4.0 Scheduling Mgmt (1)
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Transcript of 4.0 Scheduling Mgmt (1)
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4.0
SchedulingManagement
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4.1 Introduction
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4.1.1 Scheduling
Scheduling is to establish the timing ofthe use of equipment, facilities, andhuman activities in an organization.
The objective is to achieve tradeoffsamong conflicting goals, which include;
efficient utilization of staff, equipment, and
facilities, and minimization of customer waiting time,
inventories, and process times.
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Terms Used
Routing:The operations to be performed,their sequence, the work centers, & the timestandards
Bottleneck:A resource whose capacity is lessthan the demand placed on it
Due date:When the job is supposed to befinished
Slack:The time that a job can be delayed &still finish by its due date
Queue:A waiting line
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4.1.2 Benefits of Scheduling
Effective and efficient scheduling can be acompetitive advantage
Faster movement of goods through a facilitymeans better use of assets and lower costs
Additional capacity resulting from fasterthroughput improves customer service throughfaster delivery
Good schedules result in more dependabledeliveries
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4.1.3 Factors Affecting
SchedulingExternal factors;
Customer's demand
Customer's delivery dates and
Stock of goods already lying with the dealers and retailers.
Internal factors;
Stock of finished goods with the firm.
Time interval to process finished goods from raw material.
Availability of equipment and machinery. Availability of materials
Additional manufacturing facilities if required and
Feasibility of economic production runs.
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4.2 High-Volume System
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4.2.1 Scheduling in High-Volume
Systems High-volume systems are characterized by
standardized equipments and activities thatprovide identical or highly similar operations
on customers or products as they passthrough the system.
All items follow virtually the same sequenceof operations. The goal is to get a highutilization of labor and equipment. Because of
the highly repetitive nature of these systems,many of the loading and sequence decisionsare determined during the design of thesystem.
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4.2.2 The following factors determinethe success of high volume system:
Process and product design. Cost and manufacturability are important, asis achieving a smooth flow through the system.
Preventive maintenance. Keeping equipment in good operating order canminimize breakdowns that would disrupt the flow of work.
Rapid repair when breakdowns occur. This can require specialists as wellas stocks of critical spare parts.
Optimal product mixes. Techniques such as linear programming can beused to determine optimal blends of inputs to achieve desired outputs atminimal costs.
Minimization of quality problems. Quality problems can be extremelydisruptive, requiring shutdowns while problems are resolved. Moreover, whenoutput fails to meet quality standards, not only is there the loss of output butalso a waste of the labor, material, time, and other resources that went into it.
Reliability and timing of supplies. Shortage of supplies is an obvioussource of disruption and must be avoided. On the other hand, is the solutionis to stockpile supplies, that can lead to high carrying costs. Shortening
supply lead times, developing reliable supply schedules, and carefullyprojecting needs are all useful.
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4.3 Low-Volume System
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4.3.1 Scheduling in Low-
Volume Systems In low-volume systems, products are made to order,
and orders usually differ considerably in terms ofprocessing requirements, materials needed,processing time, and processing sequence and
setups. Because of these circumstances, job-shop
scheduling is usually fairly complex. This iscompounded by the impossibility of establishing firmschedules priori to receiving the actual job orders.
Job-shop processing gives rise to two basic issuesfor schedulers: loading, how to distribute the workload among work centers,
and
sequencing, what job processing sequence to use.
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Loading
Loading refers to the assignment of jobs toprocessing (work) centers and to variousmachines in the work centers.
Problems arise when two or more jobs are tobe processed and there are a number of workcenters capable of performing the requiredwork.
Managers often seek an arrangement that willminimize processing and setup costs,minimize idle time among work centers, orminimize job completion time.
4.3.2 Loading Jobs
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a) Input-Output Control
Identifies overloading andunderloading conditions
Prompts managerial action toresolve scheduling problems
The purpose is to manage work
flow so that queues and waitingtimes are kept under control.
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Input-Output ControlExample
Deviation = "actual - "planned
The backlog = "actual input" -"actual output
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Input-Output ControlExample
Options available to operationspersonnel include:
1. Correcting performances
2. Increasing capacity
3. Increasing or reducing input tothe work center
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b) Gantt Charts
Load chart shows the loading andidle times of departments, machines,or facilities
Displays relative workloads overtime
Schedule chart monitors jobs inprocess
All Gantt charts need to be updatedfrequently to account for changes
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Gantt Load Chart Example
Figure 15.3
DayMonday Tuesday Wednesday Thursday FridayWork
Center
Metalworks
Mechanical
Electronics
Painting
Job 349
Job 349
Job 349
Job 408
Job 408
Job 408
Processing Unscheduled Center not available
Job 350
Job 349
Job 295
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Gantt Schedule Chart
Example
Figure 15.4
JobDay
1Day
2Day
3Day
4Day
5Day
6Day
7Day
8
A
B
C
Now
Maintenance
Start of anactivity
End of anactivity
Scheduledactivity timeallowed
Actual workprogress
Nonproductiontime
Point in timewhen chart isreviewed
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c) Assignment Method
A special class of linearprogramming models that assign
tasks or jobs to resources Objective is to minimize cost or
time
Only one job (or worker) isassigned to one machine (orproject)
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Assignment Method
Build a table of costs or timeassociated with particularassignments
Typesetter
Job A B C
R-34 $11 $14 $ 6
S-66 $ 8 $10 $11
T-50 $ 9 $12 $ 7
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Assignment Method
1. Create zero opportunity costs byrepeatedly subtracting the lowest costsfrom each row and column
2. Draw the minimum number of verticaland horizontal lines necessary to coverall the zeros in the table. If the numberof lines equals either the number of
rows or the number of columns,proceed to step 4. Otherwise proceed tostep 3.
