# A New Approach for Optimization of Real Life Transportation Problem in Neutrosophic Environment

Authors: A. Thamaraiselvi; R. Santhi
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

Neutrosophic sets have been introduced as a generalization of crisp sets, fuzzy sets, and intuitionistic fuzzy sets to represent uncertain, inconsistent, and incomplete information about a real world problem. For the first time, this paper attempts to introduce the mathematical representation of a transportation problem in neutrosophic environment. The necessity of the model is discussed. A new method for solving transportation problem with indeterminate and inconsistent information is proposed briefly. A real life example is given to illustrate the efficiency of the proposed method in neutrosophic approach.


---

## Body

## 1. Introduction

In the present day, problems are there with different types of uncertainties which cannot be solved by classical theory of mathematics. To deal with the problems with imprecise or vague information, Zadeh [1] first introduced the fuzzy set theory in 1965, which is characterized by its membership values. But, in many situations, the results or decisions based on the available information are not enough to the level of accuracy. So several higher order fuzzy sets were introduced to deal with such problems. One was the concept of intuitionistic fuzzy set introduced by Atanassov [2] in 1986. Intuitionistic fuzzy sets are suitable to handle problems with imprecision information and are characterized by its membership and nonmembership values [3]. Hence, both the theories of fuzzy and intuitionistic fuzzy sets were applied in many real life decision making problems.In due course, any generalization of fuzzy set failed to handle problems with indeterminate or inconsistent information. To overcome this, Smarandache [4], in 1998, introduced neutrosophic sets as an extension of classical sets, fuzzy sets, and intuitionistic fuzzy sets. The components of neutrosophic set, namely, truth-membership degree, indeterminacy-membership degree, and falsity-membership degree, were suitable to represent indeterminacy and inconsistent information. Wang et al. [5] introduced the idea of single valued neutrosophic set in many practical problems. The notion of single valued neutrosophic set was more suitable for solving many real life problems like image processing, medical diagnosis, decision making, water resource management, and supply chain management.Study of optimal transportation model with cost effective manner played a predominant role in supply chain management. Many researchers [6, 7] formulated the mathematical model for transportation problem in various environments. The basic transportation model was introduced by Hitchcock [8], in 1941, in which the transportation constraints were based on crisp values. But, in the present world, the transportation parameters like demand, supply, and unit transportation cost may be uncertain due to several uncontrolled factors. In this situation, fuzzy transportation problem was formulated and solved by many researchers.Though many researchers worked on transportation problem in fuzzy environment, it was ÓhÉigeartaigh [9] who proposed a new method to solve fuzzy transportation problem with triangular fuzzy demand and supply. Chanas and Kuchta [10] proposed a method for optimal solution of fuzzy transportation problem with fuzzy cost coefficients. Jiménez and Verdegay [11] investigated the solution of fuzzy solid transportation problem in which the transportation parameters are trapezoidal fuzzy numbers. Kaur and Kumar [12] proposed a method for solving transportation problem using ranking function. Nagoor Gani and Abdul Razak [13] introduced a two-stage cost minimization method for solving fuzzy transportation problem. Pandian and Natarajan [14] introduced a new method, namely, fuzzy zero point method, to find the optimal solution for fuzzy transportation problem. Kaur and Kumar [15] solved the fuzzy transportation problem with generalized trapezoidal fuzzy numbers.Sometimes the membership function in fuzzy set theory was not a suitable one to describe an ambiguous situation of a problem. So, in 1986, Atanassov [2] introduced the concept of intuitionistic fuzzy set theory as an extension of fuzzy set theory, which included the degree of both membership and nonmembership of each element in the set. In recent research, intuitionistic fuzzy set theory plays an important role in decision making problems [16, 17]. Many researchers [18, 19] used intuitionistic fuzzy approach to solve transportation problems.In a supply chain optimization, transportation system was the most important economic activity among all the components of business logistics system. Apart from the vagueness or uncertainty in the constraints of the present day transportation model, there exists some indeterminacy due to various factors like unawareness of the problem, imperfection in data, and poor forecasting. Intuitionistic fuzzy set theory can handle incomplete information but not indeterminate and inconsistent information. Smarandache [20] proposed a new theory, namely, neutrosophic logic, by adding another independent membership function named as indeterminacy-membership I
x along with truth membership T
x and falsity F
x membership functions. Neutrosophic set is a generalization of intuitionistic fuzzy sets. If hesitancy degree H
x of intuitionistic fuzzy set and the indeterminacy-membership degree I
x of neutrosophic set are equal, then neutrosophic set will become the intuitionistic fuzzy set.Even though many scholars applied the notion of neutrosophic theories in multiattribute decision making problems [21–23] to the best of their knowledge, the existing supply chain theories of transportation model are not viewed in neutrosophic logic. For example, in a given conclusion, “The total transportation cost of delivering the goods would be 1000 units,” the supplier cannot conclude immediately that the precise cost is exactly 1000 units. There may be some neutral part, which is neither truthfulness nor falsity of the statement. This is very close to our human mind reasoning. In the neutral part, there may be some indeterminacy in deciding unit transportation cost, demand and supply units due to various causes like vehicle routing, road factors, no uniformity in traffic regulations, delivery time of goods, poor demand forecasting, demand mismatches, price fluctuations, lack of trust, and so on.The aim of this paper is to obtain the optimal transportation cost in neutrosophic environment. This paper is well organized as follows. In Section2, the basic concepts of fuzzy sets, intuitionistic fuzzy sets, and neutrosophic sets are briefly reviewed. In Section 3, the mathematical model of neutrosophic transportation problem is introduced. In Section 4, the solution algorithms are developed for solving neutrosophic transportation problem. In Section 5, the algorithms are illustrated with suitable real life problems. In Section 6, the results are interpreted. Finally, Section 7 concludes the paper with future work.


