Chapter 7 Homework - alternative limited warehouse...

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capital requirements alternative net present value year 1 year 2 year 3 limited warehouse expansion $ 4,000.00 $ 3,000.00 $ 1,000.00 $ 4,000.00 X1 Extensive warehouse expansion $ 6,000.00 $ 2,500.00 $ 3,500.00 $ 3,500.00 X2 Test market new products $ 10,500.00 $ 6,000.00 $ 4,000.00 $ 5,000.00 X3 Advertising campaign $ 4,000.00 $ 2,000.00 $ 1,500.00 $ 1,800.00 X4 Basic research $ 8,000.00 $ 5,000.00 $ 1,000.00 $ 4,000.00 X5 Puchase new equipment $ 3,000.00 $ 1,000.00 $ 500.00 $ 900.00 X6 Capital funds available $ 10,500.00 $ 7,000.00 $ 8,750.00 A>> B>> C >> A >> Develop and solve an integer programming model for maximizing the net present value B >> Assume that only one of the warehouse expansion prjects can be implemented. Modify your model of part (a) C >> Suppose that if test marketing of the new product is carried out, the advertising campsign also must be onducted. Modify our formulation of Part (b) to reflect this new situation.
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Maximize 4000X1+6000X2+10500X3+4000X4+8000X5+3000X6 S.T. 3000X1+2500X2+6000X3+2000X4+5000X5+1000X6<=10500 1000X1+3500X2+4000X3+1500X4+1000X5+500X6<=7000 4000X1+3500X2+5000X3+1800X4+4000X5+900X6<=8750 X1, X2, X3, X4, X5, X6 >=0 and Binary B >> X1+X2<=1 C >> X3-X4=0 MAX 0 0 1 1 0 1 << binary 4000 6000 10500 4000 8000 3000 OBJECTIV 17500 S.T. 3000 2500 6000 2000 5000 1000 9000 <= 10500 1000 3500 4000 1500 1000 500 6000 <= 7000 4000 3500 5000 1800 4000 900 7700 <= 8750 MAX 0 0 1 1 0 1 4000 6000 10500 4000 8000 3000 OBJECTIV 17500 S.T. 3000 2500 6000 2000 5000 1000 9000 <= 10500 1000 3500 4000 1500 1000 500 6000 <= 7000 4000 3500 5000 1800 4000 900 7700 <= 8750 1 1 0 <= 1 MAX 0 0 1 1 0 1 4000 6000 10500 4000 8000 3000 OBJECTIV 17500 S.T. 3000 2500 6000 2000 5000 1000 9000 <= 10500 1000 3500 4000 1500 1000 500 6000 <= 7000 4000 3500 5000 1800 4000 900 7700 <= 8750 1 1 0 <= 1 1 -1 0 = 0
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set up production cost manufacturing cost/unit 4 cylinder connecting rods 2000 15 6 cylinder connecting rods 3500 18 X4C= the number of 4 cylinder connecting rods produced next week X6C = the number of 6 cylinder connecting rods produced next week S4C = 1 if the production line is set up to produce the 4 cylinder connecting rod: 0 if oth S6C = 1 if the production line is set up to product the 6 cylinder connecting rod: 0 if othe D >> Write an objective funtion for minimizing the cost of production for next week A >> Using the decision variables X4 and S4 write a constraint that limits next weeks production fo the 4 cylinde connecing rod to either 0 or 8000 B >> using the decision variables X6 and S6 write a constraint tha tlimits next weeks prouction o the 6 cylinder connecting rod to either 0 or 6000 units C >> write three constraints that, taken together, limit the production of connection rods for next week
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weekly production capabilities A >> X4C = 8000S4C 8000 B >> X6C = 6000S6C 6000 C >> X4C X6C S4C S6C D >> MINIMIZE 0 6000 0 1 herwise 15 18 2000 3500 erwise 1 -8000 1 -6000 1 1 variable variable fixed fixed
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objective 111500 0 = 0 0 = 0 1 = 1
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Microsoft Excel 14.0 Answer Report Worksheet: [Chapter 7 Homework.xlsx]12 Report Created: 3/31/2016 11:27:50 AM Result: Solver found a solution. All Constraints and optimality conditions are satisfied. Solver Engine Engine: GRG Nonlinear Solution Time: 0.047 Seconds. Iterations: 1 Subproblems: 0 Solver Options Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic Scaling Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Forward, Require Bounds Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume NonNegative Objective Cell (Min) Cell Name Original Value Final Value $L$52 objective City 20 436512 436512 Variable Cells Cell Name Original Value Final Value Integer $L$55 City 1 Carrier 1 0 0 Binary $M$55 City 1 Carrier 2 0 0 Binary $N$55 City 1 Carrier 3 0 0 Binary $O$55 City 1 Carrier 4 0 0 Binary $P$55
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