# On Best Corrected Mixture Problems in Metallurgy: A Case Study

Authors: F. Dubeau; M. E. Ntigura Habingabwa
Journal: Mathematical Problems in Engineering
 (2017)

---

## Abstract

We consider three simple mixture problems occurring in metallurgy. The first problem considered is the classic minimum cost mixture problem. For the second problem, we consider finding a correction to a given mixture, a premix, without considering the cost of this premix. We only consider the cost and the weight of the quantity used as a correction. We show that the minimum cost correction does not correspond to the minimum weight correction, and we built the Pareto curve that gives all intermediate solutions between these two extreme solutions. Finally, the third problem is the correction problem for a nonfree premix. The correction is done to obtain a minimum cost corrected mixture.


---

## Body

## 1. Introduction

Using materials, such as scrap metal for recycling, in order to form blends with desired specifications on some basic chemical elements is a well-known mixture problem met in foundries. People working in these fields bring to our attention the three different but related problems that we are going to consider in this paper. The first problem, called thebasic problem, is to make a minimum cost mixture while checking the constraints on the specifications. It is a classic mixture problem solved by linear programming. For the second problem, called thecorrection problem, we start with a given mixture, called here apremix, which does not satisfy all the constraints on the specifications. Without considering the cost of this premix, we look for acorrection, which is itself a mixture of available materials, that we will add to the premix in such a way that the resultingcorrected mixture, which is the premix and the correction added together, satisfies all the constraints on the specifications. The correction can be done at minimum cost or at minimum weight. This problem is an example of a bicriteria linear programming problem, and we are going to identify its Pareto set which gives the link between minimum cost solution and minimum weight solution. Finally, the third problem is similar to the preceding correction problem, but now we take into account the cost of the given premix. We look for a minimum unit cost corrected mixture. This is an example of a linear-fractional programming problem. It is important to point out that the questions raised in this paper and the data used in the examples are inspired from a real situation.Mixture problems are important applications in Operations Research [1]. One such application is the diet problem which received a lot of attention [2–7]. A variant of this problem is the pooling problem [8]. Also other approaches could be used to analyze similar mixture problems; for example, the fuzzy approach could be used to introduce uncertainties on the data [9].


## 2. Basic Problem

### 2.1. Problem Description

For the basic problem, there areN materials that can be used to produce a mixture. Let π
n
≥
0 be the weight of the nth material in the final mixture (n
=
1
,
…
,
N). The total weight of the mixture is(1)
z
w
e
i
g
h
t
π
=
∑
n
=
1
N
π
n
.The total cost of the mixture of weight z
w
e
i
g
h
t is then(2)
z
c
o
s
t
π
=
∑
n
=
1
N
c
n
π
n
,where c
n
≥
0 is the unit costs of the nth material (for n
=
1
,
…
,
N).The specifications of the mixture and of the materials are the amount (proportion % of the content) of known basic chemical elements (aluminium, carbon, chromium, copper, iron, etc.) contained in a unit weight of the material. Typically, there may be between10 and 20 basic chemical elements that characterize a chemical mixture. Each nth material is characterized by M specifications noted by s
m
,
n
(
m
=
1
,
…
,
M
).The specifications of the mixture are determined by a linear combination of the specifications of theN elements of the mixture. If we denote by s
p
e
c
m, the mth specification of the mixture, we have(3)
s
p
e
c
m
z
w
e
i
g
h
t
π
=
∑
n
=
1
N
s
m
,
n
π
n
.The specifications of the mixture are subjected to two types of constraints.Type 1.
Each specification must be in a well-defined range; that is to say,(4)
s
p
e
c
m
m
i
n
≤
s
p
e
c
m
≤
s
p
e
c
m
m
a
xfor m
=
1
,
…
,
M, where s
p
e
c
m
m
i
n
≥
0 and s
p
e
c
m
m
a
x
≤
1 are, respectively, the minimum and maximum value allowed for the mth specification of the mixture.Type 2.
Specifications can be connected or related byJ linear relations of the form(5)
∑
m
=
1
M
α
j
,
m
s
p
e
c
m
≤
=
≥
β
jfor j
=
1
,
…
,
J. For example, in Figure 1, we illustrate a situation where there are 4 links between the specifications of two chemical elements denoted as A and B. The pair of specifications (
s
p
e
c
A
,
s
p
e
c
B
) must be within the region bounded by the 4 given straight lines.Figure 1
Related specifications.


### 2.2. Problem Formulation

The basic problem is to look for a minimum cost mixture of unit weight; that is to say,z
w
e
i
g
h
t
(
π
)
=
1, satisfying constraints of Types 1 and 2. The decision variables are π
n for n
=
1
,
…
,
N; they will then be the proportions of each ingredient in the unit weight mixture. The mathematical formulation is therefore P
min
⁡
z
c
o
s
t
π
=
∑
n
=
1
N
c
n
π
n
,
subjet to
i
∑
n
=
1
N
π
n
=
1
;
i
i
for
m
=
1
,
…
,
M
:
∑
n
=
1
N
s
m
,
n
π
n
≤
s
p
e
c
m
m
a
x
,
≥
s
p
e
c
m
m
i
n
;
i
i
i
for
j
=
1
,
…
,
J
:
∑
n
=
1
N
∑
m
=
1
M
α
j
,
m
s
m
,
n
π
n
≤
=
≥
β
j
;
i
v
π
n
≥
0
for
n
=
1
,
…
,
N
.Let S be the notation for the set of feasible solutions to this problem.


### 2.3. Numerical Example

In our numerical example coming from a real context, we haveN
=
8 materials and the mixture has specifications on M
=
11 chemical elements. The minimum and maximum values for the specifications of the mixture are listed in Table 1. We have 2 specifications on 11 basic chemical elements; then there are 22 constraints concerning the concentration of the basic chemical elements in the mixture. Specifications and costs of the 8 materials are given in Table 2.Table 1
Minimum and maximum specifications in% of weight in the mixture.
Specifications (%) for M
=
11 chemical element of the mixture
m
Chemical element
s
p
e
c
m
m
i
n
s
p
e
c
m
m
a
x
1
Al
Aluminium
0
0.2
2
C
Carbon
2.7
2.98
3
Cr
Chromium
23.25
25.25
4
Cu
Copper
0
0.2
5
Fe
Iron
65
85
6
Mn
Manganese
0.3
0.6
7
Mo
Molybdenum
0
0.3
8
Ni
Nickel
0
0.2
9
P
Phosphorus
0
0.06
10
S
Sulphur
0
0.06
11
Si
Silicone
0.3
0.8Table 2
Characteristics of available materials.
s
m
,
n specifications (%) of N
=
8 available materials
n
m
1
2
3
4
5
6
7
8
1
0
1
0
0
0
0.0096
0.0340
0
2
0.2744
0
80
7.2096
1.7184
6.5088
0.0768
0
3
0.1666
0
0
48.2880
51.9552
0
0
0
4
0.1666
0
0
0
0
0
0
0
5
96.1772
0
0
40.6650
42.1304
15.2928
21.7728
69.1200
6
0.8232
0
0
0
0
73.2768
0.1632
0
7
0.0490
0
0
0
0
0
0
0
8
0.1862
0
0
0
0
0
0
0
9
0.0392
0
0
0.0192
0.0192
0.2688
0.0096
0
10
0
0
0
0.0384
0.0192
0.0096
0
0
11
0.1176
0
0
2.6592
0.0672
0.5952
17.1616
0.0900
Material unit costc
n
0.1450
1.3500
0.1100
0.5050
0.6750
0.5760
0.3715
0.7250Constraints of the first type are completely determined using the data in Tables1 and 2. Constraints of the second type require some clarification. We consider 4 constraints of Type 2 connecting carbon (
m
=
2
) and chromium (
m
=
3
) specifications. Constraints are described geometrically in Figure 2. They indicate that the pair (
s
p
e
c
2
,
s
p
e
c
3
) should be inside the parallelogram enclosed by 4 straight lines. With the data provided, the constraints are(6)
line no. 1:
s
p
e
c
3
-
50
9
s
p
e
c
2
≤
9
.
2500
line no. 2:
s
p
e
c
3
-
50
9
s
p
e
c
2
≥
7
.
6944
line no. 3:
s
p
e
c
3
+
10
s
p
e
c
2
≤
54
.
05
line no. 4:
s
p
e
c
3
+
10
s
p
e
c
2
≥
51
.
25
.The coefficients of these constraints are given in Table 3.Table 3
Coefficients to Type2 constraints.
J
=
4 constraints of Type 2
n
j
1
2
3
4
5
6
7
8
β
j
1
-
1.3578
0
-
444.4444
8.2347
42.4085
-
36.1600
-
0.4267
0
≤9.2500
2
≥7.6944
3
2.9106
0
800
120.3840
69.1392
65.0880
0.7680
0
≤54.0500
4
≥51.2500Figure 2
Coupled specifications of carbon and chromium.For the problemP, the constraints (i) and (iv) ensure that the set of feasible solutions is bounded if it is not empty. This set is actually not empty and we get the optimal solution of Table 4, with (minimum) unit cost equal to 0.344984
$. If we take away the materials n
=
5 and n
=
6, set π
5
=
0 and π
6
=
0 in the model, this new problem has no solution, and the feasible set is empty. It means that it is impossible to make a mixture which satisfies all the constraints with the 6 remaining materials.Table 4
Optimal solution of the basic mixture problem.
π
1
π
2
π
3
π
4
π
5
π
6
π
7
π
8
0.530478
0
0.004486
0.272517
0.192517
0
0
0


