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.