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# sol3 - Homework#3 solutions IEOR4405 1 1(a Let(a1 a12 =(27...

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Homework #3 solutions / IEOR4405 1 1. (a) Let ( a 1 , ..., a 12 ) = (27 , 27 , 29 , 33 , 33 , 33 , 35 , 35 , 35 , 37 , 37 , 39) and let b = 100. Then we intro- duce the following jobs: j r j d j p j J 1 100 101 1 J 2 201 202 1 J 3 302 303 1 k 0 403 p k ( k = 1 , ..., 12) (b) Claim . The schedule: 4 , 5 , 7 , J 1 , 1 , 6 , 12 , J 2 , 2 , 11 , 10 , J 3 , 3 , 8 , 9 is optimal. Proof. For this schedule, L max = 1. To see that this is optimal, suppose that there is a schedule that has L max = ε < 1. Since the processing times and the release dates have integer values, any optimal schedule has integer-valued completion times. Therefore, in fact, L max = 0. Because p j = 403, it follows that there is no idle time in the schedule. Also, job J 1 starts exactly at time 100. But since there is no idle time in the first 100 time units, the processing times of the jobs processed in the first 100 time units add up to exactly 100. Clearly, since the processing times of the available jobs lie between 27 and 39, exactly three jobs are processed in this time interval. But now notice that all the processing times are odd, and the sum of three odd numbers is again odd. Hence they cannot add up to 100. (c) Since L max > 0, this means that the 3-partition problem does not have a solution. 2. Theorem . P 2 || L max is NP-complete. Proof. Consider an instance of the partition problem, i.e. we are given numbers p 1 , p 2 , ..., p n and we are asking the question: can we find a set S ⊆ { 1 , 2 , ..., n } such that i S p i = i 6∈ S p i ?

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