J Operations Scheduling

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Operations SchedulingOperations SchedulingJ

J – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

For For Operations Management, 9eOperations Management, 9e by by Krajewski/Ritzman/Malhotra Krajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education

PowerPoint Slides PowerPoint Slides by Jeff Heylby Jeff Heyl

The scheduling techniques cut across the various process types found in services

Scheduling Service and Scheduling Service and Manufacturing Processes Manufacturing Processes

various process types found in services and manufacturing

Front-office process with high customer contact, divergent work flows, customization, and a complex scheduling environmentBack-office process has low customer involvement uses more line work flows and


involvement, uses more line work flows, and provides standardized services

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Performance MeasuresPerformance Measures

Flow time is the amount of time a job spends in the service or manufacturing

tsystemPast due (tardiness) is the amount of time by which a job missed its due dateMakespan is the total amount of time required to complete a group of jobs


Makespan = Time of completion of last job – Starting time

of first job

Performance MeasuresPerformance Measures

Total inventory is used to measure the effectiveness of schedules for

f t imanufacturing processes.

Total Inventory = Scheduled receipts

for all items + On-hand inventories of all items

Utilization is the percentage of work time


that is productively spent by an employee or a machine

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Sequencing JobsSequencing Jobs

Operations schedules are short-term plans designed to implement the sales and operations planoperations planAn operation with divergent flows is often called a job shop

Low-to medium-volume productionUtilizes job or batch processesThe front office would be the equivalent for a


qservice providerDifficult to schedule because of the variability in job routings and the continual introduction of new jobs to be processed

Sequencing JobsSequencing Jobs

An operation with line flow is often called a flow shop

Medium- to high-volume productionUtilizes line or continuous flow processesThe back office would be the equivalent for a service provider Tasks are easier to schedule because the jobs have a common flow pattern through the


have a common flow pattern through the system

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Job Shop SequencingJob Shop Sequencing

Ship

ping

Dep

artm

ent

Raw

Mat

eria

ls


S

Legend:Batch of partsWorkstation

Figure J.1 – Diagram of a Manufacturing Job Shop Process

First-come, first-served (FCFS) Earliest due date (EDD)

Priority Sequencing RulesPriority Sequencing Rules

( )Critical ratio (CR)

A ratio less than 1.0 implies that the job is behind

CR =(Due date) – (Today’s date)Total shop time remaining


p jscheduleA ratio greater than 1.0 implies the job is ahead of scheduleThe job with the lowest CR is scheduled next

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Shortest processing time (SPT)Slack per remaining operations (S/RO)

Priority Sequencing RulesPriority Sequencing Rules

p g p ( )

S/RO =

Duedate

Today’sdate

Total shop time remaining– –

Number of operations remaining


The job with the lowest S/RO is scheduled next

Single-dimension rulesA job’s priority assignment based only on

Sequencing One WorkstationSequencing One Workstation

j p y g yinformation waiting for processing at the individual workstation (e.g., FCFS, EDD, and SPT)


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Comparing EDD and SPT RulesComparing EDD and SPT Rules

EXAMPLE J.1The Taylor Machine Shop rebores engine blocks. Currently, five engine blocks are waiting for processing. At any time, the g g p g y ,company has only one engine expert on duty who can do this type of work. The engine problems have been diagnosed, and the processing times for the jobs have been estimated. Expected completion times have been agreed upon with the shop’s customers. The accompanying table shows the current situation. Because the Taylor Machine Shop is open from 8:00 A.M. until 5:00 P.M. each weekday, plus weekend hours as needed, the customer pickup times are measured in business h f th t ti D t i th h d l f th


hours from the current time. Determine the schedule for the engine expert by using (a) the EDD rule and (b) the SPT rule. For each rule, calculate the average flow time, average hours early, and average hours past due. If average past due is most important, which rule should be chosen?