## 2. Preliminaries

Definition 1 (fuzzy set, see [1]).
LetX be a nonempty set. A fuzzy set A
- of X is defined as A
-
=
x
,
μ
A
¯
x
x
∈
X where μ
A
¯
x is called the membership function which maps each element of X to a value between 0 and 1.Definition 2 (fuzzy number).
A fuzzy numberA
~ is a convex normalized fuzzy set on the real line R such that(i)
there exist at least onex
∈
R with μ
A
~
x
=
1;
(ii)
μ
A
~
(
x
) is piecewise continuous.Definition 3 (trapezoidal fuzzy number, see [15]).
A fuzzy numberA
~
=
a
1
,
a
2
,
a
3
,
a
4 is a trapezoidal fuzzy number, where a
1, a
2, a
3, and a
4 are real numbers and its membership function is given as follows: (1)
μ
A
~
x
=
x
-
a
1
a
2
-
a
1
,
for
a
1
≤
x
≤
a
2
1
,
for
a
2
≤
x
≤
a
3
a
4
-
x
a
4
-
a
3
,
for
a
3
≤
x
≤
a
4
0
,
otherwise.Definition 4 (intuitionistic fuzzy set, see [2]).
LetX be a nonempty set. An intuitionistic fuzzy set A
-
I of X is defined as A
-
I
=
x
,
μ
A
-
I
x
,
ν
A
-
I
x
x
∈
X where μ
A
-
I
x and ν
A
-
I
x are membership and nonmembership functions such that μ
A
-
I
x
,
ν
A
-
I
x
:
X
→
[
0,1
] and 0
≤
μ
A
-
I
x
+
ν
A
-
I
x
≤
1 for all x
∈
X.Definition 5 (intuitionistic fuzzy number, see [3]).
An intuitionistic fuzzy subsetA
~
I
=
x
,
μ
A
~
I
x
,
ν
A
~
I
x
x
∈
R of the real line R is called an intuitionistic fuzzy number (IFN) if the following conditions hold:(i)
There existsm
∈
R such that μ
A
~
I
m
=
1 and ν
A
~
I
m
=
0.
(ii)
μ
A
~
I is a continuous function from R
→
[
0,1
] such that 0
≤
μ
A
~
I
x
+
ν
A
~
I
x
≤
1 for all x
∈
X.
(iii)
The membership and nonmembership functions ofA
~
I are in the following form: (2)
μ
A
~
I
x
=
0
,
for
-
∞
<
x
≤
a
1
f
x
,
for
a
1
≤
x
≤
a
2
1
,
for
x
=
a
2
g
x
,
for
a
2
≤
x
≤
a
3
0
,
for
a
3
≤
x
<
∞
,
ν
A
~
I
x
=
1
,
for
-
∞
<
x
≤
a
1
′
f
′
x
,
for
a
1
′
≤
x
≤
a
2
0
,
for
x
=
a
2
g
′
x
,
for
a
2
≤
x
≤
a
3
′
1
,
for
a
3
′
≤
x
<
∞
,where f
,
f
′
,
g
,
g
′ are functions from R
→
0,1, f and g
′ are strictly increasing functions, and g and f
′ are strictly decreasing functions with the conditions 0
≤
f
x
+
f
′
x
≤
1 and 0
≤
g
x
+
g
′
x
≤
1.Definition 6 (trapezoidal intuitionistic fuzzy number, see [16]).
A trapezoidal intuitionistic fuzzy number is denoted byA
~
I
=
a
1
,
a
2
,
a
3
,
a
4, a
1
′
,
a
2
,
a
3
,
a
4
′, where a
1
′
≤
a
1
≤
a
2
≤
a
3
≤
a
4
≤
a
4
′ with membership and nonmembership functions are defined as follows:(3)
μ
A
~
x
=
x
-
a
1
a
2
-
a
1
,
for
a
1
≤
x
≤
a
2
1
,
for
a
2
≤
x
≤
a
3
a
4
-
x
a
4
-
a
3
,
for
a
3
≤
x
≤
a
4
0
,
otherwise,
ν
A
~
x
=
a
2
-
x
a
2
-
a
1
′
,
for
a
1
′
≤
x
≤
a
2
0
,
for
a
2
≤
x
≤
a
3
x
-
a
3
a
4
′
-
a
3
,
for
a
3
≤
x
≤
a
4
′
1
,
otherwise.Definition 7 (neutrosophic set, see [4]).
LetX be a nonempty set. Then a neutrosophic set A
-
N of X is defined as A
-
N
=
x
,
T
A
-
N
x
,
I
A
-
N
x
,
F
A
-
N
x
x
∈
X
,
T
A
-
N
x
,
I
A
-
N
x
,
F
A
-
N
x
∈
0
-
,
1
+, where T
A
-
N
x, I
A
-
N
x and F
A
-
N
x are truth membership function, an indeterminacy-membership function, and a falsity-membership function and there is no restriction on the sum of T
A
-
N
x, I
A
-
N
x and F
A
-
N
x, so 0
-
≤
T
A
-
N
x
+
I
A
-
N
x
+
F
A
-
N
x
≤
3
+ and 0
-
,
1
+ is a nonstandard unit interval.
But it is difficult to apply neutrosophic set theories in real life problems directly. So Wang introduced single valued neutrosophic set as a subset of neutrosophic set and the definition is as follows.Definition 8 (single valued neutrosophic set, see [5]).
LetX be a nonempty set. Then a single valued neutrosophic set A
-
S
N of X is defined as A
-
S
N
=
{
x
,
T
A
-
N
x
,
I
A
-
N
x
,
F
A
-
N
x
x
∈
X
} where T
A
-
N
x, I
A
-
N
x and F
A
-
N
x
∈
[
0,1
] for each x
∈
X and 0
≤
T
A
-
N
x
+
I
A
-
N
x
+
F
A
-
N
x
≤
3.Definition 9 (single valued trapezoidal neutrosophic number).
Letw
a
~
,
u
a
~
,
y
a
~
∈
[
0,1
] and a
1
,
a
2
,
a
3
,
a
4
∈
R such that a
1
≤
a
2
≤
a
3
≤
a
4. Then a single valued trapezoidal neutrosophic number, a
~
=
a
1
,
a
2
,
a
3
,
a
4
;
w
a
~
,
u
a
~
,
y
a
~ is a special neutrosophic set on the realline set
R, whose truth-membership, indeterminacy-membership, and falsity-membership functions are given as follows:(4)
μ
a
~
x
=
w
a
~
x
-
a
1
a
2
-
a
1
,
for
a
1
≤
x
≤
a
2
w
a
~
,
for
a
2
≤
x
≤
a
3
w
a
~
a
4
-
x
a
4
-
a
3
,
for
a
3
≤
x
≤
a
4
0
,
otherwise,
ν
a
~
x
=
a
2
-
x
+
u
a
~
x
-
a
1
a
2
-
a
1
,
for
a
1
≤
x
≤
a
2
u
a
~
,
for
a
2
≤
x
≤
a
3
x
-
a
3
+
u
a
~
a
4
-
x
a
4
-
a
3
,
for
a
3
≤
x
≤
a
4
1
,
otherwise,
λ
a
~
x
=
a
2
-
x
+
y
a
~
x
-
a
1
a
2
-
a
1
,
for
a
1
≤
x
≤
a
2
y
a
~
,
for
a
2
≤
x
≤
a
3
x
-
a
3
+
y
a
~
a
4
-
x
a
4
-
a
3
,
for
a
3
≤
x
≤
a
4
1
,
otherwise,where w
a
~, u
a
~, and y
a
~ denote the maximum truth-membership degree, minimum-indeterminacy membership degree, and minimum falsity-membership degree, respectively. A single valued trapezoidal neutrosophic number a
~
=
a
1
,
a
2
,
a
3
,
a
4
;
w
a
~
,
u
a
~
,
y
a
~ may express an ill-defined quantity about a, which is approximately equal to [
a
2
,
a
3
].Definition 10 (arithmetic operations on single valued trapezoidal neutrosophic numbers).
Leta
~
=
a
1
,
a
2
,
a
3
,
a
4
;
w
a
~
,
u
a
~
,
y
a
~ and b
~
=
b
1
,
b
2
,
b
3
,
b
4
;
w
b
~
,
u
b
~
,
y
b
~ be two single valued trapezoidal neutrosophic numbers and k
≠
0; then(5)
i
a
~
+
b
~
=
a
1
+
b
1
,
a
2
+
b
2
,
a
3
+
b
3
,
a
4
+
b
4
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
,
ii
a
~
-
b
~
=
a
1
-
b
4
,
a
2
-
b
3
,
a
3
-
b
2
,
a
4
-
b
1
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
,
iii
b
~
=
a
1
b
1
,
a
2
b
2
,
a
3
b
3
,
a
4
b
4
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
if
a
4
>
0
,
b
4
>
0
a
1
b
4
,
a
2
b
3
,
a
3
b
2
,
a
4
b
1
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
if
a
4
<
0
,
b
4
>
0
a
4
b
4
,
a
3
b
3
,
a
2
b
2
,
a
1
b
1
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
if
a
4
<
0
,
b
4
<
0
,
iv
a
~
b
~
=
a
1
b
4
,
a
2
b
3
,
a
3
b
2
,
a
4
b
1
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
if
a
4
>
0
,
b
4
>
0
a
4
b
4
,
a
3
b
3
,
a
2
b
2
,
a
1
b
1
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
if
a
4
<
0
,
b
4
>
0
a
4
b
1
,
a
3
b
2
,
a
2
b
3
,
a
1
b
4
;
w
a
~
∧
w
b
~
,
u
a
~
∨
u
b
~
,
y
a
~
∨
y
b
~
if
a
4
<
0
,
b
4
<
0
,
v
k
a
~
=
k
a
1
,
k
a
2
,
k
a
3
,
k
a
4
;
w
a
~
,
u
a
~
,
y
a
~
if
k
>
0
k
a
4
,
k
a
3
,
k
a
2
,
k
a
1
;
w
a
~
,
u
a
~
,
y
a
~
if
k
<
0
,
vi
a
~
-
1
=
1
a
4
,
1
a
3
,
1
a
2
,
1
a
1
;
w
a
~
,
u
a
~
,
y
a
~
,
where
a
~
≠
0
.Definition 11 (score and accuracy functions of single valued trapezoidal neutrosophic number).
One can compare any two single valued trapezoidal neutrosophic numbers based on the score and accuracy functions. Leta
~
=
a
1
,
a
2
,
a
3
,
a
4
;
w
a
,
u
a
,
y
a be a single valued trapezoidal neutrosophic number; then (i)
score functionS
a
~
=
1
/
16
a
1
+
a
2
+
a
3
+
a
4
×
μ
a
~
+
1
-
ν
a
~
+
(
1
-
λ
a
~
);
(ii)
accuracy functionA
a
~
=
1
/
16
a
1
+
a
2
+
a
3
+
a
4
×
μ
a
~
+
1
-
ν
a
~
+
(
1
+
λ
a
~
)
.Definition 12 (comparison of single valued trapezoidal neutrosophic number).
Leta
~ and b
~ be any two single valued trapezoidal neutrosophic numbers; then one has the following:(i)
IfS
a
~
<
S
b
~ then a
~
<
b
~.
(ii)
IfS
a
~
=
S
b
~ and if
(1)
A
a
~
<
A
b
~ then a
~
<
b
~,
(2)
A
a
~
>
A
b
~ then a
~
>
b
~,
(3)
A
a
~
=
A
b
~ then a
~
=
b
~
.Example 13.
Leta
~
=
4,8
,
10,16
;
0.5,0.3,0.6 and b
~
=
3,7
,
11,14
;
0.4,0.5,0.6 be two single valued trapezoidal neutrosophic numbers; then(i)
a
~
+
b
~
=
7,15,21,30
;
0.4,0.5,0.6,
(ii)
a
~
-
b
~
=
-
10
,
-
3,3
,
13
;
0.4,0.5,0.6,
(iii)
a
~
b
~
=
12,56,110,224
;
0.4,0.5,0.6,
(iv)
a
~
/
b
~
=
4
/
14
,
8
/
11
,
10
/
7
,
16
/
3
;
0.4,0.5,0.6,
(v)
3
a
~
=
12,24,30,48
;
0.4,0.5,0.6,
(vi)
a
~
-
1
=
1
/
16
,
1
/
10
,
1
/
8
,
1
/
4
;
0.4,0.5,0.6,
(vii)
S
a
~
=
1
/
16
4
+
8
+
10
+
16
×
0.5
+
1
-
0.3
+
1
-
0.6
=
3.8,
(viii)
A
a
~
=
1
/
16
4
+
8
+
10
+
16
×
0.5
+
1
-
0.3
+
(
1
+
0.6
)
=
6.65
.


