On fuzzy multi-objective multi-item solid transportation problem
Deepika Rania,∗, T. R. Gulatia , Amit Kumarb
a Department
b School
of Mathematics, Indian Institute of Technology Roorkee, Roorkee-247 667, INDIA
of Mathematics and Computer Applications, Thapar University, Patiala-147 004, INDIA
Abstract
A transportation problem involving multiple objectives, multiple products and three constraints namely - source,
destination and conveyance is called the multi-objective multi-item solid transportation problem (MOMISTP).
Recently, Kundu et. al (Applied Mathematical Modelling 37 (2013) 2028-2038) proposed a method to solve an
unbalanced MOMISTP. In this paper, we suggest a method, which first converts an unbalanced problem to a
balanced one. In one case of an example, while the method proposed by Kundu et. al concludes infeasibility,
our method gives a feasible solution.
Keywords: fuzzy transportation problem, optimal-compromise solution, fuzzy programming technique,
multi-objective, multi-item
1. Introduction
In today’s business environment, competition is increasing day by day and each organization aims to find better
ways to deliver values to the customers in a cost effective manner within the specified time and fulfilling their
demands. For this they think of different ideas. One of the ideas can be using different type of transportation
modes to save time and money.
Taking into account this factor, the conventional transportation problem proposed by Hitchcock in 1941 [11] was
generalized to the transportation problem which takes into account three type of constraints namely - source,
destination and conveyance constraints instead of only source and destination constraints. This generalized
problem is called the solid transportation problem. It was introduced by Schell in 1955 [22]. Solid transportation problem plays a major role in today’s business economics as many business situations resemble with this.
Solution procedure to the solid transportation problem was given by Haley [10].
In many real-world problems, there are situations where several objectives are to be considered and optimized at
the same time. Such problems are called multi-objective problems. In multi-objective transportation problems,
instead of optimal solution, optimal compromise solution or efficient solution (feasible solution for which no
improvement in any objective function is possible without sacrificing on at least one of the objective functions)
is considered. Zimmermann [24] gave the fuzzy programming technique for the multi-objective transportation
problems. Liu and Liu [20] presented expected value model for fuzzy programming.
The geometric programming approach for multi-objective transportation problems was considered by Islam
and Roy [12]. Gupta et. al [8] proposed a method, called Mehar’s method, to find the exact fuzzy optimal
solution of unbalanced fully fuzzy multi-objective transportation problems. Gupta and Mehlawat [9] proposed
an algorithm for a fuzzy transportation problem to select a new type of coal.
∗ Corresponding
author
Email addresses: [email protected] (Deepika Rani), [email protected] (T. R. Gulati), [email protected]
(Amit Kumar)
The multi-objective transportation problems in which heterogeneous items are to be transported from different
production points to different consumer points using different modes of conveyance are called multi-objective
multi-item solid transportation problems (MOMISTPs).
Due to shortage of information, insufficient data, lack of evidence etc. the data for a transportation system
such as availabilities, demands and conveyance capacities are not always exact but can be fuzzy or arbitrary
or both. Bector and Chandra [4] have presented a systematic study of fuzzy mathematical programming. The
MOMISTPs in which atleast one of the parameters is represented by fuzzy number [23] are called the fuzzy
multi-objective multi-item solid transportation problems.
Genetic algorithm was used by Li et. al [18] to solve multi-objective solid transportation problem (MOSTP)
in which only objective function coefficients were taken as fuzzy numbers. Bit, Biswal and Alam [3] applied
the fuzzy programming technique to solve MOSTP. Fuzzy MOSTP where all the parameters except decision
variables are taken as fuzzy numbers have been solved by Gen et. al [7] and Ojha et al. [21]. Expectedconstrained programming for an uncertain solid transportation problem is given by Cui and Sheng [5]. Baidya
et. al have discussed the multi-item interval valued solid transportation problem in [1, 2]. A Method to solve
multi-objective solid transportation problems in uncertain environment has been presented in [15]. Kundu et. al
[16, 17] have proposed methods to solve fuzzy MOMISTP. They have applied the fuzzy programming technique
and global criterion method to find the optimal compromise solution.
In this paper a new method is proposed to solve the fuzzy MOMISTP. The application of proposed method
is shown by solving two numerical examples in which objective function coefficients, availabilities and demand
parameters are represented by trapezoidal fuzzy numbers. In one case of an example, while the method proposed in [16] concludes no feasible solution, our method gives an optimal compromise solution.
The present paper is organized as follows: Section 2 contains the basic definitions and ranking approach for
trapezoidal fuzzy numbers. Mathematical model for the fuzzy MOMISTP is presented in Section 3. The existing method [16] is discussed in Section 4. A new method to solve the fuzzy MOMISTP is proposed in Section
5. To illustrate the proposed method, two numerical examples have been solved in Section 6. Section 7 consists of the results of the two problems with the existing method [16] and that of the proposed method. The
interpretation of the results has been given in Section 8. Conclusions are discussed in Section 9.
2. Preliminaries [4, 13, 19]
In this section, some basic definitions, arithmetic operations and ranking approach of trapezoidal fuzzy numbers
are presented.
2.1. Basic definitions
e defined on the universal set of real numbers R, denoted as A
e = (a, b, c, d), is
Definition 1. A fuzzy number A
said to be a trapezoidal fuzzy number if its membership function µAe(x) is given by

(x−a)

 (b−a) ,


 1,
µAe(x) = (x−d)


(c−d) ,



0,
a≤x<b
b≤x≤c
c<x≤d
otherwise
e = (a, b, c, d) is said to be zero trapezoidal fuzzy number if and only
Definition 2. A trapezoidal fuzzy number A
if a = 0, b = 0, c = 0 and d = 0.
e = (a, b, c, d) is said to be non-negative trapezoidal fuzzy number
Definition 3. A trapezoidal fuzzy number A
if and only if a ≥ 0.
2
e on X is the crisp set of all x ∈ X such that µ e(x) > 0.
