Question

Consider the following transportation problem

	Destination				Availability
	D1	D2	D3	D4
O1	5	8	3	6	30
O2	4	5	7	4	50
O3	6	2	4	6	20
Requirement	30	40	20	10

Determine an initial basic feasible solution using Vogel’s approximation method

Answer 1

Let ‘a_i‘ denote the availability and ‘b_j‘ denote the requirement.

Then, `sum"a"_"i"` = 30 + 50 + 20 = 100 and `sum"b"_"j"` = 30 + 40 + 20 + 10 = 100

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

The given problem is a balanced transportation problem and we can get an initial basic feasible solution.

Vogel’s approximation method:

First allocation:

	D1	D2	D3	D4	(ai)	penalty
O1	5	8	3	6	30	(2)
O2	4	5	7	4	50	(1)
O3	6	(20)2	4	6	20/0	(2)
(bj)	30	40/20	20	10
penalty	(1)	(3)	(1)	(2)

The largest penalty = 3.

So allocate 20 units to the cell (O₃, D₂) which has the least cost in column D₂ 

Second allocation:

	D1	D2	D3	D4	(ai)	penalty
O1	5	8	(20)3	6	30/10	(2)
O2	4	5	7	4	50	(1)
(bj)	30	20	20/0	10
penalty	(1)	(3)	(4)	(2)

Largest penalty = 4

So allocate min (20, 30) units to the cell (O₁, D₃) which has the least cost in column D₃

Third allocation:

	D1	D2	D4	(ai)	penalty
O1	5	8	6	10	(1)
O2	4	(20)5	4	50/30	(1)
(bj)	30	20/10	10
penalty	(1)	(3)	(2)

The largest penalty is 3.

So allocate min (20, 50) to the cell (O₂, D₂) which has the least cost in column D₂.

Fourth allocation:

	D1	D4	(ai)	penalty
O1	5	6	10	(1)
O2	4	(10)4	30/20	(0)
(bj)	30	10/0
penalty	(1)	(2)

Largest penalty = 2.

So allocate min (10, 30) to cell (O₂, D₄) which has the least cost in column D₄

Fifth allocation:

	D1	(ai)	penalty
O1	(10)5	10/0	–
O2	(20)4	20/0	–
(bj)	30/10/0
penalty	(1)

Largest penalty = 1.

Allocate min (30, 20) to cell (O₂, D₁) which has the least cost in column D₁.

Finally allot the balance 10 units to cell (O₁, D₁)

We get the final allocation table as follows.

	D1	D2	D3	D4	Availability
O1	(10)5	8	(20)3	6	30
O2	(20)4	(20)5	7	(10)4	50
O3	6	(20)2	4	6	20
Requirement	30	40	20	10

Transportation schedule:

O₁ → D₁

O₁ → D₃

O₂ → D₁

O₂ → D₂

O₂ → D₄

O₃ → D₂

i.e x₁₁ = 10

x₁₃ = 20

x₂₁ = 20

x₂₂ = 20

x₂₄ = 10

x₃₂ = 20

Total cost is given by = (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)

= 50 + 60 + 80+ 100 + 40 + 40

= 370
Hence the minimum cost by YAM is ₹ 370.