Business Hours Since Order

Processing Time, Including Setup

Business Hours Until Due Date (customer

Engine BlockSince Order

ArrivedIncluding Setup

(hours)Due Date (customer

pickup time)

Ranger 12 8 10

Explorer 10 6 12

Bronco 1 15 20

Econoline 150 3 3 18

Thunderbird 0 12 22


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SOLUTIONa. The EDD rule states that the first engine block in the

sequence is the one with the closest due date.

Engine Block Sequence

Hours Since Order

ArrivedBegin Work

Processing Time (hr)

Finish Time (hr)

Flow Time (hr)

Scheduled Customer

Pickup Time

Actual Customer

Pickup Time

Hours Early

Hours Past Due

qConsequently, the Ranger engine block is processed first. The Thunderbird engine block, with its due date furthest in the future, is processed last. The sequence is shown in the following table, along with the flow times, the hours early, and the hours past due.


Sequence Arrived Work Time, (hr) (hr) (hr) Pickup Time Time Early Due

Ranger

Explorer

Econoline 150

Bronco

Thunderbird

12 0 + 8 = 8 20 10 10 2 —

10 8 + 6 = 14 24 12 13 — 2

3 14 + 15 = 17 20 18 18 1 —

1 17 + 3 = 32 33 20 32 — 12

0 32 + 12 = 44 44 22 44 — 22


The flow time for each job is its finish time, plus the time since the job arrived.1 For example, the Explorer engine block’s finish time will be 14 hours from now (8 hours waiting time before the engine expert started to work on it plus 6 hours processing)engine expert started to work on it plus 6 hours processing). Adding the 10 hours since the order arrived at this workstation (before the processing of this group of orders began) results in a flow time of 24 hours. You might think of the sum of flow times as the total job hours spent by the engine blocks since their orders arrived at the workstation until they were processed.


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The performance measures for the EDD schedule for the five engine blocks are

Average flow time =

Average hours early =

20 + 24 + 20 + 33 + 445 = 28.2 hrs

2 + 0 + 1 + 0 + 05 = 0.6 hrs


Average hours past due =0 + 2 + 0 + 12 + 22

5 = 7.2 hrs


b. Under the SPT rule, the sequence starts with the engine block that has the shortest processing time, the Econoline 150, and it ends with the engine block that has the longest processing time the Bronco The sequence along with the

Engine Block Sequence

Hours Since Order

ArrivedBegin Work

Processing Time, (hr)

Finish Time (hr)

Flow Time (hr)

Scheduled Customer

Pickup Time

Actual Customer

Pickup Time

Hours Early

Hours Past Due

Econoline 150 3 0 + 3 = 3 6 19 18 15 —

processing time, the Bronco. The sequence, along with the flow times, early hours, and past due hours, is contained in the following table:


Explorer

Ranger

Thunderbird

Bronco

10 3 + 6 = 9 19 12 12 3 —

12 9 + 8 = 17 29 10 17 — 7

0 17 + 12 = 29 29 22 29 — 7

1 29 + 15 = 44 45 20 44 — 24

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The performance measures are

Average flow time =

Average hours early =

6 + 19 + 29 + 29 + 455 = 25.6 hrs

15 + 3 + 0 + 0 + 05 = 3.6 hrs


Average hours past due =0 + 0 + 7 + 7 + 24

5 = 7.6 hrs

Comparing Sequencing RulesComparing Sequencing Rules

EDD rulePerforms well with respect to the percentage of jobs past due and the variance of hours past duepPopular with firms that are sensitive to achieving due dates

SPT rule Tends to minimize the mean flow and maximize shop utilizationFor single-workstations will always provide the lowest mean finish timeCould increase total inventory Tends to produce a large variance in past due hours


Tends to produce a large variance in past due hoursFCFS rule

Considered fairIt performs poorly with respect to all performance measures

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Application J.1Application J.1

Given the following information, devise an SPT schedule for the automatic routing machine.

Order

Standard Time, Including Setup (hr)

Due Date(hrs from now)

AZ135 14 14

DM246 8 20

SX435 10 6

PC088 3 18


Order Sequence

HoursSince Order Arrived

Begin Work

Finish Time (hr)

FlowTime(hr)

Scheduled Customer

Pickup Time

Actual Pickup Time

Hours Early

Hours Past Due


1.

