Open Access

Multi-objective multi-item solid transportation problem with fuzzy inequality constraints

  • Dipankar Chakraborty1,
  • Dipak Kumar Jana2Email author and
  • Tapan Kumar Roy3
Journal of Inequalities and Applications20142014:338

https://doi.org/10.1186/1029-242X-2014-338

Received: 23 April 2014

Accepted: 19 August 2014

Published: 3 September 2014

Abstract

Zimmermann (Int. J. Gen. Syst. 2:209-215, 1976) first introduced the concept of fuzzy inequality in the field of linear programming problem (LPP). But this concept is hardly used in any real life applications of LPP. So, in this paper, a multi-objective multi-item solid transportation problem (MMSTP) with fuzzy inequality constraints is modeled. Representing different preferences of the decision maker for transportation, three different types of models are formulated and analyzed. Fuzzy inequality solid transportation problem is converted to parameter solid transportation problem by an appropriate choice of flexible index, and then the crisp solid transportation problem is solved by the algorithm (Cao in Optimal Models and Methods with Fuzzy Quantities, 2010) for decision values. Fuzzy interactive satisfied method (FISM), global criterion method (GCM) and convex combination method (CCM) are applied to derive optimal compromise solutions for MMSTP by using MatLab and Lingo-11.0. The models are illustrated with numerical examples and some sensitivity analysis is also presented.

Keywords

solid transportation problemfuzzy inequalityglobal criterion methodfuzzy interactive satisfied methodconvex combination method

1 Introduction

The solid transportation problem (STP) is a generalization of the traditional transportation problem in which three-dimensional properties (supply, demand, convenience) are taken into account in the objective and constraint set instead of source and destination. The necessity of considering this special type of transportation problem arises when heterogeneous conveyances are available for shipment of products. The STP is used in public distribution systems. In many industrial problems, a homogeneous product is delivered from its origin to a destination by means of different modes of transport called conveyances, such as trucks, cargo flights, goods trains, ships, etc. These conveyances are taken as the third dimension. A solid transportation problem can be converted to a classical transportation problem by considering only a single type of conveyance.

The STP was first introduced by Haley [1] in 1962. In recent years, there have been numerous papers in this area. Some papers only minimize the total transportation cost. For example, Ojha et al. [2], Pramanik et al. [3] considered an STP for an item with fixed charge, vehicle cost and price discounted varying charge. However, in practical programming problems, the decision maker (DM) usually needs to optimize several objectives. Thus, the DM cannot obtain the optimal values of all the objectives simultaneously. The growing literature of STP focuses on multiple objective problems, that is, multiple objective solid transportation problems (MOSTPs). For example, Bit et al. [4] used a fuzzy programming approach to solve a MOSTP; Ida et al. [5] presented a neural network method to solve a MOSTP; Gao and Liu [6] developed two-phase fuzzy algorithms to solve multi-objective STP; Tao and Xu [7] developed a class of rough multiple objective programming and its application to a solid transportation problem.

If more than one objective is to be optimized in an STP, then the problem is called multi-objective solid transportation problem (MOSTP). If we consider more than one item, then it is called multi-item solid transportation problem. If we consider more than one item and more than one objective at a time in an STP, then it is called a multi-objective multi-item solid transportation problem (MMSTP). The MMSTP model was given by Kundu et al. [8]. Recently, Pramanik et al. [9] have developed a multi-objective STP in a fuzzy random environment.

Nowadays, in a very often changing market, the business of a single item does not pay much profit to a retailer. For this reason, almost all businessmen in the fields of transportation (Sancak and Salman [10]) do the business of several items. Generally, in all the cases of STP (multi-objective, multi-item and multi-objective multi-item ones), the inequality has been considered as a general inequality. But we can consider this inequality in the fuzzy environment named fuzzy inequality [11, 12]. Fuzzy inequality means it will essentially satisfy that inequality condition. Flexible index is used (Cao [13]) to convert it into the general inequality, so that it will give you a chance to choose the appropriate decision value. Two algorithms were given by Cao [13] to find the decision value. We have taken one of them to find the decision value. That decision value will give us a more general optimal solution and an optimal value to minimize the objectives.

The following developments are made in the formulation and solution of MMSTP models:

  • Various types of examples have been used to illustrate the single-objective fuzzy inequality constraints.

  • MMSTP has been solved in a fuzzy inequality constraint environment.

  • Three different soft computing techniques FISM, GCM and CCM have been used to make the comparison between optimal solutions in multi-objective problems.

  • Two different soft-computing tools (MATLAB and LINGO-11.0) have been used to solve the examples.

The rest of this paper is organized as follows. In Section 2, we recall some preliminary knowledge about fuzzy inequality constraint linear programming. Section 3 provides the notation and assumption used throughout this paper. In Section 4, we formulate multi-objective multi-item LPP with fuzzy inequality constraint and also provides general information about the fuzzy interactive satisfied method, global criterion method and convex combination method. MMSTP with fuzzy inequality constraints and its two special cases are developed in Section 5. In Section 6 we discuss the solution procedure to those models using the fuzzy interactive satisfied method. A numerical example is solved, and results obtained using the mentioned techniques are compared in Section 7. Section 8 summarizes the paper and also discusses the scope of future work.

2 Preliminaries about fuzzy inequality constraint linear programming

Let us consider the fuzzy inequality constraint linear programming (FICP)
max z = c x LP ˜ s.t.  A x b , s.t.  x 0 ,
(1)
its corresponding parameter linear programming is given by (Cao [13])
max z = c x LP α s.t.  A x b + ( 1 α ) d , s.t.  x 0 ,
(2)

where α [ 0 , 1 ] and d 0 . In the given discussion, we will use x α as an optimal solution, B α denotes an optimal basis and z α denotes an optimal value of linear programming ( LP α ).

Definition 2.1 [13]

Let B be one of the optimal basis matrices of ( LP α ). If an interval [ α 1 , α 2 ] exists, satisfying that B is an optimal basis matrix of ( LP α ) ( α [ α 1 , α 2 ] ) while B is not an optimal basis matrix for each α [ α 1 , α 2 ] , we call α 1 and α 2 critical values of ( LP α ) and [ α 1 , α 2 ] a characteristic interval.

Theorem 1 ( LP α ) has a finite characteristic interval on the interval [ 0 , 1 ] .

Theorem 2 Let B be an optimal basis matrix of ( LP α ) on a characteristic interval [ α 1 , α 2 ] . If ( B 1 b ) i 0 ( 1 i m ), then
α 1 = max [ [ B 1 ( b + d ) ] i [ B 1 b ] i , 0 | ( B 1 d ) i < 0 ( 1 i m ) ] , α 2 = min [ [ B 1 ( b + d ) ] i [ B 1 b ] i , 0 | ( B 1 d ) i > 0 ( 1 i m ) ]

are derived, where ( B 1 ( b + d ) ) i and ( B 1 d ) i are the ith components of B 1 ( b + d ) and B 1 d , respectively.