Definition 4. The support of a fuzzy number A
A
e on X is the crisp set of all x ∈ X such that µ e(x) = 1.
Definition 5. The core of a fuzzy number A
A
e = (a, b, c, d), b = c, then it is called a triangular fuzzy number
Remark 1. If for a trapezoidal fuzzy number A
and is denoted by (a, b, b, d) or (a, b, d) or (a, c, d).
2.2. Arithmetic operations
e1 = (a1 , b1 , c1 , d1 ) and A
e2 = (a2 , b2 , c2 , d2 ) be two trapezoidal fuzzy numbers. Then
Let A
e1 ⊕ A
e2 = (a1 + a2 , b1 + b2 , c1 + c2 , d1 + d2 )
(i) A
e
e2 = (a1 − d2 , b1 − c2 , c1 − b2 , d1 − a2 )
(ii) A1 A
(
e1 = (ka1 , kb1 , kc1 , kd1 ), k ≥ 0
(iii) kA
(kd1 , kc1 , kb1 , ka1 ), k ≤ 0
e1 ⊗ A
e2 = (a, b, c, d),
(iv) A
where a = min(a1 a2 , a1 d2 , d1 a2 , d1 d2 ), b = min(b1 b2 , b1 c2 , c1 b2 , c1 c2 ),
c = max(b1 b2 , b1 c2 , c1 b2 , c1 c2 ), d = max(a1 a2 , a1 d2 , d1 a2 , d1 d2 )
2.3. Liou and Wang ranking approach for trapezoidal fuzzy numbers
In this paper, we shall use the ranking approach suggested by Liou and Wang [19] to find the crisp value of
trapezoidal fuzzy numbers. According to this approach we have:
e = (a, b, c, d) ∈ F(R).
Let F(R) be a set of fuzzy numbers defined on the set of real numbers R and let A
Then
e = (a + b + c + d)
ℜ(A)
4
is called a ranking function ℜ : F(R) → R , which maps each fuzzy number into the real line.
2.4. Comparison of trapezoidal fuzzy numbers
e1 = (a1 , b1 , c1 , d1 ) and A
e2 = (a2 , b2 , c2 , d2 ) be two trapezoidal fuzzy numbers. Then
Let A
e1 A
e2 if ℜ(A
e1 ) > ℜ(A
e2 )
A
e
e
e
e
A1 ≺ A2 if ℜ(A1 ) < ℜ(A2 )
e1 ≈ A
e2 if ℜ(A
e1 ) = ℜ(A
e2 ).
A
3. Mathematical model of multi-objective multi-item solid transportation problem [16]
A multi-objective multi-item solid transportation problem with R objectives in which l different items are to be
transported from m sources (Si ) to n destinations (D j ) via K different conveyances, can be formulated as follows:
Minimize (Z1 , Z2 , ..., ZR )
subject to
n
K
∑ ∑ xipjk ≤ aeip ; 1 ≤ i ≤ m, 1 ≤ p ≤ l
j=1 k=1
m
K
∑ ∑ xipjk ≥ eb pj ; 1 ≤ j ≤ n, 1 ≤ p ≤ l
i=1 k=1
3
l
m
n
∑ ∑ ∑ xipjk ≤ eek ; 1 ≤ k ≤ K
p=1 i=1 j=1
xipjk ≥ 0, ∀ i, j, k, p,
where
aeip : the fuzzy availability of item p at Si ,
e
b pj : the fuzzy demand of item p at D j ,
ceti pjk : the fuzzy penalty for transporting one unit of item p from Si to D j via kth conveyance for tth objective Zt ,
eek : the total fuzzy capacity of kth conveyance,
xipjk : the quantity of item p to be transported from Si to D j using kth conveyance, and
l
Zt =
m
n
K
∑ ∑ ∑ ∑ (ecti pjk xipjk ), 1 ≤ t ≤ R.
p=1 i=1 j=1 k=1
For the above problem to be balanced, it should satisfy the following conditions:
(i) Total availability of an item at all sources should be equal to its demand at all the destinations.
(ii) Overall availability of all the items at all the sources, overall demand of all the items at all the destinations
and total conveyance capacity should be equal.
Mathematically, it means:
m n
b pj , 1 ≤ p ≤ l
(i) ℜ ∑ aeip = ℜ ∑ e
i=1
l
(ii) ℜ ∑
j=1
m
∑ aeip
p=1 i=1
l
=ℜ ∑
n
∑ eb pj
p=1 j=1
K = ℜ ∑ eek
k=1
4. The existing method
Recently, Kundu et al. [16] proposed a method to solve the fuzzy MOMISTPs. Their method does not require to
convert the unbalanced MOMISTP to a balanced one. The first step is to defuzzify the fuzzy parameters (availability, demand, conveyance capacity and objective function coefficients). On defuzzification, the problem gets
converted to the crisp multi-objective multi-item transportation problem. For defuzzification, the authors have
used two different methods. They observed out that the expected value model does not give a feasible solution.
It may be due to the fact that the expected value method gives ℜ(Total availability) = 122.25, ℜ(Total demand)
= 116, ℜ(Total conveyance capacity) = 104.5. Thus ℜ(Total conveyance capacity) < ℜ(Total demand), which
implies infeasibility.
In this paper, we propose a method for MOMISTPs. The method first converts an unbalanced problem to a
balanced one. Therefore the expected value model gives a feasible solution. This is then used to obtain its
optimal compromise solution. Our method gives better value of the objective function than obtained in [16].
5. The proposed method
In this section a new method has been proposed to find the optimal compromise solution of fuzzy MOMISTP
in which all the parameters except the decision variables are represented by trapezoidal fuzzy numbers. The
method consists of the following steps:
4
Step 1: Check whether the problem under consideration is balanced or not (according to the definition given in
Section 3).
For this find
m
n
i=1
j=1
∑ aeip and ∑ eb pj for each p, 1 ≤ p ≤ l.
