Question

(Transportation problem): Consider the transportation model in the table below. a) Use the northwest-corner method to find a starting solution. b) Develop the iterations that lead to the optimal solution. $1 19 $8 6 $6 19 5

Answer 1

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Solution: Your problem → north-west corner method {{0,3,5},{7,4,9},{1,8,6}},supply={4,7,19},demand={5,6,19} You have input data size 3x3, number of supply = 3, number of demand = 3 But number of supply = 3 and number of demand = 3, So we interchange this supply and demand values and trying to solve this problem, otherwise correct the input values TOTAL number of supply constraints : 3 TOTAL number of demand constraints : 3 Problem Table is D1 D2 D3 Supply or 0 3 5 5 7 4 96 1 8 6 19 od Demand 4 19 The rim values for S,=5 and D =4 are compared. The smaller of the two i.e. min(5.4) = 4 is assigned to S, Di This meets the complete demand of D, and leaves 5 - 4 = 1 units with S,

Table-1 D D2 D3 3 5 Supply 0(4) S 7 4 9 | 6 Sz | 8 6 19 Demand 19 The rim values for 5,=1 and D=7 are compared. The smaller of the two i.e. min(1,7) = 1 is assigned to S, D2 This exhausts the capacity of S, and leaves 7 - 1 = 6 units with D Table-2 D1 D2 D3 Supply 044) 311) $ 1 Sz 19 Demand O 19 The rim values for Sy=6 and D =6 are compared. The smaller of the two i.e. min(6.6) = 6 is assigned to S, D2 This exhausts the capacity of S, and leaves 6 - 6 = 0 units with D,

Table-3 D1 D2 D3 Supply s och 34 o o Sy 4167 9 0 Sz1 $ 619 Demand 0 0 19 The rim values for Sz=19 and D=0 are compared. The smaller of the two i.e. min(19,0) = 0 is assigned to S, D2 This meets the complete demand of D, and leaves 19 -0 = 19 units with Sz Table-4 0 D1 D2 D3 Supply Stopes 341 5 Sy 4461 o o Sz 1 $ 6 19 Demand The rim values for Sz=19 and Dz=19 are compared. The smaller of the two i.e. min(19,19) = 19 is assigned to Sz Dz

Table-5 D3 DI D2 st04 347 Supply | SyH 64940 Demand 0 0 0 Initial feasible solution is D1 D2 D3 0 (4) 3 (1) 5 Supply 5 7 4 (6) 9 6 6 (19) 19 Demand 4 7 19 The minimum total transportation cost = ( x 4+3 x 1 + 4 x 6 + 6 x 19 = 141 Here, the number of allocated cells = 4, which is one less than to m + n - 1 = 3 + 3 - 1 = 5 · This solution is degenerate