Parallel-Machine Models

1. Parallel-Machine Models Chapter 7 Elements of Sequencing and Schedulingby Kenneth R. Baker Byung-Hyun Ha R1

2. Outline • Introduction • Minimizing the makespan • Nonpreemptable jobs • Dependent nonpreemptable jobs • Preemptable jobs • Minimizing total flowtime • Summary

3. Introduction • Beyond single machine • Some types • Parallel systems, serial systems (flow shop), hybrid systems (job shop) • Sequencing + resource allocation • Parallel machines • Basic model (Pm) • n jobs independent and simultaneously available at time zero • m identical parallel machines • A job can be processed by at most one machine at a time • e.g., Pm || Cj • More general models • Parallel machines with different speeds, or uniform machines (Qm) • Speed of machine i -- vi • Processing time of job j on machine i --pij = pj /vi • Unrelated machines in parallel (Rm) • Speed of machine i on job j -- vij • pij = pj/ vij

4. Minimizing the Makespan • Makespan problem for parallel machines • Resource allocation problem  Sequence is not important • Preemptable jobs -- Pm | prmp | Cmax • Optimal makespan (McNaughton, 1959) • M* = max[j=1..n pj/m, max{pj}] • ALGORITHM 1 -- Minimizing M with m parallel, identical machines 1. Select some job to begin on machine 1 at time zero. 2. Choose any unscheduled job and schedule it as early as possible on the same machine. Repeat this step until the machine is occupied beyond time M* (or until all jobs are scheduled). 3. Reassign the processing scheduled beyond M* to the next machine instead, starting at time zero. Return to Step 2. • Exercise • Three machines, jobs with processing times of 1, 2, 3, 4, 5, 6, 7, 8 • M* ? Optimal schedule?

5. Minimizing the Makespan • Nonpreemptable jobs -- Pm || Cmax • NP-hard • PARTITION reduces to Pm || Cmax  PARTITION -- NP-hard in ordinary sense • Given positive integers a1, ..., at , is there a subset X {1, ..., t} such that 2iXai = a1 + ... + at ? • Branch and bound? Not easy to obtain good lower bound • Dynamic programming? Extremely large number of states, commonly • List scheduling -- a plausible way • Procedure 1. Construct a list of jobs. 2. Remove the first job from the list and place it in the schedule as early as possible. 3. Repeat Step 2 without changing the existing partial schedule. • Dispatching mechanism for real-time decision, possibly • Optimal schedule can always be produced by some list scheduling! (why?) • Search space reduction to n!

6. Minimizing the Makespan • Nonpreemptable jobs (cont’d) • Performance guarantee • A (worst-case) bound on the performance of a particular solution procedure • General form with error bound r • z rz* • where z -- objective from heuristic, z* -- that of optimal solution • Theorem 1 • List scheduling for independent, nonpreemptable jobs yields a makespan satisfying M/M*  2 – 1/m. • Proof of Theorem 1 • Consider a schedule by a list scheduling with makespan M. • Let k be a job that finishes at time M (start time of the job: M – pk). • At time (M – pk), all m machines must have been occupied since time zero. • Hence, m(M – pk)  j=1..npj – pk  M  j=1..npj/m + pk(m – 1)/m. • j=1..n pj/m  M* (from results on preemptable case) • pk  M* • Therefore, M  j=1..npj/m + pk(m – 1)/m  M*/m + M*(m – 1)/m.

7. Minimizing the Makespan • Nonpreemptable jobs (cont’d) • LPT list scheduling • List scheduling by longest processing time (LPT) • Theorem 2 (Graham, 1969) • LPT list scheduling for independent, nonpreemptable jobs yields a makespan satisfying M/M*  4/3 – 1/3m. • Proof of Theorem 2 (adopted from Pinedo, 2008) • Suppose there exist counterexamples such that 4/3 – 1/3m  M/M*. • Then, there must exist an example with the smallest number of jobs. • Consider this smallest one and assume it has n jobs and job 1 is shortest. • Then, C1 = M. Otherwise, a counterexample with fewer jobs exists. (why?) • Since all other machines are busy at start of job 1, C1 – p1 = M – p1  j=2..npj/m. • Hence, M  p1 + j=2..npj/m = p1(1 – 1/m) + j=1..npj/m. • Since M*  j=1..npj/m,4/3 – 1/3m  M/M*  p1(1 – 1/m)/M* + j=1..npj/mM*  p1(1 – 1/m)/M* + 1. • It leads to M*  3p1 , which implies that the optimal schedule may results in at most two jobs on each machine. (why?) • However, LPT scheduling is optimal in that case (why?), which is a contradiction.

