Transcript
Page 1: 1 Slide © 2005 Thomson/South-Western Chapter 10 Project Scheduling: PERT/CPM Project Scheduling with Known Activity Times Project Scheduling with Known

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© 2005 Thomson/South-Western© 2005 Thomson/South-Western

Chapter 10Chapter 10Project Scheduling: Project Scheduling:

PERT/CPMPERT/CPM Project Scheduling with Known Project Scheduling with Known

Activity TimesActivity Times

Project Scheduling with Uncertain Project Scheduling with Uncertain Activity TimesActivity Times

Considering Time-Cost Trade-OffsConsidering Time-Cost Trade-Offs

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PERT/CPMPERT/CPM PERTPERT

• Program Evaluation and Review TechniqueProgram Evaluation and Review Technique•Developed by U.S. Navy for Polaris missile Developed by U.S. Navy for Polaris missile

projectproject•Developed to handle uncertain activity timesDeveloped to handle uncertain activity times

CPMCPM•Critical Path MethodCritical Path Method•Developed by Du Pont & Remington RandDeveloped by Du Pont & Remington Rand•Developed for industrial projects for which Developed for industrial projects for which

activity times generally were knownactivity times generally were known Today’s project management software packages Today’s project management software packages

have combined the best features of both have combined the best features of both approaches.approaches.

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PERT/CPMPERT/CPM

PERT and CPM have been used to PERT and CPM have been used to plan, schedule, and control a wide plan, schedule, and control a wide variety of projects:variety of projects:

•R&D of new products and processesR&D of new products and processes

•Construction of buildings and Construction of buildings and highwayshighways

•Maintenance of large and complex Maintenance of large and complex equipmentequipment

•Design and installation of new Design and installation of new systemssystems

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PERT/CPMPERT/CPM

PERT/CPM is used to plan the PERT/CPM is used to plan the scheduling of individual scheduling of individual activitiesactivities that make up a project.that make up a project.

Projects may have as many as Projects may have as many as several thousand activities.several thousand activities.

A complicating factor in carrying A complicating factor in carrying out the activities is that some out the activities is that some activities depend on the activities depend on the completion of other activities completion of other activities before they can be started.before they can be started.

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PERT/CPMPERT/CPM Project managers rely on PERT/CPM to Project managers rely on PERT/CPM to

help them answer questions such as:help them answer questions such as:

•What is the What is the total timetotal time to complete to complete the project?the project?

•What are the What are the scheduled start and scheduled start and finish datesfinish dates for each specific activity? for each specific activity?

•Which activities are Which activities are criticalcritical and must and must be completed exactly as scheduled to be completed exactly as scheduled to keep the project on schedule?keep the project on schedule?

•How long can How long can noncritical activitiesnoncritical activities be be delayed before they cause an delayed before they cause an increase in the project completion increase in the project completion time?time?

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Project NetworkProject Network

A A project networkproject network can be can be constructed to model the constructed to model the precedence of the activities. precedence of the activities.

The The nodesnodes of the network of the network represent the activities. represent the activities.

The The arcsarcs of the network reflect the of the network reflect the precedence relationships of the precedence relationships of the activities. activities.

A A critical pathcritical path for the network is a for the network is a path consisting of activities with path consisting of activities with zero slack.zero slack.

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Frank’s Fine Floats is in the business of Frank’s Fine Floats is in the business of building elaborate parade floats. Frank and building elaborate parade floats. Frank and his crew have a new float to build and want to his crew have a new float to build and want to use PERT/CPM to help them manage the use PERT/CPM to help them manage the projectproject . .

The table on the next slide shows the The table on the next slide shows the activities that comprise the project. Each activities that comprise the project. Each activity’s estimated completion time (in days) activity’s estimated completion time (in days) and immediate predecessors are listed as well.and immediate predecessors are listed as well.

Frank wants to know the total time to Frank wants to know the total time to complete the project, which activities are complete the project, which activities are critical, and the earliest and latest start and critical, and the earliest and latest start and finish dates for each activity.finish dates for each activity.