## 2.1. Problem Description

For the basic problem, there areN materials that can be used to produce a mixture. Let π
n
≥
0 be the weight of the nth material in the final mixture (n
=
1
,
…
,
N). The total weight of the mixture is(1)
z
w
e
i
g
h
t
π
=
∑
n
=
1
N
π
n
.The total cost of the mixture of weight z
w
e
i
g
h
t is then(2)
z
c
o
s
t
π
=
∑
n
=
1
N
c
n
π
n
,where c
n
≥
0 is the unit costs of the nth material (for n
=
1
,
…
,
N).The specifications of the mixture and of the materials are the amount (proportion % of the content) of known basic chemical elements (aluminium, carbon, chromium, copper, iron, etc.) contained in a unit weight of the material. Typically, there may be between10 and 20 basic chemical elements that characterize a chemical mixture. Each nth material is characterized by M specifications noted by s
m
,
n
(
m
=
1
,
…
,
M
).The specifications of the mixture are determined by a linear combination of the specifications of theN elements of the mixture. If we denote by s
p
e
c
m, the mth specification of the mixture, we have(3)
s
p
e
c
m
z
w
e
i
g
h
t
π
=
∑
n
=
1
N
s
m
,
n
π
n
.The specifications of the mixture are subjected to two types of constraints.Type 1.
Each specification must be in a well-defined range; that is to say,(4)
s
p
e
c
m
m
i
n
≤
s
p
e
c
m
≤
s
p
e
c
m
m
a
xfor m
=
1
,
…
,
M, where s
p
e
c
m
m
i
n
≥
0 and s
p
e
c
m
m
a
x
≤
1 are, respectively, the minimum and maximum value allowed for the mth specification of the mixture.Type 2.
Specifications can be connected or related byJ linear relations of the form(5)
∑
m
=
1
M
α
j
,
m
s
p
e
c
m
≤
=
≥
β
jfor j
=
1
,
…
,
J. For example, in Figure 1, we illustrate a situation where there are 4 links between the specifications of two chemical elements denoted as A and B. The pair of specifications (
s
p
e
c
A
,
s
p
e
c
B
) must be within the region bounded by the 4 given straight lines.Figure 1
Related specifications.


## 2.2. Problem Formulation

The basic problem is to look for a minimum cost mixture of unit weight; that is to say,z
w
e
i
g
h
t
(
π
)
=
1, satisfying constraints of Types 1 and 2. The decision variables are π
n for n
=
1
,
…
,
N; they will then be the proportions of each ingredient in the unit weight mixture. The mathematical formulation is therefore P
min
⁡
z
c
o
s
t
π
=
∑
n
=
1
N
c
n
π
n
,
subjet to
i
∑
n
=
1
N
π
n
=
1
;
i
i
for
m
=
1
,
…
,
M
:
∑
n
=
1
N
s
m
,
n
π
n
≤
s
p
e
c
m
m
a
x
,
≥
s
p
e
c
m
m
i
n
;
i
i
i
for
j
=
1
,
…
,
J
:
∑
n
=
1
N
∑
m
=
1
M
α
j
,
m
s
m
,
n
π
n
≤
=
≥
β
j
;
i
v
π
n
≥
0
for
n
=
1
,
…
,
N
.Let S be the notation for the set of feasible solutions to this problem.


## 2.3. Numerical Example

In our numerical example coming from a real context, we haveN
=
8 materials and the mixture has specifications on M
=
11 chemical elements. The minimum and maximum values for the specifications of the mixture are listed in Table 1. We have 2 specifications on 11 basic chemical elements; then there are 22 constraints concerning the concentration of the basic chemical elements in the mixture. Specifications and costs of the 8 materials are given in Table 2.Table 1
Minimum and maximum specifications in% of weight in the mixture.
Specifications (%) for M
=
11 chemical element of the mixture
m
Chemical element
s
p
e
c
m
m
i
n
s
p
e
c
m
m
a
x
1
Al
Aluminium
0
0.2
2
C
Carbon
2.7
2.98
3
Cr
Chromium
23.25
25.25
4
Cu
Copper
0
0.2
5
Fe
Iron
65
85
6
Mn
Manganese
0.3
0.6
7
Mo
Molybdenum
0
0.3
8
Ni
Nickel
0
0.2
9
P
Phosphorus
0
0.06
10
S
Sulphur
0
0.06
11
Si
Silicone
0.3
0.8Table 2
Characteristics of available materials.
s
m
,
n specifications (%) of N
=
8 available materials
n
m
1
2
3
4
5
6
7
8
1
0
1
0
0
0
0.0096
0.0340
0
2
0.2744
0
80
7.2096
1.7184
6.5088
0.0768
0
3
0.1666
0
0
48.2880
51.9552
0
0
0
4
0.1666
0
0
0
0
0
0
0
5
96.1772
0
0
40.6650
42.1304
15.2928
21.7728
69.1200
6
0.8232
0
0
0
0
73.2768
0.1632
0
7
0.0490
0
0
0
0
0
0
0
8
0.1862
0
0
0
0
0
0
0
9
0.0392
0
0
0.0192
0.0192
0.2688
0.0096
0
10
0
0
0
0.0384
0.0192
0.0096
0
0
11
0.1176
0
0
2.6592
0.0672
0.5952
17.1616
0.0900
Material unit costc
n
0.1450
1.3500
0.1100
0.5050
0.6750
0.5760
0.3715
0.7250Constraints of the first type are completely determined using the data in Tables1 and 2. Constraints of the second type require some clarification. We consider 4 constraints of Type 2 connecting carbon (
m
=
2
) and chromium (
m
=
3
) specifications. Constraints are described geometrically in Figure 2. They indicate that the pair (
s
p
e
c
2
,
s
p
e
c
3
) should be inside the parallelogram enclosed by 4 straight lines. With the data provided, the constraints are(6)
line no. 1:
s
p
e
c
3
-
50
9
s
p
e
c
2
≤
9
.
2500
line no. 2:
s
p
e
c
3
-
50
9
s
p
e
c
2
≥
7
.
6944
line no. 3:
s
p
e
c
3
+
10
s
p
e
c
2
≤
54
.
05
line no. 4:
s
p
e
c
3
+
10
s
p
e
c
2
≥
51
.
25
.The coefficients of these constraints are given in Table 3.Table 3
Coefficients to Type2 constraints.
J
=
4 constraints of Type 2
n
j
1
2
3
4
5
6
7
8
β
j
1
-
1.3578
0
-
444.4444
8.2347
42.4085
-
36.1600
-
0.4267
0
≤9.2500
2
≥7.6944
3
2.9106
0
800
120.3840
69.1392
65.0880
0.7680
0
≤54.0500
4
≥51.2500Figure 2
Coupled specifications of carbon and chromium.For the problemP, the constraints (i) and (iv) ensure that the set of feasible solutions is bounded if it is not empty. This set is actually not empty and we get the optimal solution of Table 4, with (minimum) unit cost equal to 0.344984
$. If we take away the materials n
=
5 and n
=
6, set π
5
=
0 and π
6
=
0 in the model, this new problem has no solution, and the feasible set is empty. It means that it is impossible to make a mixture which satisfies all the constraints with the 6 remaining materials.Table 4
Optimal solution of the basic mixture problem.
π
1
π
2
π
3
π
4
π
5
π
6
π
7
π
8
0.530478
0
0.004486
0.272517
0.192517
0
0
0


