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 OffM 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 OffM 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 30 111 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 -45 621 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.