Property 2.1 Let B be an optimal matrix of ( LP α ) on the characteristic interval [ α i , α j ] . Then x α = B 1 ( b + ( 1 α ) d ) ( α i α α j ) is a linear vector function about variable α. The optimal value function z α = C B B 1 ( b + ( 1 α ) d ) is a linear function about variable α and decreases with the increase of variable α.

Property 2.2 The optimal value of function z α of ( LP α ) continues on the interval [ 0 , 1 ] .

2.1 Algorithm for fuzzy inequality constraint linear programming

Let z 1 be an optimal value of ( LP 1 ) and z 0 be an optimal value of ( LP 0 ), d 0 = z 0 z 1 > 0 . Based on the above conclusion, the algorithm (Cao [13]) of fuzzy linear programming is given as follows.

Step 1: Let the optimal solutions of ( LP 0 ) and ( LP 1 ) be x 0 and x 1 , the optimal values of ( LP 0 ) and ( LP 1 ) be z 0 and z 1 , and the optimal basis matrix of ( LP 0 ) be B 0 .

Step 2: Solve [ B 0 1 ( b + ( 1 α ) d ) ] i = 0 .

Assume the solution as
α 1 , α 2 , , α n 1 ( 0 < α 1 < α 2 < < α n 1 < 1 ) .

Let α 0 = 1 , α n = 1 , α = α 1 , k = 1 .

Step 3: Solve ( LP α ).

Let the optimal value be Z α . If Z α Z 1 + d 0 α , turn to Step 4, otherwise let k = k + 1 , α = α k , turn to Step 3.

Step 4: Solve the optimal decision
α = Z 1 α k Z 1 α k 1 Z α k 1 α k + Z α k α k 1 Z α k Z α k 1 α k d 0 + α k 1 d 0 .

Step 5: Solve linear programming ( LP α ), and we can obtain an optimal solution x α and an optimal value z α .

Example 2.1 Calculate
max 4 x 1 + 3 x 2 + 3 x 3 subject to LP ˜ 2 x 1 + 3 x 2 + 2 x 3 65 , LP ˜ 3 x 1 + x 2 + x 3 61 , LP ˜ x 2 + x 3 24 , LP ˜ x 2 14 , LP ˜ x 1 + x 3 10 , LP ˜ x 1 , x 2 , x 3 0 ,
(3)

where d = ( 0 , 0 , 0 , 0 , 7 ) .

The corresponding parametric linear programming problem of the above LP ˜ is presented as follows:
max 4 x 1 + 3 x 2 + 3 x 3 subject to LP α 2 x 1 + 3 x 2 + 2 x 3 65 , LP α 3 x 1 + x 2 + x 3 61 , LP α x 2 + x 3 24 , LP α x 2 14 , LP α x 1 + x 3 10 + 7 ( 1 α ) , LP α 0 α 1 , x 1 , x 2 , x 3 0 .
(4)
Now, using algorithm given in Section 2.1, we obtain Z 0 = 98.833 , Z 1 = 82 and d 0 = 16.833 by calculating ( LP 0 ) and ( LP 1 ). The inverse of the optimal matrix in ( LP 0 ) is
B 0 1 = ( 1.667 0.5 0 0 1.1667 0.333 0 0 0 0.6667 0.5 0.5 1 0 0.5 0.333 0 0 1 0.6667 0.1667 0.5 0 0 0.1667 ) .

Now, calculating the equations [ B 0 1 ( 65 , 61 , 24 , 14 , 10 7 α ) ] i = 0 ( i = 1 , 2 , , 5 ), respectively, we obtain α 1 = 0.0204 , α 2 = 0.7857 , assume α 0 = 0 and α 3 = 1 .

Now solving ( LP α 1 ) we get the optimal solution as x 1 = 16.8571 , x 2 = 10.4286 and x 3 = 0 and the optimal value as Z α 1 = Z 0.0204 = 98.714 .

Since Z 0.0204 = 98.714 > Z 1 + 0.0204 16.833 = 82.3434 , we must continue to solve the linear programming ( LP α 2 ). By solving ( LP α 2 ), we obtain the optimal solution as x 1 = 11.5 , x 2 = 14 and x 3 = 0 and the optimal value as Z α 2 = Z 0.7857 = 88 .

Now Z 0.7857 = 88 < Z 1 + 0.7857 16.833 = 95.2257 , so we stop here and calculate optimal decision α .

Now
α = Z 1 α 2 Z 1 α 1 Z α 1 α 2 + Z α 2 α 1 Z α 2 Z α 1 α 2 d 0 + α 1 d 0 = 0.5514 .

Now calculating ( LP 0.5514 ), we obtain the optimal solution as x 1 = 13.1405 , x 2 = 12.9063 and x 3 = 0 and the optimal value as Z 0.5514 = 91.2811 .

3 Notations and assumptions

3.1 Notations

In this solid transportation problem, the following notations are used:
  1. (i)

    M = number of sources of the transportation problem.

     
  2. (ii)

    N = number of destinations of the transportation problem.

     
  3. (iii)

    K = number of conveyances, i.e., different modes of transportation.

     
  4. (iv)

    a i p = amount of product available at i th origin for p th item.

     
  5. (v)

    b j p = demand at j th destination of p th item.

     
  6. (vi)

    e k = conveyances of the transportation problem.

     
  7. (vii)

    T = number of items.

     
  8. (viii)

    x i j k p = the amount to be transported from i th origin to j th destination by means of k th conveyance of p th item (decision variables).

     
  9. (ix)

    C i j k t p = per unit transportation cost from i th origin to j th destination by k th conveyance of p th item and t th objective.

     

3.2 Assumptions

In this solid transportation problem, the following assumptions are made.
  1. (i)

    Homogeneous product should be transported from sources to destinations.

     
  2. (ii)

    During transportation no items are damaged, i.e., the amount of received items in destination is the same as the one sent from sources.

     

4 Multi-objective multi-item LPP with fuzzy inequality constraint

Consider the following multi-objective linear programming problems with fuzzy constraint:
{ min x C l ( p ) x ( p ) , l = 1 , 2 , , k  and  p = 1 , 2 , , T s.t.  { A i ( p ) ( x ( p ) ) B i ( p ) , i = 1 , 2 , , r  and  p , x ( p ) 0 p ,
(5)

where x is an n-dimensional decision variable column vector.

Its corresponding parametric linear programming is given by
{ min x C l ( p ) x ( p ) , l = 1 , 2 , , k  and  p = 1 , 2 , , T s.t.  { A i ( p ) ( x ( p ) ) B i ( p ) + V i , i = 1 , 2 , , r  and  p , x ( p ) 0 p ,
(6)

where V i ( 1 i r ) is a flexible index by an appropriate choice.

Now the solution methodology of a multi-objective decision making problem by converting into a single-objective problem is discussed as follows.

