Question

Find the initial basic feasible solution of the following transportation problem:

	I	II	III	Demand
A	1	2	6	7
B	0	4	2	12
C	3	1	5	11
Supply	10	10	10

Using Least Cost method

Answer 1

Total demand (a_i) = 7 + 12 + 11 = 30 and total supply (b_j) = 10 + 10 + 10 = 30.

`sum"a"_"i" = sum"b"_"j"` ⇒ the problem is a balanced transportation problem and we can find a basic feasible solution.

Least Cost method

First allocation:

	I	II	III	(ai)
A	1	2	6	7
B	(10)0	4	2	12/2
C	3	1	5	11
(bj)	10/0	10	1

Second allocation:

	II	III	(ai)
A	2	6	7
B	4	2	2
C	(10)1	5	11/1
(bj)	10/0	10

Third allocation:

	III	(ai)
A	6	7
B	(2)2	2/0
C	5	1
(bj)	10/8

Fourth allocation:

	III	(ai)
A	(7)6	7/0
C	(1)5	1/0
(bj)	8/7/0

We first allot 1 unit to cell (C, III) and the balance 7 units to cell (A, III).

Thus we have the following allocations:

	I	II	III	Demand
A	1	2	(7)6	7
B	(10)0	4	(2)2	12
C	3	(10)1	(1)5	11
Supply	10	10	10

Transportation schedule:

A → III

B → I

B → III

C → II

C → III

i.e x₁₃ = 7

x₂₁ = 10

x₂₃ = 2

x₃₂ = 10

x₃₃ = 1

Total cost = (7 × 6) + (10 × 0) + (2 × 2) + (10 × 1) + (1 × 5)

= 42 + 0 + 4 + 10 + 5

= ₹ 61