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Unformatted text preview: IOE 543 Winter 2010 Homework 4 Solution Problem 1 By the hint, schedule 7‐6‐5 should be scheduled as the last three jobs. Job 4 precedes 3 so we can fathom node (3 x x x) Notice that ∑ P 46, so if we schedule 1 as the fourth job, cost is 38; schedule 2, we get 20; schedule 3 we get 4; schedule 4 we get 88. Furthermore 1‐4‐2‐3 incurs no tardy time in the first three jobs so the optimal schedule is 1‐4‐2‐3‐7‐6‐5 Problem 2 First order the jobs like this: 5, 3, 1, 4, and 2 Using ”loose due date” rule yields the following: Assign job 5 to set J1, set k=2; assign job 1 to set J1 and job 3 to set J2; since k+2<=5‐1, let k=4 and assign job 2 to J1 and job 4 to J2. The optimal schedule is 5‐1‐2‐4‐3, and the objective value is 25. Using “tight due date” rule: Iteration 1: t1=18>t2=10, schedule job 5 as the first job; Iteration 2: t1=8<t2=10, schedule job 3 as the fifth job; Iteration 3: t1=8>t2=4, schedule job 1 as the second job; Iteration 4: t1=3<t2=4, schedule job 4 as the fourth job. The optimal schedule is 5‐1‐2‐4‐3, and the objective value is 25. Problem 3 Cluster Jobs 1 1,2 2 3,4,5,6,7 Iteration 1: Cluster Early jobs E(r) Δ(r) 1 1,2 1 11 2 3,4 3 ‐9 Push all jobs backward for 1 unit. Iteration 2: Cluster Early jobs E(r) Δ(r) 1 1 1 ‐1 2 3,4 2 ‐9 End of iteration So the optimal completion time for these jobs is: Jobs 1 2 3 4 Cj 5 12 17 26 5 38 6 40 7 46 ...
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This note was uploaded on 03/16/2012 for the course IEOR 466 taught by Professor Richard during the Spring '12 term at Columbia.
 Spring '12
 Richard

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