Question

An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps D, E and F, whose requirements are 4500 L, 3000 L and 3500 L respectively. The distance (in~km)between the depots and the petrol pumps in given in the following table:

$\begin{array}{ccc}From/To& A& B\\ D& 7& 3\\ E& 6& 4\\ F& 3& 2\end{array}$

Assuming that the transportation cost of 10 L oil is Rs 1 per km. How should that delivery be scheduled in order the transportation cost is minimum? What is the minimum cost?

Answer 1

Ley x and y L 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 pump D is 4500 L. Since, x L are transported from depot A, the remaining (4500 - 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 follows:

$\therefore$ Transportation cost of 10 L is Rs 1 per km.

$\therefore$ Transportation of 1 L is $\frac{1}{10}$per km.

Let Z be the total cost of transportation then.

$Z=\frac{7}{10}x+\frac{6}{10}y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)$

=0.3x+0.1y+3950

So, our problem is to minimize Z=0.3x+0.1y +3950 ...(i)

$Subjecttoconstraintsare4500-x\ge 0\iff x\le 4500...\left(ii\right)3000-y\ge 0\iff y\le 3000..\left(iii\right)x+y-3500\ge 0\iff x+y\ge 3500..\left(iv\right)7000-x-y\ge 0\iff x+y\le 7000..\left(v\right)x\ge 0,y\ge 0..\left(vi\right)Firstly,drawthegraphofthelinex+y=7000\begin{array}{ccc}x& 0& 7000\\ y& 7000& 0\end{array}putting(0,0)intheinequalityx+y\le 7000,wehave0+0\le 100\Rightarrow 0\le 7000\left(whichistrue\right)So,thehalfplaneistowardstheorigin.Secondly,drawthegraphofthelinex=4500Putting(0,0)intheinequalityx\le 4500,wehave0\le 4500\left(Whichistrue\right)So,thehalfplaneistowardstheorigin.Thirdly,drawthegraphofthelinex+y=3500\begin{array}{ccc}x& 3500& 0\\ y& 0& 3500\end{array}$

Putting (0, 0) in the inequality x + y$\ge$ 3500, we have $0+0\ge 3500\Rightarrow 0\ge 3500$(which is false)So, the half plane is away from the origin. Fourthly, draw the graph of the line y = 3000 putting(0, 0) in the inequality $y\le 3000$, we have $0\le 3000$(which is true)So, the half plane is towards the origin. Since, x, y $\ge 0$ So, the feasible region lies in the first quadrant.

The intersection points of given lines are C(4500, 2500), D(4500, 3000) and E(500, 3000).

$\therefore$ Feasible region is ABCDEA.

The corner points of the feasible region are A(3500, 0), B(4500, 0),C(4500, 2500), D(4500, 3000) and E(500, 3000). The values of Z at these points are as follows.

$\begin{array}{cc}Cornerpoint& Z=0.3x+0.1y+3950\\ A(3500,0)& 5000\\ B(4500,0)& 5300\\ C(4500,2500)& 5550\\ D(4500,3000)& 5600\\ E(500,3000)& 4400\to Minimum\end{array}$

The minimm value of Z is 4400 at E(500, 3000).

Thus, the oil supplied from depot A is 500 L, 3000 L and 3500 L and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F respectively.

The minimum transportation cost is Rs 4400.