## 3. Correction to a Fixed Quantity of “Free” Premix

### 3.1. Problem Description

For the second problem we assume that we have in stock a given mixture already done called apremix, with its own specifications, but which does not satisfy some constraints for a new mixture. The problem we consider is to add to this premix acorrection, which is a combination of the N available materials, in such a way that thecorrected mixture obtained by adding the correction to the premix satisfies all the new constraints on the specifications. We are going to obtain this correction without considering the cost of the premix.Let us set(7)
π
0
a
≥
0
for the weight of the premix in the corrected mixture
,
π
n
a
≥
0
for the weight of the
n
th material in the corrected mixture
n
=
1
,
…
,
N
.As we want to correct a given weight of the premix, we set π
0
a
=
1, and the decision vector will be π
a
=
(
π
1
a
,
…
,
π
N
a
).The weight of the correction to add to the premix to obtain an acceptable mixture is given by(8)
z
w
e
i
g
h
t
π
a
=
∑
n
=
1
N
π
n
a
.If we consider a minimum weight correction, we have to minimize z
w
e
i
g
h
t
(
π
a
). The total weight of the corrected mixture is given by the following formula:(9)
z
w
e
i
g
h
t
t
o
t
a
l
π
a
=
∑
n
=
0
N
π
n
a
=
1
+
z
w
e
i
g
h
t
π
a
.The cost of the correction is(10)
z
c
o
s
t
π
a
=
∑
n
=
1
N
c
n
π
n
a
.If we consider a minimum cost correction, we have to minimize z
c
o
s
t
(
π
a
).TheM specifications of the premix are denoted by s
m
,
0
(
m
=
1
,
…
,
M
), and each material is also characterized by the M specifications previously described and denoted by s
m
,
n
m
=
1
,
…
,
M. The specifications s
p
e
c
m of the resulting corrected mixture are then defined by(11)
s
p
e
c
m
z
w
e
i
g
h
t
t
o
t
a
l
π
a
=
∑
n
=
0
N
s
m
,
n
π
n
a
=
s
m
,
n
+
∑
n
=
1
N
s
m
,
n
π
n
a
,and the constraints of Types 1 and 2 are directly applicable.


### 3.2. Problem Formulation

Since the problem we consider is to correct a unit weight of a premix(
π
0
a
=
1
) by adding to that premix a minimum cost or a minimum weight correction, the mathematical formulation is P
a
min
⁡
z
c
o
s
t
π
a
=
∑
n
=
1
N
c
n
π
n
a
,
min
⁡
z
w
e
i
g
h
t
π
a
=
∑
n
=
1
N
π
n
a
,
subjet to
i
∑
n
=
1
N
π
n
a
=
π
t
o
t
a
;
i
i
for
m
=
1
,
…
,
M
:
∑
n
=
0
N
s
m
,
n
π
n
a
-
s
p
e
c
m
m
a
x
π
t
o
t
a
≤
s
p
e
c
m
m
a
x
,
∑
n
=
0
N
s
m
,
n
π
n
a
-
s
p
e
c
m
m
i
n
π
t
o
t
a
≥
s
p
e
c
m
m
i
n
,
i
i
i
for
j
=
1
,
…
,
J
:
∑
n
=
0
N
∑
m
=
1
M
α
j
,
m
s
m
,
n
π
n
a
-
β
j
π
t
o
t
a
≤
=
≥
β
j
;
i
v
π
0
a
=
1
,
π
n
a
≥
0
for
n
=
1
,
…
,
N
.In this model, π
t
o
t
a is an intermediate variable. We denote by S
a the set of feasible solutions to this problem.Let us point out that the solution to this problem given byπ
0
a
=
1 and π
a
=
(
π
1
a
,
…
,
π
N
a
) is not proportions. The proportions of premix and materials in the optimal corrected mixture are (12)
π
~
i
a
=
π
i
a
z
w
e
i
g
h
t
t
o
t
a
l
π
a
=
π
i
a
1
+
z
w
e
i
g
h
t
π
afor i
=
0
,
…
,
N.


### 3.3. The Pareto Set

ProblemP
a is a problem with two criteria [10]: z
c
o
s
t
(
π
a
)
=
c
π
a and z
w
e
i
g
h
t
(
π
a
)
=
u
π
a, where c
=
c
1
,
…
,
c
N and u
=
1
,
…
,
1. The feasible set in the criteria space R
2 is defined by (13)
S
c
a
=
z
a
π
a
∈
R
2
∣
z
a
π
a
=
C
π
a
for
π
a
∈
S
a
=
C
S
a
,where (14)
z
a
π
a
=
z
c
o
s
t
π
a
z
w
e
i
g
h
t
π
a
=
∑
n
=
1
N
c
n
π
n
a
∑
n
=
1
N
π
n
a
=
C
π
a
with
C
=
c
1
⋯
c
N
1
⋯
1
.A feasible solution π
a
∈
S
a is anefficient feasible solution if and only if there is no other feasible solution π
¯
a
∈
S
a such that (i)
z
i
(
π
¯
a
)
≤
z
i
(
π
a
) for every i
∈
c
o
s
t
,
w
e
i
g
h
t;
(ii)
z
j
(
π
¯
a
)
<
z
j
(
π
a
) for at least one j
∈
c
o
s
t
,
w
e
i
g
h
t, that is to say that no other feasible solution improves both criteria or at least one criterion without damaging the other criterion.The set of efficient solution is theefficient set or thePareto set denoted by E
a. The efficient set or Pareto set in the criteria space is E
c
a
=
C
E
a. By considering the weighted sum of the two criteria using parameter λ
∈
(
0,1
) and the problem P
λ
min
⁡
z
λ
π
a
=
1
-
λ
z
c
o
s
t
π
a
+
λ
z
w
e
i
g
h
t
π
a
subject to
π
a
∈
S
a
,we can use the following relation for the Pareto set E
a [10]: (15)
E
a
=
⋃
λ
∈
0,1
arg
⁡
min
⁡
z
λ
π
a
.The efficient set E
a being a union of a finite number of faces of S
a follows that E
c
a is of the same form since it is the image of E
a by a linear transformation. As a subset of R
2, E
c
a is formed by a finite number of pairwise segments connected by their endpoints, also calledefficient vertices orefficient extreme points. It is a polygonal line. To each vertex of that polygonal line is associated an interval of values of λ, noted λ
_
,
λ
¯. The union of all those intervals is 0,1. Likewise, a unique value of λ is assigned to each segment which corresponds to the common bound of the intervals of the two endpoints of the segment. See [11] for a complete description of E
c
a.


