Operations Scheduling Supplement J Copyright ©2013 Pearson Education, Inc. publishing as Prentice...

Operations SchedulingSupplement J

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 01

Operations Scheduling

• Operations scheduling– A type of scheduling in which jobs are assigned

to workstations or employees are assigned to jobs for specified time periods.

• Additional performance measures

• Priority sequencing rules

• Scheduling multiple workstations

• Scheduling a two-station flow shopCopyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 02

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

– Front-office process has high customer contact, divergent work flows, customization, and a complex scheduling environment.

– Back-office process has low customer involvement, uses more line work flows, and provides standardized services.

Scheduling Service and Manufacturing Processes

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J -03

Performance Measures• Flow time – The time a job spends in the service or

manufacturing system• Past due (tardiness) – The amount of time by which a job missed its due

date• Makespan– The total amount of time required to complete a

group of jobs

Makespan =Time of completion

of last job– Starting time

of first jobCopyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 04

Performance Measures• Total inventory – A term used to measure the effectiveness of

schedules for manufacturing processes.

• Utilization– The degree to which equipment, space, or the

workforce is currently being used– The ratio of average output rate to maximum

capacity (%)

Total Inventory = Scheduled receipts

for all items +On-hand

inventories of all items


Sequencing Jobs• An operation with divergent flows is often

called a job shop

– Low-to medium-volume production

– Utilizes job or batch processes

– The front office would be the equivalent for a service provider.

– It is difficult to schedule because of the variability in job routings and the continual introduction of new jobs to be processed.


Sequencing Jobs

• An operation with line flow is often called a flow shop

– Medium- to high-volume production

– Utilizes line or continuous flow processes

– The 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 system.


Ship

ping

Dep

artm

ent

Raw

Mat

eria

ls

Legend:Batch of partsWorkstation

Job Shop Sequencing

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J- 08

• First-come, first-served (FCFS) • Earliest due date (EDD)• Critical ratio (CR)

Priority Sequencing Rules

• A ratio less than 1.0 implies that the job is behind schedule.

• A ratio greater than 1.0 implies the job is ahead of schedule.

• The job with the lowest CR is scheduled next.

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


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

Priority Sequencing Rules

• The job with the lowest S/RO is scheduled next

S/RO =

Due date

Today’sdate

Total shop time remaining– –

Number of operations remaining


• Single-dimension rules

–A set of rules that bases the priority of a job on a single aspect of the job, such as arrival time at the workstation, the due date, or the processing time.

Sequencing Jobs for One Workstation


Example J.1• The Taylor Machine Shop rebores engine blocks. • Currently, five engine blocks are waiting for

processing. • At any time, the 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.


Example J.1• 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 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?


Example J.1

Engine Block

Business Hours Since Order

Arrived

Processing Time, Including Setup

(hours)

Business Hours Until 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


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

Ranger

Explorer

Econoline 150

Bronco

Thunderbird

12 0 + 8 = 8 20 10 10 2 —

10 8 + 6 = 14 24 12 13 — 2

Example J.1

a. The EDD rule states that the first engine block in the sequence is the one with the closest due date. Consequently, the Ranger engine block is processed first. The Thunderbird engine block, with its due date furthest in the future, is processed last.

3 14 + 3 = 17 20 18 18 1 —

1 17 + 15 = 32 33 20 32 — 12

0 32 + 12 = 44 44 22 44 — 22Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 15

• The flow time for each job is its finish time, plus the time since the job arrived.

• 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.

• Sum of flow times is the total job hours spent by the engine blocks since their orders arrived at the workstation until they were processed.


Example J.1

Example J.1

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

Average flow time =

Average hours early =

Average hours past due =

20 + 24 + 20 + 33 + 445

= 28.2 hours

2 + 0 + 1 + 0 + 05

= 0.6 hour

0 + 2 + 0 + 12 + 225

= 7.2 hours


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

Explorer

Ranger

Thunderbird

Bronco

3 0 + 3 = 3 6 19 18 15 —

10 3 + 6 = 9 19 12 12 3 —

Example J.1

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.

12 9 + 8 = 17 29 10 17 — 7

0 17 + 12 = 29 29 22 29 — 7

1 29 + 15 = 44 45 20 44 — 24Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 18

Example J.1

The performance measures are:

Average flow time =

Average hours early =

Average hours past due =

6 + 19 + 29 + 29 + 455

= 25.6 hours

15 + 3 + 0 + 0 + 05

= 3.6 hours

0 + 0 + 7 + 7 + 245

= 7.6 hours


Example J.1• EDD rule

– Performs well with respect to the percentage of jobs past due and the variance of hours past due

– Is popular with firms that are sensitive to achieving due dates• SPT rule

– Tends to minimize the mean flow and maximize shop utilization– For single-workstations will always provide the lowest mean

finish time– Could increase total inventory – Tends to produce a large variance in past due hours

• FCFS rule– Is considered fair– Performs poorly with respect to all performance measures


Application J.1

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

Order

Standard Time, Including

Setup (hour)Due Date

(hrs from now)AZ135 14 14DM246 8 20SX435 10 6PC088 3 18


Order Sequence

Begin Work

Finish Time (hr)

FlowTime(hr)

Scheduled Customer

Pickup Time

Actual Pickup Time

Hours Early

Hours Past Due

1.

2.

3.

4.

TotalAverage

PC088

DM246

SX435

AZ135

0 3 3 18 18 153 11 11 20 20 9

11 21 21 6 21 1521 35 35 14 35 21

70 94 24 3617.5 23.5 6 9

Application J.1


Sequencing Jobs for One Workstation

• Multiple-dimension rules – A set of rules that apply to more than one aspect of a job.

