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quiz - CS 601(Autumn 2009 Quiz 2 solutions November 3 2009...

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CS 601 (Autumn 2009): Quiz 2 solutions November 3, 2009 Problem 1 Part (a) For each item i we have a 0 / 1 integer variable x i , where x i = 1 indicates that item i is picked while x i = 0 indicates that it is not picked. The integer LP is as follows. max n summationdisplay i =1 v i x i n summationdisplay i =1 w i x i W x i { 0 , 1 } for all i [5 marks] Part (b) The LP relaxation is as follows. max n summationdisplay i =1 v i x i n summationdisplay i =1 w i x i W x i 1 for all i x i 0 for all i To construct the dual of the above LP, we take the variable y i for the constraint x i 1, for each i , and the variable z for the constraint n i =1 w i x i W . The dual LP is as follows. min n summationdisplay i =1 y i + Wz y i + w i z v i for all i y i 0 for all i z 0 [5 marks] Common mistake: Almost everyone has ignored the constraint x i 1 in the primal. Note that the “standard” form of the primal is max c x ; Ax b , x 0 . Hence each of the constraints x i 1 will contribute a row to A and b . 1
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Part (c) If the object i is not fully picked in the optimal primal solution then we have x i < 1. By the property of complementary slackness, since the constraint x i 1 is slack, the corresponding variable y i must be 0 in the optimal dual solution. Thus z v i /w i for each item i that is not fully picked. [5 marks]
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