4.0 scheduling mgmt (1)

7/30/2019 4.0 Scheduling Mgmt (1)

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2008 Prentice Hall, Inc. 15 1

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



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.



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.



4.3 Low-Volume System



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.



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



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.



Input-Output ControlExample

Deviation = "actual - "planned

The backlog = "actual input" -"actual output



Input-Output ControlExample

Options available to operationspersonnel include:

1. Correcting performances

2. Increasing capacity

3. Increasing or reducing input tothe work center



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



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


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


Total job work time

Average job lateness= = 11/5 = 2.2 daysTotal late days

Number of jobs


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Sequencing Example

JobSequence


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


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





Total job work time


Number of jobs


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Sequencing Example

JobSequence


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


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





Total job work time


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



A 5 2

B 3 6

C 8 4

D 10 7

E 7 12


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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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