## 3. Introduction of Transportation Problem in Neutrosophic Environment

### 3.1. Mathematical Formulation

#### 3.1.1. Model I

In this model, a transportation problem is introduced in a single valued neutrosophic environment. Consider a transportation problem with “m” sources and “n” destinations in which the decision maker is indeterminate about the precise values of transportation cost from ith source to jth destination, but there is no uncertainty about the demand and supply of the product with the following assumptions and constraints.Distribution Assumptions
i is the source index for all i
=
1
,
2,3
,
…
,
m
.
j is the destination index for all j
=
1
,
2,3
,
…
,
n
.Transportation Parameters
x
i
j is the number of units of the product transported from ith source to jth destination.
C
~
i
j
N is the neutrosophic cost of one unit quantity transported from ith source to jth destination.
x
i is the total availability of the product at the source i.
x
j is the total demand of the product at the destination j.Transportation Constraints
Supply constraints:∑
j
=
0
n
x
i
j
=
a
i for all sourcesi.
Demand constraints:∑
i
=
0
m
x
i
j
=
b
j for all destinations j and ∑
i
=
0
m
a
i
=
∑
j
=
0
n
b
j.
Nonnegativity constraints:x
i
j
≥
0
∀
i
,
j
.
Now the mathematical formulation of the problem is given by(6)
Minimize
Z
~
N
=
∑
i
=
0
m
∑
j
=
0
n
x
i
j
c
~
i
j
N
subject to
∑
j
=
0
n
x
i
j
=
a
i
,
i
=
1,2
,
…
,
m
∑
i
=
0
m
x
i
j
=
b
j
,
j
=
1,2
,
…
,
n
,
x
i
j
≥
0
∀
i
,
j
.


#### 3.1.2. Model II

In this model the decision maker will not be sure about the unit transportation costs, supply, and the demand units. So the mathematical formulation of the problem becomes(7)
Minimize
Z
~
N
=
∑
i
=
0
m
∑
j
=
0
n
x
~
i
j
N
c
~
i
j
N
subject to
∑
j
=
0
n
x
~
i
j
N
=
a
~
i
N
,
i
=
1,2
,
…
,
m
∑
i
=
0
m
x
~
i
j
N
=
b
~
j
N
,
j
=
1,2
,
…
,
n
,
x
~
i
j
N
≥
0
∀
i
,
j
.


