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 pumps D,E,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 :

Assuming that the transportation cost of10 litres of oil is ₹ 1 per km , how should the delivery be scheduled in order that the transportation cost is minimum ?

Explanation

Let the supply of petrol from A to D be `x` litres and A to E be `y` litres. So the supply of petrol from A to F will be `(7000-x-y)` litres. Similarly, the supply of petrol from B to D,E,F will be `(4500-x)` litres, `(3000-y)` litres, `(x+y-3500)` litres respectively.
Therefore, minimum transportation cost.
`Z=1/10[7x+5y+3(7000-x-y)+3(4500-x)`
`+4(3000-y)+2(x+y-3500)]`
`=1/10(3x+y+39500)`
and constraints `xge0yge0`
`7000-x-yge0impliesx+yle7000`
`4500-xge0impliesxle4500`
`3000-yge0impliesyle3000`
`x+y-3500ge0impliesx+yge3500`
First, draw the graph of the lines `x+y=7000, x=4500,y=3000,x+y=3500`.

Now, we find the feasible region by constraints `x+yge7000,xle4500,yle3000,x+yge3500,xge0,yge0` and shade it whose vertices are `A(3500,0),B(4500,0),C(4500,2500),D(4000,30000,E(500,3000)`. We find the value of `Z` at these vertices.

`:. x=500,y=3000`
Therefore the supply of petrol form A to D,E,F will be 500 litres, 3000 litres, 3500 litres respectively and from B to D,E,F will be 4000 litres, 0 litres, 0 litres respectively.
Minimum transportation cos `=Rs. 4400`