt
and overtt must be integer.

14.2.5∀t ∈ TIME : startt ∈ IN

(14.2.9)∀t ∈ TIME : overtt ∈ IN

(14.2.10)

Implementation of question 2
The translation of the mathematical model to Mosel is again immediate.
model "I-2 Scheduling nurses 2"
uses "mmxprs"
declarations
NT = 12
TIME = 0..NT-1
WORK: set of integer! Number of time periods! Nurses started in other time periods! that are working during a period

REQ: array(TIME) of integer
NUM: integer! Required number of nurses per time period! Available total staff

start: array(TIME) of mpvar
overt: array(TIME) of mpvar
end-declarations! Nurses starting work in a period! Nurses working overtime

initializations from ’i2nurse.dat’
REQ NUM
end-initializations
WORK:= {0, -1, -3, -4}! Objective: total overtime worked
TotalOvert:= sum(t in TIME) overt(t)! Nurses working per time period
forall(t in TIME)
overt((t-5+NT) mod NT) + sum(i in WORK) start((t+i+NT) mod NT) >= REQ(t)! Limit on total number of nurses
Total <= NUM
forall(t in TIME) do
start(t) is_integer
overt(t) is_integer
overt(t) <= start(t)end-do! Solve the problem
minimize(TotalOvert)
end-model

Timetabling and personnel planning

216

Applications of optimization with Xpress-MP

14.2.6

Results for question 2
Forty of the eighty available nurses will have to work overtime to satisfy the personnel requirements ofthe service. Table 14.5 lists a detailed schedule (there are several equivalent solutions).
Table 14.5: Nurse schedule with overtime work

14.3

Time intervalStarting work

Total

Overtime

00am – 02am
02am – 04am
04am – 06am
06am – 08am
08am – 10am
10am – 12pm
12pm – 02pm
02pm – 04pm
04pm – 06pm
06pm – 08pm
08pm – 10pm
10pm – 12am

13
1
10
7
17
4
14
9
0
0
3
2

15
17
16
35
40
40
35
47
30
35
30
20

0
0
0
3
2
8
0
0
0
17
4
6

Establishing a college timetable
Mr. Miller is in charge of establishing the weekly timetable for two classes of the last year in a college.
The two classes have the same teachers, except for mathematics and sport. In the college all lessons havea duration of two hours. Furthermore, all students of the same class attend exactly the same courses.
From Monday to Friday, the slots for courses are the following: 8:00–10:00, 10:15–12:15, 14:00–16:00, and16:15–18:15. The following table lists the number of two-hour lessons that every teacher has to teach thestudents of the two classes per week.
Table 14.6: Number of 2-hour lessons per teacher and class
Teacher
Mr CheeseMrs Insulin
Mr Map
Mr Effofecks
Mrs Derivate
Mrs Electron
Mr WiseMr Muscle
Mrs Biceps

Subject

Lessons for class 1

Lessons for class 2

EnglishBiology
History-Geography
Mathematics
Mathematics
Physics
Philosophy
Sport
Sport

1
3
2
0
4
3
1
1
0

1
3
2
4
0
3
1
0
1

The sport lessons have to take place on Thursday afternoon from 14:00 to 16:00. Furthermore, the firsttime slot on Monday morning is reserved for supervised homework. Mr Effofecks is absent every Monday
morning because he teaches some courses at another college. Mrs Insulin does not work on Wednesday.
And finally, to prevent students from getting bored, every class may only have one two-hour lesson per
subject on a single day. Write a mathematical program that allows Mr Miller to determine the weekly
timetable for the two classes.

14.3.1Model formulation
Let TEACHERS, and CLASS respectively be the sets of teachers and classes. Let SLOTS denote the set oftime slots per week, numbered consecutively from 1 to NP · ND, where NP denotes the number of timeslots per day and ND the number of days per week. COURSEtc denotes the number of two-hour lessons
that the teacher t has to teach class c every week.
We define variables teachtcl that are 1 if the teacher t gives a lesson to class c in time slot l. We obtain

Timetabling and personnel planning

217

Applications of optimization with Xpress-MP

the following mathematical model:
minimize

X ND−1
X

X(teacht,c,d·NP+1 + teacht,c,(d+1)·NP )(14.3.1)

X(14.3.2)t∈TEACHERS c∈CLASS d=0∀t ∈ TEACHERS, c ∈ CLASS :teachtcl = COURSEtc

l∈SLOTS∀c ∈ CLASS, l ∈ SLOTS :X

teachtcl ≤ 1(14.3.3)

teachtcl ≤ 1(14.3.4)t∈TEACHERS∀t ∈ TEACHERS, l ∈ SLOTS :Xc∈CLASS(d+1)·NP∀t ∈ TEACHERS, c ∈ CLASS, d = 0, . . . , ND − 1 :Xteachtcl ≤ 1(14.3.5)

l=d·NP+1

teachMrMuscle,1,15 = 1(14.3.6)

teachMrsBiceps,2,15 = 1(14.3.7)∀t ∈ TEACHERS, c ∈ CLASS : teachtc1 = 0(14.3.8)∀l = 1, 2 : teachMrEffofecks,2,l = 0(14.3.9)∀c ∈ CLASS, l = 2 · NP + 1, . . . , 3 · NP : teachMrsInsuline,c,l = 0(14.3.10)∀t ∈ TEACHERS, c ∈ CLASS, l ∈ SLOTS : teachtcl ∈ {0, 1}(14.3.11)

The aim of this problem is simply to find a timetable that fulfills all constraints. We may however giveourselves the objective to minimize the holes in the timetable. With this aim, the courses should preferably be placed in the time slots from 10:15–12:15 and 14:00–16:00. If these time slices remain unused but
lessons are placed either at the beginning or at the end of the day, there will necessarily be holes in the
middle of the day. We therefore penalize lessons placed at the beginning or the end of the day, whichmeans we minimize the sum of courses taught during slots 1 and 4 of every day; hence the objectivefunction (14.3.1).
The constraints (14.3.2) indicate that all lessons taught by the teacher t to class c must be scheduled. The
constraints (14.3.3) specify that a class has at most one course at any time, and similarly the constraints(14.3.4) that a teacher must not teach more than one course at a time. The constraints (14.3.5) say thatat most one two-hour lesson per subject is taught on the same day.
The constraints (14.3.6) to (14.3.10) translate the specific conditions of this problem: the constraints(14.3.6) and (14.3.7) indicate that the sport lessons by Mr Muscle and Mrs Biceps have to take place inthe beginning of Thursday afternoon (time slice 15). The first time interval of the week being reserved
for supervised homework, the constraints (14.3.8) specify that no course may be scheduled during thisperiod. The constraints (14.3.9) and (14.3.10) respectively prohibit Mr Effofecks teaching on Monday
morning and Mrs Insulin teaching on Wednesday (time slices 9-12). The last set of constraints specifies
that all variables are binaries.

14.3.2

Implementation
The mathematical model of the previous section is implemented by the following Mosel program.
model "I-3 School timetable"
uses "mmxprs"
declarations
TEACHERS: set of string
CLASS = 1..2
NP = 4
ND = 5SLOTS=1..NP*ND! Set of teachers! Set of classes! Number of time periods for courses! Days per week
! Set of time slots for the entire week

COURSE: array(TEACHERS,CLASS) of integer ! Lessons per teacher and classend-declarations
initializations from ’i3school.dat’
COURSE
end-initializations
finalize(TEACHERS)

Timetabling and personnel planning

218

Applications of optimization with Xpress-MP

declarations
teach: array(TEACHERS,CLASS,SLOTS) of mpvar! teach(t,c,l) = 1 if teacher t gives a! lesson to class c during time period l
end-declarations! Objective: number of "holes" in the class timetables
Hole:=
sum(t in TEACHERS, c in CLASS, d in 0..ND-1) (teach(t,c,d*NP+1) +teach(t,c,(d+1)*NP))! Plan all courses
forall(t in TEACHERS, c in CLASS) sum(l in SLOTS) teach(t,c,l) = COURSE(t,c)! For every class, one course at a time
forall(c in CLASS, l in SLOTS) sum(t in TEACHERS) teach(t,c,l) <= 1
! Teacher teaches one course at a time
forall(t in TEACHERS, l in SLOTS) sum(c in CLASS) teach(t,c,l) <= 1
! Every subject only once per day
forall(t in TEACHERS, c in CLASS, d in 0..ND-1)
sum(l in d*NP+1..(d+1)*NP) teach(t,c,l) <= 1
! Sport Thursday afternoon (slot 15)
teach("Mr Muscle",1,15) = 1
teach("Mrs Biceps",2,15) = 1! No course during first period of Monday morning
forall(t in TEACHERS, c in CLASS) teach(t,c,1) = 0! No course by Mr Effofecks Monday morning
forall(l in 1..2) teach("Mr Effofecks",2,l) = 0! No Biology on Wednesday
forall(c in CLASS, l in 2*NP+1..3*NP) teach("Mrs Insulin",c,l) = 0
forall(t in TEACHERS, c in CLASS, l in SLOTS) teach(t,c,l) is_binary! Solve the problem
minimize(Hole)
end-model

14.3.3Results
The optimal solution found has the value 10, indicating that 10 courses are scheduled in the slots at the
beginning or the end of a day. The resulting timetables for the two classes displayed in Tables 14.7 and14.8 have no holes. Note that there are many equivalent solutions to this problem.
Table 14.7: Timetable for class 1
8:00–10:00

10:15–12:15

14:00–16:00

16:15–18:15

Mon Homework
Maths (Mrs Derivate) Biology (Mrs Insulin) Physics (Mrs Electron)
Tue
English (Mr Cheese) Maths (Mrs Derivate) Biology (Mrs Insulin) Physics (Mrs Electron)
Wed—
Maths (Mrs Derivate) Hist.-Geogr. (Mr Map)—
Thu—Hist.-Geogr. (Mr Map) Sport (Mr Muscle)
Physics (Mrs Electron)
Fri—
Biology (Mrs Insulin) Maths (Mrs Derivate) Philosophy (Mr Wise)

Table 14.8: Timetable for class 2
8:00–10:00

10:15–12:15

14:00–16:00

16:15–18:15

Mon HomeworkHist.-Geogr. (Mr Map) Maths (Mr Effofecks) Biology (Mrs Insulin)
Tue—
Maths (Mr Effofecks) Hist.-Geogr. (Mr Map) Biology (Mrs Insulin)
Wed—
Physics (Mrs Electron) Maths (Mr Effofecks)—
Thu Physics (Mrs Electron) English (Mr Cheese)Sport (Mrs Biceps)
Biology (Mrs Insulin)Fri
Physics (Mrs Electron) Philosophy (Mr Wise) Maths (Mr Effofecks)—

Timetabling and personnel planning

219

Applications of optimization with Xpress-MP

14.4

Exam scheduling
At a technical university every term the third-year students choose eight modules from the eleven modules that are taught, depending on the option they wish to choose in the fourth year (there are two
possible choices: ‘Production planning’ and ‘Quality and security management’). In the current term,
certain modules are obligatory for students who wish to continue with one of these options. The obligatory courses are Statistics (S), Graph models and algorithms (GMA), Production management (PM), andDiscrete systems and events (DSE). The optional modules are: Data analysis (DA), Numerical analysis (NA),Mathematical programming (MP), C++, Java (J), Logic programming (LP), and Software engineering (SE).
Table 14.9: Incompatibilities between different exams
DA
NA
C++
SE
PM
J
GMA
LP
MP
S
DSE

DA

NA

C++

SE

PM

J

GMA

LP

MP

S

DSE

–
X
–
–
X
–
X
–
–
X
X

X
–
–
–
X
–
X
–
–
X
X

–
–
–
X
X
X
X
–
X
X
X

–
–
X
–
X
X
X
–
–
X
X

X
X
X
X
–
X
X
X
X
X
X

–
–
X
X
X
–
X
–
X
X
X

X
X
X
X
X
X
–
X
X
X
X

–
–
–
–
X
–
X
–
–
X
X

–
–
X
–
X
X
X
–
–
X
X

X
X
X
X
X
X
X
X
X
–
X

X
X
X
X
X
X
X
X
X
X
–

Mrs Edeetee needs to schedule the exams at the end of the term. Every exam lasts two hours. Two days
have been reserved for the exams with the following time slices: 8:00–10:00, 10:15–12:15, 14:00–16:00,
and 16:15–18:15. For every exam she knows the set of incompatible exams that may not take place at thesame time because they have to be taken by the same students. These incompatibilities are summarizedin Table 14.4.1.
Help Mrs Edeetee construct a timetable so that no student has more than one exam at a time.

14.4.1Model formulation
Let planet be binary variables that are 1 if the exam e is scheduled in time slice t and 0 otherwise. We shallwrite EXAM for the set of exams and TIME for the set of time slices. The mathematical model associatedwith the problem is the following:∀e ∈ EXAM :X(14.4.1)

planet = 1

t∈TIME∀d, e ∈ EXAM, d < e ∧ INCOMPde = 1, ∀t ∈ TIME : planet + plandt ≤ 1(14.4.2)∀e ∈ EXAM, t ∈ TIME : planet ∈ {0, 1}(14.4.3)

This mathematical model does not contain any objective function since we simply wish to find a solutionthat satisfies all incompatibility constraints. This does not cause any problem for the solution algorithm
that will generate an arbitrary feasible solution (if one exists). We might however add an objectivefunction, for instance by associating costs or preferences with the assignment of certain exams to certain
time slices.
The constraints (14.4.1) indicate that every exam needs to be scheduled exactly once. The constraints(14.4.2) require that two incompatible exams may not be scheduled at the same time. The incompatibilities are stored in a Boolean matrix INCOMP. An element INCOMPde of this matrix is 1 if the exams d and e
are incompatible and 0 otherwise. Since the incompatibility matrix is symmetric for this problem, we only
need to define the constraints (14.4.2) for pairs of exams d and e with d < e. The last set of constraints(14.4.3) specifies that all variables are binaries.
The problem is completely stated by the constraints (14.4.1) to (14.4.3). However, if this problem hasa solution then it has an enormous number of equivalent (symmetric) solutions that are obtained bypermutating the time slots assigned to compatible sets of exams. We may try to ‘help’ the solution
algorithm by adding constraints breaking some of these symmetries and hence reducing the size of thesearch space. For instance, we can assign DA to slot 1, and NA to slot 2 (since it is incompatible with DA).
In fact, we can go further as PM is incompatible with DA and NA, so it can be assigned to slot 3. Similarly,
GMA, S, and DSE must be given unique slots (4, 5, and 6) since they are incompatible with everything.

