sequencing or1

7/29/2019 Sequencing OR1

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Sequencing


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Sequencing

Sequencing problems are of common occurrence in ourdaily ordering of jobs for processing in a manufacturing plant waiting aircrafts for landing clearance programmes to be run in a sequence at a computer centre

Such problems exist whenever there is a alternativechoice as to the orderin which a number of jobs can bedone.

The selection of an appropriate order or sequence inwhich to receive waiting customers (or jobs) is calledsequencing.

the objective is to optimize the use of available facilitiesto effectively process the items or the jobs.


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Scheduling Models

Flow-Shop Scheduling model :

In this model we have m machines and njobs. Any given job requires anexecution on each machine. The ordering of processing on the variousmachines is same for all jobs ; also the sequence in which the jobs gothrough the first machine has to be the same as the sequence in whichthe jobs go through any of the subsequent machine, i.e. a job may not`Pass' another job while waiting for processing on a machine. A flow-shop with this restriction is often called apermutation flow-shop.

J ob-Shop Scheduling Model :

In this model we have njobs and m machines and each job has its ownmachine order specified.


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Scheduling Models

Open shop scheduling model :

In this model we have m machines and n jobs. Any given jobrequires an execution on each of the m machines. The order in

which a job passes through the machines is immaterial.

Project Scheduling :

It is one of a type project where all the resources are brought to

the job.


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FLOW SHOP SCHEDULING

(n JOBS, m MACHINES)

n JOBS BANK OF m MACHINES (SERIES)

1

2

3

4 n

M1 M2 Mm

Complete enumeration

n! possible sequence for each machine (m

machine)


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N= Set of 1, 2, 3,... .., n jobs

M= Set of m machines M1,M2,......,Mm

tij=Processing time of i-th job on j-th machine Mj

Total idle time of last machine for n jobs in a

sequence S

n

i

iX1

n )D(Sor

Notations


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Various Performance Measures

In each one of the models it is assumed that the decision maker wishes to

minimize a given objective function. Some of the objectives are listedbelow :

Minimum Make-span : It is defined as total completion time in which theset of all jobs finish processing on all the machines. This is also called asTotal Elapsed Time or minimum flow time.

Minimum Mean Flow Time : It is defined as the average completion timeof any job. It is the average time a job spends in the Shop.

Minimum In-Process Inventory Time : It is defined as minimization oftotal in-process inventory waiting time for all jobs.

Minimum Penalty Cost : It is defined as the total penalty paid by virtue ofjobs being late in completion by their due dates.

Minimum Total Production Cost : It is defined as the total production costfor the production of a set of products on machines.


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Scheduling ProblemThe general scheduling problem is described

below

Determine the sequence and schedule forprocessing a specified number of jobs on agiven number of machines that minimizes awell-defined measure of performance. It may

be noted that the technological orders inwhich the jobs are processed on variousmachines, due dates of jobs, and the job andmachine availabilities are to be considered

while determining the sequence.


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Difference between

Sequencing and Scheduling

Sequencing simply refers to thedetermination of ORDER in which the jobs

are to be processed on various machines.

Scheduling refers to the time-table thatincludes the start time and completion

time of-jobs on machines etc.


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Flow-shop

(Assumptions)

Regarding Machines : No machine processes more than one operation at, a time.

Each operation on a machine, once started, must be performed to its

completion.

Each operation takes finite time and it must be completed before any

other operation begins. The given operation time includes set-up time.

Each machine is initially idle at the beginning of the scheduling period.

There is only one machine of each type. Machines never break down and man-power of uniform ability is always

available.

Each machine operates independently of the other.


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Flow-shop

(Assumptions)

Regarding J obs :

All jobs are available for processing at time zero

All jobs allow the same sequence of operations.

Jobs are independent of each other. No job is processed more than once on any machine.

Each job consists of a specified number of operations and each

operation is performed by only one-machine.