Check the equality of the two as given in Definition
2.4.
n
m Case (1): If ℜ ∑ aeip = ℜ ∑ e
b pj , 1 ≤ p ≤ l, i.e., the problem is balanced, then go to Step 2.
i=1
m
Case (2): If ℜ
∑
i=1
j=1
aeip
n
p
e
6= ℜ ∑ b j for any p, 1 ≤ p ≤ l, then to make the problem balanced, proceed
j=1
according to the following subcases. One or both of these subcases may apply.
n
m p
p
e
Subcase (2a): Check if ℜ ∑ aei < ℜ ∑ b j . Say this happens for one or more items. Then introduce
j=1
i=1
a dummy source with fuzzy availabilities of these items equal to any non-negative trapezoidal fuzzy number
whose ranks are equal to the difference of ranks of total demand and total availability of the corresponding
items. Assume the unit fuzzy cost of transporting these items from this dummy source to any of the destinations
via any of the conveyances as zero trapezoidal fuzzy number.
m n
p
p
e
Subcase (2b): Check if ℜ ∑ aei > ℜ ∑ b j for one or more items. Then introduce a dummy destii=1
j=1
nation similar to dummy source in Subcase (2a).
Now Proceed to Step 2.
l
Step 2: Using Step 1, we have ℜ ∑
u
∑ aeip
p=1 i=1
l
=ℜ ∑
v
∑ eb pj
= V (say) (where u = m or m + 1 and v =
p=1 j=1
n or n + 1).
K Case (1): V = ℜ ∑ eek , i.e., overall availability, overall demand and overall conveyance capacity are equal,
k=1
then go to Step 3.
K Case (2): V − ℜ ∑ eek = α 6= 0. Then one of the following conditions will hold:
k=1
Subcase (2a): α > 0. Then introduce a dummy conveyance having fuzzy capacity equal to any non-negative
fuzzy number whose rank is equal to α. Assume the unit fuzzy cost of transportation form all the source to all
the destination via this added dummy conveyance as zero trapezoidal fuzzy number.
Subcase (2b): β = −α > 0. Then check whether in Step 1 a dummy source and/or a dummy destination
were introduced or not.
Case (i): If both, a dummy source and a dummy destination were introduced, then increase the overall fuzzy
availability and fuzzy demand of the already added dummy source and dummy destination by a non-negative
fuzzy number whose rank is β .
Case (ii): If only a dummy source is introduced in Step 1, then increase its overall availability by a nonnegative fuzzy number having rank β and also add a dummy destination having this increased demand.
5
Case (iii): If only a dummy destination is introduced in Step 1, then increase its overall demand by a nonnegative fuzzy number having rank β and also add a dummy source having this increased availability.
Case (iv): If neither a dummy source nor a dummy destination were introduced in Step 1, then add a dummy
source as well as a dummy destination having availability and demand equal to any non-negative fuzzy number
whose rank is β .
Assume the unit fuzzy cost of transportation from the newly introduced source to all the destinations via any of
the conveyance as zero trapezoidal fuzzy number.
The problem is balanced now.
Step 3: The next step is to convert the balanced fuzzy MOMISTP problem, obtained by using Step 1 and
2, into a crisp MOMISTP. We will discuss two different methods for the same.
3.1. Using rank of the fuzzy number:
Formulate the crisp model by obtaining the rank of the objective function coefficients as well as the constraints,
as shown below:
Minimize(Z1 , Z2 , ..., ZR )
subject to
n
K
∑ ∑ xipjk = ℜ(eaip ), 1 ≤ i ≤ m, 1 ≤ p ≤ l
j=1 k=1
m
K
∑ ∑ xipjk = ℜ(eb pj ), 1 ≤ j ≤ n, 1 ≤ p ≤ l
i=1 k=1
l
m
n
K
∑ ∑ ∑ ∑ xipjk = ℜ(eek ), 1 ≤ k ≤ K
p=1 i=1 j=1 k=1
xipjk ≥ 0, ∀ i, j, k, p,
l
where
Zt =
m
n
K
∑ ∑ ∑ ∑ ℜ(ecti pjk xipjk ), 1 ≤ t ≤ R
p=1 i=1 j=1 k=1
This method provides a crisp optimal-compromise solution.
3.2. Using the concept of minimum of fuzzy number:
Brief overview of the method:
Since a fuzzy number Ze can not be minimized directly, Buckley et. al [6] proposed to find min Ze by first
converting it to a multi-objective problem:
min Ze = (max AL , min C, min AU ),
where AL , AU are the areas under the graph of membership function of Ze to the left and right of the center of
6
core C.
In the above multi-objective problem, we wish to
(i) maximize the possibility of getting values less than C and
(ii) minimize the possibility of getting values more than C.
This multi-objective problem is then changed to the following single objective optimization problem
min{λ1 [M − AL ] + λ2C + λ3 AU },
where M > 0 is so large such that max AL is equivalent to min[M − AL ]; λi > 0, i = 1, 2, 3 and λ1 + λ2 + λ3 = 1.
The values of λi ’s depend upon the decision maker’s choice.
To calculate the values of AL ,C and AR :
Since we have taken the objective function coefficients to be trapezoidal fuzzy numbers, let
p 2t p 3t p 4t p
ceti pjk = (c1t
i jk , ci jk , ci jk , ci jk )
Therefore, each objective function is also a trapezoidal fuzzy number
Zet = (Zt1 , Zt2 , Zt3 , Zt4 ),
l
where Ztr =
m
n
K
∑ ∑ ∑ ∑ (crti jkp xipjk ), 1 ≤ r ≤ 4. The values of AL ,C and AR can be calculated as follows:
p=1 i=1 j=1 k=1
Z 3 − Zt1
Z 4 − Zt2
Zt2 + Zt3
, AL = t
and AR = t
.