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Immediate CompletionImmediate Completion

ActivityActivity DescriptionDescription PredecessorsPredecessors Time (days)Time (days)

A Initial Paperwork A Initial Paperwork --- --- 3 3

B Build Body B Build Body A A 3 3

C Build Frame C Build Frame A A 2 2

D Finish Body D Finish Body B B 3 3

E Finish Frame E Finish Frame C C 7 7

F Final Paperwork F Final Paperwork B,C B,C 3 3

G Mount Body to Frame D,EG Mount Body to Frame D,E 6 6

H Install Skirt on Frame CH Install Skirt on Frame C 2 2

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Project NetworkProject Network

StartStart FinishFinish

BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

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Earliest Start and Finish TimesEarliest Start and Finish Times

Step 1:Step 1: Make a forward pass through the Make a forward pass through the network as follows: For each activity network as follows: For each activity i i beginning at the Start nodebeginning at the Start node, , compute:compute:

• Earliest Start TimeEarliest Start Time = the maximum of = the maximum of the earliest finish times of all activities the earliest finish times of all activities immediately preceding activity immediately preceding activity ii. (This is . (This is 0 for an activity with no predecessors.)0 for an activity with no predecessors.)

• Earliest Finish TimeEarliest Finish Time = (Earliest Start = (Earliest Start Time) + (Time to complete activity Time) + (Time to complete activity i i ).).

The project completion time is the The project completion time is the maximum of the Earliest Finish Times at maximum of the Earliest Finish Times at the Finish node.the Finish node.

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Earliest Start and Finish TimesEarliest Start and Finish Times

StartStart FinishFinish

BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

0 30 3

3 63 6 6 96 9

3 53 5

12 12 1818

6 96 9

5 75 7

5 125 12

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Latest Start and Finish TimesLatest Start and Finish Times

Step 2:Step 2: Make a backwards pass through Make a backwards pass through the network as follows: Move the network as follows: Move sequentially backwards from the Finish sequentially backwards from the Finish node to the Start node. At a given node, node to the Start node. At a given node, jj, consider all activities ending at node, consider all activities ending at node j j. . For each of these activities, For each of these activities, ii, compute:, compute:

• Latest Finish TimeLatest Finish Time = the minimum of = the minimum of the latest start times beginning at the latest start times beginning at node node jj. (For node . (For node NN, this is the project , this is the project completion time.)completion time.)

• Latest Start TimeLatest Start Time = (Latest Finish = (Latest Finish Time) - (Time to complete activity Time) - (Time to complete activity i i ).).

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Latest Start and Finish TimesLatest Start and Finish Times

StartStart FinishFinish

BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

0 30 3

3 63 6 6 96 9

3 53 5

12 12 1818

6 96 9

5 75 7

5 125 12

6 96 9 9 9 1212

0 30 3

3 53 5

12 12 1818

15 15 1818

16 16 1818

5 125 12

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Determining the Critical PathDetermining the Critical Path

Step 3:Step 3: Calculate the slack time for each Calculate the slack time for each activity by: activity by:

SlackSlack = (Latest Start) - (Earliest = (Latest Start) - (Earliest Start), or Start), or

= (Latest Finish) - = (Latest Finish) - (Earliest Finish).(Earliest Finish).

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Activity Slack TimeActivity Slack Time

ActivityActivity ESES EFEF LSLS LFLF SlackSlack A 0 3 0 3 0 A 0 3 0 3 0

(critical)(critical) B 3 6 6 9 3B 3 6 6 9 3 C 3 5 3 5 0 C 3 5 3 5 0

(critical)(critical) D 6 9 9 12 3D 6 9 9 12 3 E 5 12 5 12 0 E 5 12 5 12 0

(critical)(critical) F 6 9 15 18 9F 6 9 15 18 9 G 12 18 12 18 0 G 12 18 12 18 0

(critical)(critical) H 5 7 16 18 11H 5 7 16 18 11

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Determining the Critical PathDetermining the Critical Path

• A A critical pathcritical path is a path of activities, from the is a path of activities, from the Start node to the Finish node, with 0 slack Start node to the Finish node, with 0 slack times.times.

• Critical Path: A – C – E – GCritical Path: A – C – E – G

• The The project completion timeproject completion time equals the equals the maximum of the activities’ earliest finish maximum of the activities’ earliest finish times.times.

• Project Completion Time: 18 daysProject Completion Time: 18 days

Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Critical PathCritical Path

StartStart FinishFinish

BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

0 30 3

3 63 6 6 96 9

3 53 5

12 12 1818

6 96 9

5 75 7

5 125 12

6 96 9 9 9 1212

0 30 3

3 53 5

12 12 1818

15 15 1818

16 16 1818

5 125 12

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In the In the three-time estimate approachthree-time estimate approach, the , the time to complete an activity is assumed to time to complete an activity is assumed to follow a Beta distribution. follow a Beta distribution.

