Here the availability; ai = requirement; bj = 112, thus the T.P. is balanced.

Therefore there exists a feasible solution.

Consider the north west corner cell at (1, 1) allocate as; minimum (a,, b,)

= min (100, 30) x_{11} = 30.

Here a_{1} > b_{1} and replace b_{1} = 100 – 30 = 70.

Next, Allocate at (1,2) as: Min (70, 50) = 50; X_{12} = 50.

Here again a_{1} > b_{1} and replace, b_{1} = 70 – 50 = 20

Next allocations meat (1,3) as; min (20, 32)

= 20 and replace (32 – 20) – 12 A; So, X_{13 }= 20.

Next allocation is mode at (2, 3) as:

Here b_{3} > a_{1}

So, Min (12, 12) = 12 ; X_{23} = 12.

All allocations are made.

Thus, initial feasible solutions is:-

X_{11} = 30, X_{12} = 50, X_{13} = 20, X_{23 }= 12

The transportation cost is: Z = ΣΣCij.Xij

= 3 × 30 + 7 × 50 + 4 × 20 + 2 × 12

= 90 + 350 + 80 + 24 = Rs. 544.

**Note: **In the problem m(rows) = 2, n(Columns) = 3

∴ m + n – 1 = 2 + 3 – 1 =4

The number of allocations = 4 is same as (m + n – 1).

The initial feasible solutions a basic feasible solution.