2
2
2
Now, formulate the crisp model as shown below:
C=
Minimize(Z1 , Z2 , ..., ZR )
subject to
n
K
∑ ∑ xipjk = ℜ(eaip ), 1 ≤ i ≤ m, 1 ≤ p ≤ l
j=1 k=1
m
K
∑ ∑ xipjk = ℜ(eb pj ), 1 ≤ j ≤ n, 1 ≤ p ≤ l
i=1 k=1
l
m
n
K
∑ ∑ ∑ ∑ xipjk = ℜ(eek ), 1 ≤ k ≤ K
p=1 i=1 j=1 k=1
xipjk ≥ 0, ∀ i, j, k, p,
where
Zt = {λ1 [M − AL (Zt )] + λ2C(Zt ) + λ3 AU (Zt )}, 1 ≤ t ≤ R
λi > 0, i = 1, 2, 3 and λ1 + λ2 + λ3 = 1
Step 4: Solve the crisp multi-objective linear programming problem, obtained in Step 3, by any of the existing
multi-objective programming techniques and find the optimal compromise solution.
Remark 2. There are several defuzzification methods (e.g. [14, 19]) to defuzzify the fuzzy numbers. But in
this paper, we will use the defuzzification method given by Liou and Wang [19] as explained in Section 2.4,
7
called the rank value. One can use any of the defuzzification methods.
A flowchart of the proposed method is shown in Figure 1.
Figure 1: Flow chart for the proposed method
6. Numerical example
6.1. Example-1
Consider the fuzzy MOMISTP solved by Kundu et. al [16] in which two items are to be transported via two
different conveyances from two sources to three destinations and consisting of two different objective functions.
Solve the problems to find the amount of goods to be shipped from source(s) to destination(s) so that the total
demand at all the destinations is met at the minimum total cost and in minimum time possible. Data of the
problem is as shown below:
8
Table 1: Unit penalties of transportation for item 1 in the first objective
Conveyance k = 1
Destinations →
Sources ↓
S1
S2
D1
(5,8,9,11)
(8,10,13,15)
D2
(4,6,9,11)
(6,7,8,9)
Conveyance k = 2
Destinations →
Sources ↓
S1
S2
D3
(10,12,14,16)
(11,13,15,17)
D1
(9,11,13,15)
(10,11,13,15)
D2
(6,8,10,12)
(6,8,10,12)
D3
(7,9,12,14)
(14,16,18,20)
Table 2: Unit penalties of transportation for item 2 in the first objective
Conveyances k = 1
Destinations →
Sources ↓
S1
S2
D1
(9,10,12,13)
(11,13,14,16)
D2
(5,8,10,12)
(7,9,12,14)
Conveyance k = 2
Destinations →
Sources ↓
S1
S2
D3
(10,11,12,13)
(12,14,16,18)
D1
(11,13,14,15)
(14,16,18,20)
D2
(6,7,9,11)
(9,11,13,14)
D3
(8,10,11,13)
(13,14,15,16)
Table 3: Unit penalties of transportation for item 1 in the second objective
Conveyance k = 1
Destinations →
Sources ↓
S1
S2
D1
(4,5,7,8)
(6,8,9,11)
D2
(3,5,6,8)
(5,6,7,8)
Conveyance k = 2
Destinations →
Sources ↓
S1
S2
D3
(7,9,10,12)
(6,7,9,10)
D1
(6,7,8,9)
(4,6,8,10)
D2
(4,6,7,9)
(7,9,11,13)
D3
(5,7,9,11)
(9,10,11,12)
Table 4: Unit penalties of transportation for item 2 in the second objective
Conveyance k = 1
Destinations →
Sources ↓
S1
S2
D1
(5,7,9,10)
(10,11,13,14)
D2
(4,6,7,9)
(6,7,8,9)
Conveyance k = 2
Destinations →
Sources ↓
S1
S2
D3
(9,11,12,13)
(7,9,11,12)
D1
(7,8,9,10)
(6,8,10,12)
D2
(4,5,7,8)
(5,7,9,11)
Table 5: Availability and demand data
Fuzzy availability
Fuzzy demand
Conveyance capacity
ae11 = (21,24,26,28) e
b11 = (14,16,19,22)
ee1 = (46,49,51,53)
1
1
e
ae2 = (28,32,35,37) b2 = (17,20,22,25)
ee2 = (51,53,56,59)
ae21 = (32,34,37,39) e
b13 = (12,15,18,21)
ae22 = (25,28,30,33) e
b21 = (20,23,25,28)
e
b22 = (16,18,19,22)
e
b23 = (15,17,19,21)
From Table 5, wefind that
for the first item,
2
total availability ∑ ae1i = (21,24,26,28)⊕ (28,32,35,37) = (49,56,61,65) and
3 i=1
total demand ∑ e
b1j = (14, 16, 19, 22) ⊕ (17, 20, 22, 25) ⊕ (12, 15, 18, 21) = (43, 51, 59, 68).
j=1
Similarly for the second item,
2 3 total availability ∑ ae2i = (57,62,67,72) and total demand ∑ e
b2j = (51,58,63,71)
i=1
j=1
9
D3
(8,10,11,12)
(9,10,12,14)
After calculating the rank for all these, we find that
2 3 ℜ ∑ aeli > ℜ ∑ e
blj , l = 1, 2.
i=1
j=1
So to balance the problem add a dummy destination D4 with demands of items 1 and 2 equal to any fuzzy
number whose ranks are 2.5 and 3.75, respectively. This makes
2 2 2
ℜ ∑ ∑ aeli = ℜ ∑
l=1 i=1
3
∑ eblj
= 122.25
l=1 j=1
2 Also, ℜ ∑ eeh = 104.5 < 122.25. Therefore, insert a dummy conveyance having total fuzzy capacity equal
h=1
to any fuzzy number with rank 17.75. As a dummy destination and a dummy conveyance are introduced, so
assume ce1i4k = ce2i4k = ce1i j3 = ce2i j3 = (0, 0, 0, 0) ∀ i = 1, 2; j = 1, 2, 3, 4 and k = 1, 2, 3.
The considered problem is now balanced and consists of two objectives, two items, two sources, four destinations and three different modes of transportation. The number of constraints is 2 × 2 + 4 × 2 + 3 = 15.
