Question

There is a factory located at each of the two places $P$ and $Q$. From these location, a certain commodity is delivered to each of the three depots situated at $A, B$ and $C$. The weekly requirements of the depots are respectively 5,5 and 4 units commodity, while of the production capacity of the factories at $P$ and $Q$ are 8 and 6 units respectively. The cost of transportation per unit is given below.

To/From	Cost (in ₹)
	A	B	C
P	16	10	15
Q	10	12	10

Formulate the LPP, so that the units tranported from each factory to each depot in such an order that the transportation cost is minimum.

• A. Minimise $Z=x-7 y+190$
  subject to the constraints
  $x+y \leq 8 $
  $x+y \geq 4 $
  $x \leq 5$
  $y \leq 5$
  $x, y \geq 0$
• B. Minimise $Z=x+7 y+190$
  subject to the constraints
  $x+y \leq 8 $
  $x+y \leq 5 $
  $x, y \geq 0$
• C. Minimise $Z=x+7 y+90$
  subject to the constraints
  $x+y \leq 8 $
  $x+y \leq 4 $
  $x \leq 5 $
  $y \leq 5 $
  $x, y \geq 0$
• D. None of the above

Answer 1

Answer: (A)

The given information, can be expressed diagramatically as follows

Let the factory at $P$ transports $x$ units of commodity to depot at $A$ and $y$ units to depot at $B$. Since, the factory at $P$ has the capacity of 8 units of the commodity.
Therefore, the left out $(8-x-y)$ units will be transported to depot at $C$.
Since, the requirements are always non-negative quantities, therefore we have
$ x \geq 0, y \geq 0 \text { and } 8-x-y \geq 0$
$ \Rightarrow x \geq 0, y \geq 0 \text { and } x+y \leq 8$
Since, weekly requirements of the depot at $A$ is 5 units of the commodity and $x$ units are transported from the factory at $P$. Therefore, the remaining $5-x$ units are to be transported from the factory at $Q$.
Similarly, $5-y$ units of the commodity will be transported from the factory at $Q$ to the depot at $B$.
But the factory at $Q$ has the capacity of 6 units only, therefore the remaining $6-(5-x+5-y)=x+y-4$ units will be transported to the depot at $C$.
As, the requirements at the depots at $A, B$ and $C$ are always non-negative.
$ \therefore 5-x \geq 0,5-y \geq 0 \text { and } x+y-4 \geq 0$
$ \Rightarrow x \leq 5, y \leq 5 \text { and } x+y \geq 4$
The transportation cost from the factory at $P$ to the depots at $A, B$ and $C$ are respectively, $₹ 16 x$, ₹ $10 y$ and $₹ 15(8-x-y)$. Similarly, the transportation cost from the factory at $Q$ to the depots at $A, B$ and $C$ are respectively $₹ 10(5-x)$, ₹ $12(5-y)$ and $₹ 10(x+y-4)$.
Therefore, the total transportation $\operatorname{cost} Z$ is given by
$ Z=16 x+10 y+15(8-x-y)+10(5-x) +12(5-y)+10(x+y-4)$
$ =x-7 x+190$
Hence, the mathematical formulation of the given problem is as follows.
Minimise $Z=x-7 y+190$
Subject to constraints are
$x+y \leq 8$
$x+y \geq 4$
$x \leq 5$
$y \leq 5$
$x \geq 0, y \geq 0$