### 3.4. Numerical Example

Table5 shows the specifications of the premix. Note that these specifications satisfy the constraints of Type 1 but not the constraints of Type 2. Type 2 constraint coefficients are given in Table 6. The efficient vertices in the criteria space are given in Table 7, and the corresponding optimal solutions in the decision space appear in Table 8. In these tables, the minimum cost solution is on the first line, while the minimum weight solution is on the last line. The other solutions correspond to intermediate vertices on the Pareto curve E
c
a given in Figure 3. This curve gives us the best pairs (
z
c
o
s
t
,
z
w
e
i
g
h
t
), going from the minimum cost efficient solution to the minimum weight efficient solution. For any point (
z
c
o
s
t
,
z
w
e
i
g
h
t
)
∈
E
c
a, z
c
o
s
t is the minimum cost for all feasible correction π
a such that z
w
e
i
g
h
t
(
π
a
)
=
z
w
e
i
g
h
t, and, conversely, z
w
e
i
g
h
t is the minimum weight for all feasible correction π
a such that z
c
o
s
t
(
π
a
)
=
z
c
o
s
t.Table 5
Premix specifications in% of weight.
s
p
e
c
m
,
0 specifications (%) of the premix
m
Chemical elements
s
p
e
c
m
m
i
n
s
m
,
0
s
p
e
c
m
m
a
x
1
Al
Aluminium
0
0.15
0.2
2
C
Carbon
2.7
2.98
2.98
3
Cr
Chrome
23.25
25.25
25.25
4
Cu
Copper
0
0.19
0.2
5
Fe
Iron
65
80
85
6
Mn
Manganese
0.3
0.55
0.6
7
Mo
Molybdenum
0
0.25
0.3
8
Ni
Nickel
0
0.17
0.2
9
P
Phosphorus
0
0.05
0.06
10
S
Sulphur
0
0.05
0.06
11
Si
Silicone
0.3
0.70
0.8Table 6
Coefficients of Type2 constraints.
J
=
4 constraint of Type 2
n
j
0
1
2
3
4
5
6
7
8
β
j
1
8.6944
-
1.3578
0
-
444.4444
8.2347
42.4085
-
36.160
-
0.4267
0
≤9.2500
2
≥7.6944
3
55.0500
2.9106
0
800
120.3840
69.1392
65.088
0.7680
0
≤54.0500
4
≥51.2500Table 7
Efficient vertices in the criteria spaceR
2.
λ
_
λ
¯
z
c
o
s
t
a
z
w
e
i
g
h
t
a
0
0.8402832
0.0028353
0.0195544
0.8402832
0.9094854
0.0042987
0.0192762
0.9094854
0.9791369
0.0049710
0.0192093
0.9791369
1
0.0066641
0.0191732Table 8
Efficient vertices in the decision spaceR
8.
π
0
a
π
1
a
π
2
a
π
3
a
π
4
a
π
5
a
π
6
a
π
7
a
π
8
a
1
0.0195543
0
0
0
0
0
0
0
1
0.0126373
0
0
0
0
0
0.0066389
0
1
0.0113950
0
0
0
0
0
0.0066381
0.0011760
1
0.0124767
0
0
0
0
0
0
0.0066965Figure 3
Pareto curveE
c
a in the criteria space R
2.The solution of this problem shows that the correction at minimum cost effectively costs0.0028353
$ and weights 0.0195544 weight units. The unit cost of the correction is then (16)
0.0028353
$
0.0195544
weight unit
=
0.1449962
$
weight unit
.The corrected mixture formed thus weighs 1.0195543 weight units. On the other hand, the correction at minimum weight requires an addition of 0.0191732 weight unit, which costs 0.0066641
$. In this case, the unit cost of the correction is (17)
0.0066641
$
0.0191732
weight unit
=
0.3475736
$
weight unit
.The corrected mixture formed thus weighs 1.0191732 weight units.The proportions of premix and materials in the optimal corrected mixture are reported in Table9 for the efficient vertices.Table 9
Proportions of premix and materials for efficient vertices.
π
~
0
a
π
~
1
a
π
~
2
a
π
~
3
a
π
~
4
a
π
~
5
a
π
~
6
a
π
~
7
a
π
~
8
a
0.9808207
0.0191793
0
0
0
0
0
0
0
0.9810883
0.0123983
0
0
0
0
0
0.0065133
0
0.9811530
0.0111802
0
0
0
0
0
0.0065130
0.0011538
0.9811875
0.0122420
0
0
0
0
0
0
0.0065705Note that the correction obtained does not use materialsn
=
5 and n
=
6; they could be removed. Recall that the effect of removing these two materials eliminates the possibility of finding a feasible mixture with only the available materials, without premix.


### 3.5. A Theoretical Result

We conclude this section by establishing a relation between the feasible setsS and S
a. They are both closed subsets of R
N, and they can be empty or nonempty. The set S can be nonempty while S
a is empty. When S is nonempty, it is bounded because (18)
u
t
π
=
∑
n
=
1
N
π
n
=
1
,
π
n
≥
0
for
n
=
1
,
…
,
N
.When S
a is nonempty, it can be bounded (hence compact) or unbounded (thus not compact). The following result links these two situations. All possible situations are reported in Table 10.Table 10
Relationship between the feasible setsS and S
a.
S
S
a
Empty
Empty
empty
Nonempty, closed and bounded (compact)
Nonempty, closed and bounded (compact)
empty
Nonempty, closed and bounded (compact)
Nonempty, closed and unboundedTheorem 1.
IfS
a is nonempty, we have that (i)
S
a is bounded (thus compact) if and only if S is empty;
(ii)
S
a is unbounded (thus not compact) if and only if S is nonempty.Proof.
IfS
a is nonempty, let us take 0
≠
π
∈
S and also take π
a
∈
S
a. For t
≥
0, we set π
a
(
t
)
=
π
a
+
t
π. It is easily verified that π
a
(
t
)
∈
S
a and thus S
a is unbounded. Conversely, if S
a is unbounded, there are π
a
(
t
)
∈
S
a and π
∈
R
N such that u
t
π
=
1 and π
a
(
t
)
=
π
a
+
t
π
∈
S
a (for all t
≥
0). Then we check that π
∈
S by substituting π
a
(
t
) in the constraint of P
a, dividing by t and passing to the limit as t goes to infinity.


## 3.1. Problem Description

For the second problem we assume that we have in stock a given mixture already done called apremix, with its own specifications, but which does not satisfy some constraints for a new mixture. The problem we consider is to add to this premix acorrection, which is a combination of the N available materials, in such a way that thecorrected mixture obtained by adding the correction to the premix satisfies all the new constraints on the specifications. We are going to obtain this correction without considering the cost of the premix.Let us set(7)
π
0
a
≥
0
for the weight of the premix in the corrected mixture
,
π
n
a
≥
0
for the weight of the
n
th material in the corrected mixture
n
=
1
,
…
,
N
.As we want to correct a given weight of the premix, we set π
0
a
=
1, and the decision vector will be π
a
=
(
π
1
a
,
…
,
π
N
a
).The weight of the correction to add to the premix to obtain an acceptable mixture is given by(8)
z
w
e
i
g
h
t
π
a
=
∑
n
=
1
N
π
n
a
.If we consider a minimum weight correction, we have to minimize z
w
e
i
g
h
t
(
π
a
). The total weight of the corrected mixture is given by the following formula:(9)
z
w
e
i
g
h
t
t
o
t
a
l
π
a
=
∑
n
=
0
N
π
n
a
=
1
+
z
w
e
i
g
h
t
π
a
.The cost of the correction is(10)
z
c
o
s
t
π
a
=
∑
n
=
1
N
c
n
π
n
a
.If we consider a minimum cost correction, we have to minimize z
c
o
s
t
(
π
a
).TheM specifications of the premix are denoted by s
m
,
0
(
m
=
1
,
…
,
M
), and each material is also characterized by the M specifications previously described and denoted by s
m
,
n
m
=
1
,
…
,
M. The specifications s
p
e
c
m of the resulting corrected mixture are then defined by(11)
s
p
e
c
m
z
w
e
i
g
h
t
t
o
t
a
l
π
a
=
∑
n
=
0
N
s
m
,
n
π
n
a
=
s
m
,
n
+
∑
n
=
1
N
s
m
,
n
π
n
a
,and the constraints of Types 1 and 2 are directly applicable.


## 3.2. Problem Formulation

Since the problem we consider is to correct a unit weight of a premix(
π
0
a
=
1
) by adding to that premix a minimum cost or a minimum weight correction, the mathematical formulation is P
a
min
⁡
z
c
o
s
t
π
a
=
∑
n
=
1
N
c
n
π
n
a
,
min
⁡
z
w
e
i
g
h
t
π
a
=
∑
n
=
1
N
π
n
a
,
subjet to
i
∑
n
=
1
N
π
n
a
=
π
t
o
t
a
;
i
i
for
m
=
1
,
…
,
M
:
∑
n
=
0
N
s
m
,
n
π
n
a
-
s
p
e
c
m
m
a
x
π
t
o
t
a
≤
s
p
e
c
m
m
a
x
,
∑
n
=
0
N
s
m
,
n
π
n
a
-
s
p
e
c
m
m
i
n
π
t
o
t
a
≥
s
p
e
c
m
m
i
n
,
i
i
i
for
j
=
1
,
…
,
J
:
∑
n
=
0
N
∑
m
=
1
M
α
j
,
m
s
m
,
n
π
n
a
-
β
j
π
t
o
t
a
≤
=
≥
β
j
;
i
v
π
0
a
=
1
,
π
n
a
≥
0
for
n
=
1
,
…
,
N
.In this model, π
t
o
t
a is an intermediate variable. We denote by S
a the set of feasible solutions to this problem.Let us point out that the solution to this problem given byπ
0
a
=
1 and π
a
=
(
π
1
a
,
…
,
π
N
a
) is not proportions. The proportions of premix and materials in the optimal corrected mixture are (12)
π
~
i
a
=
π
i
a
z
w
e
i
g
h
t
t
o
t
a
l
π
a
=
π
i
a
1
+
z
w
e
i
g
h
t
π
afor i
=
0
,
…
,
N.