## 3.1. Mathematical Formulation

### 3.1.1. Model I

In this model, a transportation problem is introduced in a single valued neutrosophic environment. Consider a transportation problem with “m” sources and “n” destinations in which the decision maker is indeterminate about the precise values of transportation cost from ith source to jth destination, but there is no uncertainty about the demand and supply of the product with the following assumptions and constraints.Distribution Assumptions
i is the source index for all i
=
1
,
2,3
,
…
,
m
.
j is the destination index for all j
=
1
,
2,3
,
…
,
n
.Transportation Parameters
x
i
j is the number of units of the product transported from ith source to jth destination.
C
~
i
j
N is the neutrosophic cost of one unit quantity transported from ith source to jth destination.
x
i is the total availability of the product at the source i.
x
j is the total demand of the product at the destination j.Transportation Constraints
Supply constraints:∑
j
=
0
n
x
i
j
=
a
i for all sourcesi.
Demand constraints:∑
i
=
0
m
x
i
j
=
b
j for all destinations j and ∑
i
=
0
m
a
i
=
∑
j
=
0
n
b
j.
Nonnegativity constraints:x
i
j
≥
0
∀
i
,
j
.
Now the mathematical formulation of the problem is given by(6)
Minimize
Z
~
N
=
∑
i
=
0
m
∑
j
=
0
n
x
i
j
c
~
i
j
N
subject to
∑
j
=
0
n
x
i
j
=
a
i
,
i
=
1,2
,
…
,
m
∑
i
=
0
m
x
i
j
=
b
j
,
j
=
1,2
,
…
,
n
,
x
i
j
≥
0
∀
i
,
j
.


### 3.1.2. Model II

In this model the decision maker will not be sure about the unit transportation costs, supply, and the demand units. So the mathematical formulation of the problem becomes(7)
Minimize
Z
~
N
=
∑
i
=
0
m
∑
j
=
0
n
x
~
i
j
N
c
~
i
j
N
subject to
∑
j
=
0
n
x
~
i
j
N
=
a
~
i
N
,
i
=
1,2
,
…
,
m
∑
i
=
0
m
x
~
i
j
N
=
b
~
j
N
,
j
=
1,2
,
…
,
n
,
x
~
i
j
N
≥
0
∀
i
,
j
.


## 3.1.1. Model I

In this model, a transportation problem is introduced in a single valued neutrosophic environment. Consider a transportation problem with “m” sources and “n” destinations in which the decision maker is indeterminate about the precise values of transportation cost from ith source to jth destination, but there is no uncertainty about the demand and supply of the product with the following assumptions and constraints.Distribution Assumptions
i is the source index for all i
=
1
,
2,3
,
…
,
m
.
j is the destination index for all j
=
1
,
2,3
,
…
,
n
.Transportation Parameters
x
i
j is the number of units of the product transported from ith source to jth destination.
C
~
i
j
N is the neutrosophic cost of one unit quantity transported from ith source to jth destination.
x
i is the total availability of the product at the source i.
x
j is the total demand of the product at the destination j.Transportation Constraints
Supply constraints:∑
j
=
0
n
x
i
j
=
a
i for all sourcesi.
Demand constraints:∑
i
=
0
m
x
i
j
=
b
j for all destinations j and ∑
i
=
0
m
a
i
=
∑
j
=
0
n
b
j.
Nonnegativity constraints:x
i
j
≥
0
∀
i
,
j
.
Now the mathematical formulation of the problem is given by(6)
Minimize
Z
~
N
=
∑
i
=
0
m
∑
j
=
0
n
x
i
j
c
~
i
j
N
subject to
∑
j
=
0
n
x
i
j
=
a
i
,
i
=
1,2
,
…
,
m
∑
i
=
0
m
x
i
j
=
b
j
,
j
=
1,2
,
…
,
n
,
x
i
j
≥
0
∀
i
,
j
.


## 3.1.2. Model II

In this model the decision maker will not be sure about the unit transportation costs, supply, and the demand units. So the mathematical formulation of the problem becomes(7)
Minimize
Z
~
N
=
∑
i
=
0
m
∑
j
=
0
n
x
~
i
j
N
c
~
i
j
N
subject to
∑
j
=
0
n
x
~
i
j
N
=
a
~
i
N
,
i
=
1,2
,
…
,
m
∑
i
=
0
m
x
~
i
j
N
=
b
~
j
N
,
j
=
1,2
,
…
,
n
,
x
~
i
j
N
≥
0
∀
i
,
j
.


## 4. Procedure for Proposed Algorithms Based on Neutrosophic Numbers

### 4.1. Basic Assumptions of the Proposed Algorithms

(1) Requirement Assumption. The entire supply units from each source must be distributed to destinations.(2) Feasible Solution Assumption. The neutrosophic transportation problem will have feasible solution if and only if ∑
i
=
0
m
a
i
=
∑
j
=
0
n
b
j,i
=
1,2
,
…
,
m and j
=
1,2
,
…
,
n.(3) Cost Assumption. The total transportation cost depends only on the number of units transported and the unit transportation cost but not on other factors like distance and mode of transport.(4) Input Assumption. The parameters of the problem will be represented by either crisp or trapezoidal neutrosophic numbers.


### 4.2. Neutrosophic Initial Basic Feasible Solution for Model I

Step 1.
Calculate the score value of each neutrosophic costc
~
i
j
N and replace all the neutrosophic costs by its score value to obtain the classical transportation problem.Step 2.
For each row and column of the table obtained in Step1, calculate the difference between minimum and next to minimum of the transportation costs and denote it as Penalty.Step 3.
In the row/column, corresponding to maximum penalty, make the maximum allotment in the cell having the minimum transportation cost.Step 4.
If the maximum penalty corresponding to(i)
more than one row, select the topmost row,
(ii)
more than one column, select the extreme left column.Repeat the above procedure until all the supplies are fully exhausted and all the demands are satisfied.


### 4.3. Neutrosophic Optimal Solution for Model I

Step 1.
Convert each neutrosophic cost into crisp value by the score function and obtain the classical transportation problem.Step 2.
Choose the minimum in each row and subtract it from the corresponding row entries. Do the same procedure for each column. Now there will be at least one zero in each row and column in the resultant table.Step 3.
Verify whether the demand of each column is less than the sum of supplies whose reduced costs are zero in that column and supply of each row are less than the sum of demands whose reduced costs in that row are zero. If so, go to Step5, otherwise go to Step 4.Step 4.
Draw the minimum number of horizontal and vertical lines that cover all the zeros in the reduced table and revise the table as follows:(i)
Find the least element from the uncovered entries.
(ii)
Subtract it from all the uncovered entries and add it to the entries at the intersection of any two lines.
Again check the condition at Step3.Step 5.
Select a cell(
x
,
y
) whose reduced cost is maximum in the reduced cost table. If the maximum exists at more than one cell, then select any cell.Step 6.
Select a cell inx-row and y-column, which is the only cell whose reduced cost is zero. And then allot the maximum possible units in it. If such cell does not occur for the maximum cost, go for next maximum. If such cell does not occur for any value, then choose any cell at random whose reduced cost is zero.Step 7.
Revise the reduced table by omitting fully exhausted row and fully satisfied column and repeat Steps5 and 6 again.Repeat the procedure until all the supply units are fully used and all the demand units are fully received.


### 4.4. Neutrosophic Solution for Model II

Initial basic feasible solution and the optimum solution of Model II will be obtained by the procedure in Sections4.1 and 4.2 without altering the neutrosophic demand and supply units. Subtraction of single valued trapezoidal neutrosophic numbers is applied to modify the neutrosophic demand and supply units in each iteration.


