By matrix minimum method / least cost entry method let allocation is made at the least cost at (2, 1) as: X_{21} = Min (90, 75) = 75

Write down the reduced matrix after deleting ‘D_{1}’ – destination. Replace. b_{2}

= 90 by (90 – 75) = 15

The lowest cost in the reduced cost matrix is 4 at (1, 2) / (D_{1} D_{3}).

Allocate as X_{13 }= Min (70, 30) = 30

Delete D_{3} – destination and replace 70 by (70 – 30) = 40

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

OR (O_{2}, D_{2}) Allocate as

X_{22} – Min (15, 55) = 15 and replace 55 by (55 – 15) = 40.

So, the Final allocation is at the cell (1, 2) / (D_{1}, D_{2}):

Allocate as X_{12} = min (40, 40) = 40.

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

Observe O, availabilities are distributed to D_{2} and D_{3} and O_{2} availabilities are distributed to D_{1} and D_{2}, total availabilities are allocated to required destinations as : X_{12} = 40, X_{13} = 30, X_{21} = 75, X_{22 }= 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.