An activity’s An activity’s mean completion timemean completion time is: is:

tt = ( = (aa + 4 + 4mm + + bb)/6)/6

• aa = the = the optimisticoptimistic completion time completion time estimateestimate

• bb = the = the pessimisticpessimistic completion time completion time estimateestimate

• mm = the = the most likelymost likely completion time completion time estimateestimate

Uncertain Activity TimesUncertain Activity Times

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An activity’s An activity’s completion time variancecompletion time variance is: is:

22 = (( = ((bb--aa)/6))/6)22

• aa = the = the optimisticoptimistic completion time completion time estimateestimate

• bb = the = the pessimisticpessimistic completion time completion time estimateestimate

• mm = the = the most likelymost likely completion time completion time estimateestimate

Uncertain Activity TimesUncertain Activity Times

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Uncertain Activity TimesUncertain Activity Times

In the three-time estimate approach, In the three-time estimate approach, the critical path is determined as if the the critical path is determined as if the mean times for the activities were mean times for the activities were fixed times. fixed times.

The The overall project completion timeoverall project completion time is is assumed to have a normal distribution assumed to have a normal distribution with mean equal to the sum of the with mean equal to the sum of the means along the critical path and means along the critical path and variance equal to the sum of the variance equal to the sum of the variances along the critical path.variances along the critical path.

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Example: ABC Associates Example: ABC Associates

Consider the following project:Consider the following project:

Immed. Optimistic Most Likely PessimisticImmed. Optimistic Most Likely Pessimistic

ActivityActivity Predec.Predec. Time (Hr.Time (Hr.) ) Time (Hr.)Time (Hr.) Time (Hr.)Time (Hr.) A A -- 4 -- 4 6 6 8 8 B B -- 1 -- 1 4.5 4.5

5 5 C C A A 3 3 3 3

3 3 D D A 4 5 A 4 5 6 6 E E A 0.5 1 A 0.5 1

1.51.5 F F B,C 3 4 5 B,C 3 4 5 G G B,C B,C 1 1.5 1 1.5

5 5 H H E,F E,F 5 6 5 6

7 7 I I E,F 2 5 8 E,F 2 5 8 J J D,H D,H 2.5 2.75 2.5 2.75

4.5 4.5 K K G,I 3 5 7 G,I 3 5 7

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Example: ABC AssociatesExample: ABC Associates

Project NetworkProject Network

E

S tart

A

H

D

F

J

I

K

F in ish

B

C

G

66

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44

1166

33

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Example: ABC AssociatesExample: ABC Associates

Activity Expected Times and VariancesActivity Expected Times and Variances

tt = ( = (aa + 4 + 4mm + + bb)/6 )/6 22 = (( = ((bb--aa)/6))/6)22

ActivityActivity Expected TimeExpected Time VarianceVariance A A 6 6 4/9 4/9

B B 4 4 4/9 4/9 C C 3 3 0 0 D D 5 5 1/9 1/9 E E 1 1 1/36 1/36 F F 4 4 1/9 1/9 G G 2 2 4/9 4/9 H H 6 6 1/9 1/9 I I 5 5 1 1 J J 3 3 1/9 1/9 K K 5 5 4/9 4/9

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Example: ABC AssociatesExample: ABC Associates

Earliest/Latest Times and SlackEarliest/Latest Times and Slack

ActivityActivity ESES EF EF LSLS LFLF SlackSlack A A 0 6 0 6 0 * 0 6 0 6 0 *

B B 0 4 5 9 5 0 4 5 9 5 C 6 9 6 9 0 *C 6 9 6 9 0 * D D 6 11 15 20 9 6 11 15 20 9 E E 6 7 12 13 6 6 7 12 13 6 F F 9 13 9 13 0 * 9 13 9 13 0 * G 9 11 16 18 7G 9 11 16 18 7 H H 13 19 14 20 1 13 19 14 20 1 I I 13 18 13 18 0 * 13 18 13 18 0 * J J 19 22 20 23 1 19 22 20 23 1 K K 18 23 18 23 0 * 18 23 18 23 0 *

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Determining the Critical PathDetermining the Critical Path

• A A critical pathcritical path is a path of activities, from the is a path of activities, from the Start node to the Finish node, with 0 slack Start node to the Finish node, with 0 slack times.times.

• Critical Path: A – C – F – I – KCritical Path: A – C – F – I – K

• The The project completion timeproject completion time equals the equals the maximum of the activities’ earliest finish maximum of the activities’ earliest finish times.times.