2.

3.

4.

Total

Average

PC088

DM246

SX435

AZ135

2 0 3 5 18 18 15

4 3 11 15 20 20 9

1 11 21 22 6 21 15

5 21 35 40 14 35 25

82 94 24 40

20.5 23.5 6 10


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MultipleMultiple--Dimension RulesDimension Rules

The priority rules CR and S/RO incorporate information about the remaining workstationsS/RO i b tt th EDD ith t t thS/RO is better than EDD with respect to the percentage of jobs past due but usually worse than SPT and EDD with respect to average job flow timesCR results in longer job flow times than SPT, but CR also results in less variance in the distribution of past due hours


of past due hoursNo choice is clearly best; each rule should be tested in the environment for which it is intended

Sequencing with the CR and Sequencing with the CR and S/RO RulesS/RO Rules

EXAMPLE J.2The first five columns of the following table contain information about a set of four jobs that just arrived (end of hour 0 or j j (beginning of hour 1) at an engine lathe. They are the only ones now waiting to be processed. Several operations, including the one at the engine lathe, remain to be done on each job. Determine the schedule by using (a) the CR rule and (b) the S/RO rule. Compare these schedules to those generated by FCFS, SPT, and EDD.

Processing Time at Engine Lathe

Time Remaining Until Due Date

Number of Operations

Shop Time Remaining


Job (hours) (days) Remaining (days) CR S/RO

1 2.3 15 10 6.1 2.46 0.89

2 10.5 10 2 7.8 1.28 1.10

3 6.2 20 12 14.5 1.38 0.46

4 15.6 8 5 10.2 0.78 –0.44

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SOLUTIONa. Using CR to schedule the machine, we divide the time

remaining until the due date by the shop time remaining toremaining until the due date by the shop time remaining to get the priority index for each job. For job 1,

CR =Time remaining until the due date

Shop time remaining = = 2.46156.1

By arranging the jobs in sequence with the lowest critical i fi d i h h f j b b


ratio first, we determine that the sequence of jobs to be processed by the engine lathe is 4, 2, 3, and finally 1, assuming that no other jobs arrive in the meantime.


b. Using S/RO, we divide the difference between the time remaining until the due date and the shop time remaining by the number of remaining operations. For job 1,

S/RO =

Time remaininguntil the due date

Shop timeremaining–

Number of operations remaining = = 0.89

15 – 6.110

Arranging the jobs by starting with the lowest S/RO i ld 4 3 1 2 f j b


yields a 4, 3, 1, 2 sequence of jobs.

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Priority Rule Summary

FCFS SPT EDD CR S/RO

Average flow time 17.175 16.100 26.175 27.150 24.025

Average early time 3.425 6.050 0 0 0

Average past due 7.350 8.900 12.925 13.900 10.775



The following four jobs have just arrived at an idle drill process and must be scheduled.

Job

Processing Time at Drill Press

(wk)

Time Remaining to Due Date

(wks)

Number of Operations Remaining*

Shop Time Remaining*

(wks)

AA 4 5 3 4

BB 8 11 4 6

CC 13 16 10 9

DD 6 18 3 12


EE 2 7 5 3

* including drill press

Create the sequences for two schedules, one using the Critical Ratio rule and one using the S/RO rule.

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Job

Processing Time at Drill Press

(wk)

Time Remaining to Due Date

(wks)

Number of Operations Remaining*

Shop Time Remaining*

(wks)

AA 4 5 3 4

BB 8 11 4 6

Critical Ratio Slack/Remaining Operation

Job Priority IndexSequence on

Drill Press Job Priority IndexSequence on

Drill Press

BB 8 11 4 6

CC 13 16 10 9

DD 6 18 3 12

EE 2 7 5 3

AA 1.25 First AA 0.33 First


BBCCDDEE

1.831.441.502.33

FourthSecondThirdFifth

BBCCDDEE

1.250.702.000.80

FourthSecond

FifthThird

Identifying the best priority rule to use at a particular operation in a process is a

l bl b th t t

Multiple WorkstationsMultiple Workstations

complex problem because the output from one operation becomes the input to anotherComputer simulation models are effective tools to determine which priority rules work best in a given situation


work best in a given situation

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In single-workstation scheduling, the makespan is the same regardless of the priority rule chosenI th h d li f t k t ti i

