# chapter 8 solution - 5 a x 2 5 Otm s l to t pi a oui n o l...

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5. a. 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 Optimalsolution to LP relaxation (3.14, 2.60) x 2 x 1 2 x 1 + 3 x 2 = 14.08 The feasible mixed integer solutions are indicated by the boldface vertical lines in the graph above. b. The optimal solution to the LP relaxation is given by x 1 = 3.14, x 2 = 2.60. Its value is 14.08. Rounding the value of x 1 down to find a feasible mixed integer solution yields x 1 = 3, x 2 = 2.60 with a value of 13.8. This solution is clearly not optimal. With x 1 = 3 we can see from the graph that x 2 can be made larger without violating the constraints. c. 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 x 2 x 1 The optimal solution to the MILP is given by x 1 = 3, x 2 = 2.67. Its value is 14. 8 - 1 Optimal mixed integer solution (3, 2.67)
Chapter 8 7. a. x 1 + x 3 + x 5 + x 6 = 2 b. x 3 - x 5 = 0 c. x 1 + x 4 = 1 d. x 4 x 1 x 4 x 3 e. x 4 x 1 x 4 x 3 x 4 x 1 + x 3 - 1 9. a. x 4 8000 s 4 b. x 6 6000 s 6 c. x 4 8000 s 4 x 6 6000 s 6 s 4 + s 6 = 1 d. Min 15 x 4 + 18 x 6 + 2000 s 4 + 3500 s 6 11. a. Let P i = units of product i produced Max 25 P 1 + 28 P 2 + 30 P 3 s.t. 1.5 P 1 + 3 P 2 + 2 P 3 450 2 P 1 + 1 P 2 + 2.5 P 3 350 .25 P 1 + .25 P 2 + .25 P 3 50 P 1 , P 2 , P 3 0 b. The optimal solution is P 1 = 60 P 2 = 80 Value = 5540 P 3 = 60 This solution provides a profit of \$5540. 8 - 2
Integer Linear Programming c. Since the solution in part (b) calls for producing all three products, the total setup cost is \$1550 = \$400 + \$550 + \$600. Subtracting the total setup cost from the profit in part (b), we see that Profit = \$5540 - 1550 = \$3990 d. We introduce a 0-1 variable y i that is one if any quantity of product i is produced and zero otherwise. With the maximum production quantities provided by management, we obtain 3 new constraints: P 1 175 y 1 P 2 150 y 2 P 3 140 y 3 Bringing the variables to the left-hand side of the constraints, we obtain the following fixed charge formulation of the Hart problem. Max 25 P 1 + 28 P 2 + 30 P 3 - 400 y 1 - 550 y 2 - 600 y 3 s.t. 1.5 P 1 + 3 P 2 + 2 P 3 450 2 P 1 + 1 P 2 + 2.5 P 3 350 .25 P 1 + .25 P 2 + .25 P 3 50 P 1 - 175 y 1 0 P 2 - 150 y 2 0 P 3 - 140 y 3 0 P 1 , P 2 , P 3 0; y 1 , y 2 , y 3 = 0, 1 e. The optimal solution using The Management Scientist is P 1 = 100 y 1 = 1 P 2 = 100 y 2 = 1 Value = 4350 P 3 = 0 y 3 = 0 The profit associated with this solution is \$4350. This is an improvement of \$360 over the solution in part (c). 21. a. Let 1 if a camera is located at opening 0 if not i i x = min x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 + x 9 + x 10 + x 11 + x 12 + x 13 s.t. x 1 + x 4 + x 6 1 Room 1 x 6 + x 8 + x 12 1 Room 2 x 1 + x 2 + x 3 1 Room 3 8 - 3
Chapter 8 x 3 + x 4 + x 5 + x 7 1 Room 4 x 7 + x 8 + x 9 + x 10 1 Room 5 x 10 + x 12 + x 13 1 Room 6 x 2 + x 5 + x 9 + x 11 1 Room 7 x 11 + x 13 1 Room 8 b. x 1 = x 5 = x 8 = x 13 = 1. Thus, cameras should be located at 4 openings: 1, 5, 8, and 13.