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dr hab. inż., prof. nadzw. PWR Dorota Kuchta ioz.pwr.wroc.pl/Pracownicy/Kuchta/

dr hab. inż., prof. nadzw. PWR Dorota Kuchta http://www.ioz.pwr.wroc.pl/Pracownicy/Kuchta/. Operations Scheduling and Production – Activity Control. Job Shop Scheduling. Common Scheduling Criteria. Scheduling with Due Dates. Minimalizacja liczby opóźnionych elementów.

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dr hab. inż., prof. nadzw. PWR Dorota Kuchta ioz.pwr.wroc.pl/Pracownicy/Kuchta/

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  1. dr hab. inż., prof. nadzw. PWR Dorota Kuchtahttp://www.ioz.pwr.wroc.pl/Pracownicy/Kuchta/ Operations Scheduling and Production – Activity Control

  2. Job Shop Scheduling

  3. Common Scheduling Criteria

  4. Scheduling with Due Dates

  5. Minimalizacja liczby opóźnionych elementów Wstawić 1. zadanie do ciągu S Jeśli koniec wykonania ciągu przypada po terminie wykonania jego ostatniego elementu, wyrzucić najdłuższy element ciągu poza ciąg 3. Jeśli jeszcze są zadania nie ustawione, wstawić kolejne zadanie do ciągu, krok 2. W przeciwnym przypadku stop

  6. Two – Machine Flowshop Problem

  7. Two – Machine Flowshop Problem

  8. Two – Machine Flowshop Problem (Johnson’s Rule) Since the minimum time is on the second machine, job 2 is scheduled last: __ __ __ __ 2 Next, we pick the second – smallest processing time. This is 2, which corresponds to job 5 on machine 1. Therefore, job 5 is scheduled first: 5 __ __ ___ 2 In the next step, we have a tie between job 1 on the shear and job 3 on the punch press. When ties occur, either job can be chosen. If we pick job 1, we then have 5 1 __ __ 2 Continuing with Johnson’s rule, the last two steps yield: 5 1 __ 3 2 5 1 4 3 2

  9. Two – Machine Flowshop Problem (Johnson’s Rule) Since the minimum time is on the second machine, job 2 is scheduled last: __ __ __ __ 2 Next, we pick the second – smallest processing time. This is 2, which corresponds to job 5 on machine 1. Therefore, job 5 is scheduled first: 5 __ __ ___ 2 In the next step, we have a tie between job 1 on the shear and job 3 on the punch press. When ties occur, either job can be chosen. If we pick job 1, we then have 5 1 __ __ 2 Continuing with Johnson’s rule, the last two steps yield: 5 1 __ 3 2 5 1 4 3 2

  10. Job Data for Lynwood’s Job Shop

  11. Shop Status at Time T

  12. Open shop dla dwóch maszyn Wyznaczyć najkrótszy czas i odpowiedni element umieścić na tej maszynie, gdzie ten czas jest dłuższy (na tej drugiej) Uzupełnić szeregi na każdej maszynie w tej samej kolejności, co poprzednie

  13. Priority Dispatching Rules for Job Shops

  14. Simulation of Dispatching Rules (lwr)

  15. Simulation of Dispatching Rules

  16. Simulation of Lynwood Manufacturing Problem

  17. Simulation of Lynwood Manufacturing Problem

  18. Simulation of Lynwood Manufacturing Problem

  19. Simulation of Lynwood Manufacturing Problem

  20. Simulation of Lynwood Manufacturing Problem

  21. Bar Chart for Lynwood’s Job Shop

  22. Simulation Results Using Least Work Remaining for Lynwood’s Job Shop

  23. Scheduling Consecutive Days Off

  24. Scheduling Consecutive Days Off

  25. Scheduling Consecutive Days Off

  26. Scheduling Consecutive Days Off

  27. Vehicle Scheduling

  28. Vehicle Scheduling

  29. Vehicle Scheduling The total time required is reduced to 358 minutes, or about 6 hours, a savings of about 3.8 hours over the original schedule.

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