## 3.3. The Pareto Set

ProblemP
a is a problem with two criteria [10]: z
c
o
s
t
(
π
a
)
=
c
π
a and z
w
e
i
g
h
t
(
π
a
)
=
u
π
a, where c
=
c
1
,
…
,
c
N and u
=
1
,
…
,
1. The feasible set in the criteria space R
2 is defined by (13)
S
c
a
=
z
a
π
a
∈
R
2
∣
z
a
π
a
=
C
π
a
for
π
a
∈
S
a
=
C
S
a
,where (14)
z
a
π
a
=
z
c
o
s
t
π
a
z
w
e
i
g
h
t
π
a
=
∑
n
=
1
N
c
n
π
n
a
∑
n
=
1
N
π
n
a
=
C
π
a
with
C
=
c
1
⋯
c
N
1
⋯
1
.A feasible solution π
a
∈
S
a is anefficient feasible solution if and only if there is no other feasible solution π
¯
a
∈
S
a such that (i)
z
i
(
π
¯
a
)
≤
z
i
(
π
a
) for every i
∈
c
o
s
t
,
w
e
i
g
h
t;
(ii)
z
j
(
π
¯
a
)
<
z
j
(
π
a
) for at least one j
∈
c
o
s
t
,
w
e
i
g
h
t, that is to say that no other feasible solution improves both criteria or at least one criterion without damaging the other criterion.The set of efficient solution is theefficient set or thePareto set denoted by E
a. The efficient set or Pareto set in the criteria space is E
c
a
=
C
E
a. By considering the weighted sum of the two criteria using parameter λ
∈
(
0,1
) and the problem P
λ
min
⁡
z
λ
π
a
=
1
-
λ
z
c
o
s
t
π
a
+
λ
z
w
e
i
g
h
t
π
a
subject to
π
a
∈
S
a
,we can use the following relation for the Pareto set E
a [10]: (15)
E
a
=
⋃
λ
∈
0,1
arg
⁡
min
⁡
z
λ
π
a
.The efficient set E
a being a union of a finite number of faces of S
a follows that E
c
a is of the same form since it is the image of E
a by a linear transformation. As a subset of R
2, E
c
a is formed by a finite number of pairwise segments connected by their endpoints, also calledefficient vertices orefficient extreme points. It is a polygonal line. To each vertex of that polygonal line is associated an interval of values of λ, noted λ
_
,
λ
¯. The union of all those intervals is 0,1. Likewise, a unique value of λ is assigned to each segment which corresponds to the common bound of the intervals of the two endpoints of the segment. See [11] for a complete description of E
c
a.


## 3.4. Numerical Example

Table5 shows the specifications of the premix. Note that these specifications satisfy the constraints of Type 1 but not the constraints of Type 2. Type 2 constraint coefficients are given in Table 6. The efficient vertices in the criteria space are given in Table 7, and the corresponding optimal solutions in the decision space appear in Table 8. In these tables, the minimum cost solution is on the first line, while the minimum weight solution is on the last line. The other solutions correspond to intermediate vertices on the Pareto curve E
c
a given in Figure 3. This curve gives us the best pairs (
z
c
o
s
t
,
z
w
e
i
g
h
t
), going from the minimum cost efficient solution to the minimum weight efficient solution. For any point (
z
c
o
s
t
,
z
w
e
i
g
h
t
)
∈
E
c
a, z
c
o
s
t is the minimum cost for all feasible correction π
a such that z
w
e
i
g
h
t
(
π
a
)
=
z
w
e
i
g
h
t, and, conversely, z
w
e
i
g
h
t is the minimum weight for all feasible correction π
a such that z
c
o
s
t
(
π
a
)
=
z
c
o
s
t.Table 5
Premix specifications in% of weight.
s
p
e
c
m
,
0 specifications (%) of the premix
m
Chemical elements
s
p
e
c
m
m
i
n
s
m
,
0
s
p
e
c
m
m
a
x
1
Al
Aluminium
0
0.15
0.2
2
C
Carbon
2.7
2.98
2.98
3
Cr
Chrome
23.25
25.25
25.25
4
Cu
Copper
0
0.19
0.2
5
Fe
Iron
65
80
85
6
Mn
Manganese
0.3
0.55
0.6
7
Mo
Molybdenum
0
0.25
0.3
8
Ni
Nickel
0
0.17
0.2
9
P
Phosphorus
0
0.05
0.06
10
S
Sulphur
0
0.05
0.06
11
Si
Silicone
0.3
0.70
0.8Table 6
Coefficients of Type2 constraints.
J
=
4 constraint of Type 2
n
j
0
1
2
3
4
5
6
7
8
β
j
1
8.6944
-
1.3578
0
-
444.4444
8.2347
42.4085
-
36.160
-
0.4267
0
≤9.2500
2
≥7.6944
3
55.0500
2.9106
0
800
120.3840
69.1392
65.088
0.7680
0
≤54.0500
4
≥51.2500Table 7
Efficient vertices in the criteria spaceR
2.
λ
_
λ
¯
z
c
o
s
t
a
z
w
e
i
g
h
t
a
0
0.8402832
0.0028353
0.0195544
0.8402832
0.9094854
0.0042987
0.0192762
0.9094854
0.9791369
0.0049710
0.0192093
0.9791369
1
0.0066641
0.0191732Table 8
Efficient vertices in the decision spaceR
8.
π
0
a
π
1
a
π
2
a
π
3
a
π
4
a
π
5
a
π
6
a
π
7
a
π
8
a
1
0.0195543
0
0
0
0
0
0
0
1
0.0126373
0
0
0
0
0
0.0066389
0
1
0.0113950
0
0
0
0
0
0.0066381
0.0011760
1
0.0124767
0
0
0
0
0
0
0.0066965Figure 3
Pareto curveE
c
a in the criteria space R
2.The solution of this problem shows that the correction at minimum cost effectively costs0.0028353
$ and weights 0.0195544 weight units. The unit cost of the correction is then (16)
0.0028353
$
0.0195544
weight unit
=
0.1449962
$
weight unit
.The corrected mixture formed thus weighs 1.0195543 weight units. On the other hand, the correction at minimum weight requires an addition of 0.0191732 weight unit, which costs 0.0066641
$. In this case, the unit cost of the correction is (17)
0.0066641
$
0.0191732
weight unit
=
0.3475736
$
weight unit
.The corrected mixture formed thus weighs 1.0191732 weight units.The proportions of premix and materials in the optimal corrected mixture are reported in Table9 for the efficient vertices.Table 9
Proportions of premix and materials for efficient vertices.
π
~
0
a
π
~
1
a
π
~
2
a
π
~
3
a
π
~
4
a
π
~
5
a
π
~
6
a
π
~
7
a
π
~
8
a
0.9808207
0.0191793
0
0
0
0
0
0
0
0.9810883
0.0123983
0
0
0
0
0
0.0065133
0
0.9811530
0.0111802
0
0
0
0
0
0.0065130
0.0011538
0.9811875
0.0122420
0
0
0
0
0
0
0.0065705Note that the correction obtained does not use materialsn
=
5 and n
=
6; they could be removed. Recall that the effect of removing these two materials eliminates the possibility of finding a feasible mixture with only the available materials, without premix.


## 3.5. A Theoretical Result

We conclude this section by establishing a relation between the feasible setsS and S
a. They are both closed subsets of R
N, and they can be empty or nonempty. The set S can be nonempty while S
a is empty. When S is nonempty, it is bounded because (18)
u
t
π
=
∑
n
=
1
N
π
n
=
1
,
π
n
≥
0
for
n
=
1
,
…
,
N
.When S
a is nonempty, it can be bounded (hence compact) or unbounded (thus not compact). The following result links these two situations. All possible situations are reported in Table 10.Table 10
Relationship between the feasible setsS and S
a.
S
S
a
Empty
Empty
empty
Nonempty, closed and bounded (compact)
Nonempty, closed and bounded (compact)
empty
Nonempty, closed and bounded (compact)
Nonempty, closed and unboundedTheorem 1.
IfS
a is nonempty, we have that (i)
S
a is bounded (thus compact) if and only if S is empty;
(ii)
S
a is unbounded (thus not compact) if and only if S is nonempty.Proof.
IfS
a is nonempty, let us take 0
≠
π
∈
S and also take π
a
∈
S
a. For t
≥
0, we set π
a
(
t
)
=
π
a
+
t
π. It is easily verified that π
a
(
t
)
∈
S
a and thus S
a is unbounded. Conversely, if S
a is unbounded, there are π
a
(
t
)
∈
S
a and π
∈
R
N such that u
t
π
=
1 and π
a
(
t
)
=
π
a
+
t
π
∈
S
a (for all t
≥
0). Then we check that π
∈
S by substituting π
a
(
t
) in the constraint of P
a, dividing by t and passing to the limit as t goes to infinity.


