The makespan for this problem is 51 units of time 1010 CDS HEURISTIC For large

The makespan for this problem is 51 units of time

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The makespan for this problem is 51 units of time . 10.10 CDS HEURISTIC For large size problems, it would be difficult to get optimum solution in finite time, since the flow shop scheduling is a combinatorial problem. This means the time complexity function of flow shop problem is exponential in nature. Hence, we have to use efficient heuristics for large size problems. CDS (Campbell, Dudek and Smith) heuristic is one such heuristic used for flow shop scheduling. The CDS heuristic corresponds to multistage use of Johnson’s rule applied to a new problem formed from the original processing time. At stage 1 t 1 j 1 = t j i and t 1 j 2 = t jm In other words, Johnson’s rule is applied to the first and mth operations and intermediate operations are ignored. At stage 2 T 2 j 1 = t j i + t j 2 and t 2 j 2 = t jm + t j m –1 That is, Johnson’s rule is applied to the sum of the first two and the last two operation processing times. In general at stage i, t i ji = ji and t i j 2 = t j , m - k +1 For each stage i (i = 1, 2, ... m – 1), the job order obtained is used to calculate a makespan for the original problem. After m – 1, stages, the best makespan among the m – 1 schedule is identified. (Some of the m – 1 sequences may be identical).
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250 Operations Management ILLUSTRATION 6: Find the makespan using the CDS heuristic for the following flow shop problem: Job j t j1 t j2 t j3 t j4 14378 23725 31247 43432 SOLUTION Job (j) M/c–1 t j1 M/c–1 t j2 148 235 317 432 The optimal sequence for the above problem is as shown below: 3-2-1-4 The makespan calculation for the above schedule is shown below: Processing time (in hour) Job M/c–1 M/c–2 M/c–3 M/c–4 In Out In Out In Out In Out 3011337714 2 1 4 4 11 11 13 14 19 148111414212129 1 8 11 14 18 21 24 29 31 Makespan of this problem = 31. Stage 2 Job (j) M/c–1 M/c–2 T j1 t j2 1715 2107 3311 475 After applying Johnson's algorithm to the above problem, we get the sequence, 3-1-2-4. The makespan calculation is summarized in the following table:
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Scheduling and Controlling Production Activities 251 Processing time (in hour) Job M/c–1 M/c–2 M/c–3 M/c–4 In Out In Out In Out In Out 3011337714 1 1 5 5 8 8 15 15 23 2 5 8 8 15 15 17 23 28 4 8 11 15 19 19 22 28 30 The makes pan for the sequence 3-1-2-4 is 30. Stage 2 M/c–1 M/c–2 T j1 T j2 11418 21214 3713 4109 The application of Johnson’s algorithm to the above data yields the sequence 3-2-1-4. The determination of the corresponding makespan is shown below: M/c–1 M/c–2 M/c–3 In Out In Out In Out 30137714 2 141113 1419 14814212129 4 8 11 21 24 29 31 stage sequence Makespan 1 3-2-14 31 2 3-1-2-4 30 3 3-2-1-4 31 The best sequence is 3-1-2-4, which has the makespan of 30. 10.11 JOB-SHOP PROBLEM In job-shop problem, we assume that each job has m different operations. If some of the jobs are having less than m operations, required number of dummy operations with zero process times is assumed. By this assumption, the condition of equal number of operations for all the jobs is ensured.
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