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Question
An oil company has two depots, A and B, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps, D, E, F whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:
Figure
Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
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Solution
Let x and y litres of oil be supplied from A to the petrol pumps D and E. Then, (7000 − x − y) L 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)] L i.e. (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. 
Since, quantity of oil are non-negative quantities.Therefore,
x ≥0 , y ≥ 0, and (7000 - x - y ) ≥ 0
⇒ x ≥ 0 , y ≥ 0, and x + y ≥ 7000
4500 - x ≥ 0 , 3000 - y ≥ 0 , and x + y -3500 ≥ 0
⇒ x ≤ 4500, y ≤ 3000, and x + y ≥ 3500
Cost of transporting 10 L of petrol = Re 1
Cost of transporting 1 L of petrol = \[Rs \frac{1}{10}\]
Therefore, total transportation cost is given by,
`z = 7/10 xx x+6/10 y + 3/10 (7000 - x -y ) + 3/10 (4500 - x) + 4/10 ( 3000 - y ) + 2/10( x + y - 3500)`
= 0.3x + 0.1y + 3950
The problem can be formulated as follows.
Minimize Z = 0.3x + 0.1y + 3950
subject to the constraints,
\[x + y \leq 7000\]
\[x \leq 4500\]
\[y \leq 3000\]
\[x + y \geq 3500\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
x + y = 7000, x = 4500, y = 3000, x + y = 3500, x = 0 and y = 0
Region represented by x + y ≤ 7000:
The line x + y = 7000 meets the coordinate axes at A1(7000, 0) and \[B_1 \left( 0, 7000 \right)\] respectively. By joining these points we obtain the line x + y = 7000 . Clearly (0,0) satisfies the x + y = 7000 . So, the region which contains the origin represents the solution set of the inequation x + y ≤ 7000.
Region represented by x ≤ 4500:
The line x = 4500 is the line passes through C1(4500, 0) and is parallel to Y axis. The region to the left of the line x = 4500 will satisfy the inequation
x ≤ 4500.
Region represented by y ≤ 3000:
The line y = 3000 is the line passes through D1(0, 3000) and is parallel to X axis. The region below the the line y = 3000 will satisfy the inequation
y ≤ 3000.
Region represented by x + y ≥ 3500:
The line x + y = 7000 meets the coordinate axes at E1(3500, 0) and \[F_1 \left( 0, 3500 \right)\] respectively. By joining these points we obtain the line
x + y = 3500 . Clearly (0,0) satisfies the x + y = 3500. So, the region which contains the origin represents the solution set of the inequation x + y ≥ 3500.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 7000, x ≤ 4500, y ≤ 3000, x + y ≥ 3500, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E1(3500, 0), C1(4500, 0), I1(4500, 2500), H1(4000, 3000), and G1(500, 3000).
The values of Z at these corner points are as follows.
| Corner point | Z = 0.3x + 0.1y + 3950 |
| E1(3500, 0) | 5000 |
| C1(4500, 0) | 5300 |
| I1(4500, 2500) | 5550 |
| H1(4000, 3000) | 5450 |
| G1(500, 3000) | 4400 |
The minimum value of Z is 4400 at G1(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.
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