Question

Find an initial basic feasible solution of the following problem using the northwest corner rule.

	D1	D2	D3	D4	Supply
O1	5	3	6	2	19
O2	4	7	9	1	37
O3	3	4	7	5	34
Demand	16	18	31	25

Answer 1

Given the transportation table is

	D1	D2	D3	D4	Supply (ai)
O1	5	3	6	2	19
O2	4	7	9	1	37
O3	3	4	7	5	34
Demand (bj)	16	18	31	25	90

Total supply = Total Demand = 90.

The given problem is a balanced transportation problem.

Hence there exists a feasible solution to the given problem.

First allocation:

	D1	D2	D3	D4	(ai)
O1	(16)5	3	6	2	19/3
O2	4	7	9	1	37
O3	3	4	7	5	34
(bj)	16/0	18	31	25	90

Second allocation:

	D1	D2	D3	D4	(ai)
O1	(16)5	(3)3	6	2	19/3/0
O2	4	7	9	1	37
O3	3	4	7	5	34
(bj)	16/0	18/15	31	25	90

Third allocation:

	D1	D2	D3	D4	(ai)
O1	(16)5	(3)3	6	2	19/3/0
O2	4	(15)7	9	1	37/22
O3	3	4	7	5	34
(bj)	16/0	18/15/0	31	25	90

Fourth allocation:

	D1	D2	D3	D4	(ai)
O1	(16)5	(3)3	6	2	19/3/0
O2	4	(15)7	(22)9	1	37/22/0
O3	3	4	7	5	34
(bj)	16/0	18/15/0	31/9	25	90

Fifth allocation:

	D1	D2	D3	D4	(ai)
O1	(16)5	(3)3	6	2	19/3/0
O2	4	(15)7	(22)9	1	37/22/0
O3	3	4	(9)7	5	34/25
(bj)	16/0	18/15/0	31/9/0	25	35

Final allocation:

	D1	D2	D3	D4	(ai)
O1	(16)5	(3)3	6	2	19/3/0
O2	4	(15)7	(22)9	1	37/22/0
O3	3	4	(9)7	(25)5	34/25/0
(bj)	16/0	18/15/0	31/9/0	25/0	35

Transportation schedule: 

O₁ → D₁

O₁ → D₂

O₂ → D₂

O₂ → D₃

O₃ → D₃

O₃ → D_4.

i.e x₁₁ = 16

x₁₂ = 3

x₂₂ = 15

x₂₃ = 22

x₃₃ = 9

x₃₄ = 25.

Total transportation cost = (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)

= 80 + 9 + 105 + 198 + 63 + 125

= 580

Thus the minimum cost is ₹ 580 using the northwest comer rule.