(a)
The objective function for the given problem.
Answer to Problem 1.17P
The objective function for the given problem is to minimize
Explanation of Solution
The objective function is the relationship that is maximized (or minimized) by the decision-maker.
Here, in the given case, the firm is seeking to minimize the total cost to operate plant one and two. Therefore, the objective function will be:
(b)
The constraints for the given problem.
Answer to Problem 1.17P
The constraints for the given problem is:
Explanation of Solution
A constraint is a restriction that will be placed on the firm.
Here, in the given case the firm will have 1 million metric tons of emission. Therefore, the constraint can be given as:
(c)
The statement of constrained optimization.
Answer to Problem 1.17P
The statement of constrained optimization is:
Explanation of Solution
A statement of constrained optimization states the kind of functions a decision-maker wants to minimize or maximize along with the constraints.
In this, the perspective of the decision-maker must be considered.
In the given case, the objective function is
The given constraint is
. Therefore, the statement for the constraint optimization can be given as:
(d)
The emission level from each plant chosen by the firm.
Answer to Problem 1.17P
The emission level from each plant chosen by the firm is given as:
Emission plant 1 (X) | TC plant 1 | Emission plant 2 (Y) | TC plant 2 | TC |
0 | $490 | 1,000,000 | $10 | $500 |
250,000 | $360 | 750,000 | $40 | $400 |
500,000 | $250 | 500,000 | $90 | $340 |
750,000 | $160 | 250,000 | $160 | $320 |
1,000,000 | $90 | 0 | $250 | $340 |
Explanation of Solution
The allocation of operating cost for plant 1 and plant 2 can be represented in the following tabular form:
Emission plant 1 (X) | TC plant 1 | Emission plant 2 (Y) | TC plant 2 | TC |
0 | $490 | 1,000,000 | $10 | $500 |
250,000 | $360 | 750,000 | $40 | $400 |
500,000 | $250 | 500,000 | $90 | $340 |
750,000 | $160 | 250,000 | $160 | $320 |
1,000,000 | $90 | 0 | $250 | $340 |
From the above table, it is concluded that when X =750,000 metric tons and Y = 250,000 metric ton, then the total operating cost is minimized.
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