8. Minimizing the Makespan • Nonpreemptable jobs (cont’d) • First-fit decreasing (FFD) • To check whether makespan M is possible or not • Procedure 1. Order the jobs according to LPT 2. Attempt to assign the first job on the list to the first machine on which the job will fit (i.e., it completes on or before M). If no such machine exists, report FAIL. 3. Repeat Step 2 until all jobs have been scheduled. • Sometimes it fails, when a feasible schedule actually exists.  Determining whether M is valid is no easier than solving the makespan problem itself  It is from bin packing problem  Best-fit decreasing  LPT list scheduling • Exercise • 2 parallel machines, jobs with processing times 3, 3, 2, 2, 2, 2 • M by LPT list scheduling? Result of FFD using that M?

9. Minimizing the Makespan • Nonpreemptable jobs (cont’d) • Multifit algorithm • Determining schedule by searching for the smallest feasible value of M, using FFD • Min. trial M -- LB from preemptable case • Max. trial M -- max[2j=1..n pj/m, max{pj}], or makespan of any feasible schedule • Theorem 3 • The multifit algorithm for independent, nonpreemptable jobs yields a makespan satisfying M/M* 72/61, or about 1.18.

10. Minimizing the Makespan • Dependent nonpreemptable jobs -- Pm | prec | Cmax • Complexity • Strongly NP-hard, generally • Special cases • P1 | prec | Cmax • Easy (why?) • P | prec | Cmax • Project scheduling, polynomially solvable by CPM (critical path method) • Pm| pj=1,intree | Cmax , Pm| pj=1,outtree | Cmax • P2| pj=1,prec | Cmax intree (assembly tree) outtree

11. Minimizing the Makespan • Dependent nonpreemptable jobs (cont’d) • Intree and unit-length jobs -- Pm| pj=1,intree | Cmax • Critical path (CP) rule gives optimal schedule • To give highest priority to the job at the head of the longest string of jobs in the precedence graph (ties may be broken arbitrarily) • A.k.a. highest level first rule  Equivalent to ALGORITHM 2 + scheduling phase • Proof of optimality -- see p. 120 of Pinedo (2008)  Analogous to LPT list scheduling • Exercise • Three parallel machines, following intree jobs with unit processing time 12 17 13 8 5 2 14 9 6 3 1 15 10 7 4 16 11

12. Minimizing the Makespan • Dependent nonpreemptable jobs (cont’d) • Arbitrary precedence with two machines -- P2 | pj=1,prec | Cmax • Critical path (CP) rule using labels of direct successors • To give priority to the job with the lexicographically smallest sequence of direct successors’ labels in nonincreasing order  Equivalent to ALGORITHM 3 + scheduling phase • Exercise • Three parallel machines, unit-length jobs with following precedence 6 5 3 4 0 7 2 1

13. Minimizing the Makespan • Preemptable jobs • Arbitrary precedence with two machines -- P2 | prmp, prec | Cmax • A possible approach • Transforming to P2 | pj=1,prec | Cmax with chains • Example • Guarantee optimal? • No! The jobs must be divided into half-unit-length jobs. 5 6 7 1 1 1 2 2 2 5 6 7 5 6 7 1 1 1 4 4 1 1 1 2 3 1 2 3 1 1 1 1 1 1

14. Minimizing Total Flowtime • Minimizing F -- Pm|| Cj • Notation • pi[j] -- processing time of the jth job in sequence on the ith machine • Fi[j] -- flowtime of the jth job in sequence on the ith machine • nj -- number of jobs processed by the ith machine • Objective function • F = i=1..mj=1..niFi[j] = i=1..mj=1..ni(ni – j + 1)pi[j] • Solution approach • F as scalar product of two vectors • Matching integer coefficients (ni – j + 1) with processing times pi[j] • F cannot be minimized unless ni differ by at most one. (why?) • SPT list scheduling • Dispatching procedure and adaptable dynamic job arrivals • Exercise • Two parallel machines and six jobs with processing times of 1, 2, 3, 4, 5, 6 • Minimizing Fw -- Pm|| wjCj • NP-hard

15. Summary • More than one machine • Allocation and sequencing • In case of makespan problems, allocation may be enough • Two-phase method, commonly, because of complexity • Allocation first, sequencing next considering as single machine • LPT list scheduling • Applicable to some situations • No precedence, intree, critical path • Flowtime measures • Total tardiness problems • Very limited optimization approaches