4.1 Fuzzy interactive satisfied method

We introduce the interactive fuzzy satisfied method (FISM) proposed by Sakawa [14], Xu and Zhou [15]. We consider the following multi-objective decision-making model:
{ max [ G i ( x ) , i = 1 , 2 , , M ] s.t.  x X .
(7)
The objective function of equation (7) is to maximize G i ( x ) , so for each objective we introduce the fuzzy objective ‘ G i ( x ) approximately more than some value’, and the membership function is
μ i ( G i ( x ) ) = { 1 for  G i ( x ) > G i 1 , 1 G i ( x ) G i 0 G i 1 G i 0 for  G i 0 G i ( x ) G i 1 , 0 for  G i ( x ) < G i 0 .
(8)
In equation (8), the membership is 1 and 0 when the values of G i ( x ) are G i 1 and G i 0 , respectively,
G i 1 = max x X G i ( x ) , G i 0 = min x X G i ( x ) , i = 1 , 2 , , M .
(9)
For model min x X G i ( x ) , its optimal solution should be gotten at the boundary of the convex set X. If there exists no solution of max x X G i ( x ) or min x X G i ( x ) , or G i 1 = , G i 0 = , the decision maker may set the value of G i 1 , G i 0 subjectively. Hence, equation (9) could be transformed into the following form:
{ min [ μ 1 ( G 1 ( x ) ) , μ 2 ( G 2 ( x ) ) , , μ m ( G M ( x ) ) ] s.t.  x X .
(10)
For each objective function μ ( G i ( x ) ) , let the decision maker give the reference value of membership function μ ¯ i to reflect the ideal value of membership function. Through solving the minmax problem (10), we obtain an efficient solution of equation (7) as follows:
{ min max i = 1 , 2 , , M [ μ ¯ i μ i ( G i ( x ) ) ] s.t.  x X .
(11)
Equation (11) is equivalent to
{ min λ s.t.  { μ ¯ i μ i ( G i ( x ) ) λ , j = 1 , 2 , , p , 0 λ 1 , x X .
(12)

4.2 Global criteria method

The global criteria method gives a compromise solution for a multi-objective problem. Actually this method is a way of achieving compromise in minimizing the sum in derivations of the ideal solutions from the respective objective functions. The solution procedure is as follows.

Step-I: Solve the multi-objective problem using each time only one objective f t ( t = 1 , 2 , , R ) ignoring all the other objectives.

Step-II: From the results of Step-I, determine the ideal objective vector, say ( f 1 min , f 2 min , , f R min ) and the corresponding values of ( f 1 max , f 2 max , , f R max ) .

Step-III: Formulate the following auxiliary problem:
min G ( x ) subject to  g j ( x ) 0 , j = 1 , 2 , , M , subject to  x 0 ,

where G ( x ) = min { t = 1 R ( f t ( x ) f t min f t min ) q } 1 q or G ( x ) = min { t = 1 R ( f t ( x ) f t min f t max f t min ) q } 1 q , where 1 q < . A usual value of q is 2. This method is then called global criteria method in L 2 norms.

4.3 Convex combination method

We consider the following multi-objective model:
{ min [ f i ( x ) , i = 1 , 2 , , M ] s.t.  g j 0 , j = 1 , 2 , , N , s.t.  x X .
(13)
Then, by the convex combination method, we transfer the above problem into the following form:
{ min i = 1 M w i f i ( x ) , where  i = 1 M w i = 1 , 0 < w i < 1 s.t.  g j 0 , j = 1 , 2 , , N , s.t.  x X .
(14)

Corresponding x and f i ( x ) is solution.

5 Formulation of different models of STP

5.1 Model-1: multi-objective multi-item STP

Let p ( p = 1 , 2 , , T ) items be transported from M origins (or sources) a i ( i = 1 , 2 , , M ), N destinations (i.e., demands) b j ( j = 1 , 2 , , N ) and K conveyances e k ( k = 1 , 2 , , K ). K conveyances, i.e., different modes of transport, may be trucks, cargo flights, goods trains, ships, etc. Let a i ( p ) be the product available at i th origin for items p ( p = 1 , 2 , , T ), b j ( p ) be the demand at j th destination for items p ( p = 1 , 2 , , T ), and let e k represent the amount of product which can be carried by k th conveyance. The variable x i j k ( p ) represents the unknown quantity to be transported from origin a i ( p ) to destination b j ( p ) by means of k th conveyance for item p = 1 , 2 , , T . Then we propose the mathematical model for the fuzzy inequality constraint. Single-objective and p ( = 1 , 2 , , T ) -item problem is to minimize the total transportation cost as follows:
f t ( p ) = i = 1 M j = 1 N k = 1 K c i j k ( t p ) x i j k ( p ) p  and  t .
(15)
From the discussion above, we develop mathematical formulations of the objectives as follows:
min f t = p = 1 T i = 1 M j = 1 N k = 1 K c i j k ( t p ) x i j k ( p ) t .
(16)
As mentioned by Haley [1], the constraints are divided into three types: source constraint, destination constraint and conveyance capacity constraint. In the fuzzy environment, the quantity from a source is essentially less than equal to the supply capacity of products for different items, that is,
j = 1 N k = 1 K x i j k ( p ) a i ( p ) , i = 1 , 2 , 3 , , M p .
(17)
In the fuzzy environment, the quantity of product transported to a destination is essentially greater than equal to its demand for different items, that is,
i = 1 M k = 1 K x i j k ( p ) b j ( p ) , j = 1 , 2 , 3 , , N p .
(18)
In the fuzzy environment, the transportation quantity of conveyance is essentially less than equal to its capacity, that is,
p = 1 T i = 1 M j = 1 N x i j k ( p ) e k , k = 1 , 2 , 3 , , K .
(19)
It is natural to require the nonnegativity of decision variable x i j k ( p ) , that is,
x i j k ( p ) 0 i , j , k  and  p .
(20)
It is noted that the decision maker (DM) and the modeling analyst are often different individuals. In the transportation problem, the DM is the manager of transport enterprise, while the modeling analyst may be an expert in transportation problems or a researcher in the enterprise. With the complexity of feasible region, the DM may give an appropriately large region so that all the feasible solutions are included in it. Hence, the above MMSTP with fuzzy inequality constraint can be written as
min f t = min p = 1 T i = 1 M j = 1 N k = 1 K c i j k ( t p ) x i j k ( p ) , t = 1 , 2 , , R s.t.  j = 1 N k = 1 K x i j k ( p ) a i ( p ) , i = 1 , 2 , 3 , , M p , s.t.  i = 1 M k = 1 K x i j k ( p ) b j ( p ) , j = 1 , 2 , 3 , , N p , s.t.  p = 1 T i = 1 M j = 1 N x i j k ( p ) e k , k = 1 , 2 , 3 , , K , s.t.  x i j k ( p ) 0 i , j , k ,
(21)

where means ‘essentially smaller than equal to’ and means ‘essentially greater than equal to’.