• Project Completion Time: 23 hoursProject Completion Time: 23 hours

Example: ABC AssociatesExample: ABC Associates

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Example: ABC AssociatesExample: ABC Associates

Critical Path (A-C-F-I-K)Critical Path (A-C-F-I-K)

E

S tart

A

H

D

F

J

I

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F in ish

B

C

G

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55

22

44

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33

55

0 60 60 60 6

9 139 139 139 13

13 1813 1813 1813 18

9 119 1116 1816 18

13 1913 1914 2014 20

19 2219 2220 2320 23

18 2318 2318 2318 23

6 76 712 1312 13

6 96 96 96 9

0 40 45 95 9

6 116 1115 2015 20

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Probability the project will be completed within 24 Probability the project will be completed within 24 hrshrs

22 = = 22AA + + 22

CC + + 22FF + + 22

HH + + 22KK

= 4/9 + 0 + 1/9 + 1 + 4/9 = 4/9 + 0 + 1/9 + 1 + 4/9

= 2= 2

= 1.414= 1.414

zz = (24 - 23)/ = (24 - 23)/(24-23)/1.414 (24-23)/1.414 = .71= .71

From the Standard Normal Distribution table: From the Standard Normal Distribution table:

P(z P(z << .71) = .5 + .2612 = .7612 .71) = .5 + .2612 = .7612

Example: ABC AssociatesExample: ABC Associates

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EarthMover is a manufacturer of road EarthMover is a manufacturer of road constructionconstruction

equipment including pavers, rollers, and graders. equipment including pavers, rollers, and graders. TheThe

company is faced with a newcompany is faced with a new

project, introducing a newproject, introducing a new

line of loaders. Managementline of loaders. Management

is concerned that the project mightis concerned that the project might

take longer than 26 weeks totake longer than 26 weeks to

complete without crashing somecomplete without crashing some

activities.activities.

Example: EarthMover, Inc.Example: EarthMover, Inc.

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Immediate Immediate CompletionCompletion ActivityActivity DescriptionDescription PredecessorsPredecessors Time (wks)Time (wks)

A Study Feasibility A Study Feasibility --- --- 6 6 B Purchase Building B Purchase Building A A

4 4 C Hire Project Leader C Hire Project Leader A A

3 3 D Select Advertising StaffD Select Advertising Staff B B

6 6 E Purchase Materials E Purchase Materials B B

3 3 F Hire Manufacturing Staff B,CF Hire Manufacturing Staff B,C

10 10 G Manufacture Prototype E,FG Manufacture Prototype E,F

2 2 H Produce First 50 Units GH Produce First 50 Units G

6 6 II Advertise Product D,G Advertise Product D,G

8 8

Example: EarthMover, Inc.Example: EarthMover, Inc.

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PERT NetworkPERT Network

Example: EarthMover, Inc.Example: EarthMover, Inc.

C

S tart

D

E

I

A

F in ish

H

G

B

F

C

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D

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F in ish

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F

6644

331010

33

66

22 66

88

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Earliest/Latest TimesEarliest/Latest Times

ActivityActivity ESES EFEF LSLS LFLF SlackSlack A A 0 6 0 6 0 * 0 6 0 6 0 * B B 6 10 6 10 0 * 6 10 6 10 0 * C C 6 9 7 10 1 6 9 7 10 1 D 10 16 16 22 6D 10 16 16 22 6 E E 10 13 17 20 7 10 13 17 20 7 F F 10 20 10 20 0 * 10 20 10 20 0 * G G 20 22 20 22 0 * 20 22 20 22 0 * H H 22 28 24 30 2 22 28 24 30 2 I I 22 30 22 30 0 * 22 30 22 30 0 *

Example: EarthMover, Inc.Example: EarthMover, Inc.

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Example: EarthMover, Inc.Example: EarthMover, Inc.

Critical ActivitiesCritical Activities

C

S tart

D

E

I

A

F in ish

H

G

B

F

C

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D

E

I

A

F in ish

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B

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6644

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66

22 66

880 60 60 60 6

10 2010 20 10 2010 20

20 2220 2220 2220 22

10 1610 1616 2216 22 22 3022 30

22 3022 30

22 2822 2824 3024 30

6 96 9 7 107 10

10 1310 1317 2017 20

6 106 10 6 106 10

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Ch. 10 – 7A project involving the installation of a computer system comprises eight activities. The following table lists immediate predecessors and activity times (in weeks).

ActivityImmediate

Predecessor Time

ABCDEFGH

--A

B,CDE

B,CF,G

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a. Draw a project network.b. What are the critical activities?c. What is the expected project completion time?


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