CCE 6100 HW4

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Apr 3, 2024

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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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