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 are 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?
Suppose that, from depot A,
x litre of oil is transported to pump D
y litre of oil is transported to pump E
and 7000 - (x + y) litre of oil is transported to pump F.
Then, from depot B,
(4500 - x) litres of oil is transported to pump D
(3000 - y) litres of oil is transported to pump E
and (3500 - {7000 - (x + y)}) litres = (x + y - 3500) litres of oil is transported to pump F.
Table
We are to minimise
subject to the constraints
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
x = 4500 is a straight line AL parallel to y-axis at a distance of 4500.
y = 3000 is a straight line BM parallel to v-axis at a distance of 3000.
Now we draw the graph of x + y = 3500.
For x = 0, y = 3500
For y = 0, x = 3500
∴ line meets OX in C(3500, 0) and OY in N(0, 3500).
Again we draw the graph of x + y = 7000.
For x = 0, y = 7000
For y = 0, x = 7000
∴ line meets OX in D(7000, 0) and OY in P(0, 7000).
Since feasible region satisfies all the constraints.
∴ CASQR is the feasible region.
The corner points are
C(3500, 0), A(4500, 0), S(4500, 2500), Q(4000, 3000), R(500, 3000).
∴ the minimum cost of transportation is Rs. 4400, when from godown A, 500 litre of oil is transported to pump D, 3000 litre of oil is transported to pump E and 7000 -(500 + 3000) = 3500 litre of oil is transported to pump F.
Hence from godown B whole of 4000 litre of oil is transported to pump D.
