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Single Machine Deterministic Models

Single Machine Deterministic Models. Jobs: J 1 , J 2 , ..., J n Assumptions : The machine is always available throughout the scheduling period. The machine cannot process more than one job at a time. Each job must spend on the machine a prescribed length of time. S ( t ). 3. 2. 1.

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Single Machine Deterministic Models

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  1. Single Machine Deterministic Models • Jobs: J1, J2, ..., Jn • Assumptions: • The machine is always available throughout the scheduling period. • The machine cannot process more than one job at a time. • Each job must spend on the machine a prescribed length of time.

  2. S(t) 3 2 1 t J2 J3 J1 ...

  3. Requirements that may restrict the feasibility of schedules: • precedence constraints • no preemptions • release dates • deadlines • Whether some feasible schedule exist? NP hard Objective function f is used to compare schedules. f(S) < f(S') whenever schedule S is considered to be better than S' problem of minimising f(S) over the set of feasible schedules.

  4. 1. Completion Time Models Due date related objectives: 2. Lateness Models 3. Tardiness Models 4. Sequence-Dependent Setup Problems

  5. Completion Time Models Contents 1. An algorithm which gives an optimal schedule with the minimum total weighted completion time 1 || wjCj 2. An algorithm which gives an optimal schedule with the minimum total weighted completion time when the jobs are subject to precedence relationship that take the form of chains 1 | chain | wjCj

  6. Literature: • Scheduling, Theory, Algorithms, and Systems, Michael Pinedo, Prentice Hall, 1995, or new: Second Addition, 2002, Chapter 3.

  7. 1 || wjCj Theorem.The weighted shortest processing time first rule (WSPT) isoptimal for 1 || wjCj WSPT: jobs are ordered in decreasing order of wj/pj The next follows trivially: The problem 1 || Cj is solved by a sequence S with jobs arranged innondecreasing order of processing times.

  8. which implies that wj pk < wk pj Proof. By contradiction. S is a schedule, not WSPT, that is optimal. j and k are two adjacent jobs such that S: ... ... k j t + pj + pk t ... ... S’ k j t + pj + pk t S: (t+pj) wj + (t+pj+pk) wk = t wj + pj wj + t wk + pj wk + pk wk S’: (t+pk) wk + (t+pk+pj) wj = t wk + pk wk + t wj + pk wj + pj wj the completion time for S’ < completion time for Scontradiction!

  9. 1 | chain | wjCj chain 1: 1  2  ...  k chain 2: k+1  k+2  ...  n Lemma. If the chain of jobs 1,...,k precedes the chain of jobs k+1,...,n. Let l* satisfy factor of chain 1,...,k l* is the job that determines the  factor of the chain

  10. Lemma. If job l* determines  (1,...,k) , then there exists an optimalsequence that processes jobs 1,...,l* one after another withoutinterruption by jobs from other chains. Algorithm Whenever the machine is free, select among the remaining chainsthe one with the highest  factor. Process this chain up to and includingthe job l* that determines its  factor.

  11. Example chain 1: 1  2  3  4 chain 2: 5  6  7  factor of chain 1 is determined by job 2: (6+18)/(3+6)=2.67  factor of chain 2 is determined by job 6: (8+17)/(4+8)=2.08 chain 1 is selected: jobs 1, 2  factor of the remaining part of chain 1 is determined by job 3:12/6=2  factor of chain 2 is determined by job 6: 2.08 chain 2 is selected: jobs 5, 6

  12.  factor of the remaining part of chain 1 is determined by job 3: 2  factor of the remaining part of chain 2 is determined by job 7:18/10=1.8 chain 1 is selected: job 3  factor of the remaining part of chain 1 is determined by job 4: 8/5=1.6  factor of the remaining part of chain 2 is determined by job 7: 1.8 chain 2 is selected: job 7 job 4 is scheduled last the final schedule: 1, 2, 5, 6, 3, 7, 4

  13. 1 | prec | wjCj • Polynomial time algorithms for the more complex precedence constraints than the simple chains are developed. • The problems with arbitrary precedence relation are NP hard. • 1 | rj, prmp | wjCj preemptive version of the WSPT rule does not always lead to an optimal solution, the problem is NP hard • 1 | rj, prmp | Cj preemptive version of the SPT rule is optimal • 1 | rj | Cj is NP hard

  14. Summary • 1 || wjCj WSPT rule • 1 | chain | wjCj a polynomial time algorithm is given • 1 | prec | wjCj with arbitrary precedence relation is NP hard • 1 | rj, prmp | wjCj the problem is NP hard • 1 | rj, prmp | Cj preemptive version of the SPT rule is optimal • 1 | rj | Cj is NP hard

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