Scheduling a TwoScheduling a Two--Station Flow ShopStation Flow Shop

In the scheduling of two or more workstations in a flow shop, the makespan varies according to the sequence chosenDetermining a production sequence for a group of jobs to minimize the makespan has two advantages

The group of jobs is completed in minimum time


The group of jobs is completed in minimum timeThe utilization of the two-station flow shop is maximized

Johnson’s RuleJohnson’s Rule

Minimizes makespan when scheduling a group of jobs on two workstations

Step 1: Scan the processing time at each workstation and find the shortest processing time among the jobs not yet scheduled. If two or more jobs are tied, choose one job arbitrarily.Step 2: If the shortest processing time is on workstation 1, schedule the corresponding job as early as possible. If the shortest processing time is on workstation 2, schedule the corresponding job as late as possible


corresponding job as late as possible.Step 3: Eliminate the last job scheduled from further consideration. Repeat steps 1 and 2 until all jobs have been scheduled.

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Scheduling Jobs on Two Workstations

EXAMPLE J.3The Morris Machine Company just received an order to refurbish five motors for materials handling equipment thatrefurbish five motors for materials handling equipment that were damaged in a fire. The motors have been delivered and are available for processing. The motors will be repaired at two workstations in the following manner.

Workstation 1: Dismantle the motor and clean the parts.Workstation 2: Replace the parts as necessary, test the

motor, and make adjustments.



The customer’s shop will be inoperable until all the motors have been repaired, so the plant manager is interested in developing a schedule that minimizes the makespan and has

th i d d th l k ti til th t hauthorized around-the-clock operations until the motors have been repaired. The estimated time to repair each motor is shown in the following table:

Time (hr)Motor Workstation 1 Workstation 2

M1 12 22M2 4 5


M2 4 5

M3 6 3M4 15 16M5 10 8

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SOLUTIONThe logic for the optimal sequence is shown in the following table:table:

Establishing a Job SequenceIteration Job Sequence Comments

1 M3 The shortest processing time is 3 hours for M3 at workstation 2. Therefore, M3 is scheduled as late as possible.

2 M2 M3 Eliminate M3 from the table of estimated times. The next shortest processing time is 4 hours for M2 at workstation 1. M2 is therefore scheduled first.

3 M2 M5 M3 Eli i t M2 f th t bl Th t h t t


3 M2 M5 M3 Eliminate M2 from the table. The next shortest processing time is 8 hours for M5 at workstation 2. Therefore, M5 is scheduled as late as possible.

4 M2 M1 M5 M3 Eliminate M5 from the table. The next shortest processing time is 12 hours for M1 at workstation 1. M1 is scheduled as early as possible.

5 M2 M1 M4 M5 M3 The last motor to be scheduled is M4. It is placed in the last remaining position, in the middle of the schedule.


Workstation

M2 (4)

M1 (12)

M4 (15)

M5 (10)

M3 (5)

Idle—available for further work

Idle M2 (5)

M1 (22)

M4 (16)

M5 (8)Idle (3)

M3

0 5 10 15 20 25 30Hour

35 40 45 50 55 60 65

1

2


Figure J.2 – Gantt Chart for the Morris Machine Company Repair Schedule

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Use the following data to schedule two workstations arranged as a flow shop

Time (hr)Job Workstation 1 Workstation 2A 4 3B 10 20C 2 15D 8 7E 14 13


Sequence: C B E D A


Workstation 1 Workstation 2

Start Finish Start Finish

C

B

E

D

A

0 2

2 12

12 26

26 34

34 38

2 17

17 37

37 50

50 57

57 60


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The resource constraint is the amount of labor available, not the number of machines or workstations