Timetabling and personnel planning

220

Applications of optimization with Xpress-MP

How useful these observations are in general is debatable, but breaking symmetry, if possible, is always
a good idea.

14.4.2

Implementation
The mathematical model may be implemented with Mosel as shown below. The problem has no objectivefunction but we wish to invoke the optimization algorithm for finding a feasible solution. We therefore
simply indicate a constant (dummy) objective in the call to the optimizer — instead of minimize we
may also use maximize since we want to solve the problem without optimizing anything. We have notintroduced the symmetry-breaking ideas discussed above.
model "I-4 Scheduling exams"
uses "mmxprs"
declarations
EXAM = {"DA","NA","C++","SE","PM","J","GMA","LP","MP","S","DSE"}! Set of exams
TIME = 1..8
! Set of time slots
INCOMP: array(EXAM,EXAM) of integer ! Incompatibility between exams
plan: array(EXAM,TIME) of mpvar
end-declarations! 1 if exam in a time slot, 0 otherwise

initializations from ’i4exam.dat’
INCOMP
end-initializations! Schedule all exams
forall(e in EXAM) sum(t in TIME) plan(e,t) = 1! Respect incompatibilities
forall(d,e in EXAM, t in TIME | d<e and INCOMP(d,e)=1)plan(e,t) + plan(d,t) <= 1
forall(e in EXAM, t in TIME) plan(e,t) is_binary! Solve the problem (no objective)
minimize(0)end-model

14.4.3Results
The linear relaxation of this problem gives an integer solution (this is not true in general). In the schedule
displayed below only seven out of the eight available time intervals are used. This problem has a large
number of different feasible solutions.
Table 14.10: Exam schedule
Day 1
Day 2

8:00–10:00

10:15–12:15

14:00–16:00

16:15–18:15

NA, SE, LP, MP
—

PM
S

GMA
J

DA, C++
DSE

14.5 Production planning with personnel assignment
The company Line Production decides to plan the production of four of its products (P1, P2, P3, P4) on its
five production lines (L1 to L5). The company gains net profits of B
C 7 for the products P1 and P4, B
C 8 forP2, and B
C 9 for P3. The maximum times during which the five production lines may operate are different
during the planning period. The maximum capacities for L1 to L5 are 4500 hours, 5000 hours, 4500 hours,
1500 hours, and 2500 hours respectively. Table 14.12 lists the processing time in hours necessary for the
production of one unit of every product on every production line. Which quantities of P1 to P4 should be
produced to maximize the total profit?
If subsequently a transfer of personnel (and hence of working hours) is authorized between production

Timetabling and personnel planning

221

Applications of optimization with Xpress-MP

Table 14.11: Unitary processing times
Products

L1

L2

Lines
L3

L4

L5

P1
P2
P3
P4

1.3
1.8
1.3
0.9

0.9
1.7
1.2
1.1

2.0
1.4
1.3
1.0

0.3
0.6
1.0
0.9

0.9
1.1
1.4
1.0

Table 14.12: Possible transfers of personnel
Origin

L1

Destination
L2
L3
L4

L5

Maximum number oftransferable hours

L1
L2
L3
L4
L5

–
no
yes
no
yes

yes
–
yes
no
yes

no
yes
no
yes
–

400
800
500
200
300

yes
yes
–
no
yes

yes
no
yes
–
no

lines during the planning period as shown in Table 14.12, which is the maximum profit? How many hours
are transfered and under what conditions?

14.5.1Model formulation
Let PRODS be the set of products and LINES the set of production lines. The per unit profit of productp is given as PROFITp and the processing duration of product p on line l as DURpl . The capacity of every
production line l in working hours is CAPl . As usual in production planning problems, a variables producep
indicates the quantity of product p that is produced. The model corresponding to the first question isfairly simple. The objective is to maximize the total profit (14.5.1) whilst remaining within the capacity
limits of every production line (constraints (14.5.2)). We obtain the following model:Xmaximize(14.5.1)

PROFITp · producepp∈PRODS∀l ∈ LINES :X(14.5.2)

DURpl · producep ≤ CAPlp∈PRODS∀p ∈ PRODS : producep ≥ 0(14.5.3)

If it is now possible to transfer personnel from one line to another, variables hoursl are required that
correspond to the working hours used on production line l. We also introduce variables transferlk that
represent the number of hours transfered from line l to line k. The constraints (14.5.5) that replace
the constraints (14.5.2) of the previous model specify that the number of hours worked is equal to the
production multiplied by the per unit duration of the work.
The constraints (14.5.7) establish the balance between the working hours carried out on a line, the hours
transfered and the maximum number of working hours on this line. On one line, hoursl hours are worked
that correspond to the available number of hours for this line, increased by the hours that are transferedto this line from other and decreased by the hours transfered to other lines (TRANSFlk is 1 if it is possible
to transfer hours from line l to line k, 0 otherwise). The constraints (14.5.8) establish the limits TMAXl on
the number of hours that may be transfered from a line to the others. The last two sets of constraints(14.5.9) and (14.5.10) are the non-negativity conditions for variables transferkl and hoursl .

Xmaximize(14.5.4)

PROFITp · producepp∈PRODS∀l ∈ LINES :X(14.5.5)

DURpl · producep ≤ hourslp∈PRODS∀p ∈ PRODS : producep ≥ 0∀l ∈ LINES : hoursl = CAPl +(14.5.6)

Xtransferkl −k∈LINESTRANSFkl =1∀l ∈ LINES :Xtransferlk ≤ TMAXl

X

transferlk(14.5.7)k∈LINESTRANSFlk =1(14.5.8)k∈LINESTRANSFlk =1

Timetabling and personnel planning

222

Applications of optimization with Xpress-MP

14.5.2∀l, k ∈ LINES : transferlk ≥ 0(14.5.9)∀l ∈ LINES : hoursl ≥ 0(14.5.10)

Implementation
The following Mosel program implements and solves the models for both questions. First the model forquestion 1 is defined and solved. Then follow the additional declarations of data and variables requiredby the second model. For the second model, the capacity constraints Load of the first model (14.5.2) are
re-defined with the modified version required by the second model (14.5.5).
model "I-5 Production planning with personnel"
uses "mmxprs"
declarations
PRODS = 1..4
LINES = 1..5

! Set of products! Set of production lines

PROFIT: array(PRODS) of integer
DUR: array(PRODS,LINES) of real
CAP: array(LINES) of integer! Profit per product! Duration of production per line
! Working hours available per line

Load: array(LINES) of linctr
produce: array(PRODS) of mpvar
end-declarations! Workload constraints! Quantity produced

initializations from ’i5pplan.dat’
PROFIT DUR CAP
end-initializations! Objective: Total profit
Profit:= sum(p in PRODS) PROFIT(p)*produce(p)! Capacity constraints on lines
forall(l in LINES) Load(l):=sum(p in PRODS) DUR(p,l)*produce(p) <= CAP(l)! Solve the problem
maximize(Profit)
writeln("Total profit: ", getobjval)! **** Allow transfer of working hours between lines ****
declarations
TRANSF: dynamic array(LINES,LINES) of integer ! 1 if transfer is allowed,! 0 otherwise
TMAX: array(LINES) of integer! Maximum no. of hours to transfer
hours: array(LINES) of mpvar! Initial working hours per linetransfer: dynamic array(LINES,LINES) of mpvar ! Hours transferred
end-declarations
initializations from ’i5pplan.dat’
TRANSF TMAX
end-initializations
forall(k,l in LINES | exists(TRANSF(k,l))) create(transfer(k,l))! Re-define capacity constraints on lines
forall(l in LINES) Load(l):=sum(p in PRODS) DUR(p,l)*produce(p) <= hours(l)! Balance constraints
forall(l in LINES)
hours(l) = CAP(l) + sum(k in LINES) transfer(k,l) sum(k in LINES) transfer(l,k)! Limit on transfer
forall(l in LINES) sum(k in LINES) transfer(l,k) <= TMAX(l)! Solve the problem
maximize(Profit)
writeln("Total profit allowing transfer: ", getobjval)
end-model

Timetabling and personnel planning

223

Applications of optimization with Xpress-MP

Note that the variables transfer in the second model are defined as a dynamic array, and only after thedata indicating the allowable transfers (array TRANSF) is known do we create the transfer variables
actually required. By proceeding in this way, we may remove the tests for the values of the entries of
TRANSF from the various constraints since Mosel will automatically sum only those transfer variables
that have been created.

14.5.3Results
If no transfer of personnel (that is, working hours) between production lines is possible, the maximum
profit is B
C 18,883. 1542.55 units of product P1 and 1010.64 units of P2 are produced, and nothing of theother two products. When examining the capacity constraints for the production lines, we find that only
line 3 and 5 work at their maximum capacity. This indicates that a transfer of working hours may be
profitable.
When the transfer of working hours between lines is allowed, the total profit increases to B
C 23,431.10.
The quantities to produce are now 702.35 units of P1, 942.82 of P2, 554.83 of P3, and 854.84 of P4. On
production line 1 a total of 4100 hours are used and 400 hours are transfered to line 4. Production line
2 works for 3840.3 hours and 800 hours are transfered to line 5. Line 3 works during 4300 hours and 200
hours are transfered to line 4. On line 4, 600 hours are added to the original capacity of 1500 hours andfor L5 the 800 hours transfered from line 2 result in a total working time of 3300 hours.

14.6

Planning the personnel at a construction site
Construction workers who erect the metal skeleton of skyscrapers are called steel erectors. The following
table lists the requirements for steel erectors at a construction site during a period of six months. Transfersfrom other sites to this one are possible on the first day of every month and cost $100 per person. At theend of every month workers may leave to other sites at a transfer cost of $160 per person. It is estimatedthat understaffing as well as overstaffing cost $200 per month per post (in the case of unoccupied posts
the missing hours have to be filled through overtime work).
Table 14.13: Monthly requirement for steel erectors
March

April

May

June

July

August

4

6

7

4

6

2

Overtime work is limited to 25% of the hours worked normally. Every month, at most three workers may
arrive at the site. The departure to other sites is limited by agreements with labor unions to 1/3 of thetotal personnel of the month. We suppose that three steel erectors are already present on site at the endof February, that nobody leaves at the end of February and that three workers need to remain on-site at
the end of August. Which are the number of arrivals and departures every month to minimize the total
cost?

14.6.1Model formulation
Let MONTHS be the set of months in the time period, numbered consecutively from FIRST to LAST. The
different basic costs are denoted as follows: CARR and CLEAVE the cost of a person arriving and leaving
respectively, COVER and CUNDER the cost of over- and understaffing per person. NSTART and NFINAL
respectively are the number of workers on-site at the beginning and end of the planning period. For
every month m the personnel requirement REQm is given.
Although it would be possible to formulate this model with fewer variables, we describe a formulationthat uses five types of variables indexed by the months to facilitate the modeling and provide all thedetails for the recruitment plan. In every month m, onsitem steel erectors are present (between thearrivals at the beginning of the month and the departures at its end). There are arrivem arrivals, leavem
workers leaving, overm persons more than the required personnel and underm persons missing (14.6.1).
∀m ∈ MONTHS : onsitem , arrivem , leavem , overm , underm ∈ IN

(14.6.1)

For the first month FIRST the total staff equals the initial staff plus the arrivals at the beginning of themonth (14.6.2). After the last month LAST, the remaining personnel onsite equals the personnel present
in the last month minus the departures at the end of this month (14.6.3).

Timetabling and personnel planningonsiteFIRST = NSTART + arriveFIRST(14.6.2)

NFINAL = onsiteLAST − leaveLAST(14.6.3)

224

Applications of optimization with Xpress-MP

For the intermediate months, the personnel on-site, the arrivals and departures are linked by a relationcomparable to the stock balance constraints used for the production planning problems in Chapter 8(14.6.4).
∀m ∈ {FIRST + 1, . . . , LAST − 1} : onsitem = onsitem−1 − leavem−1 + arrivem(14.6.4)
The number of workers present on-site may be different from the number of personnel actually required.
But it becomes equal to the requirement if one deducts the overstaffing and adds the missing persons(14.6.5).
∀m ∈ MONTHS : onsitem − overm + underm = REQm
(14.6.5)
For every month m, the constraints (14.6.6) formulate the limit on the hours worked overtime, the constraints (14.6.7) limit the number of arrivals and the constraints (14.6.8) the number of persons leaving
the construction site.
∀m ∈ MONTHS : overm ≤ onsitem / 4(14.6.6)∀m ∈ MONTHS : arrivem ≤ 3(14.6.7)∀m ∈ MONTHS : leavem ≤ onsitem / 3(14.6.8)

The objective function (14.6.9) accumulates the costs of arrivals, departures, over- and understaffing.
Due to the minimization of the cost, there will never be over- and understaffing at the same time in any
month.
minimizeX(CARR · arrivem + CLEAVE · leavem + COVER · overm + CUNDER · underm )(14.6.9)m∈MONTHS

14.6.2

Implementation
In the following Mosel implementation of the mathematical model, the balance constraints (14.6.2) –(14.6.4) are grouped into a single constraint expression using the inline if. Note further the index set
MONTHS corresponding to the order numbers of the months (and not starting with 1) used for all arrays.
model "I-6 Construction site personnel"
uses "mmxprs"
declarations
FIRST = 3; LAST = 8
MONTHS = FIRST..LAST! Set of time periods (months)

CARR, CLEAVE: integer
COVER, CUNDER: integer
NSTART, NFINAL: integer
REQ: array(MONTHS) of integer! Cost per arrival/departure! Cost of over-/understaffing! No. of workers at begin/end of plan
! Requirement of workers per monthonsite: array(MONTHS) of mpvar! Workers on sitearrive,leave: array(MONTHS) of mpvar ! Workers arriving/leaving
over,under: array(MONTHS) of mpvar! Over-/understaffing
end-declarations
initializations from ’i6build.dat’
CARR CLEAVE COVER CUNDER NSTART NFINAL REQ
end-initializations! Objective: total cost
Cost:= sum(m in MONTHS) (CARR*arrive(m) + CLEAVE*leave(m) +COVER*over(m) + CUNDER*under(m))! Satisfy monthly need of workers
forall(m in MONTHS) onsite(m) - over(m) + under(m) = REQ(m)! Balances
forall(m in MONTHS)
onsite(m) = if(m>FIRST, onsite(m-1) - leave(m-1), NSTART) + arrive(m)NFINAL = onsite(LAST) - leave(LAST)! Limits on departures, understaffing, arrivals; integrality constraints
forall(m in MONTHS) do
leave(m) <= 1/3*onsite(m)under(m) <= 1/4*onsite(m)arrive(m) <= 3Timetabling and personnel planning