## 4. Correction to a Fixed Quantity of “Nonfree” Premix

### 4.1. Problem Description

It would have been more realistic to consider that the premix has a possible nonzero cost in the preceding problem. So we reconsider the preceding problem assuming a possible nonzero unit costc
0 for the premix. We consider one criterion which is the unit cost of the corrected mixture.Let us set(19)
π
0
f
≥
0
for the weight of the premix in the corrected mixture
,
π
n
f
≥
0
for the weight of the
n
th material in the corrected mixture
n
=
1
,
…
,
N
.As we want to correct a given weight of the premix, we set π
0
f
=
1, and the decision vector will be π
f
=
(
π
1
f
,
…
,
π
N
f
).For the total weight, we have(20)
z
w
e
i
g
h
t
t
o
t
a
l
π
f
=
∑
n
=
0
N
π
n
f
=
1
+
∑
n
=
1
N
π
n
f
=
1
+
z
w
e
i
g
h
t
π
fbecause we have assumed that π
0
f
=
1, and the total cost is(21)
z
c
o
s
t
t
o
t
a
l
π
f
=
∑
n
=
0
N
c
n
π
n
f
=
c
0
+
∑
n
=
1
N
c
n
π
n
f
=
c
0
+
z
c
o
s
t
π
fwhich take into account the fact that the premix might have a nonzero cost, so c
0
≥
0.For this problem, the criteria will be the unit cost of the corrected mixture, premix, and correction added together, given by the ratio(22)
z
f
π
f
=
z
c
o
s
t
t
o
t
a
l
π
f
z
w
e
i
g
h
t
t
o
t
a
l
π
f
=
c
0
+
∑
n
=
1
N
c
n
π
n
f
1
+
∑
n
=
1
N
π
n
f
=
c
0
+
z
c
o
s
t
π
f
1
+
z
w
e
i
g
h
t
π
f
.


### 4.2. Problem Formulation

The decision variables will be againπ
n
f for n
=
1
,
…
,
N, and therefore the formulation is P
f
min
⁡
z
f
π
f
=
c
0
+
∑
n
=
1
N
c
n
π
n
f
1
+
∑
n
=
1
N
π
n
f
subject to
π
f
∈
S
f
.Let S
f be the set of feasible solution to this problem. Obviously, we have S
f
=
S
a because the constraints of this problem are the constraints of the preceding problem P
a.


### 4.3. Linear-Fractional Program

The problemP
f is an example of a linear-fractional program. Under the assumption that the feasible region S
f is nonempty and bounded, the Charnes-Cooper transformation [12] (23)
η
0
=
1
1
+
∑
i
=
1
N
π
i
f
,
η
n
=
π
i
f
1
+
∑
i
=
1
N
π
i
f
f
o
r
n
=
1
,
…
,
N
, translates the linear-fractional program above to the equivalent linear program: P
~
f
min
⁡
z
~
f
η
=
∑
n
=
0
N
c
n
η
n
subject to
i
∑
n
=
0
N
η
n
=
1
;
i
i
for
m
=
1
,
…
,
M
:
∑
n
=
0
N
s
m
,
n
η
n
≤
s
p
e
c
m
m
a
x
,
∑
n
=
0
N
s
m
,
n
η
n
≥
s
p
e
c
m
m
i
n
;
i
i
i
for
j
=
1
,
…
,
J
:
∑
n
=
0
N
∑
m
=
1
M
α
j
,
m
s
m
,
n
η
n
≤
=
≥
β
j
;
i
v
η
n
≥
0
for
n
=
0
,
…
,
N
.Let S
~
f be the set of feasible solutions η
=
(
η
0
,
…
,
η
N
) of P
~
f. Let us observe that P
~
f is similar to the first problem P, and the premix can now be considered as one of the available materials. The fact that we assume S
a nonempty and bounded, Theorem 1 implies that P has no solution; hence, P
~
f has no solution η
=
(
η
0
,
…
,
η
n
)
∈
S
~
f with η
0
=
0. It follows that any solution of P
~
f, if it exists, is such that η
0
>
0 and that the corresponding solution of P
f is given by(24)
π
n
f
=
η
n
η
0
for
n
=
0
,
…
,
N
,and hence π
0
f
=
1. In case the feasible region S
a is nonempty and unbounded, P
~
f might have a solution η
=
(
η
0
,
…
,
η
n
)
∈
S
~
f with η
0
>
0; then the solution of P
f is again given by the preceding formula. But it might happen that P
~
f has a solution η
=
(
η
0
,
…
,
η
n
)
∈
S
~
f with η
0
=
0; then the resulting mixture is not a corrected mixture. It means that it is better to use original materials (n
=
1
,
…
,
N) to produce the desired mixture because it is cheaper than trying to use the premix. Hence, (25)
π
n
=
η
n
for
n
=
1
,
…
,
N
,is a solution of the original problem P. In this case, let us define π
f by the formula π
0
f
=
0, and π
n
f
=
η
n for n
=
0
,
…
,
N, as before.


### 4.4. A Parametric Analysis

We could be interested by the variation of the criteria, the unit cost of the resulting (corrected or not) mixture, with respect to the costc
0. This information is obtained from a parametric analysis of the criteria with respect to the parameter c
0. We split the criteria of P
~
f in two parts: (26)
z
~
f
η
=
∑
n
=
1
N
c
n
η
n
+
c
0
η
0
=
z
~
1
f
η
+
c
0
z
~
2
f
ηand we consider the following bicriteria problem: P
~
f
s
min
⁡
z
~
1
f
η
=
∑
n
=
1
N
c
n
η
n
min
⁡
z
~
2
f
η
=
η
0
subject to
η
∈
S
~
f
.The Pareto curve of this problem, obtained by considering the weighted sum, is (27)
z
~
λ
f
η
=
1
-
λ
z
~
1
f
η
+
λ
z
~
2
f
η and the correspondence (28)
λ
=
c
0
1
+
c
0
or
c
0
=
λ
1
-
λ
,allows us to find a piecewise linear continuous increasing concave representation for the optimal minimal unit cost z
∗
f
c
0
=
min
⁡
z
f
π
f in terms of c
0.Since(29)
z
c
o
s
t
π
f
=
∑
n
=
1
N
c
n
π
n
f
=
1
η
0
∑
n
=
1
N
c
n
η
n
=
1
η
0
z
~
1
f
η
if
η
0
>
0
or
π
0
f
=
1
,
∑
n
=
1
N
c
n
η
n
=
z
~
1
f
η
if
η
0
=
0
or
π
0
f
=
0
,
z
w
e
i
g
h
t
π
f
=
∑
n
=
1
N
π
n
f
=
1
η
0
∑
n
=
1
N
η
n
=
1
η
0
1
-
η
0
if
η
0
>
0
or
π
0
f
=
1
,
∑
n
=
1
N
η
n
=
1
if
η
0
=
0
or
π
0
f
=
0
,we have (30)
z
∗
f
c
0
=
c
0
π
0
f
+
z
c
o
s
t
π
f
π
0
f
+
z
w
e
i
g
h
t
π
f
,that is, valid not only for π
0
f
=
1 but also for π
0
f
=
0 and where π
f is the optimal solution.