5.2 Model-2: multi-objective single-item STP

We consider a multi-objective single-item solid transportation problem with fuzzy inequality constraint. Then the model may be written as
min f t = min i = 1 M j = 1 N k = 1 K c i j k t p x i j k s.t.  j = 1 N k = 1 K x i j k a i , i = 1 , 2 , 3 , , M , s.t.  i = 1 M k = 1 K x i j k b j , j = 1 , 2 , 3 , , N , s.t.  i = 1 M j = 1 N x i j k e k , k = 1 , 2 , 3 , , K , s.t.  x i j k 0 i , j , k .
(22)

5.3 Model-3: single-objective multi-item STP

We consider a single-objective multi-item solid transportation problem with fuzzy inequality constraint. Then the model may be written as
min f = min p = 1 T i = 1 M j = 1 N k = 1 K c i j k ( p ) x i j k ( p ) s.t.  j = 1 N k = 1 K x i j k ( p ) a i ( p ) , i = 1 , 2 , 3 , , M p , s.t.  i = 1 M k = 1 K x i j k ( p ) b j ( p ) , j = 1 , 2 , 3 , , N p , s.t.  p = 1 T i = 1 M j = 1 N x i j k ( p ) e k , k = 1 , 2 , 3 , , K , s.t.  x i j k ( p ) 0 i , j , k .
(23)

6 Solution of proposed models

6.1 Model-1

Let us consider that p different items are to be transported from i th origin to j th destination by means of k th conveyance. Here we have considered a two-objective function. Let max x X f 1 = f 1 U , min x X f 1 = f 1 L , max x X f 2 = f 2 U and min x X f 2 = f 2 L . Then the membership functions of f 1 and f 2 are given by
μ 1 ( f 1 ( x ) ) = { 0 for  f 1 ( x ) > f 1 U , f 1 U f 1 ( x ) f 1 U f 1 L for  f 1 L < f 1 ( x ) < f 1 U , 1 for  f 1 ( x ) < f 1 L , μ 2 ( f 2 ( x ) ) = { 0 for  f 2 ( x ) > f 2 U , f 2 U f 2 ( x ) f 2 U f 2 L for  f 2 L < f 2 ( x ) < f 2 U , 1 for  f 2 ( x ) < f 2 L .
Now, using FISM in Section 4.1, we present the equivalent crisp linear programming of (21) as follows:
{ min x λ s.t.  { f 1 f 1 U ( μ 1 λ ) ( f 1 U f 1 L ) , f 2 f 2 U ( μ 2 λ ) ( f 2 U f 2 L ) , j = 1 N k = 1 K x i j k ( p ) a i ( p ) + ( 1 α ) d i ( p ) , i = 1 , 2 , 3 , , M , i = 1 M k = 1 K x i j k ( p ) b j ( p ) + ( 1 α ) v j ( p ) , j = 1 , 2 , 3 , , N , p = 1 T i = 1 M j = 1 N x i j k ( p ) e k + ( 1 α ) u k , k = 1 , 2 , 3 , , K , 0 α 1 , x i j k ( p ) 0 i , j , k  and  p .

6.2 Model-2

Let max x X f 1 = f 1 U , min x X f 1 = f 1 L , max x X f 2 = f 2 U and min x X f 2 = f 2 L . Then the membership functions of f 1 and f 2 are given by
μ 1 ( f 1 ( x ) ) = { 0 for  f 1 ( x ) > f 1 U , f 1 U f 1 ( x ) f 1 U f 1 L for  f 1 L < f 1 ( x ) < f 1 U , 1 for  f 1 ( x ) < f 1 L , μ 2 ( f 2 ( x ) ) = { 0 for  f 2 ( x ) > f 2 U , f 2 U f 2 ( x ) f 2 U f 2 L for  f 2 L < f 2 ( x ) < f 2 U , 1 for  f 2 ( x ) < f 2 L .
Now, using FISM in Section 4.1, we present the equivalent crisp linear programming of (22) as follows:
{ min x λ s.t.  { f 1 f 1 U ( μ 1 λ ) ( f 1 U f 1 L ) , f 2 f 2 U ( μ 2 λ ) ( f 2 U f 2 L ) , j = 1 N k = 1 K x i j k a i + ( 1 α ) d i , i = 1 , 2 , 3 , , M , i = 1 M k = 1 K x i j k b j + ( 1 α ) v j , j = 1 , 2 , 3 , , N i = 1 M j = 1 N x i j k ( p ) e k + ( 1 α ) u k , k = 1 , 2 , 3 , , K , 0 α 1 , x i j k ( p ) 0 i , j , k .

6.3 Model-3

Now corresponding parametric linear programming of equation (23) is presented as follows:
min p = 1 T i = 1 M j = 1 N k = 1 K c i j k ( p ) x i j k ( p ) s.t.  j = 1 N k = 1 K x i j k ( p ) a i ( p ) + ( 1 α ) d i ( p ) , i = 1 , 2 , 3 , , M  and  p , s.t.  i = 1 M k = 1 K x i j k ( p ) b j ( p ) + ( 1 α ) v j ( p ) , j = 1 , 2 , 3 , , N  and  p , s.t.  p = 1 T i = 1 M j = 1 N x i j k ( p ) e k + ( 1 α ) u k , k = 1 , 2 , 3 , , K , s.t.  0 α 1 , x i j k 0 i , j , k ,
(24)

where d i ( p ) for i = 1 , 2 , 3 , , M , v j p for j = 1 , 2 , 3 , , N and u k for k = 1 , 2 , 3 , , K are flexible index values p.

7 Numerical experiment

7.1 Input data for Model-1 and Model-3

Let us consider a multi-objective multi-item solid transportation problem with two types of items (i.e., T = 2 ), three origins (i.e., M = 3 ), two destinations (i.e., N = 2 ) and two types of conveyances (i.e., K = 2 ). The parameters are given as follows.

Transportation cost for 1st objective and 1st item [ c i j k ( 11 ) ]

[ c 111 ( 11 ) c 112 ( 11 ) c 121 ( 11 ) c 122 ( 11 ) c 131 ( 11 ) c 132 ( 11 ) c 211 ( 11 ) c 212 ( 11 ) c 221 ( 11 ) c 222 ( 11 ) c 231 ( 11 ) c 232 ( 11 ) ] = [ 11 10 11 13 11 10 10 11 13 14 10 12 ] .

Transportation cost for 1st objective and 2nd item [ c i j k ( 12 ) ]

[ c 111 ( 12 ) c 112 ( 12 ) c 121 ( 12 ) c 122 ( 12 ) c 131 ( 12 ) c 132 ( 12 ) c 211 ( 12 ) c 212 ( 12 ) c 221 ( 12 ) c 222 ( 12 ) c 231 ( 12 ) c 232 ( 12 ) ] = [ 18 11 12 12 13 15 12 11 13 14 15 12 ] .

Transportation cost for 2nd objective and 1st item [ c i j k ( 11 ) ]

[ c 111 ( 21 ) c 112 ( 21 ) c 121 ( 21 ) c 122 ( 21 ) c 131 ( 21 ) c 132 ( 21 ) c 211 ( 21 ) c 212 ( 21 ) c 221 ( 21 ) c 222 ( 21 ) c 231 ( 21 ) c 232 ( 21 ) ] = [ 16 52 32 21 14 26 25 35 25 11 24 13 ] .

Transportation cost for 2nd objective and 2nd item [ c i j k ( 12 ) ]

[ c 111 ( 22 ) c 112 ( 22 ) c 121 ( 22 ) c 122 ( 22 ) c 131 ( 22 ) c 132 ( 22 ) c 211 ( 22 ) c 212 ( 22 ) c 221 ( 22 ) c 222 ( 22 ) c 231 ( 22 ) c 232 ( 22 ) ] = [ 22 19 15 18 19 13 22 14 12 11 8 15 ] .