## 4.1. Basic Assumptions of the Proposed Algorithms

(1) Requirement Assumption. The entire supply units from each source must be distributed to destinations.(2) Feasible Solution Assumption. The neutrosophic transportation problem will have feasible solution if and only if ∑
i
=
0
m
a
i
=
∑
j
=
0
n
b
j,i
=
1,2
,
…
,
m and j
=
1,2
,
…
,
n.(3) Cost Assumption. The total transportation cost depends only on the number of units transported and the unit transportation cost but not on other factors like distance and mode of transport.(4) Input Assumption. The parameters of the problem will be represented by either crisp or trapezoidal neutrosophic numbers.


## 4.2. Neutrosophic Initial Basic Feasible Solution for Model I

Step 1.
Calculate the score value of each neutrosophic costc
~
i
j
N and replace all the neutrosophic costs by its score value to obtain the classical transportation problem.Step 2.
For each row and column of the table obtained in Step1, calculate the difference between minimum and next to minimum of the transportation costs and denote it as Penalty.Step 3.
In the row/column, corresponding to maximum penalty, make the maximum allotment in the cell having the minimum transportation cost.Step 4.
If the maximum penalty corresponding to(i)
more than one row, select the topmost row,
(ii)
more than one column, select the extreme left column.Repeat the above procedure until all the supplies are fully exhausted and all the demands are satisfied.


## 4.3. Neutrosophic Optimal Solution for Model I

Step 1.
Convert each neutrosophic cost into crisp value by the score function and obtain the classical transportation problem.Step 2.
Choose the minimum in each row and subtract it from the corresponding row entries. Do the same procedure for each column. Now there will be at least one zero in each row and column in the resultant table.Step 3.
Verify whether the demand of each column is less than the sum of supplies whose reduced costs are zero in that column and supply of each row are less than the sum of demands whose reduced costs in that row are zero. If so, go to Step5, otherwise go to Step 4.Step 4.
Draw the minimum number of horizontal and vertical lines that cover all the zeros in the reduced table and revise the table as follows:(i)
Find the least element from the uncovered entries.
(ii)
Subtract it from all the uncovered entries and add it to the entries at the intersection of any two lines.
Again check the condition at Step3.Step 5.
Select a cell(
x
,
y
) whose reduced cost is maximum in the reduced cost table. If the maximum exists at more than one cell, then select any cell.Step 6.
Select a cell inx-row and y-column, which is the only cell whose reduced cost is zero. And then allot the maximum possible units in it. If such cell does not occur for the maximum cost, go for next maximum. If such cell does not occur for any value, then choose any cell at random whose reduced cost is zero.Step 7.
Revise the reduced table by omitting fully exhausted row and fully satisfied column and repeat Steps5 and 6 again.Repeat the procedure until all the supply units are fully used and all the demand units are fully received.


## 4.4. Neutrosophic Solution for Model II

Initial basic feasible solution and the optimum solution of Model II will be obtained by the procedure in Sections4.1 and 4.2 without altering the neutrosophic demand and supply units. Subtraction of single valued trapezoidal neutrosophic numbers is applied to modify the neutrosophic demand and supply units in each iteration.


## 5. Illustrative Example

### 5.1. Model I

Consider a transportation problem in which the peanuts are initially stored at three sources, namely, O1, O2, and O3 and are transported to peanut butter manufacturing company located at four different destinations, namely D1, D2, D3, and D4, with trapezoidal neutrosophic unit transportation cost and crisp demand and supply as given in Table1. Obtain the optimal transportation of peanuts to minimize the total transportation cost.Table 1
Input data for neutrosophic transportation problem.
D1
D2
D3
D4
Supply
O1
(3, 5, 6, 8); 0.6, 0.5, 0.4
(5, 8, 10, 14); 0.3, 0.6, 0.6
(12, 15, 19, 22); 0.6, 0.4, 0.5
(14, 17, 21, 28); 0.8, 0.2.0.6
26
O2
(0, 1, 3, 6); 0.7, 0.5, 0.3
(5, 7, 9, 11); 0.9, 0.7, 0.5
(15, 17, 19, 22); 0.4, 0.8, 0.4
(9, 11, 14, 16); 0.5, 0.4, 0.7
24
O3
(4, 8, 11, 15); 0.6, 0.3, 0.2
(1, 3, 4, 6); 0.6, 0.3, 0.5
(5, 7, 8, 10); 0.5, 0.4, 0.7
(5, 9, 14, 19); 0.3, 0.7, 0.6
30
Demand
17
23
28
12


#### 5.1.1. Neutrosophic Initial Basic Feasible Solution for Model I

Now by score function of trapezoidal neutrosophic number, calculate the score value of each neutrosophic cost to obtain the crisp transportation problem. (Here score values are rounded off to the nearest integer.) The results are given in Table2. Then calculate the penalty for each row and each column which is presented in Table 3.Table 2
Crisp transportation problem.
D1
D2
D3
D4
Supply
O1
2
3
7
10
26
O2
1
3
6
4
24
O3
5
2
3
3
30
Demand
17
23
28
12Table 3
Tabular representation with penalties.
D1
D2
D3
D4
Supply
Penalty
O1
2
3
7
10
26
1
O2
1
3
6
4
24
2
O3
5
2
3
3
30
1
Demand
17
23
28
12
Penalty
1
1
3∗
1In Table3, the highest penalty 3 (marked with ∗) occurs at column 3. Now allot the maximum possible units 28 in the minimum cost cell (
3,3
) and revise the supply units corresponding to row 3. Then the penalties are to be revised in Table 4.Table 4
First allotment with penalties.
D1
D2
D3
D4
Supply
Penalty
O1
2
3
—
10
26
1
O2
1
3
—
4
24
2
O3
5
2
28
3
2
1
Demand
17
23
—
12
Penalty
1
1
—
1Proceeding the neutrosophic initial basic feasible solution algorithm and after few iterations we get the complete allotment transportation units as given in Table5.Table 5
Table with complete allotment.
D1
D2
D3
D4
Supply
O1
—
23
—
3
—
O2
17
—
—
7
—
O3
—
—
28
2
—
Demand
—
—
—
—The initial basic feasible solution is(8)
x
12
=
23
,
x
14
=
3
,
x
21
=
17
,
x
24
=
7
,
x
33
=
28
,
x
34
=
2
.Hence the minimum total neutrosophic cost is(9)
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
i
j
c
~
i
j
N
=
23
5,8
,
10,14
;
0.3,0.6,0.6
+
3
14,17,21,28
;
0.8,0.2
.
0.6
+
17
0,1
,
3,6
;
0.7,0.5,0.3
+
7
9,11,14,16
;
0.5,0.4,0.7
+
28
5,7
,
8,10
;
0.5,0.4,0.7
+
2
5,9
,
14,19
;
0.3,0.7,0.6
=
370,543,694,938
;
0.3,0.7,0.7
.


