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Online Scheduling with Reordering

Online Scheduling with Reordering. Matthias Westermann RWTH Aachen Joint work with Matthias Englert and Deniz Özmen. Minimum makespan scheduling. makespan. Input sequence of jobs with processing times. Scheduling algorithm assigns the jobs to m machines.

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Online Scheduling with Reordering

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  1. Online Scheduling with Reordering Matthias Westermann RWTH Aachen Joint work with Matthias Englert and Deniz Özmen

  2. Minimum makespan scheduling makespan • Input sequence of jobs with processing times. • Scheduling algorithm assigns the jobs to m machines. • Objective is to minimize the makespan. M. Westermann - RWTH Aachen

  3. Online scheduling without preemption • Online algorithm gets to know the input sequence job by job without knowledge about the future. • Classic model:Each arriving job has to be assigned immediately to one of the machines. M. Westermann - RWTH Aachen

  4. Our model • Reordering buffer can be used to reorder the input sequence: • Buffer contains the first k jobs of the input that are not assigned so far. • Scheduling algorithm assigns a job contained in the buffer to a machine. • Thereafter the next job in the input takes the place of the assigned job. input machines buffer size k M. Westermann - RWTH Aachen

  5. Competitive analysis • Comparison: • Online algorithm (without knowledge about the future) • Offline algorithm (knows the whole input in advance) • Online algorithm is c-competitive, if its makespan is at most c times the makespan of an optimal offline algorithm. M. Westermann - RWTH Aachen

  6. m identical machines- main results • Lower bound of rm, if the size of the buffer does not depend on the input sequence. • r2 = 4/3 • limm!1 rm = LambertW-1(-1/e2) /(1+LambertW-1(-1/e2)) ¼ 1.4659 • Scheduling algorithm matching the lower bound with a buffer of size d(1+2/rm)¢me+2. • 1+2/r2 = 2.5 • limm!1 1+2/rm¼ 2.36 smallest real solution to x¢ex = -1/e2 M. Westermann - RWTH Aachen

  7. m identical machines- values of rm M. Westermann - RWTH Aachen

  8. m identical machines- previous work M. Westermann - RWTH Aachen

  9. m identical machines- the lower bound of rm • Assume for contradiction that algorithm A achieves a competitive ratio r < rm with a buffer of size k. • 1/+k jobs of size  arrive. wi := min{rm/m, (rm-1)/i} rm is the solution to  wi = 1 M0 Mm-1 M. Westermann - RWTH Aachen

  10. m identical machines- the lower bound of rm • There exists a machine Mjwith load ¸ wj. • If wj = rm/m,no more jobs arrive. • The optimal makespan is · (1+k¢)/m+ = (1+(k+m)¢)/m. • A’s competitive ratio is ¸ rm/(1+(k+m)¢). M. Westermann - RWTH Aachen

  11. m identical machines- the lower bound of rm • If wj = (rm-1)/j, (m-j) large jobs of size 1/j arrive. • The optimal makespan · (1+(k+j)¢)/j. • If A schedules two large jobs on the same machine, A’s competitive ratio is ¸ 2/(1+(k+j)¢) . • Otherwise, i.e., A schedules at least one of the large jobs on a machine with load ¸ (rm-1)/j, A’s competitive ratio is ¸ rm/(1+(k+j)¢) . M. Westermann - RWTH Aachen

  12. m identical machines- the upper bound of rm • When a new job arrives: • Store this job in the buffer and remove a job J of smallest size from the buffer. • Schedule J on a machine Mi with load · wi¢(T+m¢L(J))-L(J). total scheduled load size of J Assume for contradiction that such a machine does not exist. Then T >  (wi¢(T+m¢L(J))-L(J)) = T. M. Westermann - RWTH Aachen

  13. m identical machines- the upper bound of rm • After all jobs have arrived: • Schedule virtually some of the remaining jobs on m empty machines according to LPT. Abort just before more than two jobs should be assigned to the same machine. • Schedule the remaining jobs according to Greedy. M0 Mm-1 M. Westermann - RWTH Aachen

  14. m identical machines- the upper bound of rm buffer size 3m: total load · m¢OPT, since ¸ m-1 jobs remain in the buffer wi¢(T+(m-1)¢Lmin) size of each remaining job · OPT/3 · (rm-1)¢OPT average load · OPT smallest size of all remaining jobs M0 Mm-1 M. Westermann - RWTH Aachen

  15. m identical machines- further results • 2 identical machines • Scheduling algorithm achieving the optimal competitive ratio r2 with a buffer of size 2. • Small buffer sizes • Lower bound of 3/2 > rm, if the buffer size is at most bm/2c. • Lower bound of 1+1/21/2¼ 1.7071, if the buffer size is at most bm/8c and m ¸ 8. M. Westermann - RWTH Aachen

  16. m related machines • Scheduling algorithm achieving the competitive ratio 2+ with a buffer of size m. M. Westermann - RWTH Aachen

  17. Open questions • Our algorithm for identical machines achieves the optimal competitive ratio. What buffer size is necessary to obtain this result? bm/2c· … ·d(1+2/rm)¢me+2 • Can our result for related machines be improved or can a better lower bound be shown in this case? • Reordering for other scheduling problems? M. Westermann - RWTH Aachen

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