Amount of items available at origin [ a i p ]

[ a 1 ( 1 ) a 2 ( 1 ) a 1 ( 2 ) a 2 ( 2 ) ] = [ 35 30 28 30 ] .

The demand amount of items at destination [ b j ( p ) ]

[ b 1 ( 1 ) b 2 ( 1 ) b 3 ( 1 ) b 1 ( 2 ) b 2 ( 2 ) b 3 ( 1 ) ] = [ 9 7 8 7 7 6 ] .

Amount of items transported by conveyances [ c k ]

[ e 1 e 2 ] = [ 25 26 ] .

Optimum result for Model-1

With the above input data, f 1 and f 2 are calculated using GRG, and we get
f 1 U = 691.5 , f 1 L = 602.5 , f 2 U = 1 , 106 , f 2 L = 670 .
So, we can get the membership functions of f 1 and f 2 (Figure 1) as follows:
μ 1 ( f 1 ( x ) ) = { 0 for  f 1 ( x ) > 691.5 , 691.5 f 1 ( x ) 691.5 602.5 for  602.5 < f 1 ( x ) < 691.5 , 1 for  f 1 ( x ) < 602.5 , μ 2 ( f 2 ( x ) ) = { 0 for  f 2 ( x ) > 1 , 106 , 1 , 106 f 2 ( x ) 1 , 106 670 for  1 , 106 < f 2 ( x ) < 670 , 1 for  f 2 ( x ) < 670 .
Figure 1

Membership functions of f 1 and f 2 .

Then we compute the following model to get the interactive satisfied solution:
{ min x λ s.t.  { ( 691.5 f 1 ( x ) ) ( μ 1 λ ) ( 691.5 602.5 ) , ( 1 , 106 f 2 ( x ) ) ( μ 2 λ ) ( 1 , 106 670 ) , x 111 1 + x 112 1 + x 121 1 + x 122 1 + x 131 1 + x 132 1 35 , x 211 1 + x 212 1 + x 221 1 + x 222 1 + x 231 1 + x 232 1 30 , x 111 1 + x 112 1 + x 211 1 + x 212 1 9 , x 121 1 + x 122 1 + x 221 1 + x 222 1 7 , x 131 1 + x 132 1 + x 231 1 + x 232 1 8 , x 111 2 + x 112 2 + x 121 2 + x 122 2 + x 131 2 + x 132 2 28 , x 211 2 + x 212 2 + x 221 2 + x 222 2 + x 231 2 + x 232 2 30 + 7 ( 1 a ) , x 111 2 + x 112 2 + x 211 2 + x 212 2 7 + 7 ( 1 α ) , x 121 2 + x 122 2 + x 221 2 + x 222 2 7 + 7 ( 1 α ) , x 131 2 + x 132 2 + x 231 2 + x 232 2 6 + 7 ( 1 α ) , x 111 1 + x 121 1 + x 131 1 + x 211 1 + x 221 1 + x 231 1 + x 111 2 + x 121 2 + x 131 2 + x 211 2 + x 221 2 + x 231 2 25 + 7 ( 1 α ) , x 112 1 + x 122 1 + x 132 1 + x 212 1 + x 222 1 + x 232 1 + x 112 2 + x 122 2 + x 132 2 + x 212 2 + x 222 2 + x 232 2 26 + 7 ( 1 α ) , 0 α 1 , x i j k 0 i , j , k ,
where α = 0.5 has been calculated in Example 9.1. Here we solve Model-1, to get the satisfied solutions, which are listed in Table 1.
Table 1

Employ the interactive fuzzy satisfied method based on the fuzzy inequality constraint of Model-1

μ 1

μ 2

f 1

f 2

μ 1 ( f 1 )

μ 2 ( f 2 )

Values of decision variables

λ

1

1

634.3

826.09

0.641

0.641

x 111 1 = 9 , x 121 1 = 2.04 , x 131 1 = 8 , x 222 1 = 4.95 , x 121 2 = 9.45 , x 122 2 = 1.04 , x 212 2 = 10.5 , x 232 2 = 9.50

0.35

1

0.9

630.98

853.13

0.679

0.579

x 111 1 = 9 , x 121 1 = 3.17 , x 131 1 = 8 , x 222 1 = 3.82 , x 121 2 = 8.32 , x 122 2 = 2.17 , x 212 2 = 10.5 , x 232 2 = 9.5

0.32

0.9

1

637.76

799.15

0.603

0.703

x 111 1 = 9 , x 121 1 = 0.91 , x 131 1 = 8 , x 222 1 = 6.08 , x 121 2 = 10.5 , x 212 2 = 10.5 , x 232 2 = 9.5

0.296

The first line of Table 1 lists each reference value of membership function μ 1 ( f 1 ) , when the initialized membership function is 1, the value of objective function f 1 ( x ) , and its corresponding solution x. If the decision maker hopes to improve f 2 ( x ) on the basis of sacrifice f 1 ( x ) , we may consider resetting the reference value of membership function ( μ 1 , μ 2 ) , e.g., we set ( μ 1 , μ 2 ) = ( 0.9 , 1 ) or ( μ 1 , μ 2 ) = ( 1 , 0.9 ) . The corresponding results are listed in the second and third lines. Suppose that when the reference value of membership function is ( μ 1 , μ 2 ) = ( 0.9 , 1 ) , the decision maker is satisfied, then the interactive process is stopped, so we obtain that the satisfied solutions for different items are
[ x 111 ( 1 ) x 112 ( 1 ) x 121 ( 1 ) x 122 ( 1 ) x 131 ( 1 ) x 132 ( 1 ) x 211 ( 1 ) x 212 ( 1 ) x 221 ( 1 ) x 222 ( 1 ) x 231 ( 1 ) x 232 ( 1 ) ] = [ 9 0 0.91 0 8 0 0 0 0 6.08 0 0 ] , [ x 111 ( 2 ) x 112 ( 2 ) x 121 ( 2 ) x 122 ( 2 ) x 131 ( 2 ) x 132 ( 2 ) x 211 ( 2 ) x 212 ( 2 ) x 221 ( 2 ) x 222 ( 2 ) x 231 ( 2 ) x 232 ( 2 ) ] = [ 0 0 10.5 0 0 0 0 10.5 0 0 0 9.5 ]
and the corresponding optimal values for different items are
[ f 1 ( 1 ) f 1 ( 2 ) f 2 ( 1 ) f 2 ( 2 ) ] = [ 282.26 355.5 352.15 447 ] .
Applying the convex combination method stated in Section 4.3, we get Table 2 for different weights on f 1 and f 2 . The comparison between the optimum results calculated by different methods for Model-1 is given in Table 3.
Table 2

Optimum results by using the convex combination method of Model-1

w 1

w 2

f 1

f 2

Values of decision variables

0.5

0.5

691.5

670

x 111 1 = 9 , x 131 1 = 6.5 , x 222 1 = 7 , x 232 1 = 1.5 , x 212 2 = 10.5 , x 222 2 = 10.5 , x 231 2 = 9.5

