CHAPTER SOLVING MULTI-OBJECTIVE TRANSPORTATION PROBLEM USING FUZZY PROGRAMMING TECHNIQUE-PARALLEL METHOD 40 3.

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1 CHAPTER SOLVING MULTI-OBJECTIVE TRANSPORTATION PROBLEM USING FUZZY PROGRAMMING TECHNIQUE-PARALLEL METHOD INTRODUCTION MULTIOBJECTIVE TRANSPORTATION PROBLEM THEOREMS ON TRANSPORTATION PROBLEM FUZZY PRELIMINARIES PROPOSED PARALLEL METHODOLOGY EXAMPLES SUMMARIZED RESULTS AND COMMENTS 53 39

2 CHAPTER - 3 SOLVING MULTI-OBJECTIVE TRANSPORTATION PROBLEM USING FUZZY PROGRAMMING TECHNIQUE-PARALLEL METHOD 3.1 INTRODUCTION In this chapter, we propose a parallel algorithm for solving MOTP. Here, fuzzy programming technique is used with fuzzy linear membership function for different costs to solve MOTP. The proposed method is a parallel method to new row maxima method. Instead of considering average we have considered minimum of membership value. To take minimum of cost and time in first example (or in any other examples all the cost matrices may not have the same units of measurements) is not logically true because of their units. But by converting them in membership value they become unit less pure numbers and its possible to take minimum of two or more cost matrix.it gives better results for solving MOTP with less complexity. Finally, same numerical examples as in chapter 2 are illustrated to check the feasibility of the proposed study. The transportation problem (TP) is a specific problem of resource allocation, can be formulated as a linear programming problem, where the constraints have a special structure. The classical form of the transportation problem minimizes transporting cost of some commodity that is available at m sources and required at n destinations. The sources are the production facilities, warehouses or supply points and destinations are the consumption points, warehouses or demand points. The transportation problem was firstly developed by Hitchcock in Efficient method of solution derived from the simplex algorithm were primarily developed by Dantzig and then by Charnes et. al. The transportation problem can be modelled as a standard linear programming problem, which can be solved by the Simplex method. 40

3 The objective of the transportation problem is to determine the optimum schedule from different sources to different destinations so as minimize the cost of transportation. 3.2 MULTIOBJECTIVE TRANSPORTATION PROBLEM In the real life, however all transportation problems are not single objective. The transportation problem was characterized by multiple objective functions. The decision maker would like to minimize set of p objectives simultaneously. The mathematical model of MOTP can be written as follows:, 1,2, 3.1 Subject to the constraints, 1,2,.. 3.2, 1,2,, 3.3! 0, 1,2,., # 1,2, 3.4 The superscript on and are used to identify the number of objective functions, a i > 0 for all i, b j > 0 for all j, 0 for all (i, j) and (balanced condition). The balanced condition is both necessary and sufficient for solving the transportation problem in both the cases single and multiple objectives. 3.3 THEOREMS ON TRANSPORTATION PROBLEM Theorem 1. (Existence of feasible solution). A necessary and sufficient condition for the feasible solution of a transportation problem is 41

4 ( Chapter - 3 1,..,; 1,..,. 3.5 Proof. The condition is necessary.let there exist a feasible solution to the transportation problem. Then,, * + * ) - The condition is sufficient. Let./ / # :3 / or 1 1 Thus, 1, 44 #! 0,.63 ; 0, ; 0, 44 # Hence a feasible solution exists. Theorem 2. Number of basic variables in a transportation problem are at the most m+n-1. 42

5 Proof. If it is possible to prove that equations (3.2) and (3.3) are m+n-1 linearly independent equations in mn variables consequently, number of basic variables will be at the most m+n-1. To prove this, first add m row-equations (3.2) and then subtract from the sum of first (n-1) column equations (3.3), thereby getting = = < < Or > <? < > <? < 3A B which is the last column equation. Thus, out of m+n equations, one (any) is redundant and remaining m+n-1 form a linearly independent set. Hence the theorem is proved. It is concluded that a basic feasible solution will consist of at most m + n 1 positive variables, others being zero. In the degenerate case, some of the basic variables will also be zero, that is number of positive variables will become less than m + n FUZZY PRELIMINARIES Definition 2.1 Here penalties (transportation cost, delivery time etc.) as membership value defined using membership function. As membership value is higher it is closer to the optimal solution. The decision maker would like to minimize the set of p objectives. The linear membership function is used and defined as: 43

6 C D E H 1 I J * K < G * F, * K < J J I I K 0! K ) *+ 3.6 WhereJ 2 K, k= 1,2,...p. If J K then membership value is 1 for any value of k. 3.5 PROPOSED PARALLEL METHODOLOGY For solving MOTP, the proposed method is summarized in the following Steps: Step 1: Calculate the linear membership value using equation (3.6) for each cell and for each objective table. Step 2: Construct a new table in which each cell having minimum membership value of all objective tables. Step 3: Take the first row and search maximum membership value, allocate in this cell maximum possible to get rim condition. If tie occurs then compare the two columns and allocate maximum possible for the column where minimum elements are occurred in the subsequent rows. Step 4: If rim condition is in row then go to next row and repeat step 3. If rim condition is in column, then search next maximum membership value in the same row and allocate in this cell maximum possible to get rim condition. Step 5: Repeat steps 3 and 4 until supply and demand are exhausted. 44

7 3.6 EXAMPLES Let us consider the following example to illustrate proposed method [3][20]. The data for the cost and time is as follows: Table 3.1 Data for time Destination D1 D2 D3 Supply S S S Demand Table 3.2 Data for cost Destination D1 D2 D3 Supply S S S Demand As the first step we calculate membership value for time, here U k = 28 and L k = 8 then membership values are as follows: 46

8 Table 3.3 Membership value for time Destination D1 D2 D3 Supply S S S Demand Membership value for cost, here U k = 20 and L k = 6 then membership values are as follows: Table 3.4 Membership value for cost Destination D1 D2 D3 Supply S S S Demand

9 Now calculate minimum membership value: Table 3.5 Minimum membership value Destination D1 D2 D3 Supply S S S Demand After applying the proposed parallel method we get the solution as follows: X = { X 11 = 10,X 13 = 4, X 22 = 15, X 23 = 1, X 33 = 12} The corresponding objective functions values are 518 and N 374 Example : Let us consider another example of MOTP [14,7,1,36,31] having the following characteristics. Table 3.6 Penalties Destination D1 D2 D3 D4 Supply S S S Demand

10 Table 3.7 Penalties Destination D1 D2 D3 D4 Supply S S S Demand As the first step we calculate membership value for first objective table(3.6), here U k = 9 and L k = 1, then membership values are as follows: Table Membership value for penalties table 3.6 Destination D1 D2 D3 D4 Supply S S S Demand Membership value for second objective table(3.7), here U k = 10 and L k = 1 then membership values are as follows: Table Membership value for penalties table 3.7 Destination D1 D2 D3 D4 Supply S S S Demand

11 Here we calculate minimum membership value Table 3.10 Minimum membership value Destination D1 D2 D3 D4 Supply S S S Demand On applying the proposed parallel methodology we get the solution as follows: X = { X 11 = 5,X 12 = 3, X 21 = 6, X 23 = 13, X 33 = 1, X 34 = 16} The corresponding objective functions values are 156 and N 200 Example : Let us consider one more example of MOTP [14,29,20,1,36,31] having the following characteristics. Table 3.11 Penalties Destination D1 D2 D3 D4 D5 Supply S S S S Demand

12 Table 3.12 Penalties Destination D1 D2 D3 D4 D5 Supply S S S S Demand Table 3.13 Penalties Destination D1 D2 D3 D4 D5 Supply S S S S Demand As the first step we calculate membership value for first objective table(3.11), here U k = 12 and L k = 2, then membership values are as follows: Table 3.14 Membership value for table 3.11 Destination D1 D2 D3 D4 D5 Supply S S S S Demand

13 Membership value for second objective table(3.12), here U k = 9 and L k = 1, then membership values are as follows: Table Membership value for table 3.12 Destination D1 D2 D3 D4 D5 Supply S S S S Demand Membership value for third objective table(3.13), here U k = 9 and L k = 1 then membership values are as follows: Table Membership value for table 3.13 Destination D1 D2 D3 D4 D5 Supply S S S S Demand

14 Here we calculate minimum membership value Table 3.17 Minimum membership value Destination D1 D2 D3 D4 D5 Supply S S S S Demand On applying the parallel method we get the solution as follows: X = { X 11 = 3, X 14 = 2, X 25 = 4, X 32 = 2, X 41 = 1, X 42 =2, X 43 = 6,} The corresponding objective functions values are 157, N 72 and P SUMMARIZED RESULTS AND COMMENTS The results of the above three examples are summarized and shown below in Tables 3.18, 3.19, Table Comparison between different approaches for ex Name of the approach Q R S Q T S Fuzzy goal programming [20], New Row Maxima Method [3], Proposed (Parallel )method Ideal Solution

15 Table Comparison between different approaches for ex Name of the approach Q R S Q T S Fuzzy goal programming method [14], Fuzzy programming to solve bi objective[7], Trust region algorithm[36], New Row Maxima Method [3], Proposed(parallel) method Ideal Solution Table Comparison between different approaches for ex Name of the approach Q R S Q T S Q U S Interactive algorithms[29], Fuzzy goal programming method [14], Fuzzy programming goal programming [20], Trust region algorithm[36], New Row Maxima Method [3], Proposed (Parallel )method Ideal Solution Comments: Three numerical examples are illustrated and obtained results compared with some of the methods in literature. In first and third problem we get the same solution for all the objectives as in new row matrix maxima method. But in second problem, we get improved solutions by the proposed method as compared to the new row matrix maxima method. If we compare all the three problems found in literature methods, in most of the cases we got better solution. 54

OPERATIONS RESEARCH. Transportation and Assignment Problems

OPERATIONS RESEARCH. Transportation and Assignment Problems OPERATIONS RESEARCH Chapter 2 Transportation and Assignment Problems Prof. Bibhas C. Giri Department of Mathematics Jadavpur University Kolkata, India Email: bcgiri.jumath@gmail.com 1.0 Introduction In