CCE 6100 HW4
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WESTERN MICHIGAN UNIVERSITY CCE-6100-100 - Civil Systems Analysis
HW.4
Done by: Aws Tarawneh
5. Kilgore’s Deli is a small delicatessen located near a major university. Kilgore does a large
walk-in carry-out lunch business. The deli offers two luncheon chili specials, Wimpy and Dial
911. At the beginning of the day, Kilgore needs to decide how much of each special to make (he
always sells out of whatever he makes). The profit on one serving of Wimpy is $0.45, on one
serving of Dial 911, $0.58. Each serving of Wimpy requires 0.25 pound of beef, 0.25 cup of
onions, and 5 ounces of Kilgore’s special sauce. Each serving of Dial 911 requires 0.25 pound of
beef, 0.4 cup of onions, 2 ounces of Kilgore’s special sauce, and 5 ounces of hot sauce. Today,
Kilgore has 20 pounds of beef, 15 cups of onions, 88 ounces of Kilgore’s special sauce, and 60 ounces of hot sauce on hand.
A. Develop an LP model that will tell Kilgore how many servings of Wimpy and Dial 911 to make in order to maximize his profit today.
W = Wimpy chili.
D= Dial 911 chili.
Max 0.45W + 0.58 D
s.t. 0.25 W+0.25D<=20
0.25W+0.4D<=15
5W+2D<=88
5D<=60 D,W>0
Global optimal solution found.
Objective value: 12.72000
Infeasibilities: 0.000000
Total solver iterations: 1
Model Class: LP
Total variables: 2
Nonlinear variables: 0
Integer variables: 0
Total constraints: 5
Nonlinear constraints: 0
Total nonzeros: 9
Nonlinear nonzeros: 0
Variable Value Reduced Cost
W 12.80000 0.000000
D 12.00000 0.000000
Row Slack or Surplus Dual Price
1 12.72000 1.000000
2 13.80000 0.000000
3 7.000000 0.000000
4 0.000000 0.9000000E-01
5 0.000000 0.8000000E-01
B. Find an optimal solution.
max=12.72
W=12.8
D= 12
C. What is the dual value for special sauce? Interpret the dual value.
Dual value for Kilgore’s special sauce=0.09
which means increasing the sauces one once will increase the max profit 0.09 dollar D. Increase the amount of special sauce available by 1 ounce and re-solve. Does the solution confirm the answer to part (c)? Give the new solution.
Max 0.45W + 0.58 D
s.t. 0.25 W+0.25D<=20
0.25W+0.4D<=15
5W+2D<=89
5D<=60
Global optimal solution found.
Objective value: 12.81000
Infeasibilities: 0.000000
Total solver iterations: 1
Model Class: LP
Total variables: 2
Nonlinear variables: 0
Integer variables: 0
Total constraints: 5
Nonlinear constraints: 0
Total nonzeros: 9
Nonlinear nonzeros: 0
Variable Value Reduced Cost
W 13.00000 0.000000
D 12.00000 0.000000
Row Slack or Surplus Dual Price
1 12.81000 1.000000
2 13.75000 0.000000
3 6.950000 0.000000
4 0.000000 0.9000000E-01
5 0.000000 0.8000000E-01
The max increased $0.09
11. Let xij = units of component i purchased from supplier j
Min 12x11 + 13x12 + 14x13 + 10x21 + 11x22 + 10x23
s.t. x11 + x21 <= 600
x12 + x22 <= 1000
x13 + x23 <= 800
x11 + x12 + x13 = 1000
x21 + x22 + x23 = 800
x11>= 0
x12>= 0
x13>= 0
x21>= 0
x22>= 0
x23 >= 0
Global optimal solution found.
Objective value: 20400.00
Infeasibilities: 0.000000
Total solver iterations: 4
Model Class: LP
Total variables: 6
Nonlinear variables: 0
Integer variables: 0
Total constraints: 12
Nonlinear constraints: 0
Total nonzeros: 24
Nonlinear nonzeros: 0
Variable Value Reduced Cost
X11 600.0000 0.000000
X12 400.0000 0.000000
X13 0.000000 1.000000
X21 0.000000 1.000000
X22 0.000000 1.000000
X23 800.0000 0.000000
Row Slack or Surplus Dual Price
1 20400.00 -1.000000
2 0.000000 -13.00000
3 0.000000 -10.00000
4 0.000000 1.000000
5 600.0000 0.000000
6 0.000000 0.000000
7 600.0000 0.000000
8 400.0000 0.000000
9 0.000000 0.000000
10 0.000000 0.000000
11 0.000000 0.000000
12 800.0000 0.000000
Component 1 will be bought from supplier 1 & 2 600 & 400 ,respectively Component 2 will be bought from supplier 3 X23=800
Purchase Cost = $20,400
17. MF= Manufactured Frame PF=Purchased Frame MS= Manufactured Support PS= Purchased Support MT= Manufactured Strap
PT= Purchased Strap
To produce one product we need 1 frame , 2 supports, 1 strap
min 38MF+51PF+11.5MS+15PS+6.5MT+7.5PT
s.t. MF+PF>=5000
MS+PS>=10000
MT+PT>=5000
3.5MF+1.3MS+0.8MT<=21000
2.2MF+1.7MS<=25200
3.1MF+2.6MS+1.7MT<=40800
Global optimal solution found.
Objective value: 368076.9
Infeasibilities: 0.000000
Total solver iterations: 6
Model Class: LP
Total variables: 6
Nonlinear variables: 0
Integer variables: 0
Total constraints: 7
Nonlinear constraints: 0
Total nonzeros: 20
Nonlinear nonzeros: 0
Variable Value Reduced Cost
MF 5000.000 0.000000
PF 0.000000 3.576923
MS 2692.308 0.000000
PS 7307.692 0.000000
MT 0.000000 1.153846
PT 5000.000 0.000000
Row Slack or Surplus Dual Price
1 368076.9 -1.000000
2 0.000000 -47.42308
3 0.000000 -15.00000
4 0.000000 -7.500000
5 0.000000 2.692308
6 9623.077 0.000000
7 18300.00 0.000000
a.
Formulate and solve a linear programming model for this make-or-buy application. How many of each component should be manufactured and how many should be purchased?
Manufactured
Purchased
Frame
5000
0
Support
2692
7308
Strap
0
5000
b.
What is the total cost of the manufacturing and purchasing plan?
Total cost=$368,076.9
c.
How many hours of production time are used in each department?
Cutting= 350 hours used
Milling = (25200-9623)/60=259.6 hours used
Shaping=(40800-18300)/60 = 375 hours used d.
How much should Frandec be willing to pay for an additional hour of time in the shaping department?
There is still hours in the shaping department so he doesn’t need to add more hours.
e.
Another manufacturer has offered to sell frames to Frandec for $45 each. Could Frandec improve its position by pursuing this opportunity? Why or why not?
Yes, the reduced price for the Purchased frames is 3.57 so if we reduce the cost under 47.43 the solution may improve when changing PF cost to 45 the answer will become:
Global optimal solution found.
Objective value: 361500.0
Infeasibilities: 0.000000
Total solver iterations: 7
Model Class: LP
Total variables: 6
Nonlinear variables: 0
Integer variables: 0
Total constraints: 7
Nonlinear constraints: 0
Total nonzeros: 20
Nonlinear nonzeros: 0
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