Forming the crisp model by using the ranking technique as explained in Step 3.1, the model is as shown below.
Problem (P1):
1 + 12x1 + 0x1 + 7.5x1 + 9x1 + 0x1 + 13x1 + 10.5x1 + 0x1 + 0x1 + 0x1 +
Minimize Z1 = 8.25x111
112
113
121
122
123
131
132
133
141
142
1 +11.5x1 +12.25x1 +0x1 +7.5x1 +9x1 +0x1 +14x1 +17x1 +0x1 +0x1 +0x1 +0x1 +
0x143
211
212
213
221
222
223
231
232
233
241
242
243
2 + 13.25x2 + 0x2 + 8.75x2 + 8.25x2 + 0x2 + 11.5x2 + 10.5x2 + 0x2 + 0x2 + 0x2 + 0x2 +
11x111
112
113
121
122
123
131
132
133
141
142
143
2 + 17x2 + 0x2 + 10.5x2 + 11.75x2 + 0x2 + 15x2 + 14.5x2 + 0x2 + 0x2 + 0x2 + 0x2
13.5x211
212
213
221
222
223
231
232
233
241
242
243
1 + 7.5x1 + 0x1 + 5.5x1 + 6.5x1 + 0x1 + 9.5x1 + 8x1 + 0x1 + 0x1 + 0x1 +
Minimize Z2 = 6x111
112
113
121
122
123
131
132
133
141
142
1 + 8.5x1 + 7x1 + 0x1 + 6.5x1 + 10x1 + 0x1 + 8x1 + 10.5x1 + 0x1 + 0x1 + 0x1 + 0x1 +
0x143
211
212
213
221
222
223
231
232
233
241
242
243
2 + 8.5x2 + 0x2 + 6.5x2 + 6x2 + 0x2 + 11.25x2 + 10.25x2 + 0x2 + 0x2 + 0x2 + 0x2 +
7.75x111
112
113
121
122
123
131
132
133
141
142
143
2 + 9x2 + 0x2 + 7.5x2 + 8x2 + 0x2 + 9.75x2 + 11.25x2 + 0x2 + 0x2 + 0x2 + 0x2
12x211
212
213
221
222
223
231
232
233
241
242
243
subject to
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 24.75
x111
112
113
121
122
123
131
132
133
141
142
143
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 35.5
x111
112
113
121
122
123
131
132
133
141
142
143
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 33.0
x211
212
213
221
222
223
231
232
233
241
242
243
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 29.0
x211
212
213
221
222
223
231
232
233
241
242
243
1 + x1 + x1 + x1 + x1 + x1 = 17.75
x111
112
113
211
212
213
2 + x2 + x2 + x2 + x2 + x2 = 24.0
x111
211
212
213
112
113
1 + x1 + x1 + x1 + x1 + x1 = 21.0
x121
122
123
221
222
223
2 + x2 + x2 + x2 + x2 + x2 = 18.75
x121
122
123
221
222
223
10
1 + x1 + x1 + x1 + x1 + x1 = 16.5
x131
132
133
231
232
233
2 + x2 + x2 + x2 + x2 + x2 = 18.0
x131
132
133
231
232
233
1 + x1 + x1 + x1 + x1 + x1 = 2.5
x141
142
143
241
242
243
2 + x2 + x2 + x2 + x2 + x2 = 3.75
x141
142
143
241
242
243
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1
x111
121
131
141
211
221
231
241
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 49.75
+ x111
121
131
141
211
221
231
241
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1
x112
122
132
142
212
222
232
242
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 54.75
+ x112
122
132
142
212
222
232
242
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1
x113
123
133
143
213
223
233
243
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 17.75
+ x113
123
133
143
213
223
233
243
Applying the fuzzy programming technique [24] to the above crisp multi-objective linear programming problem, the results obtained are as follows:
L1 (min(Z1 )) = 938.06,U1 (max(Z1 )) = 1061.0, L2 (min(Z2 )) = 682.5,U2 (max(Z2 )) = 777.875
and the obtained optimal-compromise solution is given below.
1 = 8.25, x1 = 16.5, x1 = 9.5, x1 = 21.0, x1 = 2.5, x2 = 18.63, x2 = 16.88, x2 = 1.79, x2 = 3.58,
x111
132
212
221
242
111
122
212
213
1 = 1.88, x2 = 3.83, x2 = 14.17, x2 = 3.75, λ = 0.78 and Z = 945.3425; Z = 703.46
x221
1
2
232
233
242
Forming the crisp model using the concept of minimum of fuzzy number as explained in Step 3.2 of the proposed
method taking M = 500, λ1 = λ2 = 0.4, λ3 = 0.2, the following crisp multi-objective transportation problem is
obtained.