#### 5.1.2. Neutrosophic Optimum Solution for Model I

Consider the neutrosophic optimal solution for the transportation problem given in Table1. After applying Steps 1 and 2 of the optimal solution algorithm we obtain Table 6.Table 6
Table with zero point.
D1
D2
D3
D4
Supply
O1
0
1
4
7
26
O2
0
2
4
2
24
O3
3
0
0
0
30
Demand
17
23
28
12Now Table6 does not satisfy the optimal solution condition stated in Step 3 of Section 4.2. So proceed to Step 4, and get the revised costs as given in Table 7.Table 7
Modified table with zero point.
D1
D2
D3
D4
Supply
O1
0
0
3
6
26
O2
0
1
3
1
24
O3
4
0
0
0
30
Demand
17
23
28
12As per allotment rules given in Steps5 to 7, one can get the complete allotment schedule which is presented in Table 8.Table 8
Table with complete allocation.
D1
D2
D3
D4
Supply
O1
3
23
—
O2
14
10
—
O3
28
2
—
Demand
—
—
—
—The optimal solution is(10)
x
11
=
3
,
x
12
=
23
,
x
21
=
14
,
x
24
=
10
,
x
33
=
28
,
x
34
=
2
.Hence the minimum total neutrosophic cost is (11)
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
i
j
c
~
i
j
N
=
3
3,5
,
6,8
;
0.6,0.5,0.4
+
23
5,8
,
10,14
;
0.3,0.6,0.6
+
14
0,1
,
3,6
;
0.7,0.5,0.3
+
10
9,11,14,16
;
0.5,0.4,0.7
+
28
5,7
,
8,10
;
0.5,0.4,0.7
+
2
5,9
,
14,19
;
0.3,0.7,0.6
=
364,537,682,908
;
0.3,0.7,0.7
.


### 5.2. Model II

Consider a problem for Model II with single valued neutrosophic trapezoidal cost, demand, and supply given in Table9.Table 9
Input data for neutrosophic transportation problem.
D1
D2
D3
D4
Supply
O1
(3, 5, 6, 8);0.6, 0.5, 0.4
(5, 8, 10, 14); 0.3, 0.6, 0.6
(12, 15, 19, 22); 0.6, 0.4, 0.5
(14, 17, 21, 28); 0.8, 0.2, 0.6
(22, 26, 28, 32); 0.7, 0.3, 0.4
O2
(0, 1, 3, 6);0.7, 0.5, 0.3
(5, 7, 9, 11); 0.9, 0.7, 0.5
(15, 17, 19, 22); 0.4, 0.8, 0.4
(9, 11, 14, 16); 0.5, 0.4, 0.7
(17, 22, 27, 31); 0.6, 0.4, 0.5
O3
(4, 8, 11, 15);0.6, 0.3, 0.2
(1, 3, 4, 6); 0.6, 0.3, 0.5
(5, 7, 8, 10); 0.5, 0.4, 0.7
(5, 9, 14, 19); 0.3, 0.7, 0.6
(21, 28, 32, 37); 0.8, 0.2, 0.4
Demand
(13, 16, 18, 21);0.5, 0.5, 0.6
(17, 21, 24, 28); 0.8, 0.2, 0.4
(24, 29, 32, 35); 0.9, 0.5, 0.3
(6, 10, 13, 15); 0.7, 0.3, 0.4For Table9, calculate the score value of each neutrosophic cost to get crisp cost and consider the demand and supply units as they are. The modified cost is presented in Table 10.Table 10
Neutrosophic transportation problem with crisp cost.
D1
D2
D3
D4
Supply
O1
2
3
7
10
(22, 26, 28, 32);0.7, 0.3, 0.4
O2
1
3
6
4
(17, 22, 27, 31);0.6, 0.4, 0.5
O3
5
2
3
3
(21, 28, 32, 37);0.8, 0.2, 0.4
Demand
(13, 16, 18, 21);0.5, 0.5, 0.6
(17, 21, 24, 28);0.8, 0.2, 0.4
(24, 29, 32, 35);0.9, 0.5, 0.3
(6, 10, 13, 15);0.7, 0.3, 0.4Here, the arithmetic operations of single valued neutrosophic trapezoidal numbers are applied to modify the neutrosophic demand and supply in each iteration. Proceed the neutrosophic optimal solution method and after few iterations the optimal solution in terms of single valued neutrosophic trapezoidal numbers is obtained as follows:(12)
x
11
=
-
6,2
,
7,15
;
0.7,0.3
.
0.4
,
x
12
=
17,21,24,28
;
0.8,0.2,0.4
,
x
21
=
-
8,6
,
19,33
;
0.6,0.5,0.5
,
x
24
=
-
2,8
,
16,25
;
0.7,0.5,0.4
,
x
33
=
26,29,32,35
;
0.9,0.5,0.3
,
x
34
=
-
6,0
,
6,14
;
0.8,0.5,0.4
,
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
~
i
j
N
c
~
i
j
N
=
3,5
,
6,8
;
0.6,0.5,0.4
×
-
6,2
,
7,15
;
0.7,0.3,0.4
+
5,8
,
10,14
;
0.3,0.6,0.6
×
17,21,24,28
;
0.8,0.2,0.4
+
0,1
,
3,6
;
0.7,0.5,0.3
×
-
8,6
,
19,33
;
0.6,0.5,0.5
+
9,11,14,16
;
0.5,0.4,0.7
×
-
2,8
,
16,25
;
0.7,0.5,0.4
+
5,7
,
8,10
;
0.5,0.4,0.7
×
26,29,32,35
;
0.9,0.5,0.3
+
5,9
,
14,19
;
0.3,0.7,0.6
×
-
6,0.6,14
;
0.8,0.5,0.4
=
149,475,903,1726
;
0.3,0.7,0.7
.


## 5.1. Model I

Consider a transportation problem in which the peanuts are initially stored at three sources, namely, O1, O2, and O3 and are transported to peanut butter manufacturing company located at four different destinations, namely D1, D2, D3, and D4, with trapezoidal neutrosophic unit transportation cost and crisp demand and supply as given in Table1. Obtain the optimal transportation of peanuts to minimize the total transportation cost.Table 1
Input data for neutrosophic transportation problem.
D1
D2
D3
D4
Supply
O1
(3, 5, 6, 8); 0.6, 0.5, 0.4
(5, 8, 10, 14); 0.3, 0.6, 0.6
(12, 15, 19, 22); 0.6, 0.4, 0.5
(14, 17, 21, 28); 0.8, 0.2.0.6
26
O2
(0, 1, 3, 6); 0.7, 0.5, 0.3
(5, 7, 9, 11); 0.9, 0.7, 0.5
(15, 17, 19, 22); 0.4, 0.8, 0.4
(9, 11, 14, 16); 0.5, 0.4, 0.7
24
O3
(4, 8, 11, 15); 0.6, 0.3, 0.2
(1, 3, 4, 6); 0.6, 0.3, 0.5
(5, 7, 8, 10); 0.5, 0.4, 0.7
(5, 9, 14, 19); 0.3, 0.7, 0.6
30
Demand
17
23
28
12