225

Applications of optimization with Xpress-MP

arrive(m) is_integer; leave(m) is_integer; onsite(m) is_integer
under(m) is_integer; over(m) is_integer
end-do! Solve the problem
minimize(Cost)
end-model

14.6.3Results
The optimal solution to this problem has a cost of $1,780. The following table contains the detailed
recruitment plan (there are several equivalent solutions).
Table 14.14: Optimal plan for the construction site personnel
MonthRequirement
On siteArriveLeave
Overstaffing
Understaffing

14.7

Initial

March

April

May

June

July

August

Final

–
3
0
0
0
0

4
4
1
0
0
0

6
6
2
0
0
0

7
6
0
0
0
1

4
6
0
0
2
0

6
6
0
2
0
0

2
4
0
1
2
0

–
3
0
0
0
0

References and further material
The assignment problem seen for the allocation of personnel to machines applies to other situations, as
for example the recruitment of candidates for jobs according to their preferences and the perhaps more
surprising problem of flight connections in Chapter 11. Since the simplex algorithm automatically finds
integer-valued solutions, cases of considerable size (100 × 100 matrices for example) are easily dealt with.
However, even faster specialized algorithms exist, like the Hungarian algorithm in O(n3 ) described by
Papadimitriou [PS98] or the techniques for finding the maximum flow presented by Ahuja et al. [AMO93]
and Prins [Pri94a]. For serial workstations, the Mathematical Programming approach works less well, but
fast algorithms based on graphs also exist for this case [DZ78].
The nurse scheduling problem in Section 14.2 is a simplified version of personnel planning problems in
the hospital environment. A more complex problem for determining an individual plan for every nurse
and taking into account the rules imposed by law and the preferences of the nurses, whilst minimizing
the cost of the personnel is studied by Jaumard et al. [JSV98]. The authors suggest a column generation
technique as the solution method. Personnel scheduling problems of this and similar types are among
the most successful application areas of Constraint Programming (for instance the application describedin [HC99]). Typically the aim is ‘simply’ to generate a feasible plan, without any optimization. For nurse
scheduling, specialized CP software is available ([CHW98]).
The problems in Section 14.3 and 14.4 are classical timetabling problems frequently encountered in secondary schools, colleges and universities. They have in general many, sometimes difficult, constraints
that have not been taken into account here (assignment of class rooms, courses or exams of different
durations, balanced distribution of courses or exams over the days of the week). These problems are
NP-hard and usually solved by heuristics, metaheuristics or Constraint Programming techniques. Carterand Laporte [CL96] give an overview of the approaches used to solve this type of problem. In [Tri84]
the author suggests a model formulation with Mathematical Programming. Graph coloring models areused by de Werra [DW97]. Boufflet and Nègre describe in [BN96] a tabu search method for planningthe exams at the Technical University of Compiègne. In Boizumault [BDGP95] a Constraint Programming
approach for solving a timetabling problem and the problem of scheduling exams at the Institute ofApplied Mathematics of Angers may be found.
Section 14.5 presents a particular application of personnel management. In the production planningproblems studied in Chapter 8, certain aspects of resource management are often ignored or deliberately
not taken into account. The notion of personnel/human resource management in the form of working
hours transfered between production lines is nevertheless important and must not be forgotten.
The personnel planning problem in Section 14.6 has been studied by Clark and Hastings [CH77] who solve
it with a dynamic programming method. It resembles the production planning problems in Chapter 8. It
may be adapted to problems that consist of filling time slices by periods of work like contracts of limited

Timetabling and personnel planning

226

Applications of optimization with Xpress-MP

duration, shift planning, etc. It also has certain similarities with the problem of planning a fleet of vansin Chapter 10.

Timetabling and personnel planning

227

Applications of optimization with Xpress-MP

Chapter 15

Local authorities and public services
The military was the first to be interested in Operations Research, followed by industry. In the mostdeveloped countries, the public sector uses these techniques more and more to provide the best possible
service to consumers for a given budget.
The provision of drinking water will be a major problem for humanity in the 21st century. It will become
necessary to find new sources of water and to enlarge the old sewage systems and networks of pipes thatmight create unforeseen bottlenecks. In the classical maximum flow problem of Section 15.1 we show
how to calculate the maximum throughput of a water transport network. The methods may be applied
to other liquids, and also to road or telecommunications networks.
Section 15.2 is a study of closed circuit TV surveillance of streets. The problem consist of covering all thestreets with a minimum number of pivoting cameras. A political application, still quite relevant in certain
countries, is presented in Section 15.3: a party wishes to re-organize the electoral districts to maximize
its number of seats at the next elections. More seriously, this type of problem occurs when one tries toestablish a fair subdivision into geographical sectors.
The problem in Section 15.4 is gritting the streets of a village in case of ice, with a truck that drives atour of minimum total length. As opposed to the delivery of heating oil in Section 10.4, the tasks to be
executed are not placed on the nodes of the network but on its arcs. Section 15.5 deals with a location
problem for public services, in this case of income tax offices. The optimization criterion employed is theaverage distance that an inhabitant of the area has to travel to the closest income tax office. The chapter
terminates with the evaluation of the performances of four hospitals using the DEA method.

15.1

Water conveyance / water supply management
The graph in Figure 15.1 shows a water transport network. The nodes, numbered from 1 to 10, represent
the cities, the reservoirs, and the pumping stations connected by a network of pipes. The three cities
Gotham City, Metropolis, and Spider Ville are supplied from two reservoirs. The availabilities of waterfrom these reservoirs in thousands of m3 /h are 35 for reservoir 1 and 25 for reservoir 2. The capacity ofeach pipe connection is indicated in thousands of m3 /h in the graph.

35
1
Reservoir 1

15

3

20

10

15

25

6

2
Reservoir 2
22

6

10

10
15
15

22

7

4

5

12

18
8
Gotham City

7

15
9
Metropolis

10

10

20
10
Spider Ville

Figure 15.1: Water transport network

A study is undertaken to find out whether the existing network will be able to satisfy the demands of the

228

Applications of optimization with Xpress-MP

cities in ten years time, that is 18, 15, and 20 thousand m3 /h. Determine the maximum flow in the current
network. Will it be sufficient in ten years from now?

15.1.1Model formulation
The problem consists of finding a flow through the given graph that best satisfies the needs of the threecities (nodes 8, 9, and 10) taking into account the availabilities at the reservoirs 1 and 2. This problem isa classical maximum flow problem.
We start by transforming the graph by creating:• a first fictitious node called SOURCE (node 11 in Figure 15.2) connected to the reservoirs by two arcs
(11,1) and (11,2) with capacities corresponding to the availability of water from the two reservoirs(35 and 25 thousand m3 /h);• a second fictitious node called SINK (node 12 in Figure 15.2) to which the three cities are connectedby three arcs (8,12), (9,12), and (10,12) with capacities corresponding to the cities’ requirement forwater (18, 15, and 20 thousand m3 /h).
The resulting graph (Figure 15.2) is a transport network G = (NODES, ARCS, CAP, SOURCE, SINK) in which:• NODES is the set of nodes;• ARCS is the set of arcs;• CAPnm denotes the capacity of the arc a = (n, m);• SOURCE is the source (node 11);
• SINK is the sink (node 12).

1
Reservoir 1

35
11

15

3

20

25

10

6

2
Reservoir 2
22

6

10
15

5

12

8
Gotham City
18

10

15

7

4

9
Metropolis

7

12

10

15

22

15

10
Spider Ville

10

20

Figure 15.2: Water transport network after addition of source and sink

SUCCn denotes the set of immediate successors of a node n and PREDn the set of its immediate predecessors. A flow of a total throughput TotalFlow in this transport network fulfills the following constraints.
∀(n, m) ∈ ARCS : flownm ≤ CAPnm∀n 6= SOURCE, SINK :X

flownm =m∈SUCCn

TotalFlow =X(15.1.1)

flown,SINK

X

flowmn(15.1.2)m∈PREDn(15.1.3)n∈PREDSINK∀(n, m) ∈ ARCS : flownm ≥ 0(15.1.4)

The constraints (15.1.1) indicate that the flow flownm on every arc (n, m) must not exceed the capacity
CAPnm of this arc. The constraints (15.1.2) specify that the total flow arriving at any node n also has toleave this node (with the exception of the source and sink nodes). This condition is called Kirchhoff’s law.
The constraint (15.1.3) indicates that the total flow TotalFlow in the network equals the flow arriving
at the sink (it is also equal to the flow leaving the source). Finally, the non-negativity constraints forthe variables are given in (15.1.4). The problem that we need to solve corresponds to searching for the
maximum flow between the nodes SOURCE and SINK, that is, a flow maximizing TotalFlow. Hence thevery simple objective function (15.1.5):
maximize TotalFlow

Local authorities and public services229(15.1.5)

Applications of optimization with Xpress-MP

It is quite obviously possible to let non-integer flows pass through the arcs. However, in this type ofproblem, the simplex algorithm automatically finds integer solution values in the optimal solution to thelinear problem if all capacities are integer.

15.1.2

Implementation
In the Mosel implementation of this problem we represent the set of arcs as a list PIPE that defines everyarc a ∈ ARCS in the form a = (PIPEa1 , PIPEa2 ). For efficiency reasons, the arrays CAP and flow are indexedby the sequence number a of the arc, and not with the double index (n, m) indicating the two nodes
connected by the arc a as in the mathematical formulation. This indexation allows us to define thesearrays exactly with the required size.
model "J-1 Water supply"
uses "mmxprs"
declarations
ARCS: range
NODES=1..12! Set of arcs

PIPE: array(ARCS,1..2) of integer ! Definition of arcs (= pipes)
CAP: array(ARCS) of integer! Capacity of arcs
SOURCE,SINK: integer! Number of source and sink nodes
end-declarations
initializations from ’j1water.dat’
PIPE CAP SOURCE SINK
end-initializations
finalize(ARCS)
declarations
flow: array(ARCS) of mpvar
end-declarations! Flow on arcs

! Objective: total flow
TotalFlow:= sum(a in ARCS | PIPE(a,2)=SINK) flow(a)! Flow balances in nodes
forall(n in NODES | n<>SOURCE and n<>SINK)
sum(a in ARCS | PIPE(a,1)=n) flow(a) = sum(a in ARCS | PIPE(a,2)=n) flow(a)! Capacity limits
forall(a in ARCS) flow(a) <= CAP(a)! Solve the problem
maximize(TotalFlow)
end-model

15.1.3Results
17/20
31/35

1
Reservoir 1

15/22

8
Gotham City
17/18

10/10

6

0/10

5/15

5
6/6

2
Reservoir 2

7/7

4

10/10

14/15
0/12

11
21/25

7/15

3

9
Metropolis

7

12

10/10

15/15

15/22

15/15

5/10

10
Spider Ville

20/20

Figure 15.3: Maximum flow

The maximum flow that may pass through the network is 52,000 m3 per hour. Figure 15.3 shows an

Local authorities and public services

230

Applications of optimization with Xpress-MP

example how this flow may spread through the network (there are other equivalent solutions). In thisfigure, every arc is labeled with the flow passing through the arc followed by its capacity (both in m3 /h).
for instance, the flow on the arc (1,3) is 17 and its capacity is 20.
The total flow is not sufficient to satisfy all the demands in ten years time: with the given network layout
and capacities, Gotham City will only receive 17,000 m3 of the required 18,000 m3 .

15.2

CCTV surveillance
In the course of the last few months, the industrial zone of Billston has suffered from a series of break-ins
and thefts during the night. The zone is watched by security men but there are too few of them. The
town council in charge of security in this zone decides to install surveillance cameras to aid the security
men with their task. These cameras can be directed and pivot through 360◦ . By installing a camera at theintersection of several streets, it is possible to survey all adjoining streets. The map in Figure 15.4 shows
the industrial zone with the limits of the zone to be covered by closed circuit TV (CCTV) surveillance andthe 49 possible locations where to install the cameras. What is the minimum number of cameras thathave to be installed to survey all the streets of this zone and where should they be placed?

36

24
26
17

25

18

14
15

43

37

42

29
31

38

32

40

41

39
11

12
3

49

30

22
21

16

13

28

47
44

33

19
20

45
35

34
27

48

46

23
10

9

5

4

7
6
2
8

1

Figure 15.4: Industrial zone in Billston

15.2.1Model formulation
The most fastidious task in this problem is the encoding of the map of the zone, but this step is necessary
for any computerized solving of this problem. The set of possible locations of cameras (mostly street intersections) is denoted by NODES. The street network is modeled through a graph G = (NODES, STREETS),
where the set STREETS corresponds to the links (streets) between the possible locations NODES of thecameras.
We define binary variables placen that are 1 if a camera is put up at location n and 0 otherwise. The

Local authorities and public services

231

Applications of optimization with Xpress-MP

unique set of constraints (15.2.2) indicates that every street needs to be surveyed by at least one camera.
Therefore, if a street exists between the two locations n and m, we need to have a camera in n (placen = 1)
or m (placem = 1) or in both places. It is possible to cover a street by two cameras and this may pay incertain cases: in Figure 15.4 two cameras in locations 4 and 6 seem to be too much for the street but
they also cover the three cul-de-sacs leading to 5, 7, and 8. The objective function (15.2.1) minimizes thetotal number of cameras to install. Once the graph is defined, we obtain a remarkably simple 0-1 integer
program.
minimizeX

placen(15.2.1)∀(n, m) ∈ STREETS : placen + placem ≥ 1(15.2.2)∀n ∈ NODES : placen ∈ {0, 1}(15.2.3)n∈NODES

15.2.2

Implementation
The undirected graph is encoded by a symmetric adjacency matrix STREET. STREETnm = 1 if the street(n, m) exists and 0 otherwise. This matrix is read from the data file in sparse format, that is, only theentries that are 1 are defined. To reduce the amount of data in the file, only the arcs (n, m) with n < mare given. The other half of the adjacency matrix is calculated in the Mosel program.
model "J-2 Surveillance"
uses "mmxprs"
declarations
NODES=1..49
STREET: dynamic array(NODES,NODES) of integer

place: array(NODES) of mpvar
end-declarations! 1 if a street connects! two nodes, 0 otherwise! 1 if camera at node, 0 otherwise

initializations from ’j2bigbro.dat’
STREET
end-initializations
forall(n,m in NODES | exists(STREET(n,m)) and n<m )STREET(m,n):= STREET(n,m)! Objective: number of cameras to install
Total:= sum(n in NODES) place(n)! Flow balances in nodes
forall(n,m in NODES | exists(STREET(n,m)) ) place(n)+place(m) >= 1
forall(n in NODES) place(n) is_binary! Solve the problem
minimize(Total)
end-model

15.2.3Results
The optimization algorithm finds an integer solution with 24 cameras, the locations of which are marked
with circles in Figure 15.5 (there are several equivalent solutions to this problem).

