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

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dr hab. inz., prof. nadzw. PWR Dorota
Kuchtahttp//www.ioz.pwr.wroc.pl/Pracownicy/Kucht
a/

• Operations Scheduling and Production Activity
  Control

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

The optimal sequence is 3 - 5 - 1 - 2 - 4.
The optimal sequence is 3 - 5 - 1 - 2 - 4.
JOB PROCESSING TIME DUE 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 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
JOB DUE DATE LATEST START
1 15 11
2 16 9
3 8 6
4 21 15
5 9 6
5
Minimalizacja liczby opóznionych elementów

• Wstawic 1. zadanie do ciagu S
• Jesli koniec wykonania ciagu przypada po terminie
  
• wykonania jego ostatniego elementu, wyrzucic
• najdluzszy element ciagu poza ciag
• 3. Jesli jeszcze sa zadania nie ustawione,
  wstawic kolejne
• zadanie do ciagu, krok 2. W przeciwnym przypadku
  stop

Proc. 2 4 1 2 3 1
due 3 5 6 6 7 8
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Two Machine Flowshop Problem
JOB SHEAR PUNCH
1 4 5
2 4 1
3 10 4
4 6 10
5 2 3
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Two Machine Flowshop Problem

JOB FLOWTIME
1 9
2 10
3 22
4 34
5 37
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Two Machine Flowshop Problem (Johnsons 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 Johnsons rule, the last two steps yield 5
1 __ 3 2 5 1 4 3 2
JOB SHEAR PUNCH
1 4 5
2 4 1
3 10 4
4 6 10
5 2 3
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Two Machine Flowshop Problem (Johnsons 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 Johnsons rule, the last two steps yield 5
1 __ 3 2 5 1 4 3 2
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Job Data for Lynwoods Job Shop
JOB ARRIVAL TIME PROCESSING 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)
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Shop Status at Time T
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Open shop dla dwóch maszyn
Wyznaczyc najkrótszy czas i odpowiedni element
umiescic na tej maszynie, gdzie ten czas jest
dluzszy (na tej drugiej) Uzupelnic szeregi na
kazdej maszynie w tej samej kolejnosci, co
poprzednie
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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 Static Sum of all processing times for oparations not yet performed
6 Fewest operations remaining Static 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 Dynamic Time 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 Dynamic Time remaining until due date divided by days required to complete job
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Simulation of Dispatching Rules (lwr)
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Simulation of Dispatching Rules
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Simulation of Lynwood Manufacturing Problem
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Simulation of Lynwood Manufacturing Problem
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Simulation of Lynwood Manufacturing Problem
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Simulation of Lynwood Manufacturing Problem
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Simulation of Lynwood Manufacturing Problem
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Bar Chart for Lynwoods Job Shop
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Simulation Results Using Least Work Remaining for
Lynwoods 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
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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
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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
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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
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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
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Vehicle Scheduling
Customer Customer Customer Customer Customer Customer Customer Customer 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 -
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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 -
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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.