Question

Obtain an initial basic feasible solution to the following transportation problem by using least-cost method.

	D1	D2	D3	Supply
O1	9	8	5	25
O2	6	8	4	35
O3	7	6	9	40
Demand	30	25	45

Answer 1

Total supply = 25 + 35 + 40 = 100

Total demand = 30 + 25 + 45 = 100

Total supply = Total demand

∴ The given problem is a balanced transportation problem.

Hence there exists a feasible solution to the given problem.

Let ‘ai’ denote the supply and ‘bj’ denote the demand. We allocate the units according to the least transportation cost of each cell.

First allocation:

	D1	D2	D3	(ai)
O1	9	8	5	25
O2	6	8	(35)4	35/0
O3	7	6	9	40
(bj)	30	25	45/10

The least-cost 4 corresponds to cell (O₂, D₃).

So first we allocate to this cell.

Second allocation:

	D1	D2	D3	(ai)
O1	9	8	(10)5	25/15
O2	6	8	(35)4	35/0
O3	7	6	9	40
(bj)	30	25	45/10/0

The least-cost 5 corresponds to cell (O₁, D₃).

So we have allocated min (10, 25) to this cell.

Third allocation:

	D1	D2	D3	(ai)
O1	9	8	(10)5	25/15
O2	6	8	(35)4	35/0
O3	7	(25)6	9	40/15
(bj)	30	25/0	45/10/0

The least-cost 6 corresponds to cell (O₃, D₂).

So we have allocated min (25, 40) to this cell.

Fourth allocation:

	D1	D2	D3	(ai)
O1	9	8	(10)5	25/15
O2	6	8	(35)4	35/0
O3	(15)7	(25)6	9	40/15/0
(bj)	30/15	25/0	45/10/0

The least-cost 7 corresponds to cell (O₃, D₁).

So we have allocated min (30, 15) to this cell.

Final allocation:

Although the next least cost is 8, we cannot allocate to cells (O₁, D₂) and (O₂, D₂) because we have exhausted the demand 25 for this column.

So we allocate 15 to cell (O₁, D₁)

	D1	D2	D3	(ai)
O1	(15)9	8	(10)5	25/15/0
O2	6	8	(35)4	35/0
O3	(15)7	(25)6	9	40/15/0
(bj)	30/15/0	25/0	45/10/0

Transportation schedule:

O₁ → D₁

O₁ → D₃

O₂ → D₃

O₃ → D₁

O₃ → D₂

i.e x₁₁ = 15

x₁₃ = 10

x₂₃ = 35

x₃₁ = 15

x₃₂ = 25

Total cost is = (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)

= 135 + 50 + 140 + 105 + 150

= 580

Thus by the least cost method (LCM), the cost is ₹ 580.