IOEMFG 543

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IOE/MFG 543

• Chapter 5 Parallel machine models
• (Sections 5.3-5.6)

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Section 5.3 Total completion time PmS Cj

• On a single machine
• CjSk1 p(k)
• where p(j) is the processing time of the jth job
  processed on the machine
• Then
• SCjnp(1)(n-1)p(2)...p(n)
• gt Shortest Processing Time first rule minimizes
  SCj

j
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Total completion time PmS Cj (2)

• The same argument can be extended to parallel
  machines
• Theorem 5.3.1
• The SPT rule is optimal for PmS Cj
• In fact, a number of optimal schedules exist

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Weighted total completion time PmS wjCj

• WSPT rule is not always optimal
• In practice it usually does pretty well
• Worst case bound

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Other completion time models

• Precedence constraints PmprecSCj
• Strongly NP-hard
• Non-identical machines RmSCj
• Can be formulated as an integer program
• The solution of the corresponding linear program
  gives an optimal schedule
• gt Can be solved in polynomial time

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Section 5.4 Preemptions PmprmpS Cj

• The SPT non-preemptive rule is still optimal
• A special case of a more general result for
  QmprmpS Cj
• Recall vi is the speed on machine i in Qm models
• pj can be interpreted as the required work to
  complete job j (processing time pj/vi)

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QmprmpS Cj

• SRPT-FM rule
• Process the jobs such that the job with the
  shortest remaining processing time is put on the
  fastest machine. The job with the second shortest
  remaining processing time is put on the second
  fastest machine, etc.
• Whenever the fastest machine completes a job, all
  the remaining jobs are moved up on the machines
• Theorem 5.4.2
• The SRPT-FM rule is optimal for QmprmpS Cj

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Example 5.4.3 QmprmpS Cj

• 4 machines
• 7 jobs
• Use the SRPT-FM to solve QmprmpS Cj

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Section 5.5Due date related objectives

• Single machine problems that can be solved
  easily
• e.g., 1precLmax, 1rj,prmpLmax
• Single machine tardiness problems (1STj) are
  NP-hard
• If all due dates are 0
• Then PmLmax is equivalent to PmCmax
• gt PmLmax is NP-hard

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QmprmpLmax

• One of few parallel machine problems that can be
  solved in polynomial time
• First question Is there a schedule such that
  Lmax z ?
• This can be formulated (backwards) as
  Qmrj,prmpCmax
• Let djdjz be a hard deadline

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QmprmpLmax (2)

• Solve the problem backwards
• Find the job k with the latest deadline
• Let rk0 and rjdk-dj
• Solve Qmrj,prmpCmax by applying LRPT-FM
• The backward schedule is optimal for QmprmpLmax

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Example 5.5.1 P2prmpLmax

• 4 jobs
• Let z0
• Solve P2rj,prmpCmax by LRPT rule

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Section 5.6 Discussion

• Parallel machine models are much harder than
  single machine models!
• Later in the course we will study heuristics that
  can be used to obtain good schedules