### 5.1.1. Neutrosophic Initial Basic Feasible Solution for Model I

Now by score function of trapezoidal neutrosophic number, calculate the score value of each neutrosophic cost to obtain the crisp transportation problem. (Here score values are rounded off to the nearest integer.) The results are given in Table2. Then calculate the penalty for each row and each column which is presented in Table 3.Table 2
Crisp transportation problem.
D1
D2
D3
D4
Supply
O1
2
3
7
10
26
O2
1
3
6
4
24
O3
5
2
3
3
30
Demand
17
23
28
12Table 3
Tabular representation with penalties.
D1
D2
D3
D4
Supply
Penalty
O1
2
3
7
10
26
1
O2
1
3
6
4
24
2
O3
5
2
3
3
30
1
Demand
17
23
28
12
Penalty
1
1
3∗
1In Table3, the highest penalty 3 (marked with ∗) occurs at column 3. Now allot the maximum possible units 28 in the minimum cost cell (
3,3
) and revise the supply units corresponding to row 3. Then the penalties are to be revised in Table 4.Table 4
First allotment with penalties.
D1
D2
D3
D4
Supply
Penalty
O1
2
3
—
10
26
1
O2
1
3
—
4
24
2
O3
5
2
28
3
2
1
Demand
17
23
—
12
Penalty
1
1
—
1Proceeding the neutrosophic initial basic feasible solution algorithm and after few iterations we get the complete allotment transportation units as given in Table5.Table 5
Table with complete allotment.
D1
D2
D3
D4
Supply
O1
—
23
—
3
—
O2
17
—
—
7
—
O3
—
—
28
2
—
Demand
—
—
—
—The initial basic feasible solution is(8)
x
12
=
23
,
x
14
=
3
,
x
21
=
17
,
x
24
=
7
,
x
33
=
28
,
x
34
=
2
.Hence the minimum total neutrosophic cost is(9)
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
i
j
c
~
i
j
N
=
23
5,8
,
10,14
;
0.3,0.6,0.6
+
3
14,17,21,28
;
0.8,0.2
.
0.6
+
17
0,1
,
3,6
;
0.7,0.5,0.3
+
7
9,11,14,16
;
0.5,0.4,0.7
+
28
5,7
,
8,10
;
0.5,0.4,0.7
+
2
5,9
,
14,19
;
0.3,0.7,0.6
=
370,543,694,938
;
0.3,0.7,0.7
.


### 5.1.2. Neutrosophic Optimum Solution for Model I

Consider the neutrosophic optimal solution for the transportation problem given in Table1. After applying Steps 1 and 2 of the optimal solution algorithm we obtain Table 6.Table 6
Table with zero point.
D1
D2
D3
D4
Supply
O1
0
1
4
7
26
O2
0
2
4
2
24
O3
3
0
0
0
30
Demand
17
23
28
12Now Table6 does not satisfy the optimal solution condition stated in Step 3 of Section 4.2. So proceed to Step 4, and get the revised costs as given in Table 7.Table 7
Modified table with zero point.
D1
D2
D3
D4
Supply
O1
0
0
3
6
26
O2
0
1
3
1
24
O3
4
0
0
0
30
Demand
17
23
28
12As per allotment rules given in Steps5 to 7, one can get the complete allotment schedule which is presented in Table 8.Table 8
Table with complete allocation.
D1
D2
D3
D4
Supply
O1
3
23
—
O2
14
10
—
O3
28
2
—
Demand
—
—
—
—The optimal solution is(10)
x
11
=
3
,
x
12
=
23
,
x
21
=
14
,
x
24
=
10
,
x
33
=
28
,
x
34
=
2
.Hence the minimum total neutrosophic cost is (11)
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
i
j
c
~
i
j
N
=
3
3,5
,
6,8
;
0.6,0.5,0.4
+
23
5,8
,
10,14
;
0.3,0.6,0.6
+
14
0,1
,
3,6
;
0.7,0.5,0.3
+
10
9,11,14,16
;
0.5,0.4,0.7
+
28
5,7
,
8,10
;
0.5,0.4,0.7
+
2
5,9
,
14,19
;
0.3,0.7,0.6
=
364,537,682,908
;
0.3,0.7,0.7
.


## 5.1.1. Neutrosophic Initial Basic Feasible Solution for Model I

Now by score function of trapezoidal neutrosophic number, calculate the score value of each neutrosophic cost to obtain the crisp transportation problem. (Here score values are rounded off to the nearest integer.) The results are given in Table2. Then calculate the penalty for each row and each column which is presented in Table 3.Table 2
Crisp transportation problem.
D1
D2
D3
D4
Supply
O1
2
3
7
10
26
O2
1
3
6
4
24
O3
5
2
3
3
30
Demand
17
23
28
12Table 3
Tabular representation with penalties.
D1
D2
D3
D4
Supply
Penalty
O1
2
3
7
10
26
1
O2
1
3
6
4
24
2
O3
5
2
3
3
30
1
Demand
17
23
28
12
Penalty
1
1
3∗
1In Table3, the highest penalty 3 (marked with ∗) occurs at column 3. Now allot the maximum possible units 28 in the minimum cost cell (
3,3
) and revise the supply units corresponding to row 3. Then the penalties are to be revised in Table 4.Table 4
First allotment with penalties.
D1
D2
D3
D4
Supply
Penalty
O1
2
3
—
10
26
1
O2
1
3
—
4
24
2
O3
5
2
28
3
2
1
Demand
17
23
—
12
Penalty
1
1
—
1Proceeding the neutrosophic initial basic feasible solution algorithm and after few iterations we get the complete allotment transportation units as given in Table5.Table 5
Table with complete allotment.
D1
D2
D3
D4
Supply
O1
—
23
—
3
—
O2
17
—
—
7
—
O3
—
—
28
2
—
Demand
—
—
—
—The initial basic feasible solution is(8)
x
12
=
23
,
x
14
=
3
,
x
21
=
17
,
x
24
=
7
,
x
33
=
28
,
x
34
=
2
.Hence the minimum total neutrosophic cost is(9)
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
i
j
c
~
i
j
N
=
23
5,8
,
10,14
;
0.3,0.6,0.6
+
3
14,17,21,28
;
0.8,0.2
.
0.6
+
17
0,1
,
3,6
;
0.7,0.5,0.3
+
7
9,11,14,16
;
0.5,0.4,0.7
+
28
5,7
,
8,10
;
0.5,0.4,0.7
+
2
5,9
,
14,19
;
0.3,0.7,0.6
=
370,543,694,938
;
0.3,0.7,0.7
.


## 5.1.2. Neutrosophic Optimum Solution for Model I

Consider the neutrosophic optimal solution for the transportation problem given in Table1. After applying Steps 1 and 2 of the optimal solution algorithm we obtain Table 6.Table 6
Table with zero point.
D1
D2
D3
D4
Supply
O1
0
1
4
7
26
O2
0
2
4
2
24
O3
3
0
0
0
30
Demand
17
23
28
12Now Table6 does not satisfy the optimal solution condition stated in Step 3 of Section 4.2. So proceed to Step 4, and get the revised costs as given in Table 7.Table 7
Modified table with zero point.
D1
D2
D3
D4
Supply
O1
0
0
3
6
26
O2
0
1
3
1
24
O3
4
0
0
0
30
Demand
17
23
28
12As per allotment rules given in Steps5 to 7, one can get the complete allotment schedule which is presented in Table 8.Table 8
Table with complete allocation.
D1
D2
D3
D4
Supply
O1
3
23
—
O2
14
10
—
O3
28
2
—
Demand
—
—
—
—The optimal solution is(10)
x
11
=
3
,
x
12
=
23
,
x
21
=
14
,
x
24
=
10
,
x
33
=
28
,
x
34
=
2
.Hence the minimum total neutrosophic cost is (11)
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
i
j
c
~
i
j
N
=
3
3,5
,
6,8
;
0.6,0.5,0.4
+
23
5,8
,
10,14
;
0.3,0.6,0.6
+
14
0,1
,
3,6
;
0.7,0.5,0.3
+
10
9,11,14,16
;
0.5,0.4,0.7
+
28
5,7
,
8,10
;
0.5,0.4,0.7
+
2
5,9
,
14,19
;
0.3,0.7,0.6
=
364,537,682,908
;
0.3,0.7,0.7
.


