Question

An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distance (in km) between the depots and the petrol pumps is given in the following table:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Answer 1

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. Then, (7000 − x − y) will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 −x) L will be transported from petrol pump B.

Similarly, (3000 − y) L and 3500 − (7000 − x − y) = (x + y − 3500) L will be transported from depot B to petrol pump E and F respectively.

The given problem can be represented diagrammatically as follows.

The problem can be formulated as follows.

Minimize z = 0.3x + 0.1y + 3950 … (1)

subject to the constraints,

The feasible region determined by the constraints is as follows.

The corner points of the feasible region are A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000), and E (500, 3000).

The values of z at these corner points are as follows.

Corner point	z = 0.3x + 0.1y + 3950
A (3500, 0)	5000
B (4500, 0)	5300
C (4500, 2500)	5550
D (4000, 3000)	5450
E (500, 3000)	4400	→ Minimum

The minimum value of z is 4400 at (500, 3000).

Thus, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively.

The minimum transportation cost is Rs 4400.