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 North West Corner rule

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"` = Σb_j ⇒ the problem is a balanced transportation problem and we can find a basic feasible solution.

North West Comer rule (NWC)

First allocation:

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

Second allocation:

	I	II	III	(ai)
B	(3)0	4	2	12/9
C	3	1	5	11
(bj)	3/0	10	10

Third allocation:

	II	III	(ai)
B	(9)4	2	9/0
C	1	5	11
(bj)	10/1	10

Fourth allocation:

	II	III	(ai)
C	(1)1	(10)5	11/10/0
(bj)	10/1	10/0

We first allot 1 unit to (C, II) cell and then the balance 10 units to (C, III) cell.

Thus we have the following allocations:

	I	II	III	Demand
A	(7)1	2	6	7
B	(3)0	(9)4	2	12
C	3	(1)1	(10)5	11
Supply	10/3	10	10

Transportation schedule:

A → I

B → I

B → II

C → II

C → III

i.e x₁₁ = 7

x₂₁ = 3

x₂₂ = 9

x₃₂ = 1

x₃₃ = 10

Total cost = (7 × 1) + (3 × 0) + (9 × 4) + (1 × 1) + (10 × 5)

= 7 + 0 + 36 + 1 + 50

= ₹ 94