The processing times of the jobs are independent of the order in

which the jobs are performed.

Each job, once started, must be processed to completion.


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Flow-shop

(Assumptions)

Regarding Operating Policies:

Each job is processed as early as possible.

Each job is considered as indivisible entity even though it may be

composed of a number of individual units. Each machine is provided with sufficient waiting space for

allowing jobs to wait before starting their processing.

Each machine processes jobs in the same sequence, i.e., no

passing or overtaking of jobs is permitted.

Transportation time of a job between machines is negligible.

Set up times are sequenceindependent.


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Sequencing

(n J obs and one machine Flow shop Problem)

Priority rules: Simple heuristicsused to select the order inwhich jobs will be processed.

Job time: Time needed forsetup and processing of a job.

It includes set up time unless setup timesare sequence dependent


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Priority Rules

FCFS - first come, first served

SPT - shortest processing time

EDD - earliest due date

CR - critical ratio

=time remaining / processing time

S/O - slack per operation

=slack remaining / # of operations remaining

Rush - emergency


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First Come, First Served Rule

Process first job to arrive at a work center

first

Average performance on most scheduling

criteria

Appears fair & reasonable to customers

Important for service organizations

Example: Restaurants


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Shortest Processing Time Rule

Process job with shortest processing time

first.

Usually best at minimizing job flow and

minimizing the number of jobs in thesystem

Major disadvantage is that long jobs may

be continuously pushed back in the queue.


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Earliest Due Date Rule

Process job with earliest due date first

Widely used by many companies

If due dates important

Performs poorly on many scheduling

criteria


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Critical Ratio (CR)

Ratio of time remaining to work time

remaining

CR Time remainingWork days remaining

Due date - Today's date

Work (lead) time remaining

Process job with smallest CR first

Performs well on average lateness

4 2 Ad t f th C iti l R ti


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4.2 Advantages of the Critical Ratio

Scheduling Rule

Use of the critical ratio can help to:

determine the status of a specific job

establish a relative priority among jobs on

a common basis

relate both stock and make-to-order jobs

on a common basis

adjust priorities and revise schedulesautomatically for changes in both demand

and job progress

dynamically track job progress and

location


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Performance measures

Flow time of a job: Duration of time from ajob enters into the system until it leaves

Lateness of a job: Amount by which

completion date exceeds due date. Couldbe negative.

Tardiness=max(lateness,0)

Makespan: total time needed to finish agroup of jobs

Average number of jobs until the last isfinished:

=Total flow time / Makes an


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Criteria to Evaluate Priority Rules

Jobs#

timesFlowtimecompletionAverage

timesFlow

timesProcessnUtilizatio

timesProcess

timesFlowsystemtheinjobsofnumberAverage

jobsofNumber

timesLatelatenessjobAverage


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Example: Average number of jobs

Jobs: A and B with processing times 10 each

A finishes at 10

Number of jobs

1

2

B finishes at 20 Time

Process Time=20, Total Flow time=10+20

Average number of jobs=30/20

Average number of jobs


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Example: Sequencing rules

Jobs Processing time DD=Due date

A 11 61

B 29 45

C 31 31

D 1 33

E 2 32


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Ex: FCFS

Jobs Proc.time Flow time DD Late Tardy

A 11 11 61 -50 0

B 29 40 45 -5 0C 31 71 31 40 40

D 1 72 33 39 39

E 2 74 32 42 42

Total 268 202 66 121

Aver. 53.6 40.4 13.2 24.2


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Ex: SPT to minimize the total flow time


D 1 1 33 -32 0

E 2 3 32 -29 0A 11 14 61 -47 0

B 29 43 45 -2 0

C 31 74 31 43 43

Total 135 202 -67 43

Aver. 27.0 40.4 -13.4 8.6


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Ex: EDD to minimize the maximum lateness


