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 X_{41} = Min (a_{4}, b_{1})X_{41}= Min (14,7) = 7

D_{1} is satisfied so, delete D_{1} and replace (O_{14}– X_{41})= 14-7 = 7.

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

Allocate X_{43} = 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 X_{22} = Min (8, 9) = 8 and replace. (9 – 8) = 1 and delete O_{2}. There ducad cost matrix is:

The next allocation is made at the cell (3,3) which has least cost 7. Allocate X_{33 }= Min (7,11) = 7.

Replace 11 by (11 – 7) = 4 O_{3} is satisfied, since remaing only one origin, allocate all allocation in the same cost matrix.

Allocate, at next least cost 174, as X_{12} = Min (5,1)= 1 and replace 5 by (5- 1) = 4 as D_{2 }is satisfied next allocate as X_{13} = Min (4,4) = 4.

All availabilities and requirements are satisfied.

The initial basic feasible solution is:

The total transportation cost = Rs. 236.