### 4.5. Numerical Example

We consider two situations with the preceding data.Case 1.
We use all the materials, withoutn
=
5 and n
=
6. In this case, S is empty and S
a is nonempty and bounded; hence η
0 will always be strictly positive. Tables 11 and 12 present results for P
~
f
s and the corresponding values for P
f.Table 11
Efficient vertices ofP
~
f
s and P
f in the criteria space R
2.
(a)
λ
_
λ
¯
z
~
1
f
(
η
)
z
~
2
f
(
η
)
0
0.124887
0.002781
0.980820
0.124887
0.127242
0.005732
0.960140
0.127242
0.249832
0.011545
0.920268
0.249832
0.343079
0.024633
0.880968
0.343079
0.604078
0.045238
0.841514
0.604078
0.710795
0.045834
0.841124
0.710795
0.874033
0.056158
0.836911
0.874033
1
0.056666
0.836838
(b)
c
_
0
c
¯
0
z
c
o
s
t
f
π
f
z
w
e
i
g
n
t
f
π
f
0
0.142710
0.002835
0.019554
0.142710
0.145794
0.005970
0.041513
0.145794
0.333035
0.012545
0.086638
0.333035
0.522255
0.027961
0.135113
0.522255
1.525752
0.053758
0.188333
1.525752
2.450600
0.054491
0.188885
2.450600
6.938604
0.067101
0.194869
6.938604
+
∞
0.067714
0.194974Table 12
Optimal solutions ofP
~
f
s and P
f in the decision space R
7.
(a)
η
0
η
1
η
2
η
3
η
4
η
5
η
6
η
7
η
8
0.980820
0.019179
0
0
0
—
—
0
0
0.960140
0.038506
0
0.001352
0
—
—
0
0
0.920268
0.079282
0
0.000448
0
—
—
0
0
0.880968
0.098547
0
0
0.020483
—
—
0
0
0.841514
0.096656
0
0
0.061829
—
—
0
0
0.841124
0.094791
0
0
0.062039
—
—
0.002044
0
0.836911
0.081505
0
0
0.064288
—
—
0.001878
0.015415
0.836838
0.081852
0
0
0.064325
—
—
0
0.016983
(b)
π
0
f
π
1
f
π
2
f
π
3
f
π
4
f
π
5
f
π
6
f
π
7
f
π
8
f
1
0.019554
0
0
0
—
—
0
0
1
0.040105
0
0.001408
0
—
—
0
0
1
0.086151
0
0.000486
0
—
—
0
0
1
0.111862
0
0
0.023251
—
—
0
0
1
0.114859
0
0
0.073473
—
—
0
0
1
0.112696
0
0
0.073758
—
—
0.002430
0
1
0.097389
0
0
0.076816
—
—
0.002244
0.018420
1
0.097811
0
0
0.076867
—
—
0
0.020295Case 2.
We consider all theN
=
8 materials; then S is nonempty and S
a is nonempty and not bounded; hence it might happen that η
0 be 0 for large values of c
0. Tables 13 and 14 present results for P
~
f
s and the corresponding values for P
f.Table 13
Efficient vertices ofP
~
f
s and P
f in the criteria space R
2.
(a)
λ
_
λ
¯
z
~
1
f
(
η
)
z
~
2
f
(
η
)
0
0.124887
0.002781
0.980820
0.124887
0.127242
0.005732
0.960140
0.127242
0.249832
0.011545
0.920267
0.249832
0.261550
0.024633
0.880970
0.261550
0.270235
0.153784
0.516323
0.270235
1
0.344983
0
(b)
c
_
0
c
¯
0
z
c
o
s
t
f
(
π
f
)
z
w
e
i
g
n
t
f
(
π
f
)
0
0.142710
0.002835
0.019554
0.142710
0.145794
0.005970
0.041513
0.145794
0.333035
0.012545
0.086640
0.333035
0.354188
0.027961
0.135112
0.354188
0.370304
0.297848
0.936768
0.370304
+
∞
0.344984
1Table 14
Optimal solutions ofP
~
f
s and P
f in the decision space R
9.
(a)
η
0
η
1
η
2
η
3
η
4
η
5
η
6
η
7
η
8
0.980820
0.019179
0
0
0
0
0
0
0
0.960140
0.038506
0
0.001352
0
0
0
0
0
0.920267
0.079284
0
0.000448
0
0
0
0
0
0.880970
0.098547
0
0
0.020482
0
0
0
0
0.516323
0.277318
0
0
0.151270
0.055086
0
0
0
0
0.530479
0
0.004485
0.272517
0.192517
0
0
0
(b)
π
0
f
π
1
f
π
2
f
π
3
f
π
4
f
π
5
f
π
6
f
π
7
f
π
8
f
1
0.019554
0
0
0
0
0
0
0
1
0.040104
0
0.001408
0
0
0
0
0
1
0.086153
0
0.000486
0
0
0
0
0
1
0.111862
0
0
0.023249
0
0
0
0
1
0.537102
0
0
0.292976
0.106690
0
0
0
0
0.530479
0
0.004485
0.272517
0.192517
0
0
0Let us observe that the results forc
0
=
0 correspond to the results obtained in Section 3.


## 4.1. Problem Description

It would have been more realistic to consider that the premix has a possible nonzero cost in the preceding problem. So we reconsider the preceding problem assuming a possible nonzero unit costc
0 for the premix. We consider one criterion which is the unit cost of the corrected mixture.Let us set(19)
π
0
f
≥
0
for the weight of the premix in the corrected mixture
,
π
n
f
≥
0
for the weight of the
n
th material in the corrected mixture
n
=
1
,
…
,
N
.As we want to correct a given weight of the premix, we set π
0
f
=
1, and the decision vector will be π
f
=
(
π
1
f
,
…
,
π
N
f
).For the total weight, we have(20)
z
w
e
i
g
h
t
t
o
t
a
l
π
f
=
∑
n
=
0
N
π
n
f
=
1
+
∑
n
=
1
N
π
n
f
=
1
+
z
w
e
i
g
h
t
π
fbecause we have assumed that π
0
f
=
1, and the total cost is(21)
z
c
o
s
t
t
o
t
a
l
π
f
=
∑
n
=
0
N
c
n
π
n
f
=
c
0
+
∑
n
=
1
N
c
n
π
n
f
=
c
0
+
z
c
o
s
t
π
fwhich take into account the fact that the premix might have a nonzero cost, so c
0
≥
0.For this problem, the criteria will be the unit cost of the corrected mixture, premix, and correction added together, given by the ratio(22)
z
f
π
f
=
z
c
o
s
t
t
o
t
a
l
π
f
z
w
e
i
g
h
t
t
o
t
a
l
π
f
=
c
0
+
∑
n
=
1
N
c
n
π
n
f
1
+
∑
n
=
1
N
π
n
f
=
c
0
+
z
c
o
s
t
π
f
1
+
z
w
e
i
g
h
t
π
f
.


## 4.2. Problem Formulation

The decision variables will be againπ
n
f for n
=
1
,
…
,
N, and therefore the formulation is P
f
min
⁡
z
f
π
f
=
c
0
+
∑
n
=
1
N
c
n
π
n
f
1
+
∑
n
=
1
N
π
n
f
subject to
π
f
∈
S
f
.Let S
f be the set of feasible solution to this problem. Obviously, we have S
f
=
S
a because the constraints of this problem are the constraints of the preceding problem P
a.


## 4.3. Linear-Fractional Program

The problemP
f is an example of a linear-fractional program. Under the assumption that the feasible region S
f is nonempty and bounded, the Charnes-Cooper transformation [12] (23)
η
0
=
1
1
+
∑
i
=
1
N
π
i
f
,
η
n
=
π
i
f
1
+
∑
i
=
1
N
π
i
f
f
o
r
n
=
1
,
…
,
N
, translates the linear-fractional program above to the equivalent linear program: P
~
f
min
⁡
z
~
f
η
=
∑
n
=
0
N
c
n
η
n
subject to
i
∑
n
=
0
N
η
n
=
1
;
i
i
for
m
=
1
,
…
,
M
:
∑
n
=
0
N
s
m
,
n
η
n
≤
s
p
e
c
m
m
a
x
,
∑
n
=
0
N
s
m
,
n
η
n
≥
s
p
e
c
m
m
i
n
;
i
i
i
for
j
=
1
,
…
,
J
:
∑
n
=
0
N
∑
m
=
1
M
α
j
,
m
s
m
,
n
η
n
≤
=
≥
β
j
;
i
v
η
n
≥
0
for
n
=
0
,
…
,
N
.Let S
~
f be the set of feasible solutions η
=
(
η
0
,
…
,
η
N
) of P
~
f. Let us observe that P
~
f is similar to the first problem P, and the premix can now be considered as one of the available materials. The fact that we assume S
a nonempty and bounded, Theorem 1 implies that P has no solution; hence, P
~
f has no solution η
=
(
η
0
,
…
,
η
n
)
∈
S
~
f with η
0
=
0. It follows that any solution of P
~
f, if it exists, is such that η
0
>
0 and that the corresponding solution of P
f is given by(24)
π
n
f
=
η
n
η
0
for
n
=
0
,
…
,
N
,and hence π
0
f
=
1. In case the feasible region S
a is nonempty and unbounded, P
~
f might have a solution η
=
(
η
0
,
…
,
η
n
)
∈
S
~
f with η
0
>
0; then the solution of P
f is again given by the preceding formula. But it might happen that P
~
f has a solution η
=
(
η
0
,
…
,
η
n
)
∈
S
~
f with η
0
=
0; then the resulting mixture is not a corrected mixture. It means that it is better to use original materials (n
=
1
,
…
,
N) to produce the desired mixture because it is cheaper than trying to use the premix. Hence, (25)
π
n
=
η
n
for
n
=
1
,
…
,
N
,is a solution of the original problem P. In this case, let us define π
f by the formula π
0
f
=
0, and π
n
f
=
η
n for n
=
0
,
…
,
N, as before.


