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 4500 L, 3000 L and 3500 L 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 10litres of oil is Re1 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 tA to petrol pump F.

The requirement of petrol pump D is 4500 L.

Since xL are transported from depot A, the remaining (4500-x) L will be the 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. 

x≥0,y≥0, and (700-x-y)≥0  

⇒ x≥0,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, 

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

\(Z = 0.3x+0.1y+3950\)

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
(4000, 3000)	5450
(500, 3000)	4400	\(\to\) 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.