Problem (P2):
1 +4.4x1 +0x1 +2.5x1 +3.2x1 +0x1 +4.8x1 +3.7x1 +0x1 +0x1 +0x1 +
Minimize Z1 = 2.9x111
112
113
121
122
123
131
132
133
141
142
1 +4.1x1 +4.6x1 +0x1 +2.8x1 +3.2x1 +0x1 +5.2x1 +6.4x1 +0x1 +0x1 +0x1 +0x1 +
0x143
211
212
213
221
222
223
231
232
233
241
242
243
2 + 5x2 + 0x2 + 3x2 + 3x2 + 0x2 + 4.4x2 + 3.9x2 + 0x2 + 0x2 + 0x2 + 0x2 + 5.1x2 +
4.1x111
112
113
121
122
123
131
132
133
141
142
143
211
2 + 0x2 + 3.7x2 + 4.3x2 + 0x2 + 5.6x2 + 5.6x2 + 0x2 + 0x2 + 0x2 + 0x2 + 200
6.4x212
213
221
222
223
231
232
233
241
242
243
1 +2.8x1 +0x1 +1.9x1 +2.3x1 +0x1 +3.5x1 +2.8x1 +0x1 +0x1 +0x1 +
Minimize Z2 = 2.1x111
112
113
121
122
123
131
132
133
141
142
1 + 3.1x1 + 2.4x1 + 0x1 + 2.4x1 + 3.6x1 + 0x1 + 2.9x1 + 4x1 + 0x1 + 0x1 + 0x1 + 0x1 +
0x143
211
212
213
221
222
223
231
232
233
241
242
243
2 +3.2x2 +0x2 +2.3x2 +2.1x2 +0x2 +4.2x2 +3.8x2 +0x2 +0x2 +0x2 +0x2 +4.5x2 +
2.7x111
112
113
121
122
123
131
132
133
141
142
143
211
2 + 0x2 + 2.8x2 + 2.8x2 + 0x2 + 3.5x2 + 4.2x2 + 0x2 + 0x2 + 0x2 + 0x2 +200
3.2x212
213
221
222
223
231
232
233
241
242
243
subject to the constraints as in Problem (P1)
Solving this problem using fuzzy programming technique, obtained results are as follows:
L1 (min(Z1 )) = 540.88, AL (Z1 ) = 170.25, C(Z1 ) = 936.625, AR (Z1 ) = 174.625
U1 (min(Z1 )) = 587.75, AL (Z1 ) = 181.75, C(Z1 ) = 1051.125, AR (Z1 ) = 200.0
L2 (min(Z2 )) = 439.05, AL (Z2 ) = 170.25, C(Z2 ) = 733.875, AR (Z2 ) = 161.75
U2 (min(Z2 )) = 476.3, AL (Z2 ) = 144.5, C(Z2 ) = 768.375, AR (Z2 ) = 133.75
11
The optimal compromise solution is found to be:
1 = 8.25, x1 = 16.5, x1 = 9.5, x1 = 21.0, x1 = 2.5, x2 = 20.5, x2 = 12.56, x2 = 2.44, x2 = 1.31,
x111
132
212
221
242
111
122
132
212
2 = 2.19, x2 = 6.19, x2 = 15.56, x2 = 3.75, λ = 0.73
x213
222
233
242
Z1 = 553.722, AL (Z1 ) = 161.25, C(Z1 ) = 963.025, AR (Z1 ) = 165.06 and [872.37, 1053.68] is the core of the
objective function Z1 .
Z2 = 449.247, AL (Z2 ) = 163.815, C(Z2 ) = 710.79, AR (Z1 ) = 152.405 and [624.26, 797.32] is the core of the
objective function Z2 .
6.2. Example-2
If a MOMISTP in which availability of one or more items is less than the corresponding demand and/or total
conveyance capacity is less than total demand is solved by the method presented in [16], it terminates with the
conclusion that the problem is infeasible. However, such problems can be solved by the method proposed in
this paper, since it starts after balancing the unbalanced problem. Let the availability data in Example-1 (Table
5) is changed to Table 6, so that the total availability of some items is less and of some items is more than their
total demand, keeping the demand and the unit penalty of transportation to be same.
Table 6: Availability and demand data
Fuzzy availability
Fuzzy demand
Conveyance capacity
1
1
e
ae1 = (15,18,22,27) b1 = (14,16,19,22)
ee1 = (46,49,51,53)
ae12 = (20,25,29,32) e
b12 = (17,20,22,25)
ee2 = (51,53,56,59)
ae21 = (32,34,37,39) e
b13 = (12,15,18,21)
ae22 = (25,28,30,33) e
b21 = (20,23,25,28)
e
b22 = (16,18,19,22)
e
b23 = (15,17,19,21)
We find that
ℜ
2
∑
i=1
and
ae1i
3 2 3 1
2
e
b2j = 60.75
= 47 < ℜ ∑ b j = 55.25, ℜ ∑ aei = 64.5 > ℜ ∑ e
i=1
j=1
j=1
2 ℜ ∑ eeh = 104.5.
h=1
So to balance the problem add a dummy source S3 having availability of item 1 only and a dummy destination D4 with demand of item 2 only, equal to any non-negative fuzzy numbers whose ranks are 8.25 and 3.75,
respectively and a dummy conveyance having availability equal to any fuzzy number having rank 15.25. Assume the unit transportation penalties from S3 to all the destinations and from all the sources to D4 via any of
the conveyance as ce13 jk = ce2i4k = ce1i j3 = ce2i j3 = (0, 0, 0, 0) for all i = 1, 2, 3; j = 1, 2, 3, 4 and k = 1, 2, 3.
After balancing the problem and forming the expected value model using ranking technique the obtained problem is shown below.
