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 distances (in km) between the depots and the petrol pumps is given in the following table:

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?

4400

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 4500L. Since x L are transported from depot A, the remaining (45000−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 shown in first figure.

$x≥0,y≥0$ and $(7000−x−y)≥0$

$⇒ x≥,y≥0$ and $x+y≤7000$

$4500−x≥0,3000−y≥0$ and $x+y−3500≥0$

$⇒ x≤4500,y≤3000$ and $x+y≥3500$

Cost of transporting 10L of petrol = Re.1

Cost of transporting 1L of petrol = Rs. $\frac{1}{10}$

Therefore, total transportation cost is given by

$z=\frac{7}{10}×x+\frac{6}{10}×y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)$

$=0.3x+0.1y+3950$

The problem can be formulated as follows:

Minimize $z=0.3x+0.1y+3950$......(1)

subject to the constraints

$x+y≥7000$......(2)

$x≤4500$........(3)

$y≤3000$........(4)

$x+y≥3500$....(5)

$x,y≥0$........(6)

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

The minimum transportation cost is Rs.4400