Answer to Question #237056 in Operations Research for opr

Question

 Consider the transportation problem presented in the following table:

Destination

Origin 1 2 3 Supply

1 2 7 4 50

2 3 3 1 80

3 5 4 7 70

4 1 6 2 140

Demand 70 90 180 340

Use North West Corner Rule to determine the minimum cost of transportation hence use MODI approach to determine the transportation cost. Use Least Cost Method to determine the minimum cost of transportation. Use Vogel Approximation Method to determine the minimum cost of transportation. 

Answer 1

North West Corner Rule:

TOTAL number of supply constraints : 4

TOTAL number of demand constraints : 4

Problem Table is

Here Total Demand = 680 is greater than Total Supply = 340. So We add a dummy supply constraint with 0 unit cost and with allocation 340.

Now, The modified table is

The rim values for S1=50 and D1=70 are compared.

The smaller of the two i.e. min(50,70) = 50 is assigned to S1D1

This exhausts the capacity of S1 and leaves 70 - 50=20 units with D1

Table-1

The rim values for S2=80 and D1=20 are compared.

The smaller of the two i.e. min(80,20) = 20 is assigned to S2D1

This meets the complete demand of D1 and leaves 80 - 20=60 units with S2

Table-2

The rim values for S2=60 and D2=90 are compared.

The smaller of the two i.e. min(60,90) = 60 is assigned to S2D2

This exhausts the capacity of S2 and leaves 90 - 60=30 units with D2

Table-3

The rim values for S3=70 and D2=30 are compared.

The smaller of the two i.e. min(70,30) = 30 is assigned to S3D2

This meets the complete demand of D2 and leaves 70 - 30=40 units with S3

Table-4

The rim values for S3=40 and D3=180 are compared.

The smaller of the two i.e. min(40,180) = 40 is assigned to S3D3

This exhausts the capacity of S3 and leaves 180 - 40=140 units with D3

Table-5

The rim values for S4=140 and D3=140 are compared.

 The smaller of the two i.e. min(140,140) = 140 is assigned to S4D3

This exhausts the capacity of S4 and leaves 140 - 140=0 units with D3

Table-6

The rim values for S_dummy=340 and D3=0 are compared.

The smaller of the two i.e. min(340,0) = 0 is assigned to S_dummyD3

This meets the complete demand of D3 and leaves 340 - 0=340 units with S_dummy

Table-7

The rim values for S_dummy=340 and D4=340 are compared.

The smaller of the two i.e. min(340,340) = 340 is assigned to S_dummyD4

Table-8

Initial feasible solution is

The minimum total transportation cost =1×50+2×20+3×60+5×30+4×40+6×140+0×340=1420

Here, the number of allocated cells = 7, which is one less than to m + n - 1 = 5 + 4 - 1 = 8

∴ This solution is degenerate.

Least Cost Method:

TOTAL number of supply constraints : 4

TOTAL number of demand constraints : 4

Problem Table is:

Here Total Demand = 680 is greater than Total Supply = 340. So We add a dummy supply constraint with 0 unit cost and with allocation 340.

Now, The modified table is:

Initial feasible solution is:

The minimum total transportation cost =1×50+2×20+3×60+4×70+1×90+6×50+0×340=940

Here, the number of allocated cells = 7, which is one less than to m + n - 1 = 5 + 4 - 1 = 8

∴ This solution is degenerate.

Vogel Approximation Method:

Initial feasible solution is:

The minimum total transportation cost =2×50+1×80+3×70+1×40+2×100+0×180+0×160=630

Here, the number of allocated cells = 7, which is one less than to m + n - 1 = 5 + 4 - 1 = 8

∴ This solution is degenerate