C 31 31 31 0 0

E 2 33 32 1 1D 1 34 33 1 1

B 29 63 45 18 18

A 11 74 61 13 13

Total 235 202 33 33

Aver. 47.0 40.4 6.6 6.6


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235/74=3.176.647EDD

135/74=1.828.627SPT

268/74=3.6224.253.6FCFS

Average

Number of

Jobs at theWork Center

Average

Tardiness(days)

Average

Flow Time(days)Rule

Example summary


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CR: This is an Iterative Process using this

model:

Set Current Time (sum of time of allscheduled jobs so far)

Compute:

Model Starts with current time = 0

Current time updates after eachselection by adding scheduledProcess Time to current time

_ _

Pr._ _Re

Due Date Cur TimeCR

Work maining


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Try it:

JOBPr.

Time

D.

DateCR

Current Time = 0

1 11 615.54

6

2 29 451.55

2

3 31 31 1.00

4 1 33 33

5 2 32 16

JOBPr.Time

D.Date

CR

Current Time = 31

1 11 61 2.7272 29 45 .483

4 1 33 2

5 2 32 0.5


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Continuing (CR)

JOBPr.Time

D.Date

CR

Current Time = 60

1 11 610.091do

last

4 1 33 -27*

5 2 32 -14**

JOBFlow

Time

D.

DateTardy

Summary

3 31 31 0

2 60 45 15

4 61 33 28

5 63 32 31

1 74 61 13

Total: 289 87


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Summarizing from CR analysis:

Mean F. Time: (289)/5 = 57.8

Mean Tardiness: (87)/5 = 17.4


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Used to sequence N jobs through 2

machines in the same order

1995 Corel Corp.

1995 Corel

Corp.

Saw Drill

Job A

Job B

Job C

Jobs (N = 3)

Johnsons Rule


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Johnson's Rule - Scheduling N Jobs onTwo Machines

All jobs are to be listed, and the time eachrequires on a machine shown.

Select the job with the shortest activity

time. If the shortest time lies with the firstmachine, the job is scheduled first; if withthe second machine, the job is scheduledlast.

Once a job is scheduled, eliminate it.

Apply steps 2-3 to the remaining jobs,working toward the center of the

sequence.


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List jobs &activity times

Select job with

shortest time

Machine?

Schedule

FIRST

ScheduleLAST

Eliminate job

from list

Jobs left?Break

arbitrarily

Ties?

Yes

1

2

Yes

StopNo

No

Johnsons Rule Steps

Fl h


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Flow-shop

(n Jobs and two machine Flow shop Problem)

J ohnson's Rule : J ob i precedes job j in an

optimal sequence with regard to minimumtotal elapsed time if

min (ti1, tj2)min (ti2, tj1)

Fl h


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Flow-shop

(n Jobs and two machine Flow shop Problem)

J ohnson's Optimal Sequence Algorithm

Step 1 : Examine the processing times of n jobs on both machines andfind min [(min (ti1, ti2) where i=1, 2,...., n.]

Step 2 : (i) If the minimum tk1 is obtained in the column of firstmachine A for job k, schedule the k-th job in the first availableposition in sequence from the beginning, i.e., from left end. (ties maybe broken arbitrarily). Go to step 3.

(ii) If the minimum tr2 is obtained in the column of second machine Bfor job r, place the r-th job in the last available position in thesequence from the last i.e., from right end. (ties may be brokenarbitrarily). Go to step 3.

Step 3: Remove the assigned job from further considerationand return to step 1 until all the job are assigned.


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Eliminate M3 from consideration. The next shortest time

is M2 at Workstation 1, so schedule M2 first.Eliminate M5 from consideration. The next shortest time is

M1 at workstation #1, so schedule M1 next.

Eliminate M1 and the only job remaining to be

scheduled is M4.

Example Johnsons Ruleat theMorris Machine Co.

Time (hr)

Motor Workstation 1 Workstation 2

M1 12 22

M2 4 5M3 5 3

M4 15 16

M5 10 8

Sequence = M1M2 M3M4 M5

Shortest time is 3 hours at workstation 2, so

schedule job M3 last.

