Question

Consider the following transportation problem.

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

Determine initial basic feasible solution by VAM.

Answer 1

Let ‘ai‘ denote the availability and ’bj‘ denote the requirement

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

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

So the given problem is a balanced transportation problem.

Hence there exists a feasible solution to the given problem.

For VAM, we first find the penalties for rows and columns.

We allocate units to the maximum penalty column (or) row with the least cost.

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)

Largest penalty = 3. allocate min (40, 20) to (O₃, 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.

Allocate min (20, 30) to (O₁, D₃)

Third allocation:

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

The largest penalty is 3.

Allocate min (20, 50) to (O₂, D₂)

Fourth allocation:

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

The largest penalty is 2.

Allocate min (10, 30) to (O₂, D₄)

Fifth allocation:

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

The largest penalty is 1.

Allocate min (30, 20) to (O₂, D₁)

Balance 10 units we allot to (O₁, D₁).

Thus we have the following allocations:

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

Transportation schedule:

O₁ → D₁

O₁ → D₃

O₂ → D₁

O₂ → D₂

O₂ → D₄

O₃ → P₂

i.e x₁₁ = 10

x₁₃ = 20

x₂₁ = 20

x₂₂ = 20

x₂₄ = 10

x₃₂ = 20

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

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

= 370

Thus the least cost by YAM is ₹ 370.