Question

Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem.

		Destination
		D1	D2	D3	D4	Supply
	O1	2	3	11	7	6
Origin	O2	1	0	6	1	1
	O3	5	8	15	9	10
	Demand	7	5	3	2

Answer 1

Total supply (a_i) = 6 + 1 + 10 = 17

Total demand (b_j) = 7 + 5 + 3 + 2 = 17

`sum"a"_"i" = sum"b"_"j"`.

So given transportation problem is a balanced problem and hence we can find a basic feasible solution by VAM.

First allocation:

	D1	D2	D3	D4	(ai)	penalty
O1	2	3	11	7	6	(1)
O2	1	0	6	(1)1	1/0	(1)
O3	5	8	15	9	10	(3)
(bj)	7	5	3	2/1
penalty	(1)	(3)	(5)	(6)

The largest penalty is 6.

So allot min (2, 1) to the cell (O₂, D₄) which has the least cost in column D₄.

Second allocation:

	D1	D2	D3	D4	(ai)	penalty
O1	2	(5)3	11	7	6/1	(1)
O3	5	8	15	9	10	(3)
(bj)	7	5/0	3	1
penalty	(3)	(5)	(4)	(2)

The largest penalty is 5.

So allot min (5, 6) to the cell (O₁, D₂) which has the least cost in column D₂.

Third allocation:

	D1	D3	D4	(ai)	penalty
O1	(1)2	11	7	1/0	(5)
O3	5	15	9	10	(4)
(bj)	7/6	3	1
penalty	(3)	(4)	(2)

The largest penalty is 5.

So allot min (7, 1) to the cell (O₁, D₁) which has the least cost in row O₁.

Fourth allocation:

	D1	D3	D4	(ai)	penalty
O3	(6)5	15	9	10/4	(4)
(bj)	6/0	3	1
penalty	–	–	–

Fifth allocation:

	D3	D4	(ai)	penalty
O3	(3)15	(1)9	4/3/0	(6)
(bj)	3/0	1/0
penalty	–	–

The largest penalty is 6.

So allot min (1, 4) to the cell (O₃, D₄) which has the least cost in row O₃.

The balance 3 units is allotted to the cell (O₃, D₃).

We get the final allocation as given below.

	D1	D2	D3	D4	Supply
O1	(1)2	(5)3	11	7	6
O2	1	0	6	(1)1	1
O3	(6)5	8	(3)15	(1)9	10
Demand	7	5	3	2

Transportation schedule:

O₁ → D₁

O₁ → D₂

O₂ → D₄

O₃ → D₁

O₃ → D₃

O₃ → D₄

i.e x₁₁ = 1

x₁₂ = 5

x₂₄ = 1

x₃₁ = 6

x₃₃ = 3

x₃₄ = 1

Total cost is given by = (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)

= 2 + 15 + 1 + 30 + 45 + 9

= 102

Thus the optimal (minimal) cost of the given transportation problem by Vogel’s approximation method is ₹ 102.