Eliminate M2 from consideration. The next shortest time is

M5 at workstation #2, so schedule M5 next to last.


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Workstation

M2(4)

M1(12)

M4(15)

M5(10)

M3(5)

Idleavailablefor further work

0 5 10 15 20 25 30Day

35 40 45 50 55 60 65

Idle2 M2(5)M1(22)

M4(16)

M5(8)

M3(3)Idle

1

Gantt Chart for the Morris Machine Company Repair Schedule

The schedule minimizes the idle time of workstation 2

and gives the fastest repair time for all five motors.

No other sequence will produce a lower makespan.

Example Johnsons Ruleat theMorris Machine Co.

J h R l


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Johnsons Rule -Example

Job Work Center 1

(Drill Press)

Work Center 2

(Lathe)

A 5 2

B 3 6

C 8 4

D 10 7

E 7 12


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Johnsons Rule - Example

AStep 1

B AStep 2

B C AStep 3

B D C AStep 4

B E D C AStep 5

Graphical Depiction


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Graphical Depictionof J ob Flow

B E D C A

B E D C A

Work

center 1

Work

center 2

0 3 10 20 28 33

0 3 9 10 20 22 28 29 33 35Time =>

Time =>

B E D C A

= Job completed= Idle

E l k


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Example: Two Work Center

Job

Lowest

Workcenter 1

A 5 5

B 4 3

C 8 9

D 2 7

E 6 8

F 12 15

Workcenter 2

@ work center

1

D


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E l T W k C


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Example: Two Work Center cont.

Job

Lowest

Workcenter 1

A 5 5

B 4 3

C 8 9

D 2 7

E 6 8

F 12 15

Workcenter 2

Tie: pick arbitrarily

D BA

E l T W k C


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Job

Lowest

Workcenter 1

A 5 5

B 4 3

C 8 9

D 2 7

E 6 8

F 12 15

Workcenter 2

@ work station

1

D BAE A

E l T W k C t t


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Job

Lowest

Workcenter 1

A 5 5

B 4 3

C 8 9

D 2 7

E 6 8

F 12 15

Workcenter 2

@ work station

1

D BAE AC

D BAE AC FFinalsequence

E l T W k C t t


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Work station 1

D BE AC F0 2 8 16 28 33 37

D BE AC F2 9 17 26 28 43 48 51

Work station 2

Makespan = 51


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Example: Johnsons rule

Job Processing time on 1 Processing time on 2

A 15 25

B 8 6

C 12 4

D 20 18


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The sequence that minimizes the Makespan

A - D - B - C

15

25

20

18

8

6

12

4

15

15 35

40

43

58

55

64 68

15

13

Idle time = 28

Makespan = 68

MC1

MC2


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Job i: 1 2 3 4 5M1 5 3 1 7 8

M2 2 6 2 7 7

Stage Unscheduled Minimum tim Assignment Partialjobs m= 1, 2 schedule

1 1, 2, 3, 4, 5 t31 3=[Il 3, x, x, X, X

2 1, 2, 4, 5 t12 1=[5] 3, x, x, x, 1

3 2, 4, 5 t2l 2=[2] 3, 2, x, x, 1

4 4, 5 t52 5=[4] 3, 2, 4, 5, 1

Hence optimal sequence So=(3, 2, 4, 5, 1).


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Job i: 1 2 3 4 5M1 5 3 1 7 8

M2 2 6 2 7 7

The calculations of flow time of the given five jobs on both machines is, given below :

job Machine M1 Machine M2 Xi=Idle time of job in i-th

sequence position on 2ndmachineM2

in out in out

3 0 1 1 3 X1=1

2 1 4 4 10 X2=14 4 11 11 18 X3=1

5 11 19 19 26 X4=1

1 19 24 26 28 X5=0

sequencing or1

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