Local authorities and public services

232

Applications of optimization with Xpress-MP

46
47
44

35

24
28
26
18

42

37

33
20

15

30

22

40

32

21

13
10
3
4
6
2

Figure 15.5: Location of cameras

Local authorities and public services

233

Applications of optimization with Xpress-MP

15.3

Rigging elections
In a country far away, the party of the duke Sark Mevo has finally beaten the ruling coalition headed by
the princess Reguel Tekris. Mevo wishes to consolidate his position in the capital, the fourteen quartersof which need to be grouped to electoral districts. A schematic map of the capital is given in Figure 15.6.
The quarters are numbered from 1 to 14 (bold print). The two other numbers are the forecast numberof favorable votes for Mevo and the total number of electors per quarter. All electors must vote andthe winner needs to have the absolute majority. A valid electoral district is formed by several adjacentquarters and must have between 30,000 and 100,000 voters. Two quarters that touch each other just ata point like 12 and 13 are not considered adjacent. Electoral districts consisting of a single quarter are
permitted if it has at least 50,000 voters. Nevertheless, Mevo may not decently define an electoral district
solely with quarter 10, since this contains his residence.
Determine a partitioning into five electoral districts that maximizes the number of seats for Mevo. Should
this cause any difficulties, try a partitioning into six districts. Snirp, the mathematical jester, suggests Mevo
uses Mathematical Programming...

17617500/30000

12000/30000

9000/
40000

2

8

9

10000/30000

26000/40000

15000/50000

314200/20000

5

1118000/
20000

2500/
10000

10
34000/
60000

442000/70000

1227000/60000

13

1429000/
40000

15000/40000

Figure 15.6: Map of the capital and its quarters. Legend: quarter number, supporters/electorate

15.3.1Model formulation
The problem we are concerned with here is a partitioning problem: given the set of all possible electoral
districts, we have to choose a subset such that every quarter appears in a single chosen electoral district.
Let QUARTERS be the set of all quarters of the capital, MINSINGLE the minimum size for an electoral
district consisting of a single quarter and MINPOP and MAXPOP respectively the minimum and maximum
sizes for districts formed from several quarters. The number of electors per quarter is given in POP. We
may then calculate the complete set of electoral districts using the following algorithm:• forall q in QUARTERS:– if (electorate of q ≥ MINSINGLE and q 6= 10): add q to the list of districts– forall neighbors p of q:
if electorate of (p + q) ≤ MAXPOP: call add_neighbor(p, {q})• procedure add_neighbor(toadd, set)
– add toadd to set– if electorate of set ≥ MINPOP: add set to the list of districts– forall neighbors p of toadd:
if electorate of (set ∪ {q}) ≤ MAXPOP: call add_neighbor(p, set)
If the set of neighbors of every quarter is defined in such a way that it only contains those adjacentquarters with sequence numbers larger than the quarter itself, the algorithm calculates every possible
electoral district exactly once.
In the following we shall assume that the set RDIST of possible districts (there are 48 different ones inthis example) has been saved in the form of an array DISTR. An entry DISTRdq of this array has the value

Local authorities and public services

234

Applications of optimization with Xpress-MP

1 if quarter q is contained in district d and 0 otherwise. Based on the given forecasts of favorable votes
per quarter, we can calculate the majority indicator MAJd for every district d. MAJd is 1 if the sum offavorable votes is at least 50% of the electorate of all quarters in this district (that is, Mevo has theabsolute majority), and 0 otherwise.
The formulation of this problem requires a binary variable choosed per possible district d that takes thevalue 1 if the district is chosen for the partitioning (15.3.3). The constraint (15.3.2) means that exactly
REQD districts are formed, where REQD indicates the required number of districts. The objective function(15.3.1) consists of maximizing the number of seats in the chosen partitioning.
maximize

X

MAJd · choosed(15.3.1)

d∈RDISTX

choosed = REQD

(15.3.2)∀d ∈ RDIST : choosed ∈ {0, 1}(15.3.3)d∈RDIST

Every quarter has to appear in exactly one district of the chosen partitioning. This constraint is expressedby (15.3.4).
X∀q ∈ QUARTERS :DISTRdq · choosed = 1(15.3.4)
d∈RDIST

After the corresponding data preprocessing, the mathematical model thus has a very simple form. It has
the particularity of an entirely binary coefficient matrix and right hand side (constant terms).

15.3.2

Implementation
For clarity’s sake, the Mosel implementation of this problem is split into two files separating the data
preprocessing from the model itself. In the following, we first list the code implementing the mathematical model. At the beginning of this model, after the declaration of the data arrays, we include thefile j3elect_calc.mos that generates the data in the form required for the model. Any code included
into a Mosel program using the include statement is treated as if it were printed at this place in the
program itself. In particular, the included code must not contain model "..." and end-model . The
required number of districts REQD is defined as a parameter so as to enable the user to run this modelfor different values of REQD without having to modify the code.
model "J-3 Election districts"
uses "mmxprs"
parameters
REQD = 5
end-parameters
declarations
QUARTERS = 1..14
RDIST: range! Required number of districts! Number of quarters! Set of election districts

MAJ: array(RDIST) of integer! 1 if majority of votes, 0 otherwise
DISTR: array(RDIST,QUARTERS) of integer ! 1 if quarter is in district,! 0 otherwise
end-declarations
include "j3elect_calc.mos"declarations
choose: array(RDIST) of mpvar
end-declarations! 1 if district chosen, 0 otherwise! Objective: number of votes
Votes:= sum(d in RDIST) MAJ(d)*choose(d)! Partitioning
forall(q in QUARTERS) sum(d in RDIST) DISTR(d,q)*choose(d) = 1! Desired number of districtssum(d in RDIST) choose(d) = REQD
forall(d in RDIST) choose(d) is_binary

Local authorities and public services

235

Applications of optimization with Xpress-MP

! Solve the problem
maximize(Votes)! Solution printing
if(getprobstat<>XPRS_OPT) thenwriteln("Problem is infeasible")else
writeln("Total number of votes: ", getobjval)
end-ifend-model

In this model, we test the status of the optimizer (using function getprobstat) to decide whether to try
printing a solution or not. If no optimal solution is available (XPRS_OPT), the problem may be infeasible(XPRS_INF), unbounded (XPRS_UNB) or the algorithm has not terminated (XPRS_UNF), for instance due
to a time limit.
The following Mosel procedures implement the algorithm described in Section 15.3.1. For every quarterof the capital, the data file contains the numbers of the voters POP and of the expected favorable votes
VOTES, and the set of neighboring quarters with order numbers greater than the quarter itself. From
this input, the procedure calculate_distr calculates the list DISTR of possible electoral districts. This
procedure is called at the end of this piece of code to start the whole data preprocessing algorithm. Afterthe list of possible electoral districts DISTR has been established, the majority indicator MAJ(d) for every
entry d in this list is calculated.
declarations
NUMD: integer
RN: range! Number of possible districts! Neighboring quarters

NEIGHB: array(QUARTERS,RN) of integer ! Set of neighboring quarters
POP: array(QUARTERS) of integer! Number of electors (in 1000)
VOTES: array(QUARTERS) of real
! Number of favorable votes (in 1000)MINPOP,MAXPOP,MINSINGLE: integer! Limits on electors per district
end-declarations
initializations from ’j3elect_rev.dat’
NEIGHB POP VOTES MINPOP MAXPOP MINSINGLE
end-initializations!**** Save a new entry in the list of districts ****
procedure save_distr(sQ: set of integer)
NUMD+=1
forall(q in sQ) DISTR(NUMD,q):=1
end-procedure!**** Add a neighboring quarter to the current set of quarters ****
procedure add_neighb(toadd:integer, sQ:set of integer)
declarations
nQ: set of integer
end-declarations
nQ:=sQ+{toadd}
if(sum(q in nQ) POP(q) >= MINPOP) then
! Large enough to form distr.
save_distr(nQ)
end-ifforall(p in RN | exists(NEIGHB(toadd,p)))! Try adding every neighborif(sum(q in nQ) POP(q)+POP(NEIGHB(toadd,p))<=MAXPOP) then
add_neighb(NEIGHB(toadd,p), nQ)
end-ifend-procedure!**** Calculate the list of possible districts ****
procedure calculate_distr
NUMD:=0
forall(q in QUARTERS) do
if (POP(q) >= MINSINGLE and q<>10) then
! Single quarter districts
save_distr({q})
end-ifforall(p in RN | exists(NEIGHB(q,p)))! Try adding every neighborif(POP(q)+POP(NEIGHB(q,p))<=MAXPOP) then
add_neighb(NEIGHB(q,p),{q})end-ifLocal authorities and public services

236

Applications of optimization with Xpress-MP

end-doforall(d in 1..NUMD)! Calculate majorities
MAJ(d):= if(((sum(q in QUARTERS | DISTR(d,q)>0) VOTES(q)) /(sum(q in QUARTERS | DISTR(d,q)>0) POP(q))>= 0.5), 1, 0)
finalize(RDIST)
end-procedure!**** Start the calculation of the list of possible districts ****
calculate_distr15.3.3Results
The problem has no solution for REQD = 5. Snirp should have noticed this, as the total population is
540,000, and no district may have more than 100,000 electors; so obviously we need at least 6 districts.
Mevo executes Snirp, and appoints Che Pike as Court Mathematician.
We run the Mosel program a second time with REQD = 6. For this value we find the partitioning represented in Figure 15.7: all but one district (grey shaded area) is favorable to Mevo.

7

1
6
8

2

9

5
3

11

12

13

14

10
4

Figure 15.7: Electoral districts for REQD = 6

15.4

Gritting roads
In the case of ice, all the streets of a village need to be gritted. A schematic map of the streets is given in
Figure 15.8. The nodes correspond to street intersections, the arcs to the streets that need to be gritted.
The arcs are labeled with the length of the street in meters.

150

1

140

2

165
5

160
144

170

144
194

174

148

190
109

7

180
8

128
174

218
9

200
122

6

4

100

130

230
165

100

3

185

185

233
10

135

190
141

11

110

12

Figure 15.8: Graph of the village streets

Local authorities and public services

237

Applications of optimization with Xpress-MP

The highway maintenance depot with the gritting truck is located at intersection number 1. The truck
has a sufficiently large capacity to grit all the streets during a single tour. Due to the one-way streets, it
may be forced to pass several times through the same street that has already been gritted. Determine
a tour for the gritting truck of minimum total length that passes through all streets. For bidirectional
streets, both directions need to be gritted separately.

15.4.1Model formulation
Let G = (ISEC, ARCS, LEN) be the directed graph of the street network. ISEC denotes the set of street
intersections, ARCS the set of arcs, and LENij the length LENij of arc (i, j) in meters. The tour we aresearching for corresponds to a circuit that starts and finishes at the depot node 1, passes at least once
over every arc and has a minimum total length. This classical problem is called the Chinese postman
problem because it was first solved by a Chinese mathematician, Mei-Ko Kuan [Kua62].
A Eulerian circuit in a directed graph is a circuit that visits every arc exactly once. If a Eulerian circuit
exists, then this is an optimal tour since it is equal to the sum of all arc lengths. A simple test exists to findout whether G may contain a Eulerian circuit: every node i needs to be equilibrated, that is the numberof arcs arriving at i and leaving i must be equal, see for instance Gibbons [Gib85]. This test must not be
confused with the better known case of undirected graphs in which a Eulerian circuit exists if and only ifeither 0 or 2 nodes are of odd degree.
The graph G of this problem does not have a Eulerian circuit because among others the node 1 is notequilibrated. The circuit will thus re-use streets that are already gritted. The objective is to minimize thetotal length of such unproductive stretches. Formulating a mathematical model that directly deals withthis problem is possible, but difficult. We are going to overcome this difficulty by dealing with a simpler
problem, namely the problem of transforming G into a Eulerian graph G0 by adding copies of arcs toequilibrate the nodes, whilst minimizing the total length of these copies. It then suffices to extract a
Eulerian circuit from G0 , a task for which simple algorithms exist.
The problem of equilibrating the graph at minimum cost is modeled by a fairly simple mathematicalmodel. For every arc (i, j) in ARCS, an integer variable useij is defined that indicates the number of times
the tour uses this arc (15.4.3). This number includes one pass for gritting the arc, plus possible otherunproductive passages (copies of arcs). Since the gritting of the arc is obligatory, the useij are at least
1. The constraints (15.4.2) imply that the transformed graph G0 must have equilibrated nodes. These
constraints are similar to Kirchhoff’s law used in flow problems, such as water conveyance in Section 15.2.
The objective function (15.4.1) is to minimize the total length driven.

