sequencing (models) - lecture

7/30/2019 Sequencing (Models) - Lecture

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Sequencing (Models)

Job Sequencing

What is best sequence : Based on objective

Assumptions :

(1) One operation on one machine at a time.

(2) Processing times known and constant.

(3) Processing times not depending on sequence.

(4) Idleness of job between any two machines = 0.

(5) Operation started must be completed.

(6) Only one machine of each type is available.


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[2] n jobs 2 machines

FSP (Flow Shop Problem)

JSP (Job Shop Problem)

[1] n jobs 1 machine

[3] n jobs 3 machines

Categories :

[4] n jobs m machines

[5] 2 jobs m machines


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n job 1 machine[1]

Data for this case :

Jobs (Ji) 1 2 3 4 5 6 7 8

Processing Time (Pi) 5 8 6 3 10 14 7 3

Due days (Di) 15 10 15 25 20 40 45 50

Priority (wi) 1 2 4 5 3 8 6 7


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Ji 1 2 3 4 5 6 7 8

Pi 5 8 6 3 10 14 7 3

Fi 5 13 19 22 32 46 53 56

If sequence is 1-2-3-4-5-6-7-8 then

Mean Flow Time (MFT) = Fi / No. of jobs = 246/8 = 30.75

Hence, one of the criterion of Optimization is MFT.


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Ji 1 2 3 4 5 6 7 8

Pi 5 8 6 3 10 14 7 3

Fi

Di 15 10 15 25 20 40 45 50

Li

5 13 19 22 32 46 53 56

-10 3 4 -3 12 6 8 6


Lavr= Li / No. of jobs = 26/8 = 3.25

Lmax = 12 for 5th job, No. of late jobs = 6

Hence, the other criteria of Optimization are MaximumLateness of a Job as well as Number of Late Jobs.


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Jobs

Processing Time (Pi)

Due days (Di)

Priority (wi)

1 2 3 4 5 6 7 8

5 8 6 3 10 14 7 3

15 10 15 25 20 40 45 50

1 2 4 5 3 8 6 7

Pi * Wi 5 16 24 15 30 112 42 21

Pi * Wi is weighted processing time

Hence, the other criterion of Optimization is Weighted

Processing Time.


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Criteria of Optimization

(A) Mean Flow Time (MFT)

(B) Maximum lateness

(C) Number of late jobs

(D) Priority


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Ji 1 2 3 4 5 6 7 8

Pi 5 8 6 3 10 14 7 3

Fi

Di 15 10 15 25 20 40 45 50

Li

5 13 19 22 32 46 53 56

-10 3 4 -3 12 6 8 6


MFT = Fi / No. of jobs = 246/8 = 30.75

Lavr= Li / No. of jobs = 26/8 = 3.25



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To minimize MFT, Shortest Processing Time

(SPT) rule is applied.

Ji

Pi

1 2 3 4 5 6 7 8

5 8 6 3 10 14 7 3

Applying SPT rule, the optimal sequence to minimize

MFT will be :

4-8-1-3-7-2-5-6

To get MFT, Lavr, Lmax and No. of late jobs :

(A)


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4 8 1 3 2 7 5 6

MFT = Fi/ No. of jobs = 191/8 = 23.87

Lavr= Li / No. of jobs = 29/8 =3.62


3 6 11 17 24 32 42 56Fi

Pi

Ji

Di 15 10 15 25 20 40 45 50

-12 - 4 - 4 - 8 4 - 8 - 3 6Li

3 3 5 6 7 8 10 14


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To minimize maximum lateness, Shortest Due

Days (SDD) rule is applied.

Ji

Pi

Di

1 2 3 4 5 6 7 8

5 8 6 3 10 14 7 3

15 10 15 25 20 40 45 50

Applying SDD rule, the optimal sequence to minimize

maximum lateness will be :

2-1-3-5-4-6-7-8

To get MFT, Lavr, Lmax and No. of late jobs :

(B)


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To minimize maximum lateness, Shortest Due

Days (SDD) rule is applied.

