MSA Computer Corporation manufactures three models of computers: the Alpha4, Beta5, and the Delta6. The firm employs thirteen technicians working 160 hours each per month on its assembly line. Management insists that no more than full employment on the assembly line (i.e., all 160 hours of time) can be maintained for each worker during next month’s operations, so there will be no overtime. It requires 20 labor hours to assemble each Alpha4 computer, 25 labor hours for each Beta5 model, and 15 hours for each Delta6. Final testing and inspection requires 3 hours for each Alpha4, 1.5 hours for each Beta5, and 1.2 hours for each Delta6. In addition to the workers on the assembly line, the company has up to 80 hours available for inspection and testing each month. Demand is such that the company can and will sell all of the computers that it produces. MSA must produce at least 10 Delta6 computers. In addition, the number of Alpha4 computers produced each month must equal or exceed the combined total of Beta5 and Delta6 computers produced. Alpha4 computers generate $1,200 profit per unit, Beta5 computers yield $1,800 each, and Delta6 computers yield $1,000 per unit. If the optimal answer is in fractions of computers, leave them as fractions.
a) Formulate the linear programming problem to maximize the profit during the coming month. Provide the variables being used, the objective function and all of the constraints that are indicated by the above description of the problem.
b) Using any available software package (POM/QM, Solver, etc.), what is the maximum profit and the quantity of each type of computer to produce. Provide your computer software input and output supporting your answers to parts a) and b). Not doing this will result in loss of 75% of the credit on this problem. If the optimal answer is in fractions of computers that is acceptable.
c) Based on the computer output provided as part of part b) and not resolving the problem, provide two sensitivity analysis interpretations. One interpretation must use the objective function sensitivity analysis interpretation and one must use a constraint(s) sensitivity analysis interpretation. Provide your supporting sensitivity analysis (ranging) output.
d) Instead of the constraint that requires that MSA must produce at least 10 Delta6 computers, suppose it is now required that at least 35% of the computers produced must be Beta5 computers. Write the constraint to express this relationship. You do not need to resolve the revised linear programming formulation for the optimal answer.
e) The company now wants to ensure that for every Beta5, MSA must produce at least three Alpha4. Write the constraint to express this requirement however you do not need to resolve the revised linear programming formulation for the optimal answer.
Formulate the linear programming problem to maximize the profit during the coming month. Provide the variable being used, the objective function and all of the constraints that are indicated by the above description of the problem.
Let Alpha4 computer be X, Beta5 computer be Y, and the Delta6 computer be Z
Objective function is
Maximize Profit 1200X + 1800Y + 1000Z
Subject to :
20X + 25Y+15Z < = 13*160 (13 workers x 160 hr per month)
3X + 1.5Y + 1.2Z <=80 (No of inspection & testing hours)
Z >= 10 (Atleast 10 Delts6 Computers)
X > = Y+Z (Alpha4 must be greater than sum total of Beta5 & Delta6)
Or X – Y-Z >=0
Using any available software package (POM/QM, Solver, etc.), what is the maximum profit and the quantity of each type of computer to produce. Provide your computer software input and output supporting your answers to parts a) and b). Not doing this will result in loss of 75% of the credit on this problem. If the optimal answer is in fractions of computers that is acceptable.
Profit |
$ 1,200 |
$ 1,800 |
$ 1,000 |
|||
No of PCs |
18.44 |
8.44 |
10 |
|||
X |
Y |
Z |
Used |
Available |
||
Workers |
20 |
25 |
15 |
<= |
730 |
2080 |
I&T |
3 |
1.5 |
1.2 |
<= |
80 |
80 |
Delta6 |
0 |
0 |
1 |
>= |
10 |
10 |
Alpha4 |
1 |
-1 |
-1 |
>= |
0 |
0 |
Max Profit |
$ 47,333.33 |
Based on the computer output provided as part of part b) and not resolving the problem, provide two sensitivity analysis interpretations. One interpretation must use the objective function sensitivity analysis interpretation and one must use a constraint(s) sensitivity analysis interpretation. Provide your supporting sensitivity analysis (ranging) output.
Using Variable cell $D$4 (Y = No of Beta5 PCs), Allowable Increase/Decrease columns tell us that, provided the coefficient of Y in the objective function lies between 1800+ 0 = 1800 and 1800-642.86 =1157.14, the values of the variables in the optimal LP solution will remain unchanged.
For each constraint the column headed Shadow Price tells us exactly how much the objective function will change if we change the right hand side of the corresponding constraint within the limits given in the Allowable Increase/Decrease columns.
For example for the Inspection & Testing hrs (G7) constraint, provided the right-hand side of that constraint remains between 80 + 135 =215 and 80-38 = 42 the objective function change will be exactly 666.67 [change in right-hand side from 80].
The direction of the change in the objective function (up or down) depends upon the direction of the change in the right-hand side of the constraint and the nature of the objective (maximise or minimise).
Variable Cells |
|||||||
Final |
Reduced |
Objective |
Allowable |
Allowable |
|||
Cell |
Name |
Value |
Cost |
Coefficient |
Increase |
Decrease |
|
$C$4 |
No of PCs |
18.44 |
0 |
1200 |
2400 |
3000 |
|
$D$4 |
No of PCs |
8.44 |
0 |
1800 |
1E+30 |
642.86 |
|
$E$4 |
No of PCs |
10 |
0 |
1000 |
600 |
1E+30 |
|
Constraints |
|||||||
Final |
Shadow |
Constraint |
Allowable |
Allowable |
|||
Cell |
Name |
Value |
Price |
R.H. Side |
Increase |
Decrease |
|
$G$6 |
<= Used |
730 |
0 |
2080 |
1E+30 |
1350 |
|
$G$7 |
<= Used |
80 |
666.67 |
80 |
135 |
38 |
|
$G$6 |
<= Used |
730 |
0 |
0 |
730 |
1E+30 |
|
$G$7 |
<= Used |
80 |
0 |
0 |
80 |
1E+30 |
|
$G$8 |
>= Used |
10 |
0 |
0 |
10 |
1E+30 |
|
$G$9 |
>= Used |
0 |
0 |
0 |
0 |
1E+30 |
|
$G$8 |
>= Used |
10 |
-600 |
10 |
9.05 |
10 |
|
$G$9 |
>= Used |
0 |
-800 |
0 |
12.67 |
0 |
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