## 4.4. A Parametric Analysis

We could be interested by the variation of the criteria, the unit cost of the resulting (corrected or not) mixture, with respect to the costc
0. This information is obtained from a parametric analysis of the criteria with respect to the parameter c
0. We split the criteria of P
~
f in two parts: (26)
z
~
f
η
=
∑
n
=
1
N
c
n
η
n
+
c
0
η
0
=
z
~
1
f
η
+
c
0
z
~
2
f
ηand we consider the following bicriteria problem: P
~
f
s
min
⁡
z
~
1
f
η
=
∑
n
=
1
N
c
n
η
n
min
⁡
z
~
2
f
η
=
η
0
subject to
η
∈
S
~
f
.The Pareto curve of this problem, obtained by considering the weighted sum, is (27)
z
~
λ
f
η
=
1
-
λ
z
~
1
f
η
+
λ
z
~
2
f
η and the correspondence (28)
λ
=
c
0
1
+
c
0
or
c
0
=
λ
1
-
λ
,allows us to find a piecewise linear continuous increasing concave representation for the optimal minimal unit cost z
∗
f
c
0
=
min
⁡
z
f
π
f in terms of c
0.Since(29)
z
c
o
s
t
π
f
=
∑
n
=
1
N
c
n
π
n
f
=
1
η
0
∑
n
=
1
N
c
n
η
n
=
1
η
0
z
~
1
f
η
if
η
0
>
0
or
π
0
f
=
1
,
∑
n
=
1
N
c
n
η
n
=
z
~
1
f
η
if
η
0
=
0
or
π
0
f
=
0
,
z
w
e
i
g
h
t
π
f
=
∑
n
=
1
N
π
n
f
=
1
η
0
∑
n
=
1
N
η
n
=
1
η
0
1
-
η
0
if
η
0
>
0
or
π
0
f
=
1
,
∑
n
=
1
N
η
n
=
1
if
η
0
=
0
or
π
0
f
=
0
,we have (30)
z
∗
f
c
0
=
c
0
π
0
f
+
z
c
o
s
t
π
f
π
0
f
+
z
w
e
i
g
h
t
π
f
,that is, valid not only for π
0
f
=
1 but also for π
0
f
=
0 and where π
f is the optimal solution.


## 4.5. Numerical Example

We consider two situations with the preceding data.Case 1.
We use all the materials, withoutn
=
5 and n
=
6. In this case, S is empty and S
a is nonempty and bounded; hence η
0 will always be strictly positive. Tables 11 and 12 present results for P
~
f
s and the corresponding values for P
f.Table 11
Efficient vertices ofP
~
f
s and P
f in the criteria space R
2.
(a)
λ
_
λ
¯
z
~
1
f
(
η
)
z
~
2
f
(
η
)
0
0.124887
0.002781
0.980820
0.124887
0.127242
0.005732
0.960140
0.127242
0.249832
0.011545
0.920268
0.249832
0.343079
0.024633
0.880968
0.343079
0.604078
0.045238
0.841514
0.604078
0.710795
0.045834
0.841124
0.710795
0.874033
0.056158
0.836911
0.874033
1
0.056666
0.836838
(b)
c
_
0
c
¯
0
z
c
o
s
t
f
π
f
z
w
e
i
g
n
t
f
π
f
0
0.142710
0.002835
0.019554
0.142710
0.145794
0.005970
0.041513
0.145794
0.333035
0.012545
0.086638
0.333035
0.522255
0.027961
0.135113
0.522255
1.525752
0.053758
0.188333
1.525752
2.450600
0.054491
0.188885
2.450600
6.938604
0.067101
0.194869
6.938604
+
∞
0.067714
0.194974Table 12
Optimal solutions ofP
~
f
s and P
f in the decision space R
7.
(a)
η
0
η
1
η
2
η
3
η
4
η
5
η
6
η
7
η
8
0.980820
0.019179
0
0
0
—
—
0
0
0.960140
0.038506
0
0.001352
0
—
—
0
0
0.920268
0.079282
0
0.000448
0
—
—
0
0
0.880968
0.098547
0
0
0.020483
—
—
0
0
0.841514
0.096656
0
0
0.061829
—
—
0
0
0.841124
0.094791
0
0
0.062039
—
—
0.002044
0
0.836911
0.081505
0
0
0.064288
—
—
0.001878
0.015415
0.836838
0.081852
0
0
0.064325
—
—
0
0.016983
(b)
π
0
f
π
1
f
π
2
f
π
3
f
π
4
f
π
5
f
π
6
f
π
7
f
π
8
f
1
0.019554
0
0
0
—
—
0
0
1
0.040105
0
0.001408
0
—
—
0
0
1
0.086151
0
0.000486
0
—
—
0
0
1
0.111862
0
0
0.023251
—
—
0
0
1
0.114859
0
0
0.073473
—
—
0
0
1
0.112696
0
0
0.073758
—
—
0.002430
0
1
0.097389
0
0
0.076816
—
—
0.002244
0.018420
1
0.097811
0
0
0.076867
—
—
0
0.020295Case 2.
We consider all theN
=
8 materials; then S is nonempty and S
a is nonempty and not bounded; hence it might happen that η
0 be 0 for large values of c
0. Tables 13 and 14 present results for P
~
f
s and the corresponding values for P
f.Table 13
Efficient vertices ofP
~
f
s and P
f in the criteria space R
2.
(a)
λ
_
λ
¯
z
~
1
f
(
η
)
z
~
2
f
(
η
)
0
0.124887
0.002781
0.980820
0.124887
0.127242
0.005732
0.960140
0.127242
0.249832
0.011545
0.920267
0.249832
0.261550
0.024633
0.880970
0.261550
0.270235
0.153784
0.516323
0.270235
1
0.344983
0
(b)
c
_
0
c
¯
0
z
c
o
s
t
f
(
π
f
)
z
w
e
i
g
n
t
f
(
π
f
)
0
0.142710
0.002835
0.019554
0.142710
0.145794
0.005970
0.041513
0.145794
0.333035
0.012545
0.086640
0.333035
0.354188
0.027961
0.135112
0.354188
0.370304
0.297848
0.936768
0.370304
+
∞
0.344984
1Table 14
Optimal solutions ofP
~
f
s and P
f in the decision space R
9.
(a)
η
0
η
1
η
2
η
3
η
4
η
5
η
6
η
7
η
8
0.980820
0.019179
0
0
0
0
0
0
0
0.960140
0.038506
0
0.001352
0
0
0
0
0
0.920267
0.079284
0
0.000448
0
0
0
0
0
0.880970
0.098547
0
0
0.020482
0
0
0
0
0.516323
0.277318
0
0
0.151270
0.055086
0
0
0
0
0.530479
0
0.004485
0.272517
0.192517
0
0
0
(b)
π
0
f
π
1
f
π
2
f
π
3
f
π
4
f
π
5
f
π
6
f
π
7
f
π
8
f
1
0.019554
0
0
0
0
0
0
0
1
0.040104
0
0.001408
0
0
0
0
0
1
0.086153
0
0.000486
0
0
0
0
0
1
0.111862
0
0
0.023249
0
0
0
0
1
0.537102
0
0
0.292976
0.106690
0
0
0
0
0.530479
0
0.004485
0.272517
0.192517
0
0
0Let us observe that the results forc
0
=
0 correspond to the results obtained in Section 3.


## 5. Conclusion

Three simple mixture problems occurring in metallurgy were analyzed. We first considered the basic mixture problem. We then considered a correction problem to a “free” premix. We have shown that, in the case of a free premix, the minimal cost correction does not generally correspond to the minimum weight correction. We built the Pareto curve that gives all intermediate values between these two extremes. The same problem, with a “nonfree” premix, leads to a unique minimal unit cost solution.



---

*Source: 2742540-2017-02-27.xml*