Carla Lee, a current MBA student, decides to spend her summer designing and mark
ID: 3123044 • Letter: C
Question
Carla Lee, a current MBA student, decides to spend her summer designing and marketing bicycling maps of Western Pennsylvania. She has designed maps, corresponding to three areas around Pittsburgh. The maps differ in size, colors used, and complexity of the topographical relief (the maps are actually 3-dimensional, showing hills and valleys). She has retained a printer to produce the maps. Each map must be printed, cut, and folded. The time (in minutes) to do this for the three types of maps is:
Cut
Fold
A
1
2
3
B
2
4
2
C
3
1
5
The printer has a limited amount of time in his schedule: 1500 minutes for printing, and 2000 minutes each for cutting and folding.
The profit per map, based on the projected selling price minus printers cost and other variable cost, comes out to approximately $1 for A and B and $2 for C. In order to have a sufficiently nice display, at least 100 of each type must be produced.
Use the simplex method to determine the production quantities and projected profit.
Suppose we reduced the 100 limit on one item to 90. Which map should be decreased, and how much more would Carla make?
Cut
Fold
A
1
2
3
B
2
4
2
C
3
1
5
Explanation / Answer
Let x denote the production quantity of A, y denote the production quantity of B, and z denote the production quantity of C.
Given that unit profit from A and B is 1 and C is 2.
The objective function is Maximize P = x + y+ 2z.
Total time available for printing is 1500, cutting is 2000 and folding is 2000.
Hence the constraints are
x + y+ 3z <= 1500
2x + 4y + z <= 2000
3x + 2y + 5z <= 2000
It is also given that x >= 100, y>= 100, z>= 100
Hence the problem is
Maximize p = x+y+2z subject to
x+2y+3z <= 1500
2x+4y+z <=2000
3x+2y+5z<= 2000
x>=100
y>= 100
z>= 100
Solution
Table #1
x y z s1 s2 s3 s4 s5 s6 p
1 2 3 1 0 0 0 0 0 0 1500
2 4 1 0 1 0 0 0 0 0 2000
3 2 5 0 0 1 0 0 0 0 2000
1 0 0 0 0 0 -1 0 0 0 100
0 1 0 0 0 0 0 -1 0 0 100
0 0 1 0 0 0 0 0 -1 0 100
-1 -1 -2 0 0 0 0 0 0 1 0
Tableau #2
x y z s1 s2 s3 s4 s5 s6 p
0 2 3 1 0 0 1 0 0 0 1400
0 4 1 0 1 0 2 0 0 0 1800
0 2 5 0 0 1 3 0 0 0 1700
1 0 0 0 0 0 -1 0 0 0 100
0 1 0 0 0 0 0 -1 0 0 100
0 0 1 0 0 0 0 0 -1 0 100
0 -1 -2 0 0 0 -1 0 0 1 100
Tableau #3
x y z s1 s2 s3 s4 s5 s6 p
0 0 3 1 0 0 1 2 0 0 1200
0 0 1 0 1 0 2 4 0 0 1400
0 0 5 0 0 1 3 2 0 0 1500
1 0 0 0 0 0 -1 0 0 0 100
0 1 0 0 0 0 0 -1 0 0 100
0 0 1 0 0 0 0 0 -1 0 100
0 0 -2 0 0 0 -1 -1 0 1 200
Tableau #4
x y z s1 s2 s3 s4 s5 s6 p
0 0 0 1 0 0 1 2 3 0 900
0 0 0 0 1 0 2 4 1 0 1300
0 0 0 0 0 1 3 2 5 0 1000
1 0 0 0 0 0 -1 0 0 0 100
0 1 0 0 0 0 0 -1 0 0 100
0 0 1 0 0 0 0 0 -1 0 100
0 0 0 0 0 0 -1 -1 -2 1 400
Tableau #5
x y z s1 s2 s3 s4 s5 s6 p
0 0 0 1 0 -0.6 -0.8 0.8 0 0 300
0 0 0 0 1 -0.2 1.4 3.6 0 0 1100
0 0 0 0 0 0.2 0.6 0.4 1 0 200
1 0 0 0 0 0 -1 0 0 0 100
0 1 0 0 0 0 0 -1 0 0 100
0 0 1 0 0 0.2 0.6 0.4 0 0 300
0 0 0 0 0 0.4 0.2 -0.2 0 1 800
Tableau #6
x y z s1 s2 s3 s4 s5 s6 p
0 0 0 1 -0.222222 -0.555556 -1.11111 0 0 0 55.5556
0 0 0 0 0.277778 -0.0555556 0.388889 1 0 0 305.556
0 0 0 0 -0.111111 0.222222 0.444444 0 1 0 77.7778
1 0 0 0 0 0 -1 0 0 0 100
0 1 0 0 0.277778 -0.0555556 0.388889 0 0 0 405.556
0 0 1 0 -0.111111 0.222222 0.444444 0 0 0 177.778
0 0 0 0 0.0555556 0.388889 0.277778 0 0 1 861.111
The final solution is
Optimal solution is P= 861.11
The production quantity of A, x = 100;
Production quantity of B, y = 405.56;
Production quantity of C, z = 177.8
We should reduce the limit of item A from 100 to 90.
In this case Carla could make a profit P = 863.89.
Reduction of B and C will not make any change in the profit of P = 861.11