Question

Determine an initial basic feasible solution to the following transportation problem by using least cost method

		Destination			Supply
		D1	D2	D3
	S1	9	8	5	25
Source	S2	6	8	4	35
	S3	7	6	9	40
	Requirement	30	25	45

Answer 1

Total supply = 25 + 35 + 40 = 100 = `sum"a"_"i"`

Total requirement = 30 + 25 + 45 = 100 = `sum"b"_"j"`

Since `sum"a"_"i" = sum"b"_"j"` 

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

Least cost method (LCM)

First allocation:

	D1	D2	D3	(ai)
S1	9	8	5	25
S2	6	8	(35)4	35/0
S3	7	6	9	40
(bj)	30	25	45/10

Second allocation:

	D1	D2	D3	(ai)
S1	9	8	(10)5	25/15
S3	7	6	9	40
(bj)	30	25	10/0

Third allocation:

	D1	D2	(ai)
S1	9	8	15
S3	7	(25)6	40/15
(bj)	30	25/0

Fourth allocation:

	D1	(ai)
S1	(15)9	15/0
S3	(15)7	15/0
(bj)	30/15/0

We first allow 15 units to cell (S₃, D₁) since it has the least cost.

Then we allow the balance 15 units to cell (S₁, D₁).

The final allotment is given as follows.

	D1	D2	D3	Supply
S1	(15)9	8	(10)5	25
S2	6	8	(35)4	35
S3	(15)7	(25)6	9	40
Requirement	30	25	45

Transportation schedule:

S₁ → D₁

S₁ → D₃

S₂ → D₃

S₃ → D₁

S₃ → 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)

= 136 + 50 + 140 + 105 + 150

= 580

The optimal cost by LCM is ₹ 580.