Given the following information set up the problem in a transportation table and

Question

Given the following information set up the problem in a transportation table and solve for the minimum-cost plan:

PERIOD

Suppose that an increase in warehousing costs and other costs brings inventory carrying costs to $2 per unit per month. All other costs and quantities remain the same. Determine a revised solution to this transportation problem. (Omit the "$" sign in your response.)

Given the following information set up the problem in a transportation table and solve for the minimum-cost plan:

Explanation / Answer

Ans:Solution of Transportation Problem Using North-West Corner Method

TOTAL no. of supply constraints : 3
TOTAL no. of demand constraints : 3
Problem Table is

Here Total Demand = 2000 is greater than Total Supply = 227. So We add a dummy supply constraint with 0 unit cost and with allocation 1773.
Now, The modified table is

The rim values for S1=50 and D1=550 are compared.
The smaller of the two i.e. min(50,550) = 50 is assigned to S1 D1
This exhausts the capacity of S1 and leaves 550 - 50 = 500 units with D1
Table-1

The rim values for S2=85 and D1=500 are compared.
The smaller of the two i.e. min(85,500) = 85 is assigned to S2 D1
This exhausts the capacity of S2 and leaves 500 - 85 = 415 units with D1
Table-2

The rim values for S3=92 and D1=415 are compared.
The smaller of the two i.e. min(92,415) = 92 is assigned to S3 D1
This exhausts the capacity of S3 and leaves 415 - 92 = 323 units with D1
Table-3

The rim values for S4=1773 and D1=323 are compared.
The smaller of the two i.e. min(1773,323) = 323 is assigned to S4 D1
This meets the complete demand of D1 and leaves 1773 - 323 = 1450 units with S4
Table-4

The rim values for S4=1450 and D2=700 are compared.
The smaller of the two i.e. min(1450,700) = 700 is assigned to S4 D2
This meets the complete demand of D2 and leaves 1450 - 700 = 750 units with S4
Table-5

The rim values for S4=750 and D3=750 are compared.
The smaller of the two i.e. min(750,750) = 750 is assigned to S4 D3
Table-6

Final Allocation Table is

Here, the number of allocation is equal to m + n - 1 = 4 + 3 - 1 = 6
The solution is feasible.
Total Transportation cost = 500 × 50 + 50 × 85 + 120 × 92 + 0 × 323 + 0 × 700 + 0 × 750 = 40290

The minimized total transportation cost = $ 40290

D1 D2 D3 Supply S1 500 500 500 50 S2 50 50 50 85 S3 120 120 100 92 Demand 550 700 750

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