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 Vogel’s approximation 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.

Vogel’s approximation method (VAM)

First allocation:

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

Largest penalty = 3.

Allocate min (10, 12) to (B, III)

Second allocation:

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

Largest penalty = 4.

Allocate min (10, 2) to cell (B, I)

Third allocation:

	I	III	(ai)	Penalty
A	1	2	7	(1)
B	(2)0	4	2/0	(2)
C	3	1	11	(2)
(bj)	10/8	10
Penalty	(1)	(1)

The largest penalty is 2.

We can choose the I column or C row.

Allocate min (8, 7) to cell (A, I)

Fourth allocation:

	I	III	(ai)	Penalty
A	(7)1	2	7/0	(1)
C	3	1	11	(2)
(bj)	8/1	10
Penalty	(2)	(1)

First, we allocate 10 units to cell (C, II).

Then balance 1 unit we allot to cell (C, I)

Thus we have the following allocations:

	I	III	(ai)	Penalty
C	(1)3	(10)1	11/1/0	(2)
(bj)	1/0	10/0
Penalty	–	–

Tansportation schedule:

A → I

B → I

B → III

C → I

C → II

(i.e) x₁₁ = 7

x₂₁ = 2

x₂₃ = 10

x₃₁ = 1

x₃₂ = 10

Total cost = (7 × 1) + (2 × 0) + (10 × 2) + (1 × 3) + (10 × 1)

= 7 + 0 + 20 + 3 + 10

= ₹ 40