McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003 8.1 Table of Contents Chapter 8 (PERT/CPM...

© The McGraw-Hill Companies, Inc., 20038.1McGraw-Hill/Irwin

Table of ContentsChapter 8 (PERT/CPM Models for Project Management)

A Case Study: The Reliable Construction Co. Project (Section 8.1) 8.2–8.3Using a Network to Visually Display a Project (Section 8.2) 8.4–8.7Scheduling a Project with PERT/CPM (Section 8.3) 8.8–8.17Dealing with Uncertain Activity Durations (Section 8.4) 8.18–8.26Considering Time-Cost Tradeoffs (Section 8.5) 8.27–8.38Scheduling and Controlling Project Costs (Section 8.6) 8.39–8.46

Introduction to Project Management (UW Lecture) 8.47–8.57These slides are based upon a lecture from the course “Principles of Operations Management”, (an undergraduate business class) at the University of Washington (as taught by one of the authors).

Project Management with Uncertain Activity Durations (UW Lecture) 8.58–8.69These slides are based upon a lecture from the course “Principles of Operations Management” (an undergraduate business class) at the University of Washington (as taught by one of the authors).

Time-Cost Tradeoffs (UW Lecture) 8.70–8.81These slides are based upon a lecture from the course “Principles of Operations Management” (an undergraduate business class) at the University of Washington (as taught by one of the authors).


Reliable Construction Company Project

• The Reliable Construction Company has just made the winning bid of $5.4 million to construct a new plant for a major manufacturer.

• The contract includes the following provisions:– A penalty of $300,000 if Reliable has not completed construction within 47 weeks.

– A bonus of $150,000 if Reliable has completed the plant within 40 weeks.

Questions:1. How can the project be displayed graphically to better visualize the activities?

2. What is the total time required to complete the project if no delays occur?

3. When do the individual activities need to start and finish?

4. What are the critical bottleneck activities?

5. For other activities, how much delay can be tolerated?

6. What is the probability the project can be completed in 47 weeks?

7. What is the least expensive way to complete the project within 40 weeks?

8. How should ongoing costs be monitored to try to keep the project within budget?


Activity List for Reliable Construction

Activity Activity DescriptionImmediate

PredecessorsEstimated

Duration (Weeks)

A Excavate — 2

B Lay the foundation A 4

C Put up the rough wall B 10

D Put up the roof C 6

E Install the exterior plumbing C 4

F Install the interior plumbing E 5

G Put up the exterior siding D 7

H Do the exterior painting E, G 9

I Do the electrical work C 7

J Put up the wallboard F, I 8

K Install the flooring J 4

L Do the interior painting J 5

M Install the exterior fixtures H 2

N Install the interior fixtures K, L 6


Project Networks

• A network used to represent a project is called a project network.

• A project network consists of a number of nodes and a number of arcs.

• Two types of project networks:– Activity-on-arc (AOA): each activity is represented by an arc. A node is used to

separate an activity from its predecessors. The sequencing of the arcs shows the precedence relationships.

– Activity-on-node (AON): each activity is represented by a node. The arcs are used to show the precedence relationships.

• Advantages of AON (used in this textbook):– considerably easier to construct

– easier to understand

– easier to revise when there are changes


Reliable Construction Project Network

A

START

G

H

M

F

J

K L

N

Activity Code

A. Excavate

B. Foundation

C. Rough wall

D. Roof

E. Exterior plumbing

F. Interior plumbing

G. Exterior siding

H. Exterior painting

I. Electrical work

J. Wallboard

K. Flooring

L. Interior painting

M. Exterior fixtures

N. Interior fixtures

2

4

10

746

7

9

5

8

4 5

6

2

0

0FINISH

D IE

C

B


Microsoft Project Gantt Chart


Microsoft Project—Project Network


The Critical Path

• A path through a network is one of the routes following the arrows (arcs) from the start node to the finish node.

• The length of a path is the sum of the (estimated) durations of the activities on the path.

• The (estimated) project duration equals the length of the longest path through the project network.

• This longest path is called the critical path. (If more than one path tie for the longest, they all are critical paths.)