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Assignment Method
3. Subtract the smallest number notcovered by a line from all otheruncovered numbers. Add the same
number to any number at theintersection of two lines. Return tostep 2.
4. Optimal assignments are at zero
locations in the table. Select one, drawlines through the row and columninvolved, and continue to the nextassignment.
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Assignment Example
A B CJob
R-34 $11 $14 $ 6
S-66 $ 8 $10 $11
T-50 $ 9 $12 $ 7
Typesetter
A B CJob
R-34 $ 5 $ 8 $ 0
S-66 $ 0 $ 2 $ 3
T-50 $ 2 $ 5 $ 0
Typesetter
Step 1a - Rows
A B CJob
R-34 $ 5 $ 6 $ 0
S-66 $ 0 $ 0 $ 3
T-50 $ 2 $ 3 $ 0
Typesetter
Step 1b - Columns
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Assignment Example
Step 2 - Lines
A B CJob
R-34 $ 5 $ 6 $ 0S-66 $ 0 $ 0 $ 3
T-50 $ 2 $ 3 $ 0
Typesetter
Because only two linesare needed to cover allthe zeros, the solutionis not optimal
Step 3 - Subtraction
A B CJob
R-34 $ 3 $ 4 $ 0
S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
The smallest uncoverednumber is 2 so this issubtracted from all otheruncovered numbers andadded to numbers at theintersection of lines
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Assignment Example
Because three lines areneeded, the solution isoptimal andassignments can bemade
Step 2 - Lines
A B CJob
R-34 $ 3 $ 4 $ 0S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
Start by assigning R-34 toworker C as this is the onlypossible assignment forworker C.
Step 4 - Assignments
A B CJob
R-34 $ 3 $ 4 $ 0
S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
Job T-50 mustgo to worker A as worker C
is already assigned. Thisleaves S-66 for worker B.
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Assignment Example
From the original cost table
Minimum cost= $6 + $10 + $9 = $25
Step 4 - Assignments
A B CJob
R-34 $ 3 $ 4 $ 0S-66 $ 0 $ 0 $ 5
T-50 $ 0 $ 1 $ 0
Typesetter
A B CJob
R-34 $11 $14 $ 6S-66 $ 8 $10 $11
T-50 $ 9 $12 $ 7
Typesetter
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Sequencing
Sequencing is concerned with determiningboth;
the order in which jobs are processed at variouswork centers and
the order in which jobs are processed at individualworkstations within the work centers.
When work centers are heavily loaded, theorder of processing can be very important interms of costs associated with jobs waiting forprocessing and in terms of idle time at the
work centers.
4.3.3 Sequencing Jobs
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Sequencing Jobs
Specifies the order in which jobs shouldbe performed at work centers
Priority rules are used to dispatch or
sequence jobs
FCFS: First come, first served
SPT: Shortest processing time
EDD: Earliest due date CR: Critical Ratio
S/O: Slack of Operation
RUSH
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Sequencing Example
Job
Job Work(Processing) Time
(Days)
Job DueDate
(Days)
A 6 8
B 2 6
C 8 18
D 3 15
E 9 23
Apply the four popular sequencing rulesto these five jobs
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Sequencing Example
Job
Sequence
Job Work(Processing)
Time
Flow
Time
Job Due
Date
Job
LatenessA 6 6 8 0
B 2 8 6 2
C 8 16 18 0
D 3 19 15 4
E 9 28 23 5
28 77 11
FCFS: Sequence A-B-C-D-E
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Sequencing Example
JobSequence
Job Work(Processing)
TimeFlowTime
Job DueDate
JobLateness
A 6 6 8 0
B 2 8 6 2
C 8 16 18 0
D 3 19 15 4
E 9 28 23 5
28 77 11
FCFS: Sequence A-B-C-D-E
Average completion time= = 77/5 = 15.4 daysSum of total flow time
Number of jobs
Utilization= = 28/77 = 36.4%Total job work time
Sum of total flow time
Average number ofjobs in the system = = 77/28 = 2.75 jobs
Sum of total flow time
Total job work time
Average job lateness= = 11/5 = 2.2 daysTotal late days
Number of jobs
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Sequencing Example
JobSequence
Job Work(Processing)
TimeFlowTime
Job DueDate
JobLateness
B 2 2 6 0
D 3 5 15 0
A 6 11 8 3
C 8 19 18 1
E 9 28 23 5
28 65 9
SPT: Sequence B-D-A-C-E
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Sequencing Example
JobSequence
Job Work(Processing)
TimeFlowTime
Job DueDate
JobLateness
B 2 2 6 0
D 3 5 15 0
A 6 11 8 3
C 8 19 18 1
E 9 28 23 5
28 65 9
SPT: Sequence B-D-A-C-E
Average completion time= = 65/5 = 13 daysSum of total flow time
Number of jobs
Utilization= = 28/65 = 43.1%Total job work time
Sum of total flow time
Average number ofjobs in the system = = 65/28 = 2.32 jobs
Sum of total flow time
Total job work time
Average job lateness= = 9/5 = 1.8 daysTotal late days
Number of jobs
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Sequencing Example
JobSequence
Job Work(Processing)
TimeFlowTime
Job DueDate
JobLateness
B 2 2 6 0
A 6 8 8 0
D 3 11 15 0
C 8 19 18 1E 9 28 23 5
28 68 6
EDD: Sequence B-A-D-C-E
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Sequencing Example
JobSequence