0.6

0.4

688

673.5

x 111 1 = 9 , x 131 1 = 8 , x 222 1 = 7 , x 212 2 = 10.5 , x 221 2 = 2 , x 222 2 = 8.5 , x 231 2 = 9.5

0.7

0.3

651

748.5

x 111 1 = 9 , x 131 1 = 8 , x 222 1 = 7 , x 212 2 = 10.5 , x 221 2 = 10.5 , x 232 2 = 9.5

Table 3

Comparison of optimum results of Model-1

Method

f 1

f 2

Values of decision variables

FISM

630.98

799.15

x 111 1 = 9 , x 121 1 = 0.91 , x 131 1 = 8 , x 222 1 = 6.08 , x 121 2 = 10.5 , x 212 2 = 10.5 , x 232 2 = 9.5

Convex combination

651

748.5

x 111 1 = 9 , x 131 1 = 8 , x 222 1 = 7 , x 212 2 = 10.5 , x 221 2 = 10.5 , x 232 2 = 9.5

Global criteria

669

711.2

x 111 1 = 9 , x 131 1 = 6.75 , x 222 1 = 7 , x 232 1 = 1.24 , x 221 2 = 7.95 , x 212 2 = 10 , x 231 2 = 4.78 , x 222 1 = 2.54 , x 232 2 = 4.71 , GC = 0.1265

7.2 Input data for Model-2

Let us consider a multi-objective single-item solid transportation problem with three origins (i.e., M = 3 ), two destinations (i.e., N = 2 ) and two types of conveyances (i.e., K = 2 ). The parameters are given as follows.

Transportation cost for 1st objective [ c i j k ( 1 ) ]

[ c 111 ( 1 ) c 112 ( 1 ) c 121 ( 1 ) c 122 ( 1 ) c 131 ( 1 ) c 132 ( 1 ) c 211 ( 1 ) c 212 ( 1 ) c 221 ( 1 ) c 222 ( 1 ) c 231 ( 1 ) c 232 ( 1 ) ] = [ 10 12 11 13 9 19 12 11 13 14 13 19 ] .

Transportation cost for 2nd objective [ c i j k ( 2 ) ]

[ c 111 ( 2 ) c 112 ( 2 ) c 121 ( 2 ) c 122 ( 2 ) c 131 ( 2 ) c 132 ( 2 ) c 211 ( 2 ) c 212 ( 2 ) c 221 ( 2 ) c 222 ( 2 ) c 231 ( 2 ) c 232 ( 2 ) ] = [ 16 15 12 11 14 16 15 15 13 11 14 13 ] .

Amount of items available at origin [ a i ]

[ a 1 a 2 ] = [ 21 20 ] .

The demand amount of items at destination [ b j ]

[ b 1 b 2 b 3 ] = [ 7 5 2 ] .

Amount of items transported by conveyances [ c k ]

[ e 1 e 2 ] = [ 20 22 ] .

Optimum result for Model-2

With the above input data, f 1 and f 2 are calculated using the GRG technique, and we get
f 1 U = 345.76 , f 1 L = 252.1 , f 2 U = 344.20 , f 2 L = 323.20 .
So we can get the membership functions of f 1 and f 2 (Figure 2) as follows:
μ 1 ( f 1 ( x ) ) = { 0 for  f 1 ( x ) > 345.76 , 345.76 f 1 ( x ) 345.76 252.1 for  252.1 < f 1 ( x ) < 345.76 , 1 for  f 1 ( x ) < 252.1 , μ 2 ( f 2 ( x ) ) = { 0 for  f 2 ( x ) > 344.20 , 344.20 f 2 ( x ) 344.20 323.20 for  323.20 < f 2 ( x ) < 344.20 , 1 for  f 2 ( x ) < 323.20 .
Then we compute the following model to get the interactive satisfied solution:
{ min x λ s.t.  { ( 345.76 f 1 ( x ) ) ( μ 1 λ ) ( 345.76 252.1 ) , ( 344.20 f 2 ( x ) ) ( μ 2 λ ) ( 344.20 323.20 ) , x 111 + x 112 + x 121 + x 122 + x 131 + x 132 21 , x 211 + x 212 + x 221 + x 222 + x 231 + x 232 20 , x 111 + x 112 + x 211 + x 212 7 + 7 ( 1 α ) , x 121 + x 122 + x 221 + x 222 5 + 7 ( 1 α ) , x 131 + x 132 + x 231 + x 232 2 + 7 ( 1 α ) , x 111 + x 121 + x 131 + x 211 + x 221 + x 231 20 + 7 ( 1 α ) , x 112 + x 122 + x 132 + x 212 + x 222 + x 232 22 + 7 ( 1 α ) , x i j k 0 i , j , k , 0 α 1 ,
where α = 0.4974 has been already calculated in Example 9.2. Here we solve Model-2 to get the satisfied solutions, which are listed in Table 4.
Figure 2

Membership functions of f 1 and f 2 .

Table 4

Employ the interactive fuzzy satisfied method based on fuzzy inequality constraint for Model-2

μ 1

μ 2

f 1

f 2

μ 1 ( f 1 )

μ 2 ( f 2 )

Values of decision variables

λ

1

1

276.55

328.68

0.73

0.73

x 122 = 8.51 , x 131 = 5.47 , x 212 = 10.51 , x 232 = 0.045

0.261

0.9

1

283.03

328.03

0.66

0.76

x 122 = 8.51 , x 131 = 4.82 , x 212 = 10.51 , x 232 = 0.69

0.2302

1

0.9

273.4

330.07

0.67

0.77

x 121 = 1.34 , x 122 = 7.16 , x 131 = 5.51 , x 212 = 10.51

0.227

The first line of Table 4 lists each reference value of membership function μ 1 ( f 1 ) , when the initialized membership function is 1, the value of objective function f 1 ( x ) , and its corresponding solution x. If the decision maker hopes to improve f 2 ( x ) on the basis of sacrifice f 1 ( x ) , we may consider resetting the reference value of membership function ( μ 1 , μ 2 ) , e.g., we set ( μ 1 , μ 2 ) = ( 0.9 , 1 ) or ( μ 1 , μ 2 ) = ( 1 , 0.9 ) . The corresponding results are listed in the second and third lines. Suppose that when the reference value of membership function is ( μ 1 , μ 2 ) = ( 1 , 0.90 ) , the decision maker is satisfied, then the interactive process is stopped, so we obtain that the satisfied solutions are x 121 = 1.34 , x 122 = 7.16 , x 131 = 5.51 , x 212 = 10.51 . The corresponding optimal values are
[ f 1 f 2 ] = [ 273.40 330.67 ] .
Applying the convex combination method stated in Section 4.3, we get Table 5 for different weights on f 1 and f 2 . The comparison between optimum results calculated by different methods for Model-2 is given in Table 6.
Table 5

