Solutions-Assignment3

Solutions-Assignment3 - x1 = x7 = x11= x12 =1 Optimal value...

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Assignment 3 - solutions Problem 1. a. (20 points) Let x i = 1 if investment alternative i is selected, 0 otherwise max 4000x1 + 6000x2 + 10500x3 + 4000x4 + 8000x5 +3000x6 s.t. 3000x 1 + 2500x 2 + 6000x 3 + 2000x 4 + 5000x 5 + 1000x 6 10500 1000x 1 + 3500x 2 + 4000x 3 + 1500x 4 + 1000x 5 + 500x 6 7000 4000x 1 + 3500x 2 + 5000x 3 + 1800x 4 + 4000x 5 + 900x 6 8750 x 1, x 2, x 3, x 4, x 5, x 6 = 0,1 b) (10 points) Optimal Solution found using Excel x3=x4=x6 = 1 ; x1=x2=x5 = 0 Value = 17,500 Problem 2 a. (10 points) Let x i = 1 if a camera is located at opening i , 0 otherwise Min. x1+x2+ x3+ x4+ x5+x6+x7+ x8+ x9+ x10+ x11+ x12+x13 s.t. x1+x4+x6 1 Room 1 x6+x8+x12 1 Room 2 x1+x2+x3 1 Room 3 x3+x4+x5+x7 1 Room 4 x7+x8+x9+x10 1 Room 5 x10+x12+x13 1 Room 6 x2+x5+x9+x11 1 Room 7 x11+x13 1 Room 8 b. (10 points) Solution via Excel: x1 = x5 = x8 = x13 = 1. Thus, cameras should be located at 4 openings: 1,5,8, and 13. An alternative optimal solution is
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Unformatted text preview: x1 = x7 = x11= x12 =1 Optimal value = 4. c. (10 points) Change the constraint for room 7 to x2+x5+x9+x11 2 d. (10 points) Solution via Excel: x3 = x6 = x9 = x11 = x12 = 1. Thus, cameras should be located at openings 3,6,9,11, and 12. An alternate optimal solution is x2 = x4 = x6 = x10 = x11 = 1. Optimal Value = 5. Problem 3 (30 points) Min. 30.95x11 + 31.10x12 + 30.90x13 + 32.30x21 + 31.80x23 + 31.55x31 + 31.55x32 + 32.15x33 s.t. x11 + x12 + x13 400 x21 + x23 600 x31 + x32 + x33 300 x11 + x21+ x31 =400 x12 + x32 =400 x13 + x23 + x33 =400 x ij 0 for all i,j . Optimal Plan: Martinsville to Chicago: 300 Martinsville to Dallas: 100 Plymouth to Chicago: 100 Plymouth to New York: 400 Franklin to Dallas: 300 Total Cost = $37,810 Note: Plymouth has excess supply of 100....
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This note was uploaded on 11/18/2011 for the course BUS 220 taught by Professor Drexel during the Fall '08 term at SUNY Stony Brook.

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Solutions-Assignment3 - x1 = x7 = x11= x12 =1 Optimal value...

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