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Graphical Method of Solving Linear Programming Problems

Two godowns ...

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?

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Solution

Let's assume that godown A supplies $X$ quintals grain to ration shop D and $Y$ quintals grain to ration shop E.

So, godown A will supply remaining grains $100 - X - Y$ quintals to ration shop F.

Also, godown B will supply $60 - X$ quintals grain to ration shop D, $50 - Y$ quintals grain to ration shop E and $40 - (100 - X - Y) = X + Y - 60$ quintals grain to ration shop F.

Now total transportation cost is $Z = 6 \times X + 3 \times Y + 2.5 \times (100 - X - Y) + 4 \times (60 - X) + 2 \times (50 - Y) + 3 \times (X + Y - 60)$

$\Rightarrow Z = 6 X + 3 Y + 250 - 2.5 X - 2.5 Y + 240 - 4 X + 100 - 2 Y + 3 X + 3 Y - 180$

$\Rightarrow Z = 2.5 X + 1.5 Y + 410 . . . (1)$

Now, Since godown A can supply maximum 60 quintals to ration shop D and 50 quintals to ration shop E and have maximum 100 quintals capacity to supply.

$∴ X + Y \leq 100, X \leq 60 a n d Y \leq 50 . . . (2)$

Also, if godown A supplies all 40 quintals to ration shop F, then remaining 60 quintals will be supplied to ration shop D and E.

So, $X + Y \geq 60 . . . (3)$

Since, X and Y is amount of grains. It can never be negative.
So, $X \geq 0, Y \geq 0 . . . (4)$

We have to minimise transportation cost given by equation (1).

After plotting all the constraints given by equation (2), (3) and (4), we got the feasible region as shown in the image.

Corner points Value of $Z = 2.5 X + 1.5 Y + 410$
A (60, 0) 560
B (60, 40) 620
C (50, 50) 610
D (10, 50) 510 (Minimum)
Hence minimum transportation cost is $510 R s$ . Transporter needs to supply as follows:

Godown A to Ration shop D $= X = 10$ quintals

Godown A to Ration shop E $= Y = 50$ quintals

Godown A to Ration shop F $= 100 - X - Y = 40$ quintals

Godown B to Ration shop D $= 60 - X = 50$ quintals

Godown B to Ration shop E $= 50 - Y = 0$ quintals

Godown B to Ration shop F $= X + Y - 60 = 0$ quintals

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Transportation cost per quintal (in Rs)
From/ To	$A$	$B$
$D$	$6$	$4$
$E$	$3$	$2$
$F$	$2.50$	$3$

Corner points	Value of $Z = 2.5 X + 1.5 Y + 410$
A (60, 0)	560
B (60, 40)	620
C (50, 50)	610
D (10, 50)	510 (Minimum)