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Find the Initial basic feasible solution by matrix minimum method for the following T.P.

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By matrix minimum method / least cost entry method let allocation is made at the least cost at (2, 1) as: X21 = Min (90, 75) = 75

Write down the reduced matrix after deleting ‘D1’ – destination. Replace. b2 

= 90 by (90 – 75) = 15

The lowest cost in the reduced cost matrix is 4 at (1, 2) / (D1 D3). 

Allocate as X13 = Min (70, 30) = 30 

Delete D3 – destination and replace 70 by (70 – 30) = 40

The lowest cost in the reduced cost matrix is 5 at (2, 2) 

OR (O2, D2) Allocate as 

X22 – Min (15, 55) = 15 and replace 55 by (55 – 15) = 40. 

So, the Final allocation is at the cell (1, 2) / (D1, D2):

 Allocate as X12 = min (40, 40) = 40. 

Thus all allocations are made as below: From the original table.

Observe O, availabilities are distributed to D2 and D3 and O2 availabilities are distributed to D1 and D2, total availabilities are allocated to required destinations as : X12 = 40, X13 = 30, X21 = 75, X22 = 15 and the total transportation cost is:

∴ Total transportation cost is Rs. 660/- is the basic feasible solution for the given T.P. 

Note: In the above problem: number of rows: m = 2; number of columus ; n = 3(m + n- 1) 

= (2 + 3- 1) = 4. 

Number of allocations = 4. Both are some, the solution is degenerate.

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