Xminimize(15.4.1)

LENij · useij(i,j)∈ARCS

X∀i ∈ ISEC :

X

useij

(15.4.2)∀(i, j) ∈ ARCS : useij ≥ 1 ∧ usei,j ∈ IN

(15.4.3)(j,i)∈ARCS

useji =

(i,j)∈ARCS

It is in fact not necessary to specify that the variables are integer because the matrix of this model isa node-arc incidence matrix. The simplex algorithm will automatically find an integer solution. The
optimization provides the optimal length and the number of times useij an arc is used. We could stop
there if the exact composition of the tour was not part of the question to answer. We obtain the Euleriangraph G0 by adding useij − 1 copies to every arc (i, j). The first pass for gritting the street is alreadyrepresented by the existing example of the arc.
To obtain a Eulerian circuit, a procedure add_path(i,tour) is used. Starting from a node i, it consistsof using arcs that have not yet been used as far as possible. An arc that has been visited is marked to
avoid re-using it. At the end, the procedure returns the obtained tour tour. It is possible to show thattour always stops at the end in i and forms a circuit. The procedure add_path is first applied at the depot
node 1. If tour visits all the arcs of G0 , this is a Eulerian circuit. Otherwise, we increase tour by executing
add_path from a node in tour that still has free arcs to visit. The implementation of this algorithm isgiven after the Mosel implementation of the mathematical model in the following section. The following
is pseudo-code for the algorithm implemented with Mosel:• Call add_path(1,tour)• As long as the tour does not visit all arcs:– Search the first node i in tour from which non-visited arcs leave– Call add_path(i,tour)

Local authorities and public services

238

Applications of optimization with Xpress-MP• procedure add_path(node,tour)– Determine the insertion position pos– Find an unused path subtour starting from and returning to node
– Add the new path subtour to the main tour tour at position pos

15.4.2

Implementation
The mathematical model translates into the following Mosel program. As we have seen, a graph may
be encoded in two different ways: in the form of a list of arcs or as an adjacency matrix. The first
method is used in this book, for instance, for the opencast mining problem in Section 6.5, the assemblyline balancing problem in Section 7.6, and the composition of flight crews in Section 11.2. The second
matrix-based representation that is employed here also finds use in the network reliability problem in
Section 12.1 and the CCTV surveillance problem of Section 15.2.
The graph is defined by the matrix of arc lengths LEN, with LENij = 0 if arc (i, j) does not exist. The graphis given in sparse format in the data file (that is, it contains only the values of non-zero entries with thecorresponding index tuples). By defining LEN as a dynamic array, only the entries that are listed in thedata file will actually be created so that we can use the function exists on this array to test the existenceof a street (and at the same time, to enumerate the defined entries of LEN efficiently). The variables use
are created after reading the data, only for the existing entries of LEN. It is not required to repeat thefilter exists(LEN(i,j))) every time we define sums with these variables: Mosel automatically skips
the undefined entries.
model "J-4 Gritting circuit"uses "mmxprs"
declarations
ISEC = 1..12
LEN: dynamic array(ISEC,ISEC) of integer
use: dynamic array(ISEC,ISEC) of mpvar
end-declarations! Set of street intersections! Length of streets! Frequency of use of streets

initializations from ’j4grit.dat’LEN
end-initializations
forall(i,j in ISEC | exists(LEN(i,j))) create(use(i,j))! Objective: length of circuit
Length:= sum(i,j in ISEC | exists(LEN(i,j))) LEN(i,j)*use(i,j)! Balance traffic flow through intersections
forall(i in ISEC) sum(j in ISEC) use(i,j) = sum(j in ISEC) use(j,i)! Grit every street
forall(i,j in ISEC) use(i,j) >= 1
! Solve the problem
minimize(Length)
end-model

The variables useij indicate the number of times the truck passes through a street, but what we reallywant to know is the complete tour it has to drive to grit all streets. To print the result for this problem
in a useful form, we need to calculate a Eulerian circuit. This may be done by adding the following lines
to the program above. This code fragment implements the algorithm outlined in the previous section.
The procedure add_path has been turned into a function that returns the number of arcs in the new(sub)tour. We define a counter ct to control when all arcs of the Eulerian graph G0 calculated by the
optimization have been used. Its value is initially set to the result of the optimization and decreased by
the number of arcs added as subtours to the main tour. The algorithm stops when the counter is at 0.
forward function find_unused(J: array(range) of integer):integer
forward function add_path(node:integer, J: array(range) of integer):integer
ct:=round(getsol(sum(i,j in ISEC) use(i,j)))Local authorities and public services

239

Applications of optimization with Xpress-MP

declarations
TOUR: array(1..ct+1) of integer
COUNT: array(ISEC,ISEC) of integer
end-declarations! Main loop of the Eulerian circuit algorithm
forall(i,j in ISEC | exists(LEN(i,j))) COUNT(i,j):=round(getsol(use(i,j)))TOUR(1):=1
ct-=add_path(1,TOUR)
while(ct>0)
ct-=add_path(find_unused(TOUR),TOUR)
writeln("Tour: ", TOUR)!----------------------------------------------------------------! Find first node in list with free path(s)function find_unused(J: array(range) of integer):integeri:=1
returned:=1
while(J(i)>0 and i<getsize(J))
if (sum(j in ISEC) COUNT(J(i),j) > 0) thenreturned:=J(i)
break
else
i+=1
end-ifend-function!----------------------------------------------------------------! Add a subtour to the current tour
function add_path(node:integer, J: array(range) of integer):integer
declarations
NEWJ: array(1..getsize(J)) of integer
end-declarations! Insertion position
pos:=1
while(J(pos)<>node and pos<getsize(J)) pos+=1! Find a new path
cur:=node; newct:=0
while(sum(j in ISEC) COUNT(cur,j) > 0) do
forall(j in ISEC) if(COUNT(cur,j) > 0) then
COUNT(cur,j)-=1
newct+=1; NEWJ(newct):=j
cur:=jbreak
end-ifend-do! Add the new path to main journey
i:=getsize(J)-newct
while(i>pos) do
J(i+newct):=J(i)i-=1
end-doforall(j in 1..newct) J(pos+j):=NEWJ(j)returned:=newct
end-function

15.4.3Results
The optimization finds a tour of length 5990 meters. All arcs are used once except for the following thatare used twice: (3,4), (4,8), (5,1), (5,6), (6,9), (10,6), (11,7).
Figure 15.9 shows the resulting Eulerian graph G0 ; arcs that are doubled (used twice) are printed withthick lines. A Eulerian circuit in this graph is the tour 1 → 2 → 6 → 5 → 6 → 7 → 8 → 12 → 11 → 7 →
11 → 10 → 7 → 3 → 4 → 3 → 4 → 8 → 4 → 8 → 11 → 7 → 6 → 9 → 5 → 6 → 9 → 10 → 6 → 10 → 6 →2 → 3 → 2 → 5 → 1 → 5 → 1 (there are several equivalent tours).

Local authorities and public services

240

Applications of optimization with Xpress-MP

1

2

3

4

5

6

7

8

9

10

11

12

Figure 15.9: Eulerian graph for the extraction of the circuit

15.5

Location of income tax offices
The income tax administration is planning to restructure the network of income tax offices in a region.
The graph in Figure 15.10 shows the cities in the region and the major roads. The bold numbers printed
close to the cities indicate the population in thousands of inhabitants. The arcs are labeled with the
distances in kilometers (printed in italics). The income tax administration has determined that offices
should be established in three cities to provide sufficient coverage. Where should these offices be locatedto minimize the average distance per inhabitant to the closest income tax office?

15

10
15

1

2
5

24

18

15

16
8

22

20
24
6

12

13

30

9
19

11

22

12

24

12

19

4

16

25
10

18
18

3

5

11
7

12
22

19

22

19

21

12
20

Figure 15.10: Graph of towns and roads of the region

15.5.1Model formulation
We are dealing here with a classical problem called the p-median problem. Let CITIES denote the set ofcities in the region and NUMLOC the number of offices to install. We write POPc for the population ofcity c. The lengths of the shortest paths between cities are given by a distance matrix DIST shown in thefollowing table.
The distance matrix can be calculated from the graph in Figure 15.10 by an all-pairs shortest path algorithm like the Floyd-Warshall algorithm [AMO93]:• For all node pairs (c, d): initialize the distance label DISTcd with plus infinity (a sufficiently large
positive value)• For all nodes c: set DISTcc := 0
• For all arcs a = (c, d): set DISTcd to the length of the arc

Local authorities and public services

241

Applications of optimization with Xpress-MP

Table 15.1: Distance Matrix
1
2
3
4
5
6
7
8
9
10
11
12

1

2

3

4

5

6

7

8

9

10

11

12

0
15
37
55
24
60
18
33
48
40
58
67

15
0
22
40
38
52
33
48
42
55
61
61

37
22
0
18
16
30
43
28
20
58
39
39

55
40
18
0
34
12
61
46
24
62
43
34

24
38
16
34
0
36
27
12
24
49
37
43

60
52
30
12
36
0
57
42
12
50
31
22

18
33
43
61
27
57
0
15
45
22
40
61

33
48
28
46
12
42
15
0
30
37
25
46

48
42
20
24
24
12
45
30
0
38
19
19

40
55
58
62
49
50
22
37
38
0
19
40

58
61
39
43
37
31
40
25
19
19
0
21

67
61
39
34
43
22
61
46
19
40
21
0

• For all nodes b, c, d:
if DISTcd > DISTcb + DISTbd : set DISTcd := DISTcb + DISTbd
For the formulation of the problem two groups of binary variables are necessary: a variable buildc that is
1 if and only if a tax office is established in city c (15.5.1), and a variable dependcd that equals 1 if the city
c depends on the office in city d (15.5.2). The variables dependcd are required for calculating the average
distance per inhabitant and to find out which city depends on which office.
∀c ∈ CITIES : buildc ∈ {0, 1}(15.5.1)∀c, d ∈ CITIES : dependcd ∈ {0, 1}(15.5.2)

NUMLOC offices should be opened, which is expressed by the constraint (15.5.3). The constraints (15.5.4)
indicate that every city depends on a single office.

Xbuildc = NUMLOC(15.5.3)

X(15.5.4)c∈CITIES∀c ∈ CITIES :dependcd = 1

d∈CITIES

The objective function (15.5.5) to be minimized is the total distance weighted by the number of inhabitants of the cities. We need to divide the resulting value by the total population of the region to obtain
the average distance per inhabitant to the closest income tax office.
minimize

X

X

POPc · DISTcd · dependcd(15.5.5)c∈CITIES d∈CITIESIf we stop here, we risk assigning cities to tax offices that do not exist! We make this impossible bytranslating the implication buildc = 0 ⇒ dependcd = 0 into a linear constraint. From this result theconstraints (15.5.6): buildc at 0 forces dependcd to be 0.
∀c, d ∈ CITIES : dependcd ≤ buildd

15.5.2(15.5.6)

Implementation
The following Mosel program implements the mathematical model of the previous section. In theory it isnecessary to define all the variables as binaries, but one may try to solve the problem by omitting these
constraints for the dependcd variables to reduce the total number of integer variables.
model "J-5 Tax office location"
uses "mmxprs"
forward procedure calculate_dist
declarations
CITIES = 1..12! Set of cities

DIST: array(CITIES,CITIES) of integer ! Distance matrix

Local authorities and public services

242

Applications of optimization with Xpress-MP

POP: array(CITIES) of integer! Population of cities
LEN: dynamic array(CITIES,CITIES) of integer ! Road network
NUMLOC: integer! Desired number of tax officesbuild: array(CITIES) of mpvar! 1 if office in city, 0 otherwise
depend: array(CITIES,CITIES) of mpvar ! (c,d) 1 if city c depends on office! in city d, 0 otherwise
end-declarations
initializations from ’j5tax.dat’
LEN POP NUMLOC
end-initializations! Calculate the distance matrix
calculate_dist! Objective: weighted total distance
TotDist:= sum(c,d in CITIES) POP(c)*DIST(c,d)*depend(c,d)! Assign cities to offices
forall(c in CITIES) sum(d in CITIES) depend(c,d) = 1! Limit total number of officessum(c in CITIES) build(c) <= NUMLOC! Relations between dependencies and offices built
forall(c,d in CITIES) depend(c,d) <= build(d)forall(c in CITIES) build(c) is_binary! Solve the problem
minimize(TotDist)
end-model

The Floyd-Warshall algorithm for finding the shortest distance between every pair of nodes is implemented with procedure calculate_dist.
procedure calculate_dist! Initialize all distance labels with a sufficiently large value
BIGM:=sum(c,d in CITIES | exists(LEN(c,d))) LEN(c,d)
forall(c,d in CITIES) DIST(c,d):=BIGM! Set values on the diagonal to 0
forall(c in CITIES) DIST(c,c):=0! Length of existing road connections
forall(c,d in CITIES | exists(LEN(c,d))) do
DIST(c,d):=LEN(c,d)
DIST(d,c):=LEN(c,d)
end-do! Update shortest distance for every node triple
forall(b,c,d in CITIES | c<d )if DIST(c,d) > DIST(c,b)+DIST(b,d) then
DIST(c,d):= DIST(c,b)+DIST(b,d)
DIST(d,c):= DIST(c,b)+DIST(b,d)
end-ifend-procedure

15.5.3Results
Without specifying that the variables dependcd are binary, luckily the optimizer finds a solution with all
variables taking integer values and a total weighted distance of 2438. Since the region has a total of
185,000 inhabitants, the average distance per inhabitant is 2438/185 ≈ 13.178 km. The three offices are
established at nodes 1, 6, and 11. The first serves cities 1,2,5,7, the office in 6 cities 3,4,6,9, and the officein 11 cities 8,10,11,12.

Local authorities and public services

243

Applications of optimization with Xpress-MP15.6

Efficiency of hospitals
The administration of the hospitals in Paris decides to measure the efficiency of the surgery departments
in four major hospitals with a desire to improve the service to the public. To keep this study anonymous,the hospitals are named H1 to H4. The method suggested to measure the efficiency is DEA (Data Envelopment Analysis). This method compares the performance of a fictitious hospital with the performancesof the four hospitals.
Three initial indicators (resources) are taken into account: the number of non-medical personnel, thegeneral expenses, and the available number of beds. In addition, four final indicators (services) are
analyzed: the number of hospital admissions per day, the number of consultations in the outpatients’clinic, the number of nurses on duty every day, and the number of interns and doctors on duty every day.
The corresponding data have been analyzed over a period of two years and the numbers representing a
day of average activity in every hospital are given in the following two tables.
Table 15.2: Resource indicators
Non-medical personnel
General expenses (in kB
C)
Number of beds

H1

H2

H3

H4

90
38.89
34

87
109.48
33

51
40.43
20

66
48.41
33

Table 15.3: Service indicators
Admissions
Consultations
Interns and doctors
Nurses on duty

H1

H2

H3

H4

30.12
13.54
13
79

18.54
14.45
7
55

20.88
8.52
8
47

10.42
17.74
26
50

Justify through the DEA method how hospital H2 is performing compared to the others.

15.6.1Model formulation
15.6.1.1 General idea of the DEA method
To use the DEA method and measure the efficiency of a hospital, we need to develop a mathematicalmodel for every hospital that is to be evaluated. Hospital H2 is used here as an example, but the procedure for measuring the performance of the other hospitals would be similar.
By using Mathematical Programming we are going to construct a fictitious hospital based on the datafrom the four hospitals. The service indicators of the fictitious hospital will be weighted sums of theservice indicators of the hospitals H1 to H4. Similarly, the resource indicators will be weighted sums of
the resource indicators of the four hospitals, always using the same coefficients.
The service indicators of the fictitious hospital must be larger than those of the hospital H2. If they aresmaller, then the fictitious hospital requires less resources than H2 for a service that is at least equivalent.
In other words, hospital H2 is less performing than the fictitious hospital and hence less performing thanthe other hospitals.