Job Work(Processing)
TimeFlowTime
Job DueDate
JobLateness
B 2 2 6 0
A 6 8 8 0
D 3 11 15 0
C 8 19 18 1E 9 28 23 5
28 68 6
EDD: Sequence B-A-D-C-E
Average completion time= = 68/5 = 13.6 daysSum of total flow time
Number of jobs
Utilization= = 28/68 = 41.2%Total job work time
Sum of total flow time
Average number ofjobs in the system = = 68/28 = 2.43 jobs
Sum of total flow time
Total job work time
Average job lateness= = 6/5 = 1.2 daysTotal late days
Number of jobs
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Sequencing Example
Rule
AverageCompletion
Time (Days)
Utilization
(%)
Average Numberof Jobs in
System
AverageLateness
(Days)
FCFS 15.4 36.4 2.75 2.2
SPT 13.0 43.1 2.32 1.8
EDD 13.6 41.2 2.43 1.2
Summary of Rules
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Comparison ofSequencing Rules
No one sequencing rule excels on all criteria
SPT does well on minimizing flow time andnumber of jobs in the system
But SPT moves long jobs tothe end which may resultin dissatisfied customers
FCFS does not do especially
well (or poorly) on anycriteria but is perceivedas fair by customers
EDD minimizes lateness
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Critical Ratio (CR)
An index number found by dividing thetime remaining until the due date by thework time remaining on the job
Jobs with low critical ratios arescheduled ahead of jobs with highercritical ratios
Performs well on average job latenesscriteria
CR = =Due date -Todays date
Work (lead) time remaining
Time remaining
Workdays remaining
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Critical Ratio Example
JobDueDate
WorkdaysRemaining Critical Ratio
PriorityOrder
A 30 4 (30 - 25)/4 = 1.25 3
B 28 5 (28 - 25)/5 = .60 1
C 27 2 (27 - 25)/2 = 1.00 2
Currently Day25
With CR < 1, Job B is late. Job C is just on scheduleand Job A has some slack time.
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Critical Ratio Technique
1. Helps determine the status of specificjobs
2. Establishes relative priorities among
jobs on a common basis
3. Relates both stock and make-to-orderjobs on a common basis
4. Adjusts priorities automatically forchanges in both demand and jobprogress
5. Dynamically tracks job progress
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Slack per Operation (S/O)
Jobs are processed according toaverage slack time (time until due
date minus remaining time toprocess).
Compute by dividing slack time bynumber of remaining operations,including the current one.
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Example:S/O
Note that processing time includesthe time remaining for the current
and subsequent operations. In addition, you will need to know the
number of operations remaining,including the current one.
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Solution:1. Determine the difference between the due date and the processing time for
each operation.
2. Divide the amount by the number of remaining operations, and3. Rank them from low to high. This yields the sequence of jobs:
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Rush
Emergency or preferred customersfirst.
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Summary Priority Rules
LOCAL GLOBAL
Description
Local rules take into
account information
pertaining only to asingle workstation.
Global rules take into
account information
pertaining to multipleworkstations.
Rules
FCFS, SPT, and EDD. CR and S/O
Rush
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4.3.3 Sequencing N Jobs on
Two Machines: Johnsons Rule
Works with two or more jobs that
pass through the same twomachines or work centers
Minimizes total production time and
idle time
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Johnsons Rule
1. List all jobs and times for each workcenter
2. Choose the job with the shortest activity
time. If that time is in the first work center,schedule the job first. If it is in the secondwork center, schedule the job last.
3. Once a job is scheduled, it is eliminatedfrom the list
4. Repeat steps 2 and 3 working toward thecenter of the sequence
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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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B E D C A
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
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
Time 0 3 10 20 28 33
Time 0 1 3 5 7 9 10 11 12 13 17 19 21 22 2325 27 29 31 33 35
B ACDE
B ACDE
WC
1
WC2
B E D C A
B ACDE
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4.3.4 Sequence Jobs When SetupTimes Are Sequence-Dependent
The simplest wayto determinewhichsequence will result in thelowest total setuptimeis;
to list each possible sequence and
determine its total setup time.
As the number of jobs increases, a managerwould use a computer to generate the list andidentify the best alternative(s).
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