Two godowns A and B have grain capacities 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:

$\begin{matrix} T r a n s p o r t a t i o n c o s t p e r q u i n t a l (i n R s) \end{matrix} \begin{matrix} F r o m / T o & A & B D & 6 & 4 E & 3 & 2 F & 2.50 & 3 \end{matrix}$

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

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Solution

Ley godown A supplies 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 A. Therefore, the remaining (60 - 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 shops E and F respectively. The given problem can be represented diagrammatically as follows:

Ley Z be the total cost of transportation then,

Z = 6x + 3y + 2.50(100 - x - y)+4(60 - x)+2(50 - y) +3(x + y) - 60

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

=2.50x + 1.50y + 410

Subject to constraints are $60 - x \geq 0 \Leftrightarrow x \leq 60 (i i) 50 - y \geq 0 \Leftrightarrow y \leq 50 (i i i) 100 - (x + y) \geq 0 \Leftrightarrow x + y \leq 100 (i v) x + y - 60 \geq\Leftrightarrow x + y \geq 60 (v) x \geq 0, y \geq 0 (v i) F i r s t l y, d r a w t h e g r a p h o f t h e l i n e x + y = 100, w e h a v e 0 + 0 \leq 100 \Rightarrow 0 \leq 100 (w h i c h i s t r u e) S o, t h e h a l f p l a n e i s t o w a r d s t h e o r i g i n . S e c o n d l y, d r a w t h e g r a p h o f t h e l i n e x = 60 P u t t i n g (0, 0) i n t h e i n e q u a l i t y x + y \leq 100, w e h a v e 0 \leq 60 (W h i c h i s t r u e) S o, t h e h a l f p l a n e i s t o w a r d s t h e o r i g i n . T h i r d l y, d r a w t h e g r a p h o f t h e l i n e x + y = 60 \begin{matrix} x & 0 & 60 y & 60 & 0 \end{matrix}$

Putting (0, 0) in the inequality x + y $\geq$ 60, we have $0 + 0 \geq 60 \Rightarrow 0 \geq 60$ (which is false)So, the half plane is away from the origin. Fourthly, draw the graph of the line y = 50 putting(0, 0) in the inequality $y \leq 50$ , we have $0 \leq 50$ (which is true)So, the half plane is towards the origin. Since, x, y $\geq 0$ So, the feasible region lies in the first quadrant. The intersection points of the given lines are A (60, 0), B(60, 40), C(50, 50) and D(10, 50).

$∴$ ~~Feasible region is ABCDA.

The corner points of the feasible region are A(60, 0) B(60, 40), C(50, 50) and D(10, 50). The values of Z at these points are as follows:

$\begin{matrix} C o r n e r p o i n t & Z = 2.5 x + 1.5 y + 410 A (60, 0) & 560 B (60, 40) & 620 C (50, 50) & 610 D (10, 50) & 510 \to m i n i m u m \end{matrix}$

The minimum value of Z is 510 at D(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 quintal and 0 quintal respectively.

The minimum cost is Rs. 510.

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Similar questions

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

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A

and

B

have grain capacity of

100

quintals and

50

quintals respectively. They supply to

3

ration shops,

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and

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whose requirements are

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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	$A$	$B$
$D$	$6$	$4$
$E$	$3$	$2$
$F$	$2.50$	$3$

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