IOEMFG 543 Chapter 6 Flow shops Sections 6

IOE/MFG 543 Chapter 6: Flow shops Sections 6. 1 and 6. 2 (skip section 6. 3) 1

Flow shop (Fm) n n m machines, n jobs Jobs are processed on the machines in series – Processing time of job j on machine i is pi, j n Buffers between machines – Unlimited – Limited => blocking 2

Section 6. 1. Unlimited storage - Permutation rule n Permutation rule – All jobs are processed in the same order on the machines – Equivalent to a FCFS rule For F 2||Cmax and F 3||Cmax there exists a permutation schedule that is optimal n It is much harder to minimize the makespan when the sequencing is not restricted to the permutation rule n 3

Computing the makespan for a given permutation n Let j 1, …, jn be a given permutation – i. e. , job jk is the kth job on all the machines n Ci, j=completion time of job j on machine i i Ci, j 1= S pl, j 1 i=1, …, m C 1, jk= S p 1, jl k=1, …, n l=1 k l=1 Ci, jk= max(Ci-1, jk, Ci, jk-1)+pi, jk i=2, …, m; k=2, …, n 4

Computing the makespan for a given permutation (2) n n Instead of solving the recursive equations on the previous slide the makespan can be computed by a critical path method Example 6. 1. 1 job j p 1 j p 2 j p 3 j 1 5 4 4 2 5 4 4 3 3 2 3 4 6 4 4 5 3 4 1 p 4 j 3 6 3 2 5 5

Johnson’s rule for F 2||Cmax n n Set I: All jobs such that p 1 j<p 2 j Set II: All jobs such that p 1 j>p 2 j Jobs with p 1 j=p 2 j can be put in either set SPT(1) – LPT(2) schedule (Johnson’s rule): – Jobs in Set I go first and in an increasing (nondecreasing) order of p 1 j => SPT(1) – Jobs in Set II go last and in a decreasing (nonincreasing) order of p 2 j => LPT(2) n Theorem 6. 1. 4 – Any SPT(1)-LPT(2) schedule is optimal for F 2||Cmax 6

Fm|prmu|Cmax n Theorem 6. 1. 7 – F 3|prmu|Cmax is strongly NP-hard – 3 -Partition reduces to F 3|prmu|Cmax 7

Mixed integer programming formulation of Fm|prmu|Cmax n Notation – xjk=1 if job j is the kth job in the sequence and 0 otherwise – Iik is the idle time on machine i between jobs in the kth and (k+1)th position – Wik is the waiting time after it has finished on the ith machine of the job in the kth position – Dik is the difference between the time when the job in the (k+1)th position starts on machine i+1 and the time the job in the kth position finishes on machine i – pi(k) is the processing time on machine i of the job in the kth position 8

Proportionate flow shops The processing time (work) for job j is pij=pj n Theorem 6. 1. 8 n – The makespan of Fm|prmu, pij=pj|Cmax is Cmax=Spj+(m-1)max(p 1, …, pn) and is independent of the schedule 9

Single machine models and proportionate flow shops Rule/algorithm Single machine Proportionate flow shop SPT rule 1||SCj Fm|pij=pj|SCj Algorithm 3. 3. 1 1||SUj Fm|prmu, pij=pj|SUj Algorithm 3. 2. 1 1||hmax Fm|prmu, pij=pj|hmax Algorithm 3. 4. 4 1||STj Fm|prmu, pij=pj|STj Lemma 3. 5. 1 1||Swj. Tj Fm|prmu, pij=pj|Swj. Tj Note: WSPT is not always optimal for Fm|prmu, pij=pj|Swj. Cj 10

Slope heuristic for Fm|prmu|Cmax n Slope index of job j m Aj= - S (m-(2 i-1))pij j=1, …, n i=1 n The slope index is large if the processing times on the downstream machines are large relative to the processing times on the upstream machines Heuristic rule n Example 6. 1. 10 n – Sequence jobs in decreasing order of the slope index 11

Section 6. 2 Limited storage flow shops Only need to consider the case where the storage between machines is zero n New notation n – Dij is the time when job j departs machine i – D 0 j is the time when job j starts processing on machine 1 – Note that Cij≤Dij 12

Computing the makespan of a sequence i Di, j 1= S pl, j 1 i=1, …, m l=1 Di, jk= max(Di-1, jk+pi, jk, Di+1, jk-1) i=1, …, m-1; k=2, …, n Dm, jk= Dm-1, jk+pm, jk The makespan of a given sequence can also be computed by a critical path method n The problem F 3|block|Cmax is strongly NP-hard n 13

Profile fitting (PF) heuristic for Fm|block|Cmax 1. 2. A job j 1 is selected to go first Try all the other jobs as the next job – – – 3. Use the equations on the previous slide to compute the departure times Compute a penalty as the sum of idle times and blocked times on all machines Choose the job with the lowest penalty to go next If all jobs have been scheduled=> STOP Otherwise go to Step 2. 14

Example 6. 2. 5 job j 1 2 3 4 5 p 1 j 5 5 3 6 3 p 2 j 4 4 2 4 4 p 3 j 4 4 3 4 1 p 4 j 3 6 3 2 5 15