The Paths for Reliable’s Project Network

Path Length (Weeks)

StartA B C D G H M Finish 2 + 4 + 10 + 6 + 7 + 9 + 2 = 40

Start A B C E H M Finish 2 + 4 + 10 + 4 + 9 + 2 = 31

Start A B C E F J K N Finish 2 + 4 + 10 + 4 + 5 + 8 + 4 + 6 = 43

Start A B C E F J L N Finish 2 + 4 + 10 + 4 + 5 + 8 + 5 + 6 = 44

Start A B C I J K N Finish 2 + 4 + 10 + 7 + 8 + 4 + 6 = 41

Start A B C I J L N Finish 2 + 4 + 10 + 7 + 8 + 5 + 6 = 42


Earliest Start and Earliest Finish Times

• The starting and finishing times of each activity if no delays occur anywhere in the project are called the earliest start time and the earliest finish time.

– ES = Earliest start time for a particular activity

– EF = Earliest finish time for a particular activity

• Earliest Start Time Rule: ES = Largest EF of the immediate predecessors.

• Procedure for obtaining earliest times for all activities:1. For each activity that starts the project (including the start node), set its ES = 0.

2. For each activity whose ES has just been obtained, calculate EF = ES + duration.

3. For each new activity whose immediate predecessors now have EF values, obtain its ES by applying the earliest start time rule. Apply step 2 to calculate EF.

4. Repeat step 3 until ES and EF have been obtained for all activities.


ES and EF Values for Reliable Constructionfor Activities that have only a Single Predecessor

A

START

G

H

M

F

J

FINISH

K L

N

D IE

C

B

2

4

10

746

7

9

5

8

4 5

6

2

ES = 0 EF = 2

ES = 2 EF = 6

ES = 16 EF = 22

ES = 16 EF = 20

ES = 16 EF = 23

ES = 20 EF = 25

ES = 22 EF = 29

ES = 6 EF = 16

0

0


ES and EF Times for Reliable Construction

A

START

G

H

M

F

J

FINISH

K L

N

D IE

C

B

2

4

10

746

7

9

5

8

4 5

6

2

ES = 0 EF = 2

ES = 2 EF = 6

ES = 16 EF = 22

ES = 16 EF = 20

ES = 16 EF = 23

ES = 20 EF = 25

ES = 22 EF = 29

ES = 6 EF = 16

ES = 0 EF = 0

ES = 25 EF = 33

ES = 33 EF = 38

ES = 38 EF = 44

ES = 33 EF = 37

ES = 29 EF = 38

ES = 38 EF = 40

ES = 44 EF = 44

0

0


Latest Start and Latest Finish Times

• The latest start time for an activity is the latest possible time that it can start without delaying the completion of the project (so the finish node still is reached at its earliest finish time). The latest finish time has the corresponding definition with respect to finishing the activity.

– LS = Latest start time for a particular activity

– LF = Latest finish time for a particular activity

• Latest Finish Time Rule: LF = Smallest LS of the immediate successors.

• Procedure for obtaining latest times for all activities:1. For each of the activities that together complete the project (including the finish

node), set LF equal to EF of the finish node.

2. For each activity whose LF value has just been obtained, calculate LS = LF – duration.

3. For each new activity whose immediate successors now have LS values, obtain its LF by applying the latest finish time rule. Apply step 2 to calculate its LS.

4. Repeat step 3 until LF and LS have been obtained for all activities.


LS and LF Times for Reliable’s Project

A

START

G

H

M

F

J

FINISH

K L

N

D IE

C

B

2

4

10

746

7

9

5

8

4 5

6

2

LS = 0 LF = 2

LS = 2 LF = 6

LS = 20 LF = 26

LS = 16 LF = 20

LS = 18 LF = 25

LS = 20 LF = 25

LS = 26 LF = 33

LS = 6 LF = 16

LS = 0 LF = 0

LS = 25 LF = 33

LS = 33 LF = 38

LS = 38 LF = 44

LS = 34 LF = 38

LS = 33 LF = 42

LS = 42 LF = 44

LS = 44 LF = 44

0

0


The Complete Project Network

A

START

G

H

M

F

J

FINISH

K L

N

D IE

C

B

2

4

10

746

7

9

5

8

4 5

6

2

S = (0, 0) F = (2, 2)

S = (2, 2) F = (6, 6)

S = (16, 20) F = (22, 26)

S = (16, 16) F = (20, 20)

S = (16, 18) F = (23, 25)

S = (20, 20) F = (25, 25)

S = (22, 26) F = (29, 33)

S = (6, 6) F = (16, 16)

S = (0, 0) F = (0, 0)

S = (25, 25) F = (33, 33)

S = (33, 33) F = (38, 38)

S = (38, 38) F = (44, 44)

S = (33, 34) F = (37, 38)

S = (29, 33) F = (38, 42)

S = (38, 42) F = (40, 44)

S = (44, 44) F = (44, 44)

0

0


Slack for Reliable’s Activities

Activity Slack (LF–EF) On Critical Path?

