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 E F 7 6 3 3 4 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, depot A transport x litres to petrol pump D and depot B transport y litres to petrol pump E and transportation cost of 1 km is equal to Rs 0.1.

To/from	A	cost	B	cost
D	7	A−D 7×0.1=0.7	3	B−D 3×0.1=0.3
E	6	A−E 6×0.1=0.6	4	B−E 4×0.1=0.4
F	3	A−F 3×0.1=0.3	2	B−F 2×0.1=0.2

The equations from the above diagram can be written as,

x+y≤7000 x≤4500 y≤3000 x+y≥3500

We need to minimize the cost of transportation so we can use function which will minimize Z.

Minimize Z=0.7x+0.6y+0.3[ 7000−( x+y ) ]+0.3( 4500−x )+0.4( 3000−y ) +0.2( x+y−3500 ) Z=0.3x+0.1y+3950

All constraints are given as,

x+y≤7000 x+y≥3500 y≤3500,x≤4500 x≥0,y≥0

x+y≤7000

x	0	7000
y	7000	0

x+y≥3500

x	0	3500
y	3500	0

Plot the points of all the constraint lines,

Substitute these points in the given objective function to find the maximum value of Z.

Corner points	Value of Z
( 500,3000 )	4400 (Minimum)
( 4000,3000 )	5450
( 4500,2500 )	5550
( 4500,0 )	5300
( 3500,0 )	5000

Thus, oil transported from A is 500,3000,3500 litres to petrol pump D,E,F respectively and from B is 4000,0,0 litres to petrol pump D,E,F respectively and minimum transportation cost is Rs. 4400.