Dr hab in prof nadzw pwr dorota kuchta http www ioz pwr wroc pl pracownicy kuchta
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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 in prof nadzw pwr dorota kuchta http www ioz pwr wroc pl pracownicy kuchta

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

Operations Scheduling and Production – Activity Control





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 johnson s rule
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 rule1
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




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












Scheduling consecutive days off
Scheduling Consecutive Days Off Lynwood’s Job Shop


Scheduling consecutive days off1
Scheduling Consecutive Days Off Lynwood’s Job Shop


Scheduling consecutive days off2
Scheduling Consecutive Days Off Lynwood’s Job Shop


Scheduling consecutive days off3
Scheduling Consecutive Days Off Lynwood’s Job Shop


Vehicle scheduling
Vehicle Scheduling Lynwood’s Job Shop


Vehicle scheduling1
Vehicle Scheduling Lynwood’s Job Shop


Vehicle scheduling2
Vehicle Scheduling Lynwood’s Job Shop

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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