15.6.1.2 Modeling our problem
Let HOSP be the set of hospitals in this study and h0 the hospital the performance of which we wish tomeasure. Let coefh be the DEA coefficient (decision variable) associated with hospital h. The DEA method
imposes a requirement that the sum of these coefficients is equal to 1 (15.6.1).

X

coeffh = 1(15.6.1)h∈HOSP

The constants INDSERVsh represent the service indicator s of every hospital h (given in Table 15.3). The
index s takes its value in the set SERV of service indicator types. Similarly, INDRESrh represents the indicator for resource r of hospital h (data given in Table 15.2), where r ranges over the set RES of resource
indicator types. To simplify the model formulation, we introduce variables fservs and fresr for the sums

Local authorities and public services

244

Applications of optimization with Xpress-MP

weighted by the coefficients coefh for every service and resource indicator type, constraints (15.6.2) and(15.6.3).
∀s ∈ SERV :XINDSERVsh · coefh = fservs(15.6.2)h∈HOSP∀r ∈ RES :XINDRESrh · coefh = fresr(15.6.3)h∈HOSP

The variables fservs (service indicators of the fictitious hospital) must be larger than the service indicatorsof the hospital h0 for which we wish to evaluate the performance. The constraints (15.6.4) establish these
relations for all service types in SERV.
∀s ∈ SERV : fservs ≥ INDSERVsh0(15.6.4)

The weighted sum of the resource indicators fresr (that is, the resource indicators of the fictitious hospital)
will not be compared directly with the value of the resource indicators of hospital h0 , but with a fraction
eff of these indicators. The constraints (15.6.5) link the coefficients coefh and this new variable eff. The
sense of the inequality has to be “less than or equal to” since we are trying to show that the hospital
h0 uses less resources than the fictitious hospital. To give a numerical example, the number of beds in
hospital h0 is 33. In this case 33 · eff corresponds to the number of beds in the fictitious hospital. If eff = 1
then the number of beds available in the fictitious hospital is identical to that of hospital h0 . If eff islarger than 1, then the fictitious hospital uses more beds than the hospital h0 and in the opposite case,the hospital h0 is performing worse than the fictitious hospital for this resource.
∀r ∈ RES : fresr ≤ INDRESrh0 · eff(15.6.5)

If a solution exists with eff < 1, this means that the fictitious hospital needs less resources than thehospital h. The objective function (15.6.6) for the DEA method minimizes the value of eff, that is, theresources required by the fictitious hospital. To complete the mathematical model, we need to add thenon-negativity constraints (15.6.7) to (15.6.10) for all variables.

15.6.2minimize eff(15.6.6)∀h ∈ HOSP : coefh ≥ 0(15.6.7)

eff ≥ 0(15.6.8)∀s ∈ SERV : fservs ≥ 0(15.6.9)∀r ∈ RES : fresr ≥ 0(15.6.10)

Implementation
The following Mosel implementation of the mathematical model solves the problem for every hospitalin the set HOSP. The constraint on the sum of the DEA coefficients (15.6.1) and the relations defining
the service and resource indicators of the fictitious hospital (constraints (15.6.2) and (15.6.3)) are statedonly once. The definition of all constraints referring to a specific hospital (namely constraints (15.6.4) and15.6.5) is replaced in every new execution of the loop by the corresponding constraints for the hospitalthat is to be evaluated. To be able to replace (and hence delete) a constraint in the problem held in Mosel,
it needs to be named as shown in the following program for LimServ and LimRes. (The declaration of
constraints is optional, but it is recommended for efficiency reasons if the array sizes are known like inthis example).
model "J-6 Hospital efficiency"
uses "mmxprs"
declarations
HOSP = 1..4
SERV: set of string
RES: set of string
INDSERV: array(SERV,HOSP) of real
INDRES: array(RES,HOSP) of real
end-declarations! Set of hospitals! Service indicators! Resource indicators! Service indicator values! Resource indicator values

initializations from ’j6hospit.dat’
INDSERV INDRES
end-initializations

Local authorities and public services

245

Applications of optimization with Xpress-MP

finalize(SERV); finalize(RES)
declarations
eff: mpvar
coef: array(HOSP) of mpvar
fserv: array(SERV) of mpvar
fres: array(RES) of mpvar
LimServ: array(SERV) of linctr
LimRes: array(RES) of linctr
end-declarations! Efficiency value! Coefficients for DEA method! Service indicator of fict. hospital! Resource indicator of fict. hospit.
! Hospital-specific service constr.
! Hospital-specific resource constr.

! DEA coefficientssum(h in HOSP) coef(h) = 1! Relations between service and resource indicators
forall(s in SERV) fserv(s) = sum(h in HOSP) INDSERV(s,h)*coef(h)forall(r in RES) fres(r) = sum(h in HOSP) INDRES(r,h)*coef(h)! Solve the problem for every hospital
forall(h in HOSP) do! Limits on services and resources for the hospital currently looked atforall(s in SERV) LimServ(s):= fserv(s) >= INDSERV(s,h)
forall(r in RES) LimRes(r):= fres(r) <= INDRES(r,h)*eff! Minimize efficiency index
minimize(eff)writeln("Evaluation of hospital ", h, ": ", getobjval)
end-doend-model

15.6.3Results
The program finds efficiency values 1 for the hospitals 1, 3, and 4, so these hospitals have the same
performance as the fictitious hospital. For hospital 2, the algorithm returns a value of 0.921. This meansthat the fictitious hospital reaches the same level of services with only approximately 92% of the resources
used by hospital 2. In other words, hospital 2 performs less well than the other hospitals in the study. For
every evaluation run, the coefficients coefh represent the proportions of the four hospitals that form thefictitious hospital.

15.7

References and further material
The mathematical model in Section 15.1 allows us to deal with large maximum flow problems. However,even more efficient graph algorithms exist. Historically the first one was suggested by Ford and Fulkerson[FF67]. Its complexity is O(nm2 ), where n denotes the number of nodes in the graph and m the numberof arcs. An even faster algorithm in O(n2 m) is due to Ahuja and Orlin [AMO93]. In the latter reference
an improved version in O(mn · log U) is also described, where U denotes the maximum value of the arc
capacities.
The problem in Section 15.2 is a problem of covering the edges by the nodes, node cover or vertex coverin graph theory. This problem is NP-hard [CLR90]. Heuristics are described by Papadimitriou and Steiglitz[PS98].
Partitioning problems are defined over a set S of m elements and a set U of n pre-selected subsets of S.
They consist of choosing in U the subsets forming a partitioning of S to optimize the total cost of theselection. Every subset may be encoded as a binary vector and from these vectors results a binary matrix
A (m × n). The partitioning constraints may then be written as A · x = s, s denoting a vector formed bym entries of 1. Covering problems are quite similar, but an element of S may belong to several chosen
subsets: the constraints are therefore written as A·x ≥ s. These problems frequently occur for subdividing
geographical regions or for the coverage of zones by services.
Besides the partitioning problem in Section 15.3, this book contains two other covering problems: thecutting of sheet metal in Chapter 9 and the location of telephone transmitters in Chapter 12. Covering problems are easier, since they always have a trivial solution, namely the one using all the available
subsets. Partitioning problems may be infeasible like the subdivision into five electoral districts. In bothproblem types, the simplex algorithm finds a majority of the variables with integer values in the optimum which facilitates the subsequent tree search for the remaining fractional variables, see for instance

Local authorities and public services

246

Applications of optimization with Xpress-MP

Beasley for the covering problem [Bea87] and Fisher [FK90] for both types of problem.
The book by Syslo describes heuristics and exact tree search methods for both problem types, with Pascal
source code [SDK83]. Very efficient heuristics guided by the Linear Programming relaxations have recently
been suggested for very large size covering problems [CFT99].
The Chinese Postman problem seen in Section 15.4 for gritting the streets of a village also exists forundirected graphs. Edmonds and Johnson give efficient polynomial algorithms for the directed andundirected case [EJ73]. Pascal source code for the extraction of Eulerian circuits and the directed Chinese
Postman are provided in the book by Prins [Pri94a]. The Chinese Postman problem becomes a Rural
Postman problem if only a subset of the arcs needs to be visited.
If quantities are added that need to be distributed over the arcs and if trucks of limited capacity areavailable and several tours are necessary we obtain a Capacitated Arc Routing Problem (CARP). The rural
postman and the CARP are NP-hard problems. Numerical examples and heuristics are given in the book by
Evans and Minieka [EM92]. Hertz et al. describe local search procedures for the rural postman [HLNH99]
and a tabu search methods for the CARP [HLM00].
Location problems like the placement of income tax offices (Section 15.5) form a rich class of combinatorial problems to which the book by Daskin is entirely dedicated [Das95]. The problem of p-centers is
another classical one that differs from the p-median problem in its objective that consists of minimizing
the maximum distance between an office and a city. This book contains other problems of this type,
like the depot location one in Chapter 10 and the choice of locations for mobile phone transmitters in
Chapter 12.
The DEA (Data Envelopment Analysis) method from Section 15.6 is widely used for comparing the performances related to several almost identical environments. The first applications were indeed to measure
the efficiency of a hospital. Other applications and examples for this methods may be found in Sherman[She84] and Lewin et al. [LM81]. A recent book by Cooper et al. gives a very complete overview of the
DEA method and its numerous applications [CST99].

Local authorities and public services

247

Applications of optimization with Xpress-MP

Bibliography[AA95]

E. D. Andersen and K. D. Andersen. Presolving in Linear Programming. Mathematical Programming,
71(2):221–245, 1995.

[AC91]

D. Applegate and W. Cook. A Computational Study of the Job-Shop Scheduling Problem. ORSA Journal on
Computing, 3(2):149–156, 1991.

[AD93]

R. W. Ashford and R. C. Daniel. Mixed-Integer Programming in Production Scheduling: A Case Study. In
T. Ciriani and R. C. Leachman, editors, Optimization in Industry, pages 231–239, New York, 1993. John Wiley& Sons.

[AMO93]R. K.. Ahuja, T. L. Magnanti, and J. B. Orlin. Network Flows. Theory, Algorithms and Applications. Prentice
Hall, Englewood Cliffs, NJ, 1993.

[Arb93]

A. Arbel. Exploring Interior-Point Linear Programming: Algorithms and Software. The MIT Press, 1993.

[ARVK89] I. Adler, M. Resende, G. Veiga, and N. Karmarkar. An Implementation of Karmarkar’s Algorithm for Linear
Programming. Mathematical Programming, 44:297–335, 1989.
[Baa88]

S. Baase. Computer Algorithms. Addison-Wesley, 1988.

[BC96]

J. E. Beasley and B. Cao. A Tree Search Algorithm for the Crew cheduling Problem. European Journal ofOperational Research, 94:517–526, 1996.

[BDB95]

H. Beringer and B. De Backer. Combinatorial Problem Solving in Constraint Logic Programming with Cooperating Solvers. In C. Beierle and L. Plümer, editors, Logic Programming: Formal Methods and Practical
Applications, pages 245–272, Amsterdam, 1995. Elsevier Science B. V./North-Holland.

[BDGP95] P. Boizumault, Y Delon, C. Guéret, and L. Péridy. Résolution de problèmes en Programmation Logique avec
Contraintes. Revue d’Intelligence Artificielle, 9(3):383–406, 1995.
[Bea59]E. M. L. Beale. On Quadratic Programming. Naval. Res. logist. Q., 6:227–243, 1959.

[Bea87]J. E. Beasley. An Algorithm for Set Covering Problems. European Journal of Operational Research, 31:85–93,
1987.

[BESW93] J. Błazewicz, K. Ecker, G. Schmidt, and J. Wȩglarz. Scheduling in Computer and Manufacturing Systems.
Springer, Berlin , Heidelberg, 1993.
[BF76]

E. M. L. Beale and J. J. H. Forrest. Global Optimization Using Special Ordered Sets. Mathematical Programming, 10(1):52–69, 1976.

[BHJS95]

C. Barnhart, C. A. Hane, E. L. Johnson, and G. Sigismondi. A Column Generation and Partitioning Approachfor Multi-commodity Flow Problems. Telecommunication Systems, 3:239–258, 1995.

[BJ90]

M. S. Bazaraa and J. J. Jarvis. Linear Programming and Network Flows. Wiley, 1990.

[BJN+ 98]

C. Barnhart, E. L. Johnson, G. L. Nemhauser, M. W. P. Savelsbergh, and P. H. Vance. Branch-and-Price: Column
Generation for Solving Huge Integer Programs. Operations Research, 46(3):316–329, 1998.

[BK98]

A. Bockmayr and T. Kasper. Branch and Infer: A Unifying Framework for Integer and Finite Domain Constraint Programming. INFORMS Journal on Computing, 10(3):287–300, 1998.

[BKSA00]

J. E. Beasley, M. Krishnamoorthy, Y. M. Sharaiha, and D. Acramson. Scheduling Aircraft Landings – the Static
Case. Transportation Science, 34:180–197, 2000.

[BL83]

E. Balas and P. R. Landweer. Traffic Assignment in Communication Satellites. Operation Research Letters,
2(4):141–147, 1983.

[BMR94]

L. Bianco, A. Mingozzi, and S. Ricciardelli. A Set Partitioning Approach to the Multiple Depot Vehicle
Scheduling Problem. Optimization Methods and Software, 3:163–194, 1994.

[BN96]J. P. Boufflet and S. Nègre. Three Methods Used to Solve an Examination Timetabling Problem. In E.K.
Burke and P. Ross, editors, Practice and Theory of Automated Timetabling, pages 327–344. Springer-Verlag
Lecture Notes in Computer Science, 1996.

[Bow56]

E. H. Bowman. Production Scheduling by the Transportation Method of Linear Programming. Operations
Research, 4(1):100–103, 1956.

[BS87]

E. S. Buffa and R. K. Sarin. Modern Production/Operations Management. Wiley, 1987.

[BT70]

E. M. L. Beale and J. A. Tomlin. Special Facilities in a General Mathematical Programming System for
Nonconvex Problems Using Ordered Sets of Variables. In J. Lawrence, editor, Operational Research 69,
pages 447–454. Tavistock Publishing, London, 1970.

[BT79]

E. S. Buffa and W. Taubert. Production-Inventory Systems: Planning and control (3rd Ed.). Homewood III,R.D. Irwin, 1979.

248

Applications of optimization with Xpress-MP

[CA99]C. Chu and J. Antonio. Approximation Algorithms to Solve Real-Life Multicriteria Cutting Stock Problems.
Operations Research, 47(4):495–508, 1999.

