The production department for an aluminum valve plant is scheduling its work for
ID: 406022 • Letter: T
Question
The production department for an aluminum valve plant is scheduling its work for next month. Each valve must go through three separate machines during the fabrication process. After fabrication, each valve is inspected by a human being, who spends 15 minutes per valve. There are 525 inspection hours available for the month. The time required (in hours) by each machine to work on each valve is shown in the following table. Also shown are the minimum number of valves that must be produced for the month and the unit profit for each valve.
PRODUCT
V231
V242
V784
V906
CAPACITY
(hours)
Drilling
0.40
0.30
0.45
0.35
700
Milling
0.60
0.65
0.52
0.48
890
Lathe
1.20
0.60
0.5
0.70
1200
Unit profit
$16
$12
$13
$8
Minimum needed
200
250
600
450
Determine the optimal production mix for the valve plant to make the best use of its profit potential.
PRODUCT
V231
V242
V784
V906
CAPACITY
(hours)
Drilling
0.40
0.30
0.45
0.35
700
Milling
0.60
0.65
0.52
0.48
890
Lathe
1.20
0.60
0.5
0.70
1200
Unit profit
$16
$12
$13
$8
Minimum needed
200
250
600
450
Explanation / Answer
This problem is a linear programming problem with four decision variables which are the number of four different valves to be manufactured. We have to maximize the total Profits, Subjected to the constraints. This problem can be formulated in excel and can be solved using the solver module.
Decision Variables:
X1 = Number of V231 valves.
X2 = Number of V242 valves.
X3 = Number of V784 valves.
X4 = Number of V906 valves.
Constraints:
(15/60)*(X1 + X2 + X3 + X4)<= 525
Objective Function:
Maximize Z = 16*X1 + 12*X2 + 13*X3 + 8*X4
V231 = 332, V242 = 250, V784 = 600, V906 = 450.
Maximum profit = $19720
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.