Optimum results by using the convex combination method for Model-2

w 1

w 2

f 1

f 2

Values of decision variables

0.5

0.5

259.06

337.24

x 121 = 8.51 , x 131 = 5.51 , x 212 1 = 10.51 , x 121 1 = 8.51

0.3

0.7

276.1

328.72

x 122 = 8.51 , x 131 = 5.51 , x 212 1 = 10.51

0.6

0.4

252.1

344.20

x 111 = 6.96 , x 121 = 8.51 , x 131 1 = 5.51 , x 212 1 = 3.55

Table 6

Comparison of optimum results of Model-2

Method

f 1

f 2

Values of decision variables

FISM

273.40

330.07

x 121 = 1.34 , x 122 = 7.16 , x 131 = 5.51 , x 212 = 10.51

Convex combination

252.1

344.20

x 111 = 6.96 , x 121 = 8.51 , x 131 = 5.51 , x 212 = 3.55

Global criteria

259.06

337.24

x 121 = 8.51 , x 131 = 5.51 , x 212 = 10.51 , GC = 0.0514

7.3 Optimum result for Model-3

To solve Model-3, we will solve MSSTP α (Example 9.1) and the optimum solution has come for decision α = 0.5 , and the optimal solutions for different items are
[ x 111 ( 1 ) x 112 ( 1 ) x 121 ( 1 ) x 122 ( 1 ) x 131 ( 1 ) x 132 ( 1 ) x 211 ( 1 ) x 212 ( 1 ) x 221 ( 1 ) x 222 ( 1 ) x 231 ( 1 ) x 232 ( 1 ) ] = [ 0 2.70 7 0 0 2.47 6.29 0 0 0 5.52 0 ] , [ x 111 ( 2 ) x 112 ( 2 ) x 121 ( 2 ) x 122 ( 2 ) x 131 ( 2 ) x 132 ( 2 ) x 211 ( 2 ) x 212 ( 2 ) x 221 ( 2 ) x 222 ( 2 ) x 231 ( 2 ) x 232 ( 2 ) ] = [ 0 4.81 8.32 2.17 0 0 0 5.68 0 0 0 9.5 ]
and the corresponding optimal values for different items are
[ f 1 ( 1 ) f 1 ( 2 ) f 2 ( 1 ) f 2 ( 2 ) ] = [ 247 355.5 719.10 477.59 ] .

8 Conclusion

The multi-objective multi-item solid transportation problem in fuzzy inequality constraints has been explored in this paper. Three different models have been derived. First, a fuzzy inequality solid transportation problem has been converted to a parametric solid transportation problem using flexible index, and then the fuzzy inequality solid transportation problem has been solved by using the decision making technique. The fuzzy interactive satisfied method, global criterion method and convex combination method have been applied to calculate the optimal compromise solutions of multi-objective STP problem, and then it was solved by using MatLab and Lingo-11.0. The models are illustrated with numerical examples and corresponding results are compared. This paper only researches the problem under fuzzy inequality constraints, and the problem in other more complex environments or multi-objective uncertain transportation problem may become new topics in further research. The present formulation and solution procedures can be applied to other fuzzy transportation models with different fuzzy numbers.

Appendix

Example 9.1 Consider the following FICP:
MMSTP ˜ min 11 x 111 1 + 10 x 112 1 + 11 x 121 1 + 13 x 122 1 + 11 x 131 1 + 10 x 132 1 + 10 x 211 1 + 11 x 212 1 MMSTP ˜ min + 13 x 221 1 + 14 x 222 1 + 10 x 231 1 + 12 x 232 1 + 18 x 111 2 + 11 x 112 2 + 12 x 121 2 + 12 x 122 2 MMSTP ˜ min + 13 x 131 2 + 15 x 132 2 + 12 x 211 2 + 11 x 212 2 + 13 x 221 2 + 14 x 222 2 + 15 x 231 2 + 12 x 232 2 MMSTP ˜ s.t.  x 111 1 + x 112 1 + x 121 1 + x 122 1 + x 131 1 + x 132 1 35 , MMSTP ˜ s.t.  x 211 1 + x 212 1 + x 221 1 + x 222 1 + x 231 1 + x 232 1 30 , MMSTP ˜ s.t.  x 111 1 + x 112 1 + x 211 1 + x 212 1 9 , MMSTP ˜ s.t.  x 121 1 + x 122 1 + x 221 1 + x 222 1 7 , MMSTP ˜ s.t.  x 131 1 + x 132 1 + x 231 1 + x 232 1 8 , MMSTP ˜ s.t.  x 111 2 + x 112 2 + x 121 2 + x 122 2 + x 131 2 + x 132 2 28 , MMSTP ˜ s.t.  x 211 2 + x 212 2 + x 221 2 + x 222 2 + x 231 2 + x 232 2 30 , MMSTP ˜ s.t.  x 111 2 + x 112 2 + x 211 2 + x 212 2 7 , MMSTP ˜ s.t.  x 121 2 + x 122 2 + x 221 2 + x 222 2 7 , MMSTP ˜ s.t.  x 131 2 + x 132 2 + x 231 2 + x 232 2 6 , MMSTP ˜ s.t.  x 111 1 + x 121 1 + x 131 1 + x 211 1 + x 221 1 + x 231 1 + x 111 2 MMSTP ˜ s.t.  + x 121 2 + x 131 2 + x 211 2 + x 221 2 + x 231 2 25 , MMSTP ˜ s.t.  x 112 1 + x 122 1 + x 132 1 + x 212 1 + x 222 1 + x 232 1 + x 112 2 MMSTP ˜ s.t.  + x 122 2 + x 132 2 + x 212 2 + x 222 2 + x 232 2 26 , MMSTP ˜ s.t.  x i j k p 0 i , j , k  and  p = 1 , 2 , d = ( 0 , 0 , 0 , 0 , 0 , 0 , 7 , 7 , 7 , 7 , 7 , 7 ) .
The corresponding parametric linear programming problem of MMSTP ˜ is presented as follows:
MMSTP α min 11 x 111 1 + 10 x 112 1 + 11 x 121 1 + 13 x 122 1 + 11 x 131 1 + 10 x 132 1 MMSTP α min + 10 x 211 1 + 11 x 212 1 + 13 x 221 1 + 14 x 222 1 + 10 x 231 1 + 12 x 232 1 MMSTP α min + 18 x 111 2 + 11 x 112 2 + 12 x 121 2 + 12 x 122 2 + 13 x 131 2 + 15 x 132 2 MMSTP α min + 12 x 211 2 + 11 x 212 2 + 13 x 221 2 + 14 x 222 2 + 15 x 231 2 + 12 x 232 2 MMSTP α s.t.  x 111 1 + x 112 1 + x 121 1 + x 122 1 + x 131 1 + x 132 1 35 , MMSTP α s.t.  x 211 1 + x 212 1 + x 221 1 + x 222 1 + x 231 1 + x 232 1 30 , MMSTP α s.t.  x 111 1 + x 112 1 + x 211 1 + x 212 1 9 , MMSTP α s.t.  x 121 1 + x 122 1 + x 221 1 + x 222 1 7 , MMSTP α s.t.  x 131 1 + x 132 1 + x 231 1 + x 232 1 8 , MMSTP α s.t.  x 111 2 + x 112 2 + x 121 2 + x 122 2 + x 131 2 + x 132 2 28 , MMSTP α s.t.  x 211 2 + x 212 2 + x 221 2 + x 222 2 + x 231 2 + x 232 2 30 + 7 ( 1 a ) , MMSTP α s.t.  x 111 2 + x 112 2 + x 211 2 + x 212 2 7 + 7 ( 1 α ) , MMSTP α s.t.  x 121 2 + x 122 2 + x 221 2 + x 222 2 7 + 7 ( 1 α ) , MMSTP α s.t.  x 131 2 + x 132 2 + x 231 2 + x 232 2 6 + 7 ( 1 α ) , MMSTP α s.t.  x 111 1 + x 121 1 + x 131 1 + x 211 1 + x 221 1 + x 231 1 + x 111 2 MMSTP α s.t.  + x 121 2 + x 131 2 + x 211 2 + x 221 2 + x 231 2 25 + 7 ( 1 α ) , MMSTP α s.t.  x 112 1 + x 122 1 + x 132 1 + x 212 1 + x 222 1 + x 232 1 + x 112 2 MMSTP α s.t.  + x 122 2 + x 132 2 + x 212 2 + x 222 2 + x 232 2 26 + 7 ( 1 α ) , MMSTP α s.t.  x i j k p 0 i , j , k , p = 1 , 2 and 0 α 1 .
(25)
Now, using the algorithm given in Section 2.1, we obtain Z 0 = 725 , Z 1 = 480 and d 0 = 245 by calculating ( MMSTP 0 ) and ( MMSTP 1 ) corresponding to (25). Let the inverse of the optimal matrix in ( MMSTP 0 ) be
B 0 1 = ( 1 0 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 ) .