[Cam96]

J. F. Campbell. Hub Location and the p-hub Median Problem. Operations Research, 44(6):923–935, 1996.

[Can77]

D. Candea. Issues of Hierarchical Planning in Multistage Production Systems. Technical Report 134, Operations Research Center, MIT, Cambridge, Mass., USA, 1977.

[CC88]

J. Carlier and P. Chrétienne. Problèmes d’ordonnancement : modélisation / complexité / algorithmes. Masson, 1988.

[CCE+ 93]

J. Carlier, Chrétienne, J. Erschler, C. Hanen, P. Lopez, A. Munier, E. Pinson, M.-C. Portmann, C. Prins, C. Proust,and P. Villon. Les problèmes d’ordonnancement. RAIRO-Recherche Opérationnelle, 27(1):77–150, 1993.

[CD99]

G. Cornuejols and M. Dawande. A Class of Hard Small 0-1 Programs. INFORMS Journal on Computing,11(2):205–210, 1999.

[CFT99]A. Caprara, M. Fischetti, and P. Toth. A Heuristic Method for the Set Covering Problem. Operations Research,
47(5):730–743, 1999.

[CGJ96]E. G. Coffman, M. R. Garey, and D. S. Johnson. Approximation Algorithms for Bin-packing: a Survey.
In D. Hochbaum, editor, Approximation Algorithms for NP-Hard Problems, pages 46–93. PWS Publishing,
Boston, 1996.

[CH77]J. A. Clark and N. A. J. Hastings. Decision Networks. Operations Research Quaterly, 20:51–68, 1977.

[CH02]

Y. Colombani and S. Heipcke. Mosel: An Extensible Environment for Modeling and Programming Solutions.
In N. Jussien and F. Laburthe, editors, Proceedings of CP-AI-OR’02, pages 277–290, Le Croisic, March 2002.

[Chv83]

V. Chvàtal. Linear Programming. W.H. Freeman, 1983.

[CHW98]

P. Chan, K. Heus, and G. Weil. Nurse Scheduling with Global Constraints in CHIP: GYMNASTE. In Proceedingsof the Fourth International Conference on the Practical Application of Constraint Technology, PACT98,
pages 157–169, London, UK, March 1998. The Practical Application Company.

[CL96]

M. W. Carter and G. Laporte. Recent Development in Practical Examination Timetabling. In E.K. Burke and
P. Ross, editors, Practice and Theory of Automated Timetabling, pages 3–21. Springer-Verlag Lecture Notes
in Computer Science, 1996.

[CLR90]T. H. Cormen, C. L. Leiserson, and R. L. Rivest. Introduction to Algorithms. MIT Press and McGraw-Hill, New
York, 1990.

[CMT79a] N. Christofides, A. Mingozzi, and P. Toth. Loading Problems. In N. Christofides and al., editors, Combinatorial Optimization, pages 339–369. Wiley, 1979.
[CMT79b] N. Christofides, A. Mingozzi, and P. Toth. The Vehicle Routing Problem. In N. Christofides and al., editors,
Combinatorial Optimization, pages 315–338. Wiley, 1979.
[CST99]W. W. Cooper, L. M. Seiford, and K. Tone. Data Envelopment Analysis. Kluwer, 1999.

[CW64]

G. Clarke and J. W. Wright. Xheduling of Vehicles from a Central Depot to a Number of Delivery Points.
Operations Research, 12(4):568–581, 1964.

[Dan63]G. B. Dantzig. Linear Programming and Extensions. Princeton University Press, Princeton, NJ, 1963.

[Das95]

M. S. Daskin. Network and Discrete Location. Wiley, 1995.

[DDS92]

M. Desrochers, J. Desrosiers, and M. Solomon. A New Optimization Algorithm for the Vehicle Routing
Problem with Time Windows. Operations Research, 40(2):342–354, 1992.

[DE73]

E. M. Dar-El. MALB, a Heuristic Technique for Balancing Large Scale Single-Model Assembly Lines. AIIE
Transactions, 5(4):–, 1973.

[DK99]

A. Dutta and P. Kubat. Design of Partially Survivable Networks for Cellular Telecommunications Systems.
European Journal of Operational Research, 118(1):52–64, 1999.

[DPL94]

S. Dauzère-Pérès and J. B. Lasserre. An Integrated Approach in Production Planning and Schedulin. In
Lecture Notes in Economics and Mathematical Systems 411. Springer-Verlag, 1994.

[DW97]

D. De Werra. The Combinatorics of Timetabling. European Journal of Operational Research, 96(3):504–513,
1997.

[DZ78]

U. Derigs and U. Zimmerman. An Augmenting Path Method for Solving Linear Bottleneck Assignment
Problems. Computing, 19:285–295, 1978.

[EJ73]

J. Edmonds and E. L. Johnson. Matching, Euler Tours, and the Chinese Postman. Mathematical Programming, 5:88–124, 1973.

[EM92]

J. E. Evans and E. Minieka. Optimization Algorithms for Networks and Graphs. Marcel Dekker Inc., 1992.

[Erl78]

D. Erlenkotter. A Dual-Based Procedure for Uncapacited Facility Location. Operations Research, 26:992–
1009, 1978.

[Eva64]

H. F. Evart. Introduction to PERT. Allyn & Bacon, 1964.

[FF67]

L. Ford and D. Fulkerson. Flots dans les graphes. Gauthier-Villars, Paris, 1967.

[FJVW86] M. L. Fisher, R. Jaikumar, and L. N. Van Wassenhove. A Multiplier Adjustment Method for the Generalized
Assignment Problem. Management Science, 39(2):1095–1103, 1986.
[FK90]

M. L. Fisher and P. Kedia. Optimal Solutions of Set Covering/Partitioning Problems Using Dual Heuristics.
Management Science, 36:674–688, 1990.

[Fle87]

R. Fletcher. Practical Methods of Optimization (2nd ed.). John Wiley, Chichester UK, 1987.

Bibliography

249

Applications of optimization with Xpress-MP

[Fre82]

S. French. Sequencing and Scheduling. An Introduction to the Mathematics of the Job-Shop. John Wileyand Sons, New York, 1982.

[Gar63]L. L. Garver. Power Scheduling by Integer Programming. IEEE Transactions on Power Apparatus and Systems,
81:730–735, 1963.

[Gar00]D. T. Gardner. Efficient Formulation of Electric Utility Resource Planning Models. Journal of the Operational
Society, 51(2):231–236, 2000.

[GG61]

P. C. Gilmore and R. E. Gomory. A Linear Programming Approach to the Cutting Stock Problem. Part I.
Operations Research, 9:849–859, 1961.

[GG63]

P. C. Gilmore and R. E. Gomory. A Linear Programming Approach to the Cutting Stock Problem. Part II.
Operations Research, 11:863–888, 1963.

[GH91]

M. Grötschel and O. Holland. Solution of Large-Scale Symmetric Traveling Salesman Problems. Mathematical Programming, 51:141–202, 1991.

[GHL94]

M. Gendreau, A. Hertz, and G. Laporte. A Tabu Search Algorithm for the Vehicle Routing Problem. Management Science, 40:1276–1290, 1994.

[Gib85]

A. Gibbons. Algorithmic Graph Theory. Cambridge University Press, Cambridge, New York, 1985.

[GJ79]

M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of NP-completeness.
W.H. Freeman & Company, San Francisco, 1979.

[Gle80]J. J. Glen. A Parametric Programming Method for Beef Cattle Ration Formulation. Journal of the Operational Research Society, 31:689–690, 1980.

[GM90]

M. Gondran and M. Minous. Graphes et algorithmes (2nd Ed.). Eyrolles, 1990.

[GP98]

C. Guéret and C. Prins. Classical and New Heuristics for the Open-Shop Scheduling Problem: a Computational Evaluation. European Journal of Operational Research, 107(2):306–314, 1998.

[GP01]

Gunluk and Y. Pochet. Mixing mixed-integer inequalities. Mathematical Programming, 90:429–457, 2001.

[Gra69]R. L. Graham. Bounds on Multiprocessing Timing Anomalies. Siam Journal on Applied Mathematics, 17:416–
429, 1969.

[GS76]

T. Gonzalez and S. Sahni. Open-Shop Scheduling to Minimize Finish Time. Journal of the ACM, 23:665–679,
1976.

[GS87]

A. Gersht and A. Shulman. A New Algorithm for the Solution of the Minimum Cost Multicommodity Flow
Problem. In IEEE 26th Conference on Decision and Control (CDC’87), pages 748–758, Los Angeles, December
1987.

[GTdW93] F. Glover, E. Taillard, and D. de Werra. A User’s Guide to Tabu Search. Annals of Operations Research,
41(1–4):3–28, 1993.
[HC84]A. C. Hax and D. Candea. Production and Inventory Management. Prentice Hall, 1984.

[HC99]S. Heipcke and Y. Colombani. Julian’s Problem. Work Assignment at a Disabled Care Centre. ORInsight,
12(1):16–20, 1999.

[Hei99a]

S. Heipcke. Combined Modelling and Problem Solving in Constraint Programming and Mathematical Programming. PhD thesis, University of Buckingham, 1999.

[Hei99b]

S. Heipcke. Comparing Constraint Programming and Mathematical Programming Approaches to Discrete
Optimisation. The Change Problem. Journal of the Operational Research Society, 50(6):581–595, 1999.

[HH60]

F. Hanssmann and S. W. Hess. A Linear Programming Approach to Production and Employment Scheduling.
Management Technology, 1:–, 1960.

[Hit41]F. L. Hitchcock. The Distribution of a Product from Several Sources to Numerous Localities. J. Math. Phys.,20(2):224–230, 1941.

[HK70]

M. Held and R. Karp. The Traveling Salesman Problem and Minimum Spanning Trees. Operations Research,
18:1138–1162, 1970.

[HLM00]

A. Hertz, G. Laporte, and M. Mittaz. A Tabu Search Heuristic for the Capacited Arc Routing Problem.
Operations Research, 48(1):129–135, 2000.

[HLNH99] A. Hertz, G. Laporte, and P. Nanchen Hugo. Improvement Procedures for the Undirected Rural Postman
Problem. INFORMS Journal on Computing, 11(1):53–62, 1999.
[HW95]J. L. Ho and J. S. Wong. Makespan Minimization for m Parallel Identical Processors. Naval Research Logistics,
42(6):935–948, 1995.

[HZ96]

M. Hifi and V. Zissimopoulos. A Recursive Exact Algorithm for Weighted Two-Dimensional Cutting. European Journal of Operational Research, 91:553–564, 1996.

[Jak96]

S. Jakobs. On Genetic Algorithms for the Packing of Polygons. European Journal of Operational Research,
88:165–181, 1996.

[JM74]

L. A. Johnson and D. C. Montgomery. Operations Research in Production Planning, Scheduling, and Inventory Control. Wiley, New York, 1974.

[JNS00]E. L. Johnson, G. L. Nemhauser, and M. W. P. Savelsbergh. Progress in Linear Programming Based Algorithms
for Integer Programming: An Exposition. INFORMS Journal on Computing, 12(1):2–23, 2000.

[JSV98]B. Jaumard, F. Semet, and T. Vovor. A Generalized Linear Programming Model for Nurse Scheduling. European Journal of Operational Research, 107(1):1–18, 1998.

Bibliography

250

Applications of optimization with Xpress-MP

[Kao79]

E. P. C. Kao. A Multi-Product Dynamic Lot-Size Problem with Individual and Joint Set-up Costs. Operations
Research, 27(2):279–289, 1979.

[Kar84]

N. Karmarkar. A New Polynomial Time Algorithm for Linear Programming. Combinatorica, 4:373–395,
1984.

[Kas98]R. G. Kasilingam. Logistics and Transportation. Kluwer Academic Publishers, 1998.

[KGV83]

S. Kirkpatrick, C. D. Gelatt, and M. P. Vecchi. Optimization by Simulated Annealing. Science, 220:671–680,
1983.

[Kua62]M. K. Kuan. Graphic Programming Using Odd and Even Points. Chinese Mathematics, 1:273–277, 1962.

[LD60]A. H. Land and A. G. Doig. An Automatic Method for Solving Discrete Programming Problems. Econometrica, 28:497–520, 1960.

[LDN84]

G. Laporte, M. Desrochers, and Y. Nobert. Two Exact Algorithms for the Distance-Constrained Vehicle
Routing Problem. Networks, 14:161–172, 1984.

[LM81]A. Y. Lewin and R. C. Morey. Measuring the Relative Efficiency and Output Potential of Public Sector
Organizations: an Application of Data Envelopment Analysis. International Journal of Policy Analysis andInformation Systems, 5(4):267–285, 1981.

[LMV99]A. Lodi, S. Martello, and D. Vigo. Heuristics and Metaheuristics for a class of two-Dimensional Bin Packing
Problems. INFORMS Journal on Computing, 11(4):345–357, 1999.

[LQ92]

J. B. Lasserre and M. Queyranne. Generic Scheduling Polyedra and a new Mixed-Integer Formulation for aSingle Machine Scheduling. In Second IPCO Conference, Pittsburg, 1992. Carnegie-Mellon University.

[LR99]

P. Lopez and F. Roubellat. Ordonnancement. Economica, 1999.

[LS98]J. T. Linderoth and M. W. P. Savelsbergh. A Computational Study of Search Strategies for Mixed Integer
Programming. INFORMS Journal on Computing (to appear), 1998.

[LT71]

L. S. Lasdon and R. C. Terjung. An Efficient Algorithm for Multi-Item Scheduling. Operations Research,
19(4):946–969, 1971.

[Lue84]D. G. Luenberger. Linear and Non-Linear Programming. Addison-Wesley, 1984.

[Man86]

A. S. Manne. GAMS/MINOS: Three Examples. Technical report, Department of Operations Research, Stanford University, 1986.

[MB86]

M. Minoux and G. Bartnik. Graphes, algorithmes, logiciels. Dunod, 1986.

[McC69]W. H. S. McColl. Management and Operations in an Oil Company. Operations Research Quaterly, 20:64–65,
1969.

[Min75]

M. Minoux. Résolution des problèmes de multiflots en nombres entiers dans les grands réseaux. RAIRO Recherche Opérationnelle, 3:–, 1975.

[MS98]

K. Marriott and P. J. Stuckey. Programming with Constraints. MIT Press, Cambridge MA, 1998.

[MT79]

S. Martello and P. Toth. The 0-1 Knapsack Problem. In N. Christofides and al., editors, Combinatorial
Optimization. Wiley, 1979.

