Q. An oil company required 12000, 20000 and 15000 barrels of high-grade, medium grade and low grade oil, respectively. Refinery A produces 100, 300 and 200 barrels per day of high-grade, medium-grade and low-grade oil, respectively, while refinery B produces 200, 400 and 100 barrels per day of high-grade, medium-grade and low-grade oil, respectively. If refinery A costs ₹ 400 per day and refinery B costs ₹ 300 per day to operate, then the days should each be run to minimize costs while satisfying requirements are
Linear Programming
Solution:
The given data may be put in the following tabular form
Refinery High grade Medium grade Low grade Cos t per day A 100 300 200 Rs 400 B 200 400 100 Rs 3100 Minimum
Requirement 12000 20000 15000
Suppose refineries A and B should run for x and y days respectively to minimize the total cost.
The mathematical form of the above is Minimize $Z = 400x + 300y$
Subject to
$100x + 200y \ge 12000$
$300x + 400y \ge 20000$
$200x + 100y \ge 15000$
and $x, y \ge 0$
The feasible region of the above LPP is represented by the shaded region in the given figure. The corner points of the feasible region are $A_2(120, 0), P(60, 30)$ and $B_3(0, 1 50)$. The value of the objective function at these points are given in the following table
Point (x , y ) Value of the objective function
Z = 400x + 300y $A_2 (120, 0)$ $Z = 400 × 120 + 300 × 0 = 48000$ $P (60, 30)$ $Z = 400 × 60 + 300 × 30 = 33000$ $B_3 (0, 150)$ $Z = 400 × 0 + 300 × 150 = 45000$
Clearly, $Z$ is minimum when $x = 60, y = 30.$ Hence, the machine $A$ should run for $60$ days and the machine$ B$ should run for $30$ days to minimize the cost while satisfying the constraints.
| Refinery | High grade | Medium grade | Low grade | Cos t per day |
|---|---|---|---|---|
| A | 100 | 300 | 200 | Rs 400 |
| B | 200 | 400 | 100 | Rs 3100 |
| Minimum Requirement | 12000 | 20000 | 15000 |
| Point (x , y ) | Value of the objective function Z = 400x + 300y |
|---|---|
| $A_2 (120, 0)$ | $Z = 400 × 120 + 300 × 0 = 48000$ |
| $P (60, 30)$ | $Z = 400 × 60 + 300 × 30 = 33000$ |
| $B_3 (0, 150)$ | $Z = 400 × 0 + 300 × 150 = 45000$ |