Ji

Pi

Fi

2 1 3 5 4 6 7 8

8 5 6 10 3 14 7 3

8 13 19 29 32 46 53 56

MFT = Fi / No. of jobs = 256/8 = 32

Lavr= Li / No. of jobs = 36/8 = 4.50


Di 10 15 15 20 25 40 45 50

Li 2 2 4 9 7 6 8 6

T i i i b f l t j b(C)


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To minimize number of late jobs :

Ji

Pi

Fi

Di

Li

2 1 3 5 4 6 7 8

8 5 6 10 3 14 7 3

8 13 19 29 32 46 53 56

10 15 15 20 25 40 45 50

2 2 4 9 7 6 8 6

Sequence for minimum (maximum lateness)


For Minimizing late jobs remove Job2 (Pi = 8)

(C)

The first job with lateness is 3rd Job (Job No. 3)

Out of the first three jobs Pij is Max. (=8)for 1st job

(Job No. 2)


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Ji

Pi

Fi

Di

Li

1 3 5 4 6 7 8

5 6 10 3 14 7 3

5 11 21 24 38 45 48

15 15 20 25 40 45 50

10 4 1 1 2 0 2


For Minimizing late jobs remove Job5 (Pi = 10)

The first job with lateness is 3rd Job (Job No. 5)

Out of the first three jobs Pij is Max. (=10)for 3rd job

(Job No. 5)


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Ji

Pi

Fi

Di

Li

1 3 4 6 7 8

5 6 3 14 7 3

5 11 14 28 35 38

15 15 25 40 45 50

10 4 11 12 10 12

Sequence for minimum (maximum lateness)

Hence, the optimal sequence to minimize late jobs will

be 1-3-4-6-7-8-2-5 or 1-3-4-6-7-8-5-2

No job is late.


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Ji

Pi

Fi

Di

Li

1 3 4 6 7 8 2 5

5 6 3 14 7 3 8 10

5 11 14 28 35 38 46 56

15 15 25 40 45 50 10 20

10 4 11 12 10 12 36 36

Sequence for minimizing No. of late jobs

Lmax = 36 for 2nd & 5th job, No. of late jobs = 2

Lavg= Li / No. of jobs = 13/8 = 1.62


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Based on priority(D)

Jobs

Processing Time (Pi)

Due days (Di)

Priority (wi)

1 2 3 4 5 6 7 8

5 8 6 3 10 14 7 3

15 10 15 25 20 40 45 50

1 2 4 5 3 8 6 7

Pi * Wi 5 16 24 15 30 112 42 21

Pi * Wi is weighted processing time

Based on weighted processing time the optimal sequence will be :

1 4 2 8 3 5 7 6

E i


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Jobs 1 2 3 4 5 6

Pi 5 8 3 10 4 7di 4 25 6 36 10 28

wi 2 3 4 6 1 5

Exercise

Jobs Pi di wi Jobs Pi di wi

1 10 11 2 1 3 6 4

2 7 9 4 2 6 20 13 3 35 1 3 9 30 5

4 15 29 3 4 5 12 3

5 6 42 6 5 7 25 2

6 9 50 5

[ 1 ]

[ 2 ] [ 3 ]


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n jobs 2 machines

Flow Shop Problem (FSP) Job Shop Problem (JSP)

[2]


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(a) Flow Shop Problem (FSP)

Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4

Elapsed Time for a Sequence is the duration

from starting of 1st Job to the end of last Job.

The criterion of Optimization is Elapsed Time.

T S Mi i i El d Ti


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Jobs M1

M2

1 8 2

2 5 10

3 3 9

4 11 6

5 7

6 12 4

Applying Johnsons rule :

11

5

To get Sequence to Minimize Elapsed Time,Johnsons Rule is applied.

Flow Shop Problem (FSP)


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Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4

5

2

1



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Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4

5 1

3

3



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Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4

5 3 16

4



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Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4

5 3 6 12

5



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Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4

5 3 2 6 1

Hence, Optimal Job Sequence is : 5 3 2 4 6 1

4


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How to get Elapsed Time (T) for any sequence ?

Elapsed Time for a Sequence is the total time fromstarting of first job to the completion of last job in asequence.

