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

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Unformatted text preview: 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 ∈ { , 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 ≥ [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 ≥ . Hence each of the constraints x i ≤ 1 will contribute a row to A and b . 1 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] Comment: This should suggest the idea that the items should be ordered by...
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This note was uploaded on 01/04/2010 for the course CSE CS601 taught by Professor Prof.ranade during the Summer '09 term at IIT Bombay.

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

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