Question

Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively.They supply to 3 ration shops,D,E and F whose requirements are 60,50 and 40 quintals respectively.The cost of transportation per quintal from the godowns to the shops are given in the following table:

Transportation cost per quintal (in Rs)

From/To| A| B
D| 6| 4
E| 3| 2
F| 2.50| 3

How should the supplies be transported in order that the transportation cost is minimum?What is the minimum cost?

Answer 1

Let godown A supply x and y quintals of grain to the shops D and E respectively.

Then, (100-x-y) will be supplied to shop F.
The requirement at shop D is 60 quintals since x quintals are transported from godown B.

Similarly, (50-y) quintals and 40-(100-x-y) = (x+y-60) quintals will be transported from godown B to shop E and F respectively.

The given problem can be represented diagrammatically as follows.

x≥0,y≥0, and 100-x-y ≥ 0

⇒ x≥0, y≥0, and x+y≤100

60-x≥0,50-y≥0, and x+y-60 ≥ 0

⇒ x≤60, y≤50, and x+y≥60

Total transportation cost Z is given by,

Z = 6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)

Z = 6x+3y+250-2.5x-2.5y+240-4x+100-2y+3x+3y-180

Z = 2.5x+1.5y+410

The given problem can be formulated as

Minimize Z = 2.5x+1.5y+410 ...(1)

Subject to the constraints,

x+y≤100 .....(2)
x≤60 .....(3)
y≤50 ..…(4)
x+y≥60 .....(5)
x,y≥0 .....(6)

The reasonable region determined by the system of constraints is as follows.

The corner points are A(60,0), B(60,40), C(50,50), and D(10,50). The value of Z at these corner points is as follows.

Corner point| z = 2.5x + 1.5y + 410|
---|---|---
A (60, 0)| 560|
B (60, 40)| 620|
C (50, 50)| 610|
D (10, 50)| 510| → Minimum

The minimum value of Z is 510 at (10,50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively, and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.