• Choosing a rule

– S/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 times.

– CR results in longer job flow times than SPT, but CR also results in less variance in the distribution of past due hours.

– No choice is clearly best; each rule should be tested in the environment for which it is intended.Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

J - 23

• The first five columns of the following table contain information about a set of four jobs that just arrived (end of hour 0 or 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.

Job

Processing Time at Engine Lathe

(hours)

Time Remaining Until Due Date

(days)

Number of Operations Remaining

Shop Time 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 –0.44Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 24

Example J.2

a. Using CR to schedule the machine, we divide the time remaining until the due date by the shop time remaining to get the priority index for each job.

CR = Time remaining until the due dateShop time remaining

= = 2.46156.1

By arranging the jobs in sequence with the lowest critical 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.


Job 1

Example J.2

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.

S/RO =

Time remaininguntil the due date

Shop timeremaining–

Number of operations remaining = = 0.8915 – 6.1

10

Arranging the jobs by starting with the lowest S/RO yields a 4, 3, 1, 2 sequence of jobs.


Job 1

Example J.2

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


Example J.2

Application J.2

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 4BB 8 11 4 6CC 13 16 10 9DD 6 18 3 12EE 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.


Critical Ratio Slack/Remaining Operation

Job Priority Index

Sequence on

Drill PressJob Priority

Index

Sequence on

Drill Press

Application J.2

AABBCCDDEE

1.251.831.771.502.33

FirstFourthThird

SecondFifth

AABBCCDDEE

0.331.250.702.000.80

FirstFourthSecond

Fifth

Third


• Identifying the best priority rule to use at a particular operation in a process is a complex problem because the output from one operation becomes the input to another.

• Computer simulation models are effective tools to determine which priority rules work best in a given situation.

Scheduling Jobs for Multiple Workstations


• In the scheduling of two or more workstations in a flow shop, the makespan varies according to the sequence chosen.

• Determining 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 utilization of the two-station flow shop is

maximized.

Scheduling Jobs for a Two-Station Flow Shop


Johnson’s Rule• Minimizes makespan when scheduling a group of jobs

on two workstations

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.

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.

3. Eliminate the last job scheduled from further consideration. Repeat steps 1 and 2 until all jobs have been scheduled.Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall J - 32

Example J.3• The Morris Machine Company just received an

order to refurbish 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.


Example J.3• 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 authorized 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 5M3 6 3M4 15 16M5 10 8


Example J.3Establishing a Job Sequence

Iteration 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 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.


Example J.3

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

Workstation

0 5 10 15 20 25 30Hour

35 40 45 50 55 60 65

1

2


Application J.3

Use the following data to schedule two workstations arranged as a flow shop using Johnson’s Rule:

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

Sequence: C B E D A


Application J.3

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


• Labor-Limited Environment – An environment in which the resource constraint is the

amount of labor available, not the number of machines or workstations

• Some possible labor assignment rules:– Assign personnel to the workstation with the job that

has been in the system longest.– Assign personnel to the workstation with the most

jobs waiting for processing.– Assign personnel to the workstation with the largest

standard work content.– Assign personnel to the workstation with the job that

has the earliest due date.

Labor-Limited Environment


• 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.

• Currently, five engines with varying problems are awaiting service.

• Customers usually do not pick up their engines early.

Solved Problem 1


Solved Problem 1Using the table below:a. Develop separate schedules by using the SPT and EDD

rules.b. Compare the two schedules on the basis of average flow

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

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


Solved Problem 1

a. 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 — 5100-hp Mercury 8 10 23 31 12 23 — 11

Total 83


Solved Problem 1

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


Solved Problem 1b. 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 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.


Solved Problem 2• 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 total 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.


Solved Problem 2Using the Table below: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.

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


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

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

remaining operations of Number

remaining time Shopdate sToday'date DueS/RO

154 150 8E555:S/RO 1

5

-0.80

158 150 6B272:S/RO 2

9

0.22

170 150 18D707:S/RO 3

8

0.25

162 150 9A101:S/RO 4

10

0.30

152 150 1C105:S/RO 5

1

1.00


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

remaining time Shopdate sToday'date Due

CR

154 150E555:CR 1

8

0.50

158 150B272:CR 3

6

1.33

170 150D707:CR 2

18

1.11

162 150A101:CR 4

9

1.33

152 150C105:CR 5

1

2.00


Solved Problem 2

b. The average flow times at this single machine are:

8 15 19 29 44S/RO : hours

5

23.30

8 12 19 29 44CR: hours

5

22.4

• In this example, the average flow time per job is lower for the CR rule, which is not always the case.

• If we arbitrarily assigned A101 before B272, the average flow time would increase to (8 + 12 + 22 + 29 + 44)/5 = 23.0 hours.


Solved Problem 3The 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 storage 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):

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


Solved Problem 3

• 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 makespan.

• 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.


Solved Problem 3

• We can use Johnson’s rule to find the schedule that minimizes 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:


Solved Problem 3

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

E H — — — — — — — —

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

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

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

E — — — — — — — — —

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


Solved Problem 3Several optimal solutions are available to this problem because of the ties at the start of the scheduling procedure. However, all have the same makespan.

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 18J 11 15 18 26I 15 23 26 28C 23 26 28 30D 26 32 32 33A 32 35 35 36

Total 200


E H G F B J I C D A

E H G F B J I C D A

Solved Problem 3• The makespan is 36 days.

• The average flow time is the sum of incineration finish times divided by 10, or 200/10 = 20 days.

Dredge

Incinerate



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