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) x11 = 30.
Here a1 > b1 and replace b1 = 100 – 30 = 70.
Next, Allocate at (1,2) as: Min (70, 50) = 50; X12 = 50.
Here again a1 > b1 and replace, b1 = 70 – 50 = 20
Next allocations meat (1,3) as; min (20, 32)
= 20 and replace (32 – 20) – 12 A; So, X13 = 20.
Next allocation is mode at (2, 3) as:
Here b3 > a1
So, Min (12, 12) = 12 ; X23 = 12.
All allocations are made.
Thus, initial feasible solutions is:-
X11 = 30, X12 = 50, X13 = 20, X23 = 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.