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
JOBPROCESSING TIME
(HOURS)
1 8
2 4
3 7
4 3
5 6
6 5
JOBPROCESSING TIME
(HOURS)
4 3
2 3 + 4 = 7
6 7 + 5 = 12
5 12 + 6 = 18
3 18 + 7 = 25
1 25 + 8 = 33
Common Scheduling Criteria
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
Scheduling with Due Dates
JOBPROCESSING
TIMEDUE
DATE
1 4 15
2 7 16
3 2 8
4 6 21
5 3 9
JOB FLOWTIME TARDINESS
1 4 0
2 4 + 7 = 11 0
3 11 + 2 = 13 5
4 13 + 6 = 19 0
5 19 + 3 = 22 13
average 13.8 3.6
JOB DUE DATELATESTSTART
1 15 11
2 16 9
3 8 6
4 21 15
5 9 6
JOB FLOWTIME DUE DATE TARDINESS
3 2 8 0
5 2 + 3 = 5 9 0
1 5 + 4 = 9 15 0
2 9 + 7 = 16 16 0
4 16 + 6 = 22 21 1
Minimalizacja liczby opóźnionych elementów
1. Wstawić 1. zadanie do ciągu S2. 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 kolejnezadanie do ciągu, krok 2. W przeciwnym przypadku stop
Proc. 2 4 1 2 3 1
due 3 5 6 6 7 8
Two – Machine Flowshop Problem
JOB SHEAR PUNCH
1 4 5
2 4 1
3 10 4
4 6 10
5 2 3
10 20 30 40
1 2 3 4 5
1 2 3 4 5Shear
Punch
Days
Job number
Two – Machine Flowshop Problem
JOB FLOWTIME
1 9
2 10
3 22
4 34
5 37
Two – Machine Flowshop Problem (Johnson’s Rule)
JOB SHEAR PUNCH
1 4 5
2 4 1
3 10 4
4 6 10
5 2 3
Since the minimum time is on the second machine, job 2 is scheduled last:__ __ __ __ 2Next, 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 __ __ ___ 2In 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 have5 1 __ __ 2Continuing with Johnson’s rule, the last two steps yield:5 1 __ 3 25 1 4 3 2
Two – Machine Flowshop Problem (Johnson’s Rule)
10 20 30 40
Shear
Punch
Days
5 1 4 3 2
5 1 4 3 2
Since the minimum time is on the second machine, job 2 is scheduled last:__ __ __ __ 2Next, 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 __ __ ___ 2In 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 have5 1 __ __ 2Continuing with Johnson’s rule, the last two steps yield:5 1 __ 3 25 1 4 3 2
Job Data for Lynwood’s Job Shop
JOB ARRIVAL
TIMEPROCESSING SEQUENCE
(PROCESSING TIME)
1 0 L(10) - D(20) - G(35)
2 0 D(25) - L(20) - G(30) - M(15)
3 20 D(10) - M(10)
4 30 L(15) - G(10) - M(20)
Shop Status at Time T
kk
j
35
Drill
Current time =T
Grinder Milling machine
Lathe
i
t1 t2
Completion time of job i Completion time of job j
Job i being processed Job j being processed Job k waiting Machine idle
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
LP. RULE TYPE DESCRIPTION
1 Earliest release date Static Time job is released to the shop
2 Shortest processing time Static Processing time of operation for which job is waiting
3 Total work Static Sum of all processing times
4 Earliest due date Static Due date of job
5 Least work remaining StaticSum of all processing times for oparations not yet performed
6Fewest operations remaining Static Number of operations yet to be performed
7 Work in next queue DynamicAmount of work awaiting the next machine in a job's processing time
8 Slack time DynamicTime remaining until due date minus remaining processing time
9 Slack/ remaining operations Dynamic Slack time divided by the number of operations remaining
10 Critical ratio DynamicTime remaining until due date divided by days required to complete job
Simulation of Dispatching Rules (lwr)
21
10 25
Drill
Current time = 0
Grinder Milling
machine
11
2
25
Drill
Current time = 10
Grinder Milling
machine Lathe
Simulation of Dispatching Rules
11
2
25
Drill
Current time = 20
Grinder Milling
machine Lathe
33
11
3
35
Drill
Current time = 25
Grinder Milling
machine Lathe
2
45
Simulation of Lynwood Manufacturing Problem
11
3
35
Current time = 30
2
44
45
1
55
Current time = 35
2
44
45 45
3
Simulation of Lynwood Manufacturing Problem
1
55
Current time = 45
4
60 75
2
2
35
Current time = 55
4
60
11
75
Simulation of Lynwood Manufacturing Problem
2
Current time = 60
11 44
75
11
4
85
Drill
Current time = 75
Grinder Milling machine
Lathe
2
90
Simulation of Lynwood Manufacturing Problem
44
1
120
Current time = 85
2
90
1
120
Current time = 90
4
110
Simulation of Lynwood Manufacturing Problem
1
120
Current time = 110
Drill
Current time = 120
Grinder Milling
machine Lathe
Bar Chart for Lynwood’s Job Shop
10 30 50 70 90 110
Mill
Grind
Drill
Lathe
3 2 4
2 4 1
2 3 1
1 2 4
Simulation Results Using Least Work Remaining for Lynwood’s Job Shop
JOB WAITING TIME COMPLETION TIME MACHINE IDLE TIME (MAKESPAN - PROCESSING TIME)
1 55 120 Lathe 75
2 0 90 Drill 65
3 25 45 Grind 45
4 65 110 Mill 75
Scheduling Consecutive Days Off
M T W T F S S
8 6 6 6 9 5 3
EMPLOYEE NO. M T W T F S S
1 8 6 6 6 9 5 3
M T W T F S S
7 5 5 5 8 5 3
EMPLOYEE NO. M T W T F S S
2 7 5 5 5 8 5 3
Scheduling Consecutive Days Off
M T W T F S S
6 4 4 4 7 5 3
EMPLOYEE NO. M T W T F S S
3 6 4 4 4 7 5 3
M T W T F S S
5 4 4 3 6 4 2
Scheduling Consecutive Days Off
EMPLOYEE NO. M T W T F S S
4 5 4 4 3 6 4 2
5 4 3 3 2 5 4 2
6 3 2 3 2 4 3 1
7 2 1 2 1 3 3 1
8 2 1 1 0 2 2 0
9 1 0 1 0 1 1 0
10 1 0 0 0 0 0 0
Scheduling Consecutive Days Off
EMPLOYEE M T W T F S S
1 x x x x x
2 x x x x x
3 x x x x x
4 x x x x x
5 x x x x x
6 x x x x x
7 x x x x x
8 x x x x x
9 x x x x x
10 x x x x x
Total 8 6 6 8 10 6 6
Vehicle Scheduling
Customer
0 1 2 3 4 5 6 7
0 -
1 20 -
2 57 51 -
3 51 10 50 -
4 50 55 20 50 -
5 10 25 30 11 70 -
6 15 80 10 90 60 50 -
7 90 53 47 38 10 90 12 -
Vehicle Scheduling
1 2 3 4 5 6 7
1 -
2 26 -
3 61 58 -
4 15 87 51 -
5 5 37 50 -10 -
6 -45 62 -24 5 -25 -
7 57 100 103 130 10 93 -
Vehicle Scheduling
ROUTE TIME
0 - 4 - 7 - 0 150
0 - 3 - 1 - 0 81
0 - 2 - 5 - 0 97
0 - 6 - 0 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.
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