A 0 Yes

B 0 Yes

C 0 Yes

D 4 No

E 0 Yes

F 0 Yes

G 4 No

H 4 No

I 2 No

J 0 Yes

K 1 No

L 0 Yes

M 4 No

N 0 Yes


Spreadsheet to Calculate ES, EF, LS, LF, Slack

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B C D E F G H I JActivity Description Time ES EF LS LF Slack Critical?

A Excavate 2 0 2 0 2 0 YesB Foundation 4 2 6 2 6 0 YesC Rough Wall 10 6 16 6 16 0 YesD Roof 6 16 22 20 26 4 NoE Exterior Plumbing 4 16 20 16 20 0 YesF Interior Plumbing 5 20 25 20 25 0 YesG Exterior Siding 7 22 29 26 33 4 NoH Exterior Painting 9 29 38 33 42 4 NoI Electrical Work 7 16 23 18 25 2 NoJ Wallboard 8 25 33 25 33 0 YesK Flooring 4 33 37 34 38 1 NoL Interior Painting 5 33 38 33 38 0 YesM Exterior Fixtures 2 38 40 42 44 4 NoN Interior Fixtures 6 38 44 38 44 0 Yes

Project Duration 44


The PERT Three Estimate Approach

Most likely estimate (m) = Estimate of most likely value of the durationOptimistic estimate (o) = Estimate of duration under most favorable conditions.Pessimistic estimate (p) = Estimate of duration under most unfavorable conditions.

Elapsed time

p0

Beta distribution

mo


Mean and Standard Deviation

An approximate formula for the variance (2) of an activity is

2

p o6

2

An approximate formula for the mean () of an activity is

o 4m p6


Time Estimates for Reliable’s Project

Activity o m p Mean Variance

A 1 2 3 2 1/9

B 2 3.5 8 4 1

C 6 9 18 10 4

D 4 5.5 10 6 1

E 1 4.5 5 4 4/9

F 4 4 10 5 1

G 5 6.5 11 7 1

H 5 8 17 9 4

I 3 7.5 9 7 1

J 3 9 9 8 1

K 4 4 4 4 0

L 1 5.5 7 5 1

M 1 2 3 2 1/9

N 5 5.5 9 6 4/9


Pessimistic Path Lengths for Reliable’s Project

Path Pessimistic Length (Weeks)

StartA B C D G H M Finish 3 + 8 + 18 + 10 + 11 + 17 + 3 = 70

Start A B C E H M Finish 3 + 8 + 18 + 5 + 17 + 3 = 54

Start A B C E F J K N Finish 3 + 8 + 18 + 5 + 10 + 9 + 4 + 9 = 66

Start A B C E F J L N Finish 3 + 8 + 18 + 5 + 10 + 9 + 7 + 9 = 69

Start A B C I J K N Finish 3 + 8 + 18 + 9 + 9 + 4 + 9 = 60

Start A B C I J L N Finish 3 + 8 + 18 + 9 + 9 + 7 + 9 = 63


Three Simplifying Approximations of PERT/CPM

1. The mean critical path will turn out to be the longest path through the project network.

2. The durations of the activities on the mean critical path are statistically independent. Thus, the three estimates of the duration of an activity would never change after learning the durations of some of the other activities.

3. The form of the probability distribution of project duration is the normal distribution. By using simplifying approximations 1 and 2, there is some statistical theory (one version of the central limit theorem) that justifies this as being a reasonable approximation if the number of activities on the mean critical path is not too small.


Calculation of Project Mean and Variance

Activities on Mean Critical Path Mean Variance

A 2 1/9

B 4 1

C 10 4

E 4 4/9

F 5 1

J 8 1

L 5 1

N 6 4/9

Project duration p = 44 2p = 9


Probability of Meeting Deadline

44 (Mean)

47 (Deadline)

Project duration (in weeