Question

a) Solve the following transportation problem using any suitable method. The cell| entries are unit costs D D2 D D Ds 15 available 0 68 35 4 18 02 57 88 91 Irn 8 17 03 91 60 75 43 60 19 04 52 53 7 13 Os 13 7 15 51 16 18 18 Required 20 14 14 b) Reduce the following game by Dominance and find the game value. TABLE 1 PLAYERB II III IV 2 It PLAYER A II 4 2 III 4 2 4 0 IV 4 ba

Answer 1

Find Solution using Row minima method D1 D2 D3 D4 D5 Supply S1 68 354 74 15 18 58 354 S2 57 88 91 3 8 17 S3 91 60 75 45 60 19 52 53 24 7 82 13 51 18 82 137 15 Demand 16 18 20 14 14 S4 S5 Solution: TOTAL number of supply constraints : 5 TOTAL number of demand constraints : 5 Problem Table is D1 1 D2 D3 D4 D5 Supply S1 68 35 4 74 15 18 S2 S3 57 88 91 3 8 17 91 60 75 45 60 19 2 53 247 82 18 82 137 S4 13 S. 51 18 82 137 15 Demand 16 18 20 14 14 In 1st row, The smallest transportation cost is 4 in cell S1D3. The allocation to this cell is min(18,20) = 18. This exhausts the capacity of Sy and leaves 20- 18 = 2 units with D3

Table-1 D1 D2 D3 D4 D5 Supply ST 68 35 4(18) 74 15 0 S2 57 88 91 38 S3 91 60 75 45 60 19 S4 52 53 24 7 82 13 S. 51 18 82 13 7 15 Demand 16 18 2 14 14 In 2nd row, The smallest transportation cost is 3 in cell SD The allocation to this cell is min(17,14) = 14. This satisfies the entire demand of D4 and leaves 17 - 14 = 3 units with S2 Table-2 D1 D2 D3 D4 D5 Supply S 68 35 4(18) 74 151 0 S2 57 88 88 91 3(4) 8 3 S3 91 60 75 415 60 19 S4 52 53 24 82 13 S5 51 18 82 13 7 15 Demand 16 18 2 0 14 In 2nd row, The smallest transportation cost is 8 in cell S2D5. The allocation to this cell is min(3,14) = 3.

Table-3 D1 D2 D3 DS Supply ST 68 35 4(18) 714 15 0 Sz 57 88 97 34|4) 843) 0 S3 91 60 75 45 60 19 S4 52 53 24 7 82 13 S5 51 51 18 82 13 7 15 Demand 16 18 2 0 11 In 3rd row, The smallest transportation cost is 60 in cells The allocation to this cell is min(19,18) = 18. This satisfies the entire demand of Dą and leaves 19 - 18 = 1 units with S3 Table-4 D1 D2 D3 D4 D5 Supply S 68 315 4(18) 74 15 0 Sz 57 91 344) 813) 0 S3 91 60(18) 75 45 60 1 A S4 52 53 24 82 13 S5 51 18 82 13 7 15 Demand 16 0 2 0 11

In 3rd row, The smallest transportation cost is 60 in cell SzD5 The allocation to this cell is min(1,11) = 1. This exhausts the capacity of Sz and leaves 11 - 1 = 10 units with D Table-5 D1 D2 D3 de 4 D5 Supply S 68 315 4(18) 74 15 0 S 57 818 91 344) 813) 0 91 60(18) 75 45 60(17 0 S4 52 53 24 7 82 13 S5 51 18 82 13 7 15 Demand 16 0 2. 0 10 In 4th row, The smallest transportation cost is 24 in cell S4D3 The allocation to this cell is min(13,2) = 2. This satisfies the entire demand of D3 and leaves 13 -2 = 11 units with S4

Table-6 D1 02 D3 D4 D5 Supply S 68 448) 15 0 Sz 57 94234741 843) 0 91 60418) 75 4560(1) 0 S4 52 53 242) 7 82 11 Sg 51 18 82 13 7 15 Demand 16 0 0 0 10 In 4th row, The smallest transportation cost is 52 in cell SĄD. The allocation to this cell is min(11,16) = 11. This exhausts the capacity of S4 and leaves 16 - 11 = 5 units with Di Table-7 D1 D2 D3 Da D5 Supply S 4618) 74 15 0 S 57 8b 92 344) 813) 0 S 91 6048) 715 4560(1) 0 52(11) 24/2) 0 SS 51 18 82 13 7 15 Demand 5 0 0 0 10 In 5th row, The smallest transportation cost is 7 in cell SD5. The allocation to this cell is min(15,10) = 10. This satisfies the entire demand of D, and leaves 15 - 10 = 5 units with S5

Table-8 D1 D2 D3 D4 D5 Supply ST 68 315 448) 714 115 0 Sz 57 91 344 843) 0 Sz 91 60(18) 715 45 O 6011) SA 52(11) 53 24/2) 822 0 S5 51 18 822 13 7(10) 5 Demand 5 0 0 O 0 In 5th row, The smallest transportation cost is 51 in cell szDz. The allocation to this cell is min(5,5) = 5. Table-9 11 D2 D3 D. 4 Supply s 68 448) 744 75 0 S 57 91 344 843) 0 3 911 60418) 75 45 60117 0 य 512 242) 82 0 51/5) 15 0 Demand 0 0 0 0 0 Initial feasible solution is D1 D2 D3 D4 D5 Supply Si 68 35 4 (18) 74 15 18 S2 57 88 91 3 (14) 8 (3) 17 S3 91 60 (18) 75 45 60 (1) 19 S4 52 (11) 53 24 (2) 7 82 13 Sg 51 (5) 18 82 13 7 (10) | 15 Demand 16 18 20 14 14 The minimum total transportation cost = 4 x 18 + 3 * 14 + 8* 3 + 60 x 18 + 60 x 1 + 52 11 + 24 x 2 + 51 x 5 + 7 x 10 = 2223 Here, the number of allocated cells = 9 is equal to m+ n - 1 = 5 + 5 - 1 = 9 .. This solution is non-degenerate

Find Solution of game theory problem using dominance method Player B B1 B2 B3 B4 Player A A1 3 2 4 0 A2 |3 4 2 4 A3 4 2 4 0 A4 0 4 08 Solution: dominance method Steps (Rule) Step-1:1. If all the elements of Column-i are greater than or equal to the corresponding elements of any other Column-j, then the Column-i is dominated by the Column-j and it is removed from the matrix. eg. If Column-2 > Column-4, then remove Column-2 Step-2: 1. If all the elements of a Row-i are less than or equal to the corresponding elements of any other Row-j, then the Row-i is dominated by the Row-j and it is removed from the matrix. eg. If Row-3 < Row-4, then remove Row-3 Step-3: Again repeat Step-1 & Step-2, if any Row or Column is dominated, otherwise stop the procedure.

2. Dominance rule to reduce the size of the payoff matrix Using dominance property Player B B1 B2 B3 B4 A1 3 2 4 0 A2 3 4 2 4 Player A Аз 4 2 4 0 A4 0 4 0 8 row-1 s row-3, so remove row-1, (A4 s A3:3 5 4,2 5 2,4 5 4,0 < 0) Player B B1 B2 B3 B4 A2 3 4 2 4 Player A A3 4 2 4 0 A4 0 4 08 column-1 2 column-3, so remove column-1. (B4 2 B3:3 2 2,4 2 4,0 20) Player B B2 B3 B4 A2 4 2 4 Player A A3 2 4 0 4 0 8