LaborLabor--Limited EnvironmentLimited Environment

The scheduler must also assign workers to their next workstationsSome possible labor assignment rules

Assign personnel to the workstation with the job that has been in the system longestAssign personnel to the workstation with the most jobs


g p jwaiting for processingAssign personnel to the workstation with the largest standard work contentAssign personnel to the workstation with the job that has the earliest due date

The Neptune’s Den Machine Shop specializes in overhauling outboard marine engines. Some engines require replacement of broken parts, whereas others need a complete overhaul. C tl fi i ith i bl iti

Solved Problem 1Solved Problem 1

Currently, five engines with varying problems are awaiting service. The best estimates for the labor times involved and the promise dates (in number of days from today) are shown in the following table. Customers usually do not pick up their engines early.

EngineTime Since Order

Arrived (days)

Processing Time, Including Setup

(days)Promise Date

(days from now)


50-hp Evinrude 4 5 8

7-hp Johnson 6 4 15

100-hp Mercury 8 10 12

50-hp Honda 1 1 20

75-hp Nautique 15 3 10

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a. Develop separate schedules by using the SPT and EDD rulesb. Compare the two schedules on the basis of average flow

time, percentage of past due jobs, and maximum past duetime, percentage of past due jobs, and maximum past due days for any engine



SOLUTIONa. Using the SPT rule, we obtain the following schedule:

Repair Sequence

Days Since Order

ArrivedProcessing

TimeFinish Time

Flow Time

Promise Date

Actual Pickup Date

Days Early

Days Past Due

50-hp Honda 1 1 1 2 20 20 19 —

75-hp Nautique 15 3 4 19 10 10 6 —

7-hp Johnson 6 4 8 14 15 15 7 —


50-hp Evinrude 4 5 13 17 8 13 — 5

100-hp Mercury 8 10 23 31 12 23 — 11

Total 83

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Using the EDD rule we obtain this schedule:

Repair Sequence

Days Since Order

ArrivedProcessing

TimeFinish Time

Flow Time

Promise Date

Actual Pickup Date

Days Early

Days Past Due

50-hp Evinrude 4 5 5 9 8 8 3 —

75-hp Nautique 15 3 8 23 10 10 2 —

100-hp Mercury 8 10 18 26 12 18 — 6


7-hp Johnson 6 4 22 28 15 22 — 7

50-hp Honda 1 1 23 24 20 23 — 3

Total 110


b. Performance measures are as follows:Average flow time is 16.6 (or 83/5) days for SPT and 22.0 (or 110/5) days for EDD The percentage of past due jobs is 40110/5) days for EDD. The percentage of past due jobs is 40 percent (2/5) for SPT and 60 percent (3/5) for EDD. For this set of jobs, the EDD schedule minimizes the maximum days past due but has a greater flow time and causes more jobs to be past due.


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The following data were reported by the shop floor control system for order processing at the edge grinder. The current date is day 150. The number of remaining operations and the t t l k i i i l d th ti t th d i dtotal work remaining include the operation at the edge grinder. All orders are available for processing, and none have been started yet. Assume the jobs were available for processing at the same time.

Current OrderProcessing

Time (hr)Due Date

(day)Remaining Operations

Shop Time Remaining

(days)

A101 10 162 10 9


B272 7 158 9 6

C106 15 152 1 1

D707 4 170 8 18

E555 8 154 5 8


a. Specify the priorities for each job if the shop floor control system uses slack per remaining operations (S/RO) or critical ratio (CR).

b. For each priority rule, calculate the average flow time per job at the edge grinder.


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Solved Problem 2Solved Problem 2SOLUTIONa. We specify the priorities for each job using the two

sequencing rules. The sequence for S/RO is shown in the brackets.brackets.