Now, calculating the equations [ B 0 1 ( 35 , 30 , 9 , 7 , 8 , 28 , 30 + 7 ( 1 α ) , 7 + 7 ( 1 α ) , 7 + 7 ( 1 α ) , 6 + 7 ( 1 α ) , 25 + 7 ( 1 α ) , 26 + 7 ( 1 α ) ) ] i = 0 ( i = 1 , 2 , , 12 ), respectively, we obtain α 1 = 0.8571 assume α 0 = 0 and α 2 = 1 .

Now, solving ( MMSTP α 1 ), we get an optimal solution and an optimal value as Z α 1 = Z 0.8571 = 515.01 .

Now, Z 0.8571 = 515.01 < Z 1 + 0.8571 × 245 = 689.98 , so we stop here and calculate optimal decision α . Now
α = Z 1 α 1 Z 1 α 0 Z α 0 α 1 + Z α 1 α 0 Z α 1 Z α 0 α 1 d 0 + α 0 d 0 = 0.5 .

Now, solving ( MMSTP 0.5 ), we obtain the optimal value Z 0.5 = 602.5 .

Example 9.2 Consider the following FICP:
MSSTP ˜ min 10 x 111 + 12 x 112 + 11 x 121 + 13 x 122 MSSTP ˜ min + 9 x 131 + 19 x 132 + 12 x 211 + 11 x 212 MSSTP ˜ min + 13 x 221 + 14 x 222 + 13 x 231 + 19 x 232 MSSTP ˜ s.t.  x 111 + x 112 + x 121 + x 122 + x 131 + x 132 21 , MSSTP ˜ s.t.  x 211 + x 212 + x 221 + x 222 + x 231 + x 232 20 , MSSTP ˜ s.t.  x 111 + x 112 + x 211 + x 212 7 , MSSTP ˜ s.t.  x 121 + x 122 + x 221 + x 222 5 , MSSTP ˜ s.t.  x 131 + x 132 + x 231 + x 232 2 , MSSTP ˜ s.t.  x 111 + x 121 + x 131 + x 211 + x 221 + x 231 20 , MSSTP ˜ s.t.  x 112 + x 122 + x 132 + x 212 + x 222 + x 232 22 , MSSTP ˜ s.t.  x i j k 0 i , j , k , d = ( 0 , 0 , 7 , 7 , 7 , 7 , 7 ) .
(26)
The corresponding parametric linear programming problem of MSSTP ˜ is presented as follows:
MSSTP α min 10 x 111 + 12 x 112 + 11 x 121 + 13 x 122 MSSTP α min + 9 x 131 + 19 x 132 + 12 x 211 + 11 x 212 MSSTP α min + 13 x 221 + 14 x 222 + 13 x 231 + 19 x 232 MSSTP α s.t.  x 111 + x 112 + x 121 + x 122 + x 131 + x 132 21 , MSSTP α s.t.  x 211 + x 212 + x 221 + x 222 + x 231 + x 232 20 , MSSTP α s.t.  x 111 + x 112 + x 211 + x 212 7 + 7 ( 1 α ) , MSSTP α s.t.  x 121 + x 122 + x 221 + x 222 5 + 7 ( 1 α ) , MSSTP α s.t.  x 131 + x 132 + x 231 + x 232 2 + 7 ( 1 α ) , MSSTP α s.t.  x 111 + x 121 + x 131 + x 211 + x 221 + x 231 20 + 7 ( 1 α ) , MSSTP α s.t.  x 112 + x 122 + x 132 + x 212 + x 222 + x 232 22 + 7 ( 1 α ) , MSSTP α s.t.  x i j k 0 i , j , k , 0 α 1 .
(27)
Now, using the algorithm given in Section 2.1, we obtain Z 0 = 367 , Z 1 = 143 and d 0 = 224 by calculating ( MSSTP 0 ) and ( MSSTP 1 ) corresponding to (27). The inverse of the optimal matrix in ( MSSTP 0 ) is
B 0 1 = ( 1 0 1 1 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 1 ) .

Now, calculating the equations [ B 0 1 ( 21 , 20 , 7 + 7 ( 1 α ) , 5 + 7 ( 1 α ) , 2 + 7 ( 1 α ) , 20 + 7 ( 1 α ) , 22 + 7 ( 1 α ) ) ] i = 0 ( i = 1 , 2 , , 7 ), respectively, we obtain α 1 = 0.8571 assume α 0 = 0 and α 2 = 1 .

Now, solving ( MSSTP α 1 ), we get an optimal solution as x 122 = 9 , x 311 = 9 and x 321 = 2 and an optimal value as Z α 1 = Z 0.8571 = 173.009 .

Now Z 0.8571 = 173.009 < Z 1 + 0.8571 × 224 = 334.99 , so we stop here and calculate optimal decision α . Now
α = Z 1 α 1 Z 1 α 0 Z α 0 α 1 + Z α 1 α 0 Z α 1 Z α 0 α 1 d 0 + α 0 d 0 = 0.4974 .

Now, solving ( MSSTP 0.4974 ), we obtain an optimal value as Z 0.4974 = 252.1 .

Declarations

Acknowledgements

The authors sincerely thank the anonymous reviewers and editor-in-chief for their careful reading, constructive comments and fruitful suggestions. The first two authors are also thankful to Ms. Priyanka Dey, Assistant Professor, Haldia Institute of Technology for advices on grammatical errors and organization of the paper.

Authors’ Affiliations

(1)
Department of Mathematics, Heritage Institute of Technology
(2)
Department of Applied Science, Haldia Institute of Technology
(3)
Department of Mathematics, Bengal Engineering and Science University

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© Chakraborty et al.; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.