Question

Obtain an initial basic feasible solution to the following transportation problem by north west corner method.

	D	E	F	C	Available
A	11	13	17	14	250
B	16	18	14	10	300
C	21	24	13	10	400
Required	200	225	275	250

Answer 1

Total availability is 250 + 300 + 400 = 950

Total requirement is 200 + 225 + 275 + 250 = 950

Since Σa_i = Σb_j the problem is a balanced transportation problem and we can find an initial basic feasible solution.

First Allocation:

	D	E	F	C	(ai)
A	(200)11	13	17	14	250/5
B	16	18	14	10	300
C	21	24	13	10	400
(bj)	200/0	225	275	250

Second allocation:

	E	F	C	(ai)
A	(50)13	17	14	50/5
B	18	14	10	300
C	24	13	10	400
(bj)	225/175	275	250

Third allocation:

	E	F	C	(ai)
B	(175)18	14	10	300/125
C	24	13	10	400
(bj)	175/0	275	250

Fourth allocation:

	F	C	(ai)
B	(125)14	10	125/0
C	13	10	400
(bj)	275/150	250

Fifth allocation:

	F	C	(ai)
C	(150)13	(250)10	400/250/0
(bj)	150/0	250/0

We first allocate 150 units to cell (C, F).

Then we allocate balance of 250 units to cell (C, G)

Thus we have the following allocation:

	D	E	F	C	Available
A	(200)11	(50)13	17	14	250
B	16	(175)18	(125)14	10	300
C	21	24	(150)13	(250)10	400
Required	200	225	275	250

Transportation schedule:

A → D

A → E

B → E

B → F

C → F

C → G

i.e x₁₁ = 200

x₁₂ = 50

x₂₂ = 175

x₂₃ = 125

x₃₃ = 150

x₃₄ = 250

Total cost = (200 × 11) + (50 × 13) + (175 × 18) + (125 × 14) + (150 × 13) + (250 × 10)

= 2200 + 650 + 3150 + 1750 + 1950 + 2500

= ₹ 12,200