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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
Jobs Proc.time Flow time DD Late Tardy
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
Jobs Proc.time Flow time DD Late Tardy
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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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
@ work station
1
D BAE A
E l T W k C t t
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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
@ work station
1
D BAE AC
D BAE AC FFinalsequence
E l T W k C t t
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Example: Two Work Center cont.
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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Example: Johnsons rule
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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Example: Johnsons rule
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