[MT90]

S. Martello and P. Toth. Knapsack Problems: Algorithms and Computer Implementations. John Wiley, New
York, 1990.

[MW01]

H. Marchand and L. Wolsey. Aggregation and mixed-integer rounding to solve MIPs. Operations Research,
49:363–371, 2001.

[NEH83]

M. Nawaz, E.E. Enscore, and I. Ham. A Heuristic Algorithm for the m-Machine, n-Job Flow-Shop Sequencing
Problem. Omega, 11(1):91–95, 1983.

[Orl75]J. A. Orlicky. Material Requirement Planning: the New Way of Life in Production and Inventory Management. McGraw-Hill, New York, 1975.

[PFTVed]W. H. Press, B. P. Flannery, S. A. Teukolsky, and W. T. Vetterling. Numerical Recipes in C – The Art of Scientific
Computing. Cambridge University Press, Cambridge, 1992 (2nd ed.).

[Pin95a]

M. Pinedo. Scheduling: Theory, Algorithms and Systems. Prentice Hall, 1995.

[Pin95b]

E. Pinson. The Job Shop Scheduling Problem: A Concise Survey and Some Recent Developments. In Ph.
Chrétienne, E. G. Coffman, J. K. Lenstra, and Z. Liu, editors, Scheduling Theory and its Applications, pages
277–294, Chichester, New York, 1995. John Wiley.

[Pri94a]C. Prins. Algorithmes de graphes avec programmes en Pascal. Eyrolles, 1994.

[Pri94b]

C. Prins. An Overview of Scheduling Problems Arising in Satellite Communications. J. Opl. Res. Soc.,
45(6):611–623, 1994.

[PS98]

C. H. Papadimitriou and K. Steiglitz. Combinatorial Optimization. Dover, 1998.

[Roo97]

K. Roos. Theory and Algorithms for Linear Optimization. An Interior Point Approach. John Wiley, Chichester, New York, 1997.

[Ros85]

Roseaux. Exercices et problèmes résolus de Recherche Opérationnelle – Tome 3. Masson, 1985.

[RWH99]R. Rodošek, M. G. Wallace, and M. T. Hajian. A New Approach to Integrating Mixed Integer Programmingand CLP. Annals of Operations Research, 86:63–87, 1999.

[Sav94]M. W. P. Savelsbergh. Preprocessing and Probing Techniques for Mixed Integer Programming Problems.
ORSA Journal on Computing, 6(4):445–454, 1994.

Bibliography

251

Applications of optimization with Xpress-MP

[SC81]D. W. Sutton and P. A. Coates. On-Line Mixture Calculation System for Stainless Steel Production by BSC
Stainless: the Least Through Cost Mix System (LTCM). Journal of the Operational Research Society, 32:165–
169, 1981.

[Sch86]

A. Schrijver. Theory of Linear and Integer Programming. Wiley, 1986.

[Sch97]

L. Schrage. Optimization Modeling with LINDO (5th ed.). Duxbury, 1997.

[Sch99]A. Scholl. Balancing and Sequencing of Assembly Lines. Physica-Verlag, 1999.

[SDK83]M. M. Syslo, N. Deo, and J. S. Kowalik. Discrete Optimization Algorithms with Pascal Programs. Prentice
Hall, 1983.

[Sev98]

M. Sevaux. Étude de deux problèmes d’optimisation en planification et ordonnancement. PhD thesis, Thèse
de l’université Pierre-et-Marie-Curie, Paris VI, 1998.

[She84]

H. D. Sherman. Hospital Efficiency Measurement and Evaluation. Medical Care, 22(10):922–938, 1984.

[SM74]A. M. Sasson and H. M. Merrill. Some Applications of Optimization Techniques to Power Systems Problems.
In IEEE,62, pages 959–972, 1974.

[SN96]

A. Sofer and S. G. Nash. Linear and Non-Linear Programming. McGraw-Hill, 1996.

[SPP98]

E. Silver, D. F. Pyke, and R. Peterson. Inventory Management and Production Planning and Scheduling.
Wiley, 1998.

[SRP92]P. E. Sweeney and E. Ridenour Paternoster. Cutting and Packing Problems: a Categorized, ApplicationOriented Research Bibliography. Journal of the Operational Research Society, 43(7):691–706, 1992.

[SS98]

B. Sansò and P. Soriano. Telecommunication Network Planning. Kluwer, 1998.

[SSR00]C. H. Scott, O. G. Skelton, and E. Rolland. Tactical and Strategic Models for Satellite Customer Assignment.
Journal of the Operational Research Society, 51(1):61–71, 2000.

[Tri84]

A. Tripathy. School Timetabling: a Case in Large Binary Integer Linear Programming. Management Science,30(12):1473–1489, 1984.

[VH98]

P. Van Hentenryck. The OPL Optimization Programming Language. MIT Press, Cambridge, MA, 1998.

[VW02]

S. Voß and D. L. Woodruff. Introduction to Computational Optimization Methods for Production Planningin a Supply Chain. Springer, Berlin, 2002.

[WA88]

E. A. Wasil and A. A. Assad. Project Management on the PC: Software, Applications and Trends. Interfaces,
18(2):75–84, 1988.

[Wag59]H. M. Wagner. An Integer Programming Model for Machine Scheduling. Naval Research Logistics Wuaterly,
6:131–140, 1959.

[Wal76]W. E. Walker. A Heuristic Adjacent Extreme Point Algorithm for the Fixed Charge Problem. Management
Science, 22(5):587–596, 1976.

[WH89]

M. Widmer and A. Hertz. A New Heuristic Method for the Flow Shop Sequencing Problem. European
Journal of Operational Research, 41:186–193, 1989.

[Wil93]

H. P. Williams. Model Building in Mathematical Programming. (3rd Rev. Ed.). John Wiley, Chichester, 1993.

[Win94]W. L. Winston. Operations Research: Applications and Algorithms (3rd ed.). Duxbury Press, Belmont, 1994.

[Win98]W. L. Winston. Financial Models Using Simulation and Optimization. Palisade, 1998.

[Wol98]L. A. Wolsey. Integer Programming. Wiley Interscience, 1998.

[WW58]H. M. Wagner and T. M. Whitin. A Dynamic Version of the Economic Lot Size Model. Management Science,50(1):89–96, 1958.

[Zen96]Z. A. Zenios. Financial Optimization. Cambridge University Press, 1996.

Bibliography

252

Applications of optimization with Xpress-MP

Index+, 136+=, 171:=, 171
ˆ, 141{}, 171

convexity row, 41
covering problem, 193, 246
create, 50, 51, 53, 54, 111
critical machine, 93
cut generation, 45
cutoff value, 125
cutting stock, 121
cycle, 168
cycle time, 99

accounting constraints, 25
accounting variables, 25
activity, 12
adjacency matrix, 101, 175
all item discount pricing, 43
and, 52, 61
arclist, 175
arc-paths formulation, 180
array, 61as, 50, 51, 53, 61, 111
assignment problem, 157, 211

decision variable, 7declarations, 58, 61, 72, 92, 111, 208
dense format, 83
deterministic, 11
disjunctive constraints, 90, 162
disjunctive graph, 91
div, 61
divisibility, 11do, 61dual value, 12
dummy objective, 221
dynamic, 61, 93
dynamic array, 111
dynamic array, 50, 54, 92, 166

B&B, see Branch and Boundbin packing, 122
binary variable, 30
bipartite graph, 172
blending, 62
boolean, 61bottleneck, 93, 211bottleneck assignment problem, 186
boundlower, 19
upper, 19
Branch and Bound, 31
Branch and Cut, 172
Branch and Price, 45
branching direction strategy, 33
break, 51, 54, 61, 124
break-even point, 13

elif, 61else, 61
empty set, 171
end, 61, 96
end-model, 12, 58, 235
Eulerian circuit, 238
exam, 56
exclusion constraints, 162
exists, 50, 51, 53, 54, 93, 116, 175, 239
exit, 51, 53, 124, 178
facility location problem, 155false, 61
fathoming, 32
feasibility problem, 220
feasible, 8
finalize, 50–54, 72
fix costs, 119
flow conservation law, 174
flow shop, 102
forall, 59, 61, 93, 124
forall-do, 50–54, 93
forward, 50, 61, 84
free variable, 20, 25
from, 61function, 124
function, 51, 61, 124case, 61
ceil, 50, 51, 74
Chinese postman problem, 238
column generation, 45
comment, 59
complement, 35
conjunctive constraints, 90
constraint, 8
conjunctive, 90
disjunctive, 90, 162
named, 84, 245
non-negativity, 8, 10
soft, 27strong, 151
constraint activity, 12Constraint Programming, 45
continuation line, 59

Gantt chart, 93
generalized assignment problem, 117

253

Applications of optimization with Xpress-MP

getlast, 51, 116
getobjval, 84
getprobstat, 236
getsize, 51–54, 124, 171
getsol, 50, 65, 84, 176
global optimum, 31
Gozinto graph, 109
hard constraint, 27
if, 50, 51, 53, 54, 61, 78, 96, 105, 108, 201, 225
if-then, 50, 51, 53, 54, 64, 96
if-then-elif, 52
if-then-elif-then-else, 65
if-then-else, 51, 54, 65
in, 61include, 61, 235
index set, 59, 214
infeasible, 8, 15, 236
initialisations, 61
initializations, 60, 61, 68
integer, 50, 61, 99
Integer Programming, 30
integer variable, 30
inter, 61
Interior Point algorithm, 18
inventory balance equation, 104
IP, see Integer Programming
is_binary, 50, 61
is_continuous, 61
is_free, 61
is_integer, 50, 61
is_partint, 61
is_semcont, 61
is_semint, 61
is_sos1, 61
is_sos2, 61
isodd, 51, 116
JIT, see Just-in-Time
job shop, 102
Just-in-Time, 119
Kirchhoff’s law, 142, 174, 229
knapsack constraints, 117
knapsack problem, 58, 122, 128

Newton-Barrier algorithm, 18
next, 61
node cover, 246
node selection strategy, 33
non-negativity constraint, 8, 10
not, 52, 61
NP-hard, 101
objective function, 8, 10
dummy, 221of, 61
optimization, 8
options, 61
or, 52, 61, 160
outputvisualize, 125
output printing, 59

linctr, 61, 147line break, 59
linear equation, 10
linear expression, 9
linear inequality, 10
Linear Programming, 9Linear Programming problem, 10
loading, 121
local optimum, 31
long-term planning, 103
loss equations, 21
lot sizing, 119
lower bound, 19
lower limit, 19
LP, see Linear Programming
LP solution, 77, 205

p-median problem, 241
packing, 121, 122
panic variable, 27
parameter, 96, 199
parameters, 61
partial integer variable, 30
partitioning problem, 234, 246
Parts explosion, 109
pathmakespan, 93

Index

matching, 158
matching with maximum total weight, 158
Material Requirement Planning, 119max, 52, 53, 61, 154
maximin, 122, 211
maximin assignment problem, 187
maximize, 59, 221
maximum cardinality matching, 158
maxlist, 52, 154
MCNF, see multi-commodity network flow
Method of Potentials, 102
mid-term planning, 103
min, 52, 53, 61, 154
minimax, 122
minimize, 221
minimum cost flow problem, 74, 141, 155minlist, 52, 154
MIP, see Mixed Integer Programming
MIQP, see Mixed Integer Quadratic Programming
Mixed Integer Programming, 30
Mixed Integer Quadratic Program, 207
Mixed Integer Quadratic Programming, 46
mmquad, 206
mmxprs, 206
mod, 53, 54, 61, 203model, 58
parameter, 199
model, 58, 61, 84
model cuts, 147, 150
modules, 56
mpvar, 59, 61, 72
MRP, see Material Requirement Planning
MRP method, 119
multi-commodity network flow, 192
multi-stage production planning, 119
multi-time period model, 23, 119

254

Applications of optimization with Xpress-MP

node-disjunctive, 174
pattern selection, 121
PERT method, 102
placement, 121
planning, 103
portfolio optimization, 46
presolving, 18
procedure, 50–52, 54, 61, 124prod, 61, 108
product mix, 62
production planning, 103
public, 61

symmetry breaking, 133, 221

QP, see Quadratic Programming
Quadratic Program, 206
Quadratic Programming, 46

unbounded, 15, 236
union, 61, 144
until, 61
upper bound, 19
upper limit, 19
uses, 61

technological coefficients, 21
then, 61
to, 61transport network, 229
transportation problem, 140transshipment flow formulation, 114
traveling salesman problem, 102
trim loss, 121
true, 61, 160
TSP, see traveling salesman problem

range, 50–53, 61, 83, 92, 137
range set, 59
ratio objective function, 28
real, 61, 99, 124, 176
reduced cost, 12
reference row, 40
relaxation, 31
repeat, 61
repeat-until, 50–52, 98, 124
returned, 124
RHS, see right hand sideright hand side, 10
round, 50, 53, 54, 99, 176
Rural Postman problem, 247

variable, 7
0/1, 30
binary, 30
fixed, 19
free, 20
integer, 30
partial integer, 30
semi-continuous, 30
semi-continuous integer, 30
variable selection strategy, 33
vehicle routing problem, 155vertex cover, 246
visualize output, 125
VRP, see vehicle routing problem

SC, see semi-continuous variable
schedulingflow shop, 102
job shop, 102
open shop, 193
scheduling with project crashing, 83
semi-continuous integer variable, 30
semi-continuous variable, 30, 199
separation, 32
sequence-dependent setup times, 102set, 61
set of integer, 51, 54, 137
setparam, 51
setup costs, 119
shadow price, 12, 13
short-term planning, 103
Simplex method, 18
soft constraint, 27
solving, 59
SOS1, see Special Ordered Set of type 1
SOS2, see Special Ordered Set of type 2
sparse arrays, 93
sparse format, 83
Special Ordered Set of type 1, 30
Special Ordered Set of type 2, 31
sqrt, 52, 141
Stochastic Programming, 11
strfmt, 50, 68
string, 61
sub-cycles, 168
subroutine, 84
sum, 61

Index

while, 61, 124, 188
while-do, 51, 53, 54, 124
XPRS_INF, 236
XPRS_LIN, 77, 205
XPRS_MIPADDCUTOFF, 125
XPRS_OPT, 236
XPRS_UNB, 236
XPRS_UNF, 236
XPRS_VERBOSE, 125

255

Applications of optimization with Xpress-MP