## 5.2. Model II

Consider a problem for Model II with single valued neutrosophic trapezoidal cost, demand, and supply given in Table9.Table 9
Input data for neutrosophic transportation problem.
D1
D2
D3
D4
Supply
O1
(3, 5, 6, 8);0.6, 0.5, 0.4
(5, 8, 10, 14); 0.3, 0.6, 0.6
(12, 15, 19, 22); 0.6, 0.4, 0.5
(14, 17, 21, 28); 0.8, 0.2, 0.6
(22, 26, 28, 32); 0.7, 0.3, 0.4
O2
(0, 1, 3, 6);0.7, 0.5, 0.3
(5, 7, 9, 11); 0.9, 0.7, 0.5
(15, 17, 19, 22); 0.4, 0.8, 0.4
(9, 11, 14, 16); 0.5, 0.4, 0.7
(17, 22, 27, 31); 0.6, 0.4, 0.5
O3
(4, 8, 11, 15);0.6, 0.3, 0.2
(1, 3, 4, 6); 0.6, 0.3, 0.5
(5, 7, 8, 10); 0.5, 0.4, 0.7
(5, 9, 14, 19); 0.3, 0.7, 0.6
(21, 28, 32, 37); 0.8, 0.2, 0.4
Demand
(13, 16, 18, 21);0.5, 0.5, 0.6
(17, 21, 24, 28); 0.8, 0.2, 0.4
(24, 29, 32, 35); 0.9, 0.5, 0.3
(6, 10, 13, 15); 0.7, 0.3, 0.4For Table9, calculate the score value of each neutrosophic cost to get crisp cost and consider the demand and supply units as they are. The modified cost is presented in Table 10.Table 10
Neutrosophic transportation problem with crisp cost.
D1
D2
D3
D4
Supply
O1
2
3
7
10
(22, 26, 28, 32);0.7, 0.3, 0.4
O2
1
3
6
4
(17, 22, 27, 31);0.6, 0.4, 0.5
O3
5
2
3
3
(21, 28, 32, 37);0.8, 0.2, 0.4
Demand
(13, 16, 18, 21);0.5, 0.5, 0.6
(17, 21, 24, 28);0.8, 0.2, 0.4
(24, 29, 32, 35);0.9, 0.5, 0.3
(6, 10, 13, 15);0.7, 0.3, 0.4Here, the arithmetic operations of single valued neutrosophic trapezoidal numbers are applied to modify the neutrosophic demand and supply in each iteration. Proceed the neutrosophic optimal solution method and after few iterations the optimal solution in terms of single valued neutrosophic trapezoidal numbers is obtained as follows:(12)
x
11
=
-
6,2
,
7,15
;
0.7,0.3
.
0.4
,
x
12
=
17,21,24,28
;
0.8,0.2,0.4
,
x
21
=
-
8,6
,
19,33
;
0.6,0.5,0.5
,
x
24
=
-
2,8
,
16,25
;
0.7,0.5,0.4
,
x
33
=
26,29,32,35
;
0.9,0.5,0.3
,
x
34
=
-
6,0
,
6,14
;
0.8,0.5,0.4
,
Minimize
Z
~
N
=
∑
i
=
0
3
∑
j
=
0
4
x
~
i
j
N
c
~
i
j
N
=
3,5
,
6,8
;
0.6,0.5,0.4
×
-
6,2
,
7,15
;
0.7,0.3,0.4
+
5,8
,
10,14
;
0.3,0.6,0.6
×
17,21,24,28
;
0.8,0.2,0.4
+
0,1
,
3,6
;
0.7,0.5,0.3
×
-
8,6
,
19,33
;
0.6,0.5,0.5
+
9,11,14,16
;
0.5,0.4,0.7
×
-
2,8
,
16,25
;
0.7,0.5,0.4
+
5,7
,
8,10
;
0.5,0.4,0.7
×
26,29,32,35
;
0.9,0.5,0.3
+
5,9
,
14,19
;
0.3,0.7,0.6
×
-
6,0.6,14
;
0.8,0.5,0.4
=
149,475,903,1726
;
0.3,0.7,0.7
.


## 6. Results and Discussions

In Section5.1, the neutrosophic optimum solution 364,537,682,908
;
0.3,0.7,0.7 is better than the neutrosophic initial basic feasible solution. In the optimum solution, the total minimum transportation cost will be greater than 364 and less than 908. And as the total minimum transportation cost lies between 537 and 682, the overall level of acceptance or satisfaction or the truthfulness is 30%. Also for the remaining values of total minimum transportation cost, the degree of truthfulness is μ
(
x
)
×
100 where x denotes the total cost and μ
(
x
) is given by(13)
μ
x
=
0.3
x
-
364
537
-
364
,
for
364
≤
x
≤
537
0.3
,
for
537
≤
x
≤
682
0.3
908
-
x
908
-
682
,
for
682
≤
x
≤
908
0
,
otherwise.In the optimum solution, the degrees of indeterminacy and falsity are the same. Hence, degree of indeterminacy and falsity for the minimum transportation cost are(14)
ν
x
=
537
-
x
+
0.7
x
-
364
537
-
364
,
for
364
≤
x
≤
537
0.7
,
for
537
≤
x
≤
682
x
-
682
+
0.7
908
-
x
908
-
682
,
for
682
≤
x
≤
908
1
,
otherwise
(15)
λ
x
=
537
-
x
+
0.7
x
-
364
537
-
364
,
for
364
≤
x
≤
537
0.7
,
for
537
≤
x
≤
682
x
-
682
+
0.7
908
-
x
908
-
682
,
for
682
≤
x
≤
908
1
,
otherwise,respectively. Hence, a decision maker can conclude the total neutrosophic cost from the range 364 to 908, with its truth degree, indeterminacy degree, and falsity degree. Based on the above result, he may schedule the transportation and budget constraints.


## 7. Conclusion

Neutrosophic sets being a generalization of intuitionistic fuzzy sets provide an additional possibility to represent the indeterminacy along with the uncertainty. Though there are many transportation problems that have been studied with different types of input data, this research has investigated the solutions of transportation problems in neutrosophic environment. Two different models in neutrosophic environment were considered in the study. The arithmetic operations on single valued neutrosophic trapezoidal numbers are employed to find the solutions. The solution procedures are illustrated with day-to-day problems. Though the proposed algorithms concretely analyze the solutions of neutrosophic transportation problems, there are some limitations in predicting the solutions of qualitative and complex data. The computational complexity in handling higher dimensional problems will be overcome by genetic algorithm approach. In future, the research will be extended to deal with multiobjective solid transportation problems in environment. The researchers will be interested to overcome the above stated limitations. Further, the approaches of transportation problems on fuzzy and intuitionistic fuzzy logic may be extended to neutrosophic logic.



---

*Source: 5950747-2016-03-20.xml*