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.