( )remaining operations of Number

remainingtimeShopdatesToday'date DueS/RO −−=

( ) [ ]18005

8150154S/RO:E555 .−=−−

=

( ) [ ]22209

6150158S/RO:B272 .=−−

=


( ) [ ]32508

18150170S/RO:D707 .=−−

=

( ) [ ]430010

9150162S/RO:A101 .=−−

=

( ) [ ]50011

1150152S/RO:C105 .=−−

=

Solved Problem 2Solved Problem 2The sequence of production for CR is shown in the brackets.

datesToday'dateDueCR −=

remainingtimeShopCR

[ ]15008

150154CR:E555 .=−

=

[ ]3331150158CR:B272 −

[ ]211118

150170CR:D707 .=−

=


[ ]33316

CR:B272 .==

[ ]43319

150162CR:A101 .=−

=

[ ]50021

150152CR:C105 .=−

=

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b. We are sequencing a set of jobs at a single machine, so each job’s finish time equals the finish time of the job just prior to it in sequence plus its own processing time. Further, all jobs

il bl f i t th ti h j b’were available for processing at the same time, so each job’s finish time equals its flow time. Consequently, the average flow times at this single machine are

hours 23.305

442919158S/RO =++++:

hours 22.45

442919128:CR =++++


5

In this example, the average flow time per job is lower for the CR rule, which is not always the case. For example, the critical ratios for B272 and A101 are tied at 1.33. If we arbitrarily assigned A101 before B272, the average flow time would increase to (8 + 12 + 22 + 29 + 44)/5 = 23.0 hours.


The Rocky Mountain Arsenal, formerly a chemical warfare manufacturing site, is said to be one of the most polluted locations in the United States. Cleanup of chemical waste

t b i ill i l t tistorage basins will involve two operations.Operation 1: Drain and dredge basin.Operation 2: Incinerate materials.Management estimates that each operation will require the following amounts of time (in days):

St B i


Storage Basin

A B C D E F G H I J

Dredge 3 4 3 6 1 3 2 1 8 4

Incinerate 1 4 2 1 2 6 4 1 2 8

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Management’s objective is to minimize the makespan of the cleanup operations. All storage basins are available for processing right now. First, find a schedule that minimizes the

k Th l l t th fl ti f tmakespan. Then calculate the average flow time of a storage basin through the two operations. What is the total elapsed time for cleaning all 10 basins? Display the schedule in a Gantt machine chart.

SOLUTIONWe can use Johnson’s rule to find the schedule that minimizes th t t l k F j b ti d f th h t t


the total makespan. Four jobs are tied for the shortest process time: A, D, E, and H. E and H are tied for first place, while A and D are tied for last place. We arbitrarily choose to start with basin E, the first on the list for the drain and dredge operation. The 10 steps used to arrive at a sequence are as follows:


2. Select basin H next; put it toward the front

E H — — — — — — — —

1. Select basin E first (tied with basin H); put it at the front.

E — — — — — — — — —

toward the front.3. Select basin A next (tied with

basin D); put it at the end.E H — — — — — — — A

4. Put basin D toward the end. E H — — — — — — D A

5. Put basin G toward the front. E H G — — — — — D A

7. Put basin I toward the end. E H G — — — I C D A

6. Put basin C toward the end. E H G — — — — C D A


8. Put basin F toward the front. E H G F — — I C D A

9. Put basin B toward the front. E H G F B — I C D A

10. Put basin J in the remaining space.

E H G F B J I C D A

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Several optimal solutions are available to this problem because of the ties at the start of the scheduling procedure. However, all have the same makespan. The schedule would be as follows:

Operation 1 Operation 2Basin Start Finish Start Finish

E 0 1 1 3H 1 2 3 4G 2 4 4 8F 4 7 8 14B 7 11 14 18


J 11 15 18 26I 15 23 26 28C 23 26 28 32D 26 32 32 35A 32 35 35 36

Total 200


The makespan is 36 days. The average flow time is the sum of incineration finish times divided by 10, or 200/10 = 20 days. The Gantt machine chart for this schedule is given in Figure J.3.

E H G F B J I C D A

E H G F B J I C D A

Storage Basin

Dredge

Incinerate


Figure J.3

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J Operations Scheduling