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For the following transportation problem, obtain the Initial basic feasible solution by:-  Matrix minima method (M.M.M)

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Solution by Matrix -minima method: From the given T.P:

By M.M.M / least OR lowest cost entry Method: The lowest cost in the cost matrix is 1 at (4, 1): 

The allocation is X41 = Min (a4, b1)X41= Min (14,7) = 7 

D1 is satisfied so, delete D1 and replace (O14– X41)= 14-7 = 7.

The lowest cost is 2 at (4,3) in the first table. 

Allocate X43 = Min (7,18) = 7 and replace 18 by (18 – 7) = 11. 04 is satisfied and delete and the reduced table is:

In the above reduced cost matrix the least cost is 3 at (2,2) of first table. 

So, allocate as X22 = Min (8, 9) = 8 and replace. (9 – 8) = 1 and delete O2. There ducad cost matrix is:

The next allocation is made at the cell (3,3) which has least cost 7. Allocate X33 = Min (7,11) = 7. 

Replace 11 by (11 – 7) = 4 O3 is satisfied, since remaing only one origin, allocate all allocation in the same cost matrix. 

Allocate, at next least cost 174, as X12 = Min (5,1)= 1 and replace 5 by (5- 1) = 4 as D2 is satisfied next allocate as X13 = Min (4,4) = 4. 

All availabilities and requirements are satisfied. 

The initial basic feasible solution is:

The total transportation cost = Rs. 236.

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