Download presentation

Presentation is loading. Please wait.

Published byOlivia Chandler Modified over 3 years ago

1
**Operations Scheduling and Production – Activity Control**

dr hab. inż., prof. nadzw. PWR Dorota Kuchta Operations Scheduling and Production – Activity Control

2
**PROCESSING TIME (HOURS)**

Job Shop Scheduling JOB PROCESSING TIME (HOURS) 1 8 2 4 3 7 5 6 JOB PROCESSING TIME (HOURS) 4 3 2 3 + 4 = 7 6 7 + 5 = 12 5 = 18 = 25 1 = 33

3
**Common Scheduling Criteria**

DEFINITION OBJECTIVES 1. Makespan Time to process a set of jobs Minimize makespan 2. Flowtime Time a job spends in the shop Minimize average flowtime 3. Tardiness The amount by which completion Minimize number of tardy jobs time exceeds the due date of a job Minimize the maximum tardiness

4
**Scheduling with Due Dates**

The optimal sequence is The optimal sequence is Scheduling with Due Dates JOB PROCESSING TIME DUE DATE 1 4 15 2 7 16 3 8 6 21 5 9 JOB FLOWTIME TARDINESS 1 4 2 4 + 7 = 11 3 = 13 5 = 19 = 22 13 average 13.8 3.6 JOB FLOWTIME DUE DATE TARDINESS 3 2 8 5 2 + 3 = 5 9 1 5 + 4 = 9 15 9 + 7 = 16 16 4 = 22 21 JOB DUE DATE LATEST START 1 15 11 2 16 9 3 8 6 4 21 5

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 Proc. 2 4 1 3 due 5 6 7 8

6
**Two – Machine Flowshop Problem**

JOB SHEAR PUNCH 1 4 5 2 3 10 6

7
**Two – Machine Flowshop Problem**

JOB FLOWTIME 1 9 2 10 3 22 4 34 5 37

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 JOB SHEAR PUNCH 1 4 5 2 3 10 6

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

10
**Job Data for Lynwood’s Job Shop**

ARRIVAL TIME PROCESSING SEQUENCE (PROCESSING TIME) 1 L(10) - D(20) - G(35) 2 D(25) - L(20) - G(30) - M(15) 3 20 D(10) - M(10) 4 30 L(15) - G(10) - M(20)

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**

LP. RULE TYPE DESCRIPTION 1 Earliest release date Static Time job is released to the shop 2 Shortest processing time Processing time of operation for which job is waiting 3 Total work Sum of all processing times 4 Earliest due date Due date of job 5 Least work remaining Sum of all processing times for oparations not yet performed 6 Fewest operations remaining Number of operations yet to be performed 7 Work in next queue Dynamic Amount of work awaiting the next machine in a job's processing time 8 Slack time Time remaining until due date minus remaining processing time 9 Slack/ remaining operations Slack time divided by the number of operations remaining 10 Critical ratio Time remaining until due date divided by days required to complete job

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**

WAITING TIME COMPLETION TIME MACHINE IDLE TIME (MAKESPAN - PROCESSING TIME) 1 55 120 Lathe 75 2 90 Drill 65 3 25 45 Grind 4 110 Mill

23
**Scheduling Consecutive Days Off**

M T W F S 8 6 9 5 3 EMPLOYEE NO. M T W F S 1 8 6 9 5 3 M T W F S 7 5 8 3 EMPLOYEE NO. M T W F S 2 7 5 8 3

24
**Scheduling Consecutive Days Off**

M T W F S 6 4 7 5 3 EMPLOYEE NO. M T W F S 3 6 4 7 5 M T W F S 5 4 3 6 2

25
**Scheduling Consecutive Days Off**

EMPLOYEE NO. M T W F S 4 5 3 6 2 1 7 8 9 10

26
**Scheduling Consecutive Days Off**

EMPLOYEE M T W F S 1 x 2 3 4 5 6 7 8 9 10 Total

27
**Vehicle Scheduling Customer 1 2 3 4 5 6 7 - 20 57 51 10 50 55 25 30 11**

1 2 3 4 5 6 7 - 20 57 51 10 50 55 25 30 11 70 15 80 90 60 53 47 38 12

28
**Vehicle Scheduling 1 2 3 4 5 6 7 - 26 61 58 15 87 51 37 50 -10 -45 62**

1 2 3 4 5 6 7 - 26 61 58 15 87 51 37 50 -10 -45 62 -24 -25 57 100 103 130 10 93

29
**Vehicle Scheduling ROUTE TIME 0 - 4 - 7 - 0 150 0 - 3 - 1 - 0 81**

97 30 The total time required is reduced to 358 minutes, or about 6 hours, a savings of about 3.8 hours over the original schedule.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google