12
Problem (P3):
1 + 12x1 + 0x1 + 7.5x1 + 9x1 + 0x1 + 13x1 + 10.5x1 + 0x1 + 11.5x1 +
Minimize Z1 = 8.25x111
112
113
121
122
123
131
132
133
211
1 + 0x1 + 7.5x1 + 9x1 + 0x1 + 14x1 + 17x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 +
12.25x212
213
221
222
223
231
232
233
311
312
313
321
322
1 + 0x1 + 0x1 + 0x1 + 11x2 + 13.25x2 + 0x2 + 8.75x2 + 8.25x2 + 0x2 + 11.5x2 + 10.5x2 +
0x323
331
332
333
111
112
113
121
122
123
131
132
2 + 0x2 + 0x2 + 0x2 + 13.5x2 + 17x2 + 0x2 + 10.5x2 + 11.75x2 + 0x2 + 15x2 + 14.5x2 +
0x133
141
142
143
211
212
213
221
222
223
231
232
2 + 0x2 + 0x2 + 0x2
0x233
241
242
243
1 + 7.5x1 + 0x1 + 5.5x1 + 6.5x1 + 0x1 + 9.5x1 + 8x1 + 0x1 + 8.5x1 + 7x1 +
Minimize Z1 = 6x111
112
113
121
122
123
131
132
133
211
212
1 + 6.5x1 + 10x1 + 0x1 + 8x1 + 10.5x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 +
0x213
221
222
223
231
232
233
311
312
313
321
322
323
1 + 0x1 + 0x1 + 7.75x2 + 8.5x2 + 0x2 + 6.5x2 + 6x2 + 0x2 + 11.25x2 + 10.25x2 + 0x2 +
0x331
332
333
111
112
113
121
122
123
131
132
133
2 + 0x2 + 0x2 + 12x2 + 9x2 + 0x2 + 7.5x2 + 8x2 + 0x2 + 9.75x2 + 11.25x2 + 0x2 + 0x2 +
0x141
142
143
211
212
213
221
222
223
231
232
233
241
2 + 0x2
0x242
243
subject to
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 20.5
x111
112
113
121
122
123
131
132
133
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 35.5
x111
112
113
121
122
123
131
132
133
141
142
143
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 26.5
x211
212
213
221
222
223
231
232
233
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 29.0
x211
212
213
221
222
223
231
232
233
241
242
243
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 8.25
x311
312
313
321
322
323
331
332
333
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 17.75
x111
112
113
211
212
213
311
312
313
2 + x2 + x2 + x2 + x2 + x2 = 24.0
x111
112
113
211
212
213
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 21.0
x121
122
123
221
222
223
321
322
323
2 + x2 + x2 + x2 + x2 + x2 = 18.75
x121
122
123
221
222
223
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 = 16.5
x131
132
133
231
232
233
331
332
333
2 + x2 + x2 + x2 + x2 + x2 = 18.0
x131
132
133
231
232
233
2 + x2 + x2 + x2 + x2 + x2 = 3.75
x141
142
143
241
242
243
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1
x111
121
131
211
221
231
311
321
331
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 49.75
+ x111
121
131
141
211
221
231
241
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1
x112
122
132
212
222
232
312
322
332
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 54.75
+ x112
122
132
142
212
222
232
242
1 + x1 + x1 + x1 + x1 + x1 + x1 + x1 + x1
x113
123
133
213
223
233
313
323
333
2 + x2 + x2 + x2 + x2 + x2 + x2 + x2 = 11.5
+ x113
123
133
143
213
223
233
243
Applying the fuzzy programming technique, we obtain:
L1 = 872.81, U1 = 993.875, L2 = 640.63, U2 = 728.125.
The optimal-compromise solution is
13
1 = 12.25, x1 = 8.25, x1 = 5.5, x1 = 21, x1 = 8.25, x2 = 16.5, x2 = 18.75, x2 = 0.25,
λ = 0.74, x111
132
212
221
332
111
122
132
2
2 = 4.47, x2 = 6.97, x2 = 10.78, x2 = 3.75 and Z = 903.95, Z = 663.12
x212 = 3.03, x213
1
2
232
233
242
Converting the same problem to crisp problem using the concept of minimum of fuzzy number, the model
obtained is as follows.
Problem(P4):
1 + 4.4x1 + 0x1 + 2.5x1 + 3.2x1 + 0x1 + 4.8x1 + 3.7x1 + 0x1 + 4.1x1 +
Minimize Z1 = 2.9x111
112
113
121
122
123
131
132
133
211
1 + 0x1 + 2.8x1 + 3.2x1 + 0x1 + 5.2x1 + 6.4x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 +
4.6x212
213
221
222
223
231
232
233
311
312
313
321
322
1 + 0x1 + 0x1 + 0x1 + 4.1x2 + 5x2 + 0x2 + 3x2 + 3x2 + 0x2 + 4.4x2 + 3.9x2 + 0x2 +
0x323
331
332
333
111
112
113
121
122
123
131
132
133
2 +0x2 +0x2 +5.1x2 +6.4x2 +0x2 +3.7x2 +4.3x2 +0x2 +5.6x2 +5.6x2 +0x2 +0x2 +
0x141
142
143
211
212
213
221
222
223
231
232
233
241
2 + 0x2 + 200
0x242
243
1 + 2.8x1 + 0x1 + 1.9x1 + 2.3x1 + 0x1 + 3.5x1 + 2.8x1 + 0x1 + 3.1x1 +
Minimize Z1 = 2.1x111
112
113
121
122
123
131
132
133
211
1 + 0x1 + 2.4x1 + 3.6x1 + 0x1 + 2.9x1 + 4x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 + 0x1 +
2.4x212
213
221
222
223
231
232
233
311
312
313
321
322
1 +0x1 +0x1 +0x1 +2.7x2 +3.2x2 +0x2 +2.3x2 +2.1x2 +0x2 +4.2x2 +3.8x2 +0x2 +
0x323
331
332
333
111
112
113
121
122
123
131
132
133
2 +0x2 +0x2 +4.5x2 +3.2x2 +0x2 +2.8x2 +2.8x2 +0x2 +3.5x2 +4.2x2 +0x2 +0x2 +
0x141
142
143
211
212
213
221
222
223
231
232
233
241
2 + 0x2 + 200
0x242
243
subject to the constraints as in Problem (P3)
Solving this problem using fuzzy programming technique, obtained results are as follows:
L1 (min(Z1 )) = 519.05, AL (Z1 ) = 150.375, C(Z1 ) = 871.75, AR (Z1 ) = 152.5
U1 (min(Z1 )) = 563.925, AL (Z1 ) = 168.125, C(Z1 ) = 984.75, AR (Z1 ) = 186.375
L2 (min(Z2 )) = 424.10, AL (Z2 ) = 160.75, C(Z2 ) = 645.5, AR (Z2 ) = 151
U2 (min(Z2 )) = 463.20, AL (Z2 ) = 131.75, C(Z2 ) = 728.625, AR (Z2 ) = 122.25
The optimal compromise solution is found to be:
1 = 12.25, x1 = 8.25, x1 = 5.5, x1 = 21.0, x1 = 8.25, x2 = 16.5, x2 = 13.55, x2 = 5.45, x2 = 4.8,
x111
132
212
221
332
111
122
132
212
2 = 2.7, x2 = 5.2, x2 = 12.55, x2 = 3.75
x213
222
233
242
Z1 = 532.785, AL (Z1 ) = 147.625, C(Z1 ) = 905.375, AR (Z1 ) = 148.425 and [828.1, 982.65] is the core of the
objective function Z1 .
