Ch09rp - 1 SOLUTION TO CHAPTER 9 REVIEW PROBLEMS 1. Let zt...

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SOLUTION TO CHAPTER 9 REVIEW PROBLEMS 1. Let z t = 1 if any production occurs during quarter t and z t = 0 otherwise. Then add terms 200(z 1 + z 2 + z 3 + z 4 ) to the objective function and add the constraints x 1 + y 1 Mz 1 x 2 + y 2 Mz 2 x 3 + y 3 Mz 3 x 4 + y 4 Mz 4 M = 200 will do. 2. We approximate y 2 by z 1 (0) 2 + z 2 (.25) 2 + z 3 (.50) 2 + z 4 (.75) 2 + z 5 (1) 2 and x 2 by z 1 '(0) 2 + z 2 '(.25) 2 + z 3 '(.50) 2 + z 4 '(.75) 2 + z 5 '(1) 2 . Then an appropriate IP is max z = 3(z 1 '(0) 2 + z 2 '(.25) 2 + z 3 '(.50) 2 + z 4 '(.75) 2 + z 5 '(1) 2 ) + z 1 (0) 2 + z 2 (.25) 2 + z 3 (.50) 2 + z 4 (.75) 2 + z 5 (1) 2 st. z 1 y 1 , z 2 y 1 + y 2 , z 3 y 2 + y 3 , z 4 y 3 + y 4 , z 5 y 4 z 1 + z 2 + z 3 + z 4 + z 5 = 1, y 1 + y 2 + y 3 + y 4 = 1 z 1 ' y 1 ', z 2 ' y 1 ' + y 2 ', z 3 ' y 2 ' + y 3 ', z 4 ' y 3 ' + y 4 ', z 5 ' y 4 ', z 1 ' + z 2 ' + z 3 ' + z 4 ' + z 5 ' = 1, y 1 ' + y 2 ' + y 3 ' + y 4 ' = 1, All y i and y i ' = 0 or 1, all other variables 0 From the optimal solution to this IP we approximate the optimal values of x and y as y = .25z 2 + .50z 3 + .75z 4 + z 5 and x = .25z 2 ' + .50z 3 ' + .75z 4 ' + z 5 '. 3. Let z i = 1 if gymnast i enters both events z i = 0 otherwise x i = 1 if gymnast i enters only BB x i = 0 otherwise y i = 1 if gymnast i enters just FE y i = 0 Then the appropriate IP is max z = 16.7z 1 + 17.7z 2 + . ..17.7z 6 + 8.8x 1 + . ..9.1x 6 + 7.9y 1 + . .. 8.6y 6 s.t. z 1 + z 2 + . ..+ z 6 = 3 x 1 + x 2 + . .. + x 6 = 1 y 1 + y 2 + . .. + y 6 = 1 x 1 + y 1 + z 1 1 x 2 + y 2 + z 2 1 x 3 + y 3 + z 3 1 x 4 + y 4 + z 4 1 x 5 + y 5 + z 5 1 1
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6 + y 6 + z 6 1 All variables equal 0 or 1. 4. Let x ij = 1 if students from district i are sent to school j x ij = 0 otherwise Then the appropriate IP is Min z = 110x 11 + 220x 12 + 37.5x 21 + 127.5x 22 + 80x 31 + 80x 32 + 117x 41 + 36x 42 + 135x 51 + 54x 52 s.t. 110x 11 + 75x 21 + 100x 31 + 90x 41 + 90x 51 150 110x 12 + 75x 22 + 100x 32 + 90x 42 + 90x 52 150 30x 11 + 5x 21 + 10x 31 + 40x 41 + 30x 51 .20 -------------------------------------- 110x 11 + 75x 21 + 100x 31 + 90x 41 + 90x 51 or 0 8x 11 - 10x 21 - 10x 31 + 22x 41 + 12x 51 30x 12 + 5x 22 + 10x 32 + 40x 42 + 30x 52 .20 --------------------------------------- 110x 12 + 75x 22 + 100x 32 + 90x 42 + 90x 52 or 0 8x 12 - 10x 22 - 10x 32 + 22x 42 + 12x 52 x 11 + x 12 = 1, x 21 + x 22 = 1, x 31 + x 32 = 1 x 41 + x 42 = 1, x 51 + x 52 = 1 All variables = 0 or 1 5. Let x 1 = 1 if RS is signed x 1 = 0 otherwise x 2 = 1 if BS is signed x 2 = 0 otherwise x 3 = 1 if DE is signed x 3 = 0 otherwise x 4 = 1 if ST is signed x 4 = 0 otherwise x 5 = 1 if TS is signed x 5 = 0 otherwise max z = 6x 1 + 5x 2 + 3x 3 + 3x 4 + 2x 5 s.t. 6x 1 + 4x 2 + 3x 3 + 2x 4 + 2x 5 12 (1) x 2 + x 3 + x 4 2 (2) x 1 + x 2 + x 3 + x 5 2 (3) x 1 + x 2 1 (4) All variables = 0 or 1 6. Let y i = 1 if we buy any computers from vendor i y i = 0 otherwise x i
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This homework help was uploaded on 04/02/2008 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at University of California, Berkeley.

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Ch09rp - 1 SOLUTION TO CHAPTER 9 REVIEW PROBLEMS 1. Let zt...

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