1 / 29

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

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' dr hab. inż., prof. nadzw. PWR Dorota Kuchta ioz.pwr.wroc.pl/Pracownicy/Kuchta/' - jorden-kennedy

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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

Operations Scheduling and Production – Activity Control

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

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

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

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 Lynwood’s Job Shop

Scheduling Consecutive Days Off Lynwood’s Job Shop

Scheduling Consecutive Days Off Lynwood’s Job Shop

Scheduling Consecutive Days Off Lynwood’s Job Shop

Vehicle Scheduling Lynwood’s Job Shop

Vehicle Scheduling Lynwood’s Job Shop

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.