Question

An oil company has two depots $A$ and $B$ with capacities of $7000L$ and $4000L$ 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	$A$	$B$
$D$	$7$	$3$
$E$	$6$	$4$
$F$	$3$	$2$

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 $x$ and $y$ litres of oil be supplied from $A$ to the petrol pumps, $D$ and $E$. Then $(7000-x-y)$ will be supplied from $A$ to petrol pump $F$.
The requirement at petrol pum[ $D$ is $4500L$. Since $x$ $L$ are transported from depot $A$, the remaining $(45000-x)$ $L$ will be transported from petrol pump $B$$.
Similarly, $(3000-y)L$ and $3500(7000-x-y)=(x+y-3500)L$ will be transported from depot $B$ to petrol pump $E$ and $F$ respectively.
The given problem can be represented diagrammatically as shown in first figure.
$x\ge 0,y\ge 0$ and $(7000-x-y)\ge 0$
$\Rightarrow$ $x\ge ,y\ge 0$ and $x+y\le 7000$
$4500-x\ge 0,3000-y\ge 0$ and $x+y-3500\ge 0$
$\Rightarrow$ $x\le 4500,y\le 3000$ and $x+y\ge 3500$
Cost of transporting $10L$ of petrol $=$ Re.$1$
Cost of transporting $1L$ of petrol $=$ Rs.$\frac{1}{10}$
Therefore, total transportation cost is given by
$z=\frac{7}{10}\times x+\frac{6}{10}\times y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)$
$=0.3x+0.1y+3950$
The problem can be formulated as follows:
Minimize $z=0.3x+0.1y+3950......\left(1\right)$
subject to the constraints
$x+y\ge 7000......\left(2\right)$
$x\le 4500........\left(3\right)$
$y\le 3000........\left(4\right)$
$x+y\ge 3500....\left(5\right)$
$x,y\ge \mathrm{0........}\left(6\right)$
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are $A(3500,0),B(4500,0),C(4500,2500),D(4000,3000)$ and $E(500,3000)$
The values of $z$ at these corner points are as follows

Corner point	$z=0.3x+0.1y+3950$
$A(3500,0)$	$5000$
$B(4500,0)$	$5300$

$C(4500,2500)$	$5550$
$D(4000,3000)$	$5450$
$E(500,3000)$	$4400$	$\to$ Minimum

The minimum value of $z$ is $4400$ at $(500,3000)$
Thus, the oil supplied from depot $A$ is $500L,3000L$ and $3500L$ and from depot $B$ is $4000L,0L$ and $0L$ to petrol pumps $D,E$ and $F$ respectively.
The minimum transportation cost is Rs.$4400$