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Scheduling

Chapter 6. Scheduling. Production Planning & Control (PPC). Produktionsplanung & Steuering (PPS) Production planning MPS MRP Lotsizing Capacity checks … Production control Job release Scheduling. Job Release. Flow time = completion time – release time

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Scheduling

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  1. Chapter 6 Scheduling Summarizedfromcourse „EK ABWL Produktion & Logistik“ and „KFK Production Analysis“

  2. ProductionPlanning & Control (PPC) • Produktionsplanung & Steuering (PPS) • Productionplanning • MPS • MRP • Lotsizing • Capacitychecks… • Productioncontrol • Job release • Scheduling

  3. Job Release Flow time = completion time – release time • If jobs are released too early (e.g. based on historical data on flow times + some safety times) • Long queues before machines (esp. bottlenecks) • Long observed flow times • Jobs are released even earlier • Even longer queues before machines • Even longer observed flow times • Etc. … viciouscircle

  4. Job release • To break this vicious circle  systematic job release • Basic idea: „release job only when the utimization of the sysstem is below a certain threshold“ • 2 simple strategies: • CONWIP = CONstant Work In Processsimple, used in US literature„keep „work in process“ (WIP) constant, i.e. release a new job eas soon assome job is finished • „belastungsorientierteAuftragsfreigabe“ (BOA, BORA)(„utilitazion based job release“)more complicated measure for utilization of the systemmainly German literature

  5. Scheduling • After job release there are queues of jobs in front of the various machiens • Decide in which order these are processed (and also the starting times)  scheduling • Various objectives are reasonable, e.g. • Minimize total tardiness, • Minimize cycle time (max completion time) Z • Minimize average (or total) flow time D.  Unfortunately most objectives are conflicting!

  6. Single MachineScheduling • Scheduling is typically np-hard • Single machine scheduling is relatively easy • Many problems can be solved to optimality (exact algorithm) using some simple priority rules (list scheduling) Examples: • Minimize maximum tardiness • Minimize cycle timeZ • Minimize average flow time D

  7. Tardiness • Given: due date (desired completion time) • Observed after scheduling: actual completion time • Lateness = completion time - due date • Can be positive or negative • If lateness > 0 … job is late Tardiness = max {0, lateness} • If lateness < 0 … job is earlyEarliness = max {0, -lateness} • Simple rule good for all tardiness related objectives: • Earliest Due Date (EDD) rule QEM- MgmtSci Chapter 6/7

  8. Minimize Maximum Tardiness overview • EDD rule, alwaysschedulejobwithearliest due datefirst • Is an exactalgorithmformaximumtardiness • Example: 2. 8 - 2 1. - 19 1 4. 3. 11 - 5. 26 5 Completiontimes?Tardiness?  Gantt chart Optimal sequence: BADCE QEM- MgmtSci Chapter 6/8

  9. EDD Example – Gantt Chart C E B A D 2 11 26 19 8 8 6 15 18 21 V = maximumtardiness= 5 5+1 = 6 T = total tardiness = #T = numberoftardyjobs = 2 = 13,2 (8+2+19+11+26)/5 D = averageflow time = cycle time

  10. Other ObjectivesRelatedtoTardiness • Minimizesumof all tardinesses(total tardiness)in aboveexample: total tardiness = 1 + 5 = 6typicallynp-hard (at least ifobjectiveisweighted)EDD ruleis a goodheuristic • Minimizenumberoftardyjobsin aboveexample: numberoftardyjobs= 2 • Can besolvedtooptimalitywithHodgson‘salgorithm: • Apply EDD rule • If a jobislate removescheduledtaskwiththelongestoperation time • All removedtaskareschedules last and will betardy • All scheduled (and not removed) tasks will be on time

  11. Minimize Cycle Time Z • Cycle Time (makespan) = maximum completion time of all jobs • „Minimize Cycle Time“ is typically conflicting with all other objectives • Single machine scheduling:problem is trivial: every schedule without idle times is optimal • Above Example: Z is always 26, unless idle times are scheduled

  12. MinimizeAverage Flow Time overview • Optimal Solution: SPT rule (shortestprocessingtime) 3. 11 3 - 2 1. 5. 26 8 2. 5 - 4. 18 - BDAE C Completiontimes?Tardiness?  Gantt chart Optimal sequence: QEM Chapter 6/12

  13. SPT Example – Gantt Chart E D B A C 18 26 2 11 5 V = maximumtardiness = 8 ( > 5 with EDD) = 12,4 D = averageflow time = ( < 13,2) (11+2+26+5+18)/5 • maxtardinessandaverageflow time areconflicting

  14. SchedulingWith Multiple Machines Main classes of problems: • Flow shop … same sequence of machines for all jobs (but sequence of jobs at machines can be different – to be optimized) • Permutation Flow shop … no overtaking is possible (same sequence of jobs at all machines – to be optimized) • Job shop … each job can have a different required sequence of machines • Open shop …required sequence of machines is free

  15. SchedulingWith Multiple Machines • Each job must be processed on several machines • Typically np-hard  heuristic solution • SPT: shortest (total) processing timegood for average flow time (optimal for single machine) • SOT: shortest operation timegood for average flow time (optimal for single machine) • SRPT: shortest remaining processing timetotal processing time on remaining machinesoften best for average flow time • LPT, LOT: longest processing/operation time sometimes good for cycle time Z • EDD: (earliest due date) – all due date related objectives • Critical ratio: clever variant of EDD rule(remaining time to due date)/(remaining processing time)

  16. SchedulingWithTwo Machines • 2 machines • flow shop (same sequence of machines for all jobs) • Objective “minimize cycle time”  Optimal solution is a permutation schedule: Johnson Algorithm: 1. Find smallest element in table of operation times. If this occurs with machine 1, schedule on first available position, otherwise schedule on last available position. 2. Delete schedules jobs from list of open jobs Example

  17. Example - Johnson Algorithm Given: 5 jobs ans 2 machines M1 and M2 Minimize cycle time! 2 35 7 7 14 3 9 9 33 12 20 4 29 17 33 19 22 45 7 [  ] Optimal sequence: B E D C A flowtimes, cycle time? EK Produktion & Logistik Kapitel 10/17

  18. Example – Gantt Chart 3 9 10 33 35 22 29 E A B C D B D A E C 20 33 28 3 10 Job B canstart on machine M2, assoonasfinished on M1 (M2 idle) Job E canstart on machine M2, assoonasfinished on M1 (M2 idle) Job D cannotstart on M2 assoonasfinished on M1 fertig (M2 occupied) Z = cycle time = 35 ZE = 25,6 D = averageflow time = (9+22+29+33+35)/5

  19. Example – SRPT Rule • Apply SRPT rule ( trytoget a shorteraverageflow time) B sequence: [  ] A C D E 45 8 20 5 7 16 14 26 33 B C E D A A B D C E 5 8 16 26 33 ( > 35 Johnson) Z = cycle time = 45 ZE ( < 25,6 Johnson) D = (7+14+20+33+45)/5 = 23,8

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