Z2 = 436.06, AL (Z2 ) = 148.375, C(Z2 ) = 669.825, AR (Z1 ) = 137.4 and [590.55, 749.1] is the core of the
objective function Z2 .
Similarly, other problems in which availability of all the items is less/more than its demand and/or the total
conveyance capacity is less/more than the total availability or total demand, can be solved by the proposed
method.
14
7. Results
The above models are solved with the help of MAPPLE software. Kundu et. al [16] obtained the solution of the
MOMISTP considered in Section 6.1 using the defuzzification technique given by Kikuchi [14]. The results for
both the problems are shown in the Tables below and are interpreted in Section 8.
Table 7: Results using the existing method [16]
Example
Optimal compromise solution
Z1
Z2
1139.536
837.4808
–
–
Z1 = 615.4325
core of Z1 =[1039.232,1227.548]
Z2 = 497.6407
core of Z2 =[753.1357,935.7518]
–
–
Ranking Method
1
1 = 3.1686, x1 = 5.8313, x1 = 13.8686, x1 = 14.7, x1 = 12.6314,
x111
121
221
132
212
2 = 22.7, x2 = 5.6314, x2 = 11.1, x2 = 1.0686, x2 = 16.8, λ = 0.6989
x111
221
122
222
232
2
Infeasible solution
Minimum of Fuzzy Number
1
1 = 9, x1 = 19.7, x1 = 14.7, x1 = 6.8, x2 = 18.3839, x2 = 4.1160,
x111
221
132
212
111
231
2 = 2.7321, x2 = 12.68394, x2 = 4.3160, x2 = 15.0678, λ = 0.753
x122
132
212
222
2
Infeasible solution
Table 8: Results using the proposed method
Example
Optimal compromise solution
Z1
Z2
Ranking Method
1
1 = 8.25, x1 = 16.5, x1 = 9.5, x1 = 21.0, x1 = 2.5, x2 = 18.63,
x111
132
212
221
242
111
2 = 16.88, x2 = 1.79, x2 = 3.58, x1 = 1.88, x2 = 3.83, x2 = 14.17,
x122
212
213
221
232
233
2
x242 = 3.75, λ = 0.78
945.34
703.46
2
1 = 12.25, x1 = 8.25, x1 = 5.5, x1 = 21, x1 = 8.25, x2 = 16.5,
x111
132
212
221
332
111
2 = 18.75, x2 = 0.25, x2 = 3.03, x2 = 4.47, x2 = 6.97, x2 = 10.78,
x122
132
212
213
232
233
2
x242 = 3.75, λ = 0.74
903.95
663.12
1
1 = 8.25, x1 = 16.5, x1 = 9.5, x1 = 21.0, x1 = 2.5, x2 = 20.5,
x111
132
212
221
242
111
2 = 12.56, x2 = 2.44, x2 = 1.31, x2 = 2.19, x2 = 6.19, x2 = 15.56,
x122
132
212
213
222
233
2 = 3.75, λ = 0.73
x242
Z1 = 553.722
core of Z1 =[872.37,1053.68]
Z2 = 449.247
core of Z2 =[624.26,797.32]
2
1 = 12.25, x1 = 8.25, x1 = 5.5, x1 = 21.0, x1 = 8.25, x2 = 16.5,
x111
132
212
221
332
111
2 = 13.55, x2 = 5.45, x2 = 4.8, x2 = 2.7, x2 = 5.2, x2 = 12.55,
x122
132
212
213
222
233
2
x242 = 3.75
Z1 = 532.785
core of Z1 =[828.1,982.65]
Z2 = 436.06
core of Z2 =[590.55,749.1]
Minimum of Fuzzy Number
8. Interpretation of results
The objective values found using the concept of minimum of fuzzy numbers for Example-1 (Section 6.1) can
be shown by the figures below.
Figure 2: Optimal value of Z1
15
Figure 3: Optimal value of Z2
From the membership function of the objective function Z1 shown in Figure 2, the following information about
the minimum transportation cost for the objective function Z1 may be interpreted:
(i) According to the decision maker minimum transportation cost for the transportation will be greater than
731.18 units and will be less than 1202.49 units.
(ii) The maximum chances are that the minimum transportation cost will lie in the range 872.37–1053.68
units.
(iii) The overall level of satisfaction for other values of the minimum transportation cost (say x) is µZe1 (x)×100,
where
 (x−731.18)

141.19 ,


 1,
µZe1 (x) = (1202.49−x)

,


 148.81
0,
731.18 ≤ x < 872.37
872.37 ≤ x ≤ 1053.68
1053.68 < x ≤ 1202.49
otherwise
Similarly, the results about the minimum transportation cost for the objective function Z2 can be interpreted.
The decision maker has more information about the objective function using this concept. It has been found
that the objective values obtained using the rank of fuzzy numbers Z1 = 945.3425; Z2 = 703.46 lie within the
core [872.37, 1053.68] and [624.26, 797.32] of the objective functions Z1 and Z2 , respectively and are close to
their centre of cores.
The results of Example-2 (Section 6.2) can also be interpreted similarly.
9. Conclusions
Our results show that, unlike [16], the expected value model gives the feasible, and hence the optimal solution
of Example 6.1. It has been found that the obtained optimal compromise solution is better than that obtained
by Kundu et al. [16] using different ranking approach for the objective function and the constraints (Subsection
5.1 and 5.2, pp. 2032-2033). Also, to solve the problem we do not require any condition on availability or on
total conveyance capacity. Since the proposed method is for fuzzy MOMISTP, it is also applicable for multiobjective transportation problems, solid transportation problems, multi-objective solid transportation problem,
multi-item transportation problems and multi-objective solid transportation problems in fuzzy environment.
Acknowledgement
The first author thanks CSIR, Government of India for providing financial support.
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