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Unformatted text preview: Homework #2 solutions / IEOR4405 1 3.1. (a) By Theorem 3.1.1 in the textbook, the WSPT (weighted shortest processing time) rule is optimal for 1  ∑ w j C j . In order to apply this rule, we have to calculate the values of w j /p j . These values are given below: Job 1 2 3 4 5 6 7 w j /p j 3 2 1.6 2 2.2125 1.7778 As a result, both the sequences 2635741 and 2653741 are optimal with objective value ∑ w j C j = 1610. (b) If p 2 is increased from 6 to 7, the ratio w 2 /p 2 decreases from 3 to 2.57. Since w 2 /p 2 is still the largest among all the ratios, this change will has no effect on the optimal sequences. (c) The new objective value under (b) is ∑ w j C j = 1689. 3.3 To find the optimal solution we use the WDSPT rule. Define h ( j ) = w j exp ( rp j ) 1 exp ( rp j ) . Now calculate h ( j ) for every job. h(1) = 0, h(2) = 51.4, h(3) = 34.3, h(4) = 28.2, h(5) = 42.9, h(6) = 34.6, h(7)=28.15 Hence, the optimal schedule is 2563471. (b) Using, WDSPT rule again h(1) = 0, h(2) = 0.94, h(3) = 0.63, h(4) = 0.72, h(5) = 1.25, h(6) = 0.32, h(7) = 0.18 Optimal schedule is 5243671. The schedule has changed from before....
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This note was uploaded on 05/07/2011 for the course INDUSTRIAL IE 208 taught by Professor Serolbulkan during the Fall '11 term at Marmara Üniversitesi.
 Fall '11
 serolbulkan
 Optimization

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