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


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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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dr hab. inż., prof. nadzw. PWR Dorota Kuchtahttp://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

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


Two – Machine Flowshop Problem


Two – Machine Flowshop Problem


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


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


Job Data for Lynwood’s Job Shop


Shop Status at Time T


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


Priority Dispatching Rules for Job Shops


Simulation of Dispatching Rules (lwr)


Simulation of Dispatching Rules


Simulation of Lynwood Manufacturing Problem


Simulation of Lynwood Manufacturing Problem


Simulation of Lynwood Manufacturing Problem


Simulation of Lynwood Manufacturing Problem


Simulation of Lynwood Manufacturing Problem


Bar Chart for Lynwood’s Job Shop


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


Scheduling Consecutive Days Off


Scheduling Consecutive Days Off


Scheduling Consecutive Days Off


Scheduling Consecutive Days Off


Vehicle Scheduling


Vehicle Scheduling


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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