To get Elapsed (Makespan)Time for sequence 1-2-3-4-5-6

8

M1

1

10

M2

0

Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4


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8

M1

1

10

13

2

23

Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4


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8

M1

1

10

13

2

23

16

3

32

Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4


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8

M1

1

10

13

2

23

16

3

32

27

4

38

Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4


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8

M1

1

10

13

2

23

16

3

32

27

4

38

5

28

45

Jobs M1 M2

1 8 2

2 5 10

3 3 9

4 11 6

5 1 7

6 12 4


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8

M1

1

10

13

2

23

16

3

32

27

4

38

5

28

45

40

6

49

T = 49

Jobs M1 M2

1 8 2

2 5 10

3 3 94 11 6

5 1 7

6 12 4

Idleness of M1 = 9

Idleness of M2 = 11

Elapsed Time = 49


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For Optimal Job Sequence : 5 3 2 4 6 1

Find elapsed time for this optimal sequence.

Ans. : T = 42 days

Idleness of M1 = 2 days, Idleness of M2 = 4 days

E i


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Find the optimal sequence of jobs and elapsedtime for optimal sequence for the following n jobs

2 machines (FSP) problems. All jobs pass through

M1 M2 sequence. Timings are given in days.

Jobs 1 2 3 4 5 6

M1 7 5 1 10 6 3

M2 2 4 3 7 11 8

Exercise

(b) Job Shop Problem (JSP)


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(b) Job Shop Problem (JSP)

Job Machine for Processing time for

O1 O2 O1 O2

10 M2 M1 5 8

1 M1 M2 2 8

2 M1 -- 4 -

3 M2 M1 3 6

4 M1 M2 6 3

5 M2 -- 3 -

6 M2 M1 7 4

7 M1 M2 9 5

8 M2 -- 2 -

9 M1 M2 4 7

[ M1 ] [ 2 ]

[ M2 ] [ 5, 8 ]

[ M1

M2

] [ 1, 4, 7, 9 ]

[ M2 M1 ] [ 3, 6, 10 ]

Jacksons Rule :

On M1 [ M1 M2 ] [ M1 ] [ M2 M1 ] ..

On M2 [ M2 M1 ] [ M2 ] [ M1 M2 ] ..

To get Optimal Sequence of [ M M ] jobs :


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O1 O2 O1 O2

1 M1 M2 2 8

4 M1 M2 6 3

7 M1 M2 9 5

9 M1 M2 4 7

[ M1 M2 ] [ 1, 4, 7, 9 ] 1, 9, 7, 4

Applying Johnsons rule :

[ M1 M2 ] [ 1, 4, 7, 9 ]

To get Optimal Sequence of [ M1 M2 ] jobs :

1 9 7 4

To get Optimal Sequence of [ M2 M1 ] jobs


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To get Optimal Sequence of [ M2 M1 ] jobs :


O1 O2 O1 O2

10 M2 M1 5 8

3 M2 M1 3 6

6 M2 M1 7 4

[ M2 M1 ] [ 3, 6, 10 ] 3, 10, 6

Applying Johnsons rule : 3 610

[ M2 M1 ] [ 3, 6, 10 ]


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[ M1 ] [ 2 ] 2

[ M2 ] [ 5, 8 ] 5, 8

Jacksons Rule :

On M1 [ M1 M2 ] [ M1 ] [ M2 M1 ] ..

On M2

[ M2 M1 ] [ M2 ] [ M1 M2 ]

..

[ M1 M2 ] [ 1, 4, 7, 9 ] 1, 9, 7, 4

[ M2 M1 ] [ 3, 6, 10 ] 3, 10, 6

Hence Sequence on M1 : 1-9-7-4-2-3-10-6

Sequence on M2 : 3-10-6-5-8-1-9-7-4

Exercise


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Find the optimal solution for the following Job-Shop problem.Timings are given in days.

Jobs Machine forOperation1 Operation2

Operation1 Operation2

1 M1 M2 4 6

2 M1 - 3

3 M1 - 4

4 M1 M2 5 2

5 M2 M1 1 2

6 M2 - 1 7 M2 M1 7 8

8 M2 - 3

9 M2 M1 6 7

10 M1 M2 2 4

Exercise

j b 3 hi Fl Sh P bl[3]


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Jobs A B C

12

3

45

5 4 37 3 4

9 1 2

6 2 38 5 3

n jobs 3 machines : Flow Shop Problem[3]

Min = 5 Max = 5 Min = 2

At least any one condition out of the following has to besatisfied to convert this problem into 2 m/cs problem.

(1) Maximum of middle Minimum of first column

(2) Maximum of middle Minimum of last column

J b A B C


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Jobs A B C

1

23

4

5

5 4 3

7 3 49 1 2

6 2 3

8 5 3

Jobs G H

1

23

4

5

9 7

10 710 3

8 5

13 8

J b G H


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Jobs G H

1

23

4

5

9 7

10 710 3

8 5

13 8

Applying Johnsons rule : 1 34

2

5

2 34

1

5

OR

Hence, optimal solution of original problem is 5-1-2-4-3

OR 5-2-1-4-3

[4] j b hi Fl Sh P bl


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Jobs M1 M2 M3 M4 M5

1

2

3

4

9 7 4 5 11

8 8 6 7 12

7 6 7 8 10

10 5 5 4 8

Min = 7 Max = 8 Min = 8

At least any one condition out of the following has to be

satisfied to convert this problem into 2 m/cs problem.

(1) Maximum of middle Minimum of first column

(2) Maximum of middle Minimum of last column

[4] n jobs m machines : Flow Shop Problem


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Jobs M1 M2 M3 M4 M5

12

3

4

9 7 4 5 118 8 6 7 12

7 6 7 8 10

10 5 5 4 8

Jobs G H

12

3

4

25 2729 33

28 31

24 22

G


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Jobs G H

1

23

4

25 27

29 3328 31

24 22

Applying Johnsons rule : 41 3 2

Hence, optimal solution of original problem is 1-3-2-4

Exercise


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Get the optimal solution for following n-jobs m-machinesproblem. Timings are given in days.

Jobs M1 M2 M3 M4 M5 M6

1 18 8 7 2 10 25

2 17 6 9 6 8 19

3 11 5 8 5 7 15

4 20 4 3 4 8 12

Exercise

[5] 2 jobs m machines Problems :


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Job1 C(3), A(3), D(4), B(2), E(6)

Job2 B(2), A(4), C(4), D(2), E(4)




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Job1 C(3), A(3), D(4), B(2), E(6)

Job2 B(2), A(4), C(4), D(2), E(4)


16LJ2


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C A D B E

4 8 16

12

4

8

16

12B

A

C

D

E

E

D

C

A

B

J1

T = 18+16 = 34

16LJ2


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C A D B E

4 8 16

12

4

8

16

12B

A

C

D

E

E

D

C

A

B

J1

T = 16+18 = 34

16L18 16


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C A D B E

4 8 16

12

4

8

16

12B

A

C

D

E

E

D

C

A

12,12

B

3,3

3,6

7,10

6,3

10,10

12,1618,16

T = 18+7 = 25

16L

18 16


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C A D B E

4 8 16

12

4

8

16

12B

A

C

D

E

E

D

C

A

18,16

12,12

B6,2

3,2

6,6

10,10

12,16

T = 18+8 = 26

16L

18 16


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C A D B E

4 8 16

12

4

8

16

12B

A

C

D

E

E

D

C

A

18,16

12,12

B6,2

3,2

6,6

10,10

12,16

T = 18+9 = 27

10,12

16L


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C A D B E

4 8 16

12

4

8

16

12B

A

C

D

E

E

D

C

A

18,12

B6,2

3,2

16,12

T = 18+4 = 22


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Hence, minimum time in which both the jobs

can be completed = 22 days.

Scheduling is as per the blue path shown

w.r.t. 22 days.

Exercise


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Solve the following 2-jobs, m-machines problems. Findgraphically minimum time required to complete both the

jobs. Timings are given in days.

Job1Seq. of m/c

Time

A

3

B

4

C

2

D

6

E

2

Job2Seq. of m/c

TimeB

5C

4A

3D

2E

6

Job1Seq. of m/c

TimeA

2B

3C

4D

6E

2

Job

2

Seq. of m/c

Time

C

4

A

5

D

3

E

2

B

6

e c se

[ 1 ]

[ 2 ]


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Thank you

For any Query or suggestion :

Contact :

Dr. D. B. NaikProfessor & Head, Training & Placement,S. V. National Institute of Technology,

Ichchhanath, Surat

395 007 (Gujarat)

Email : [email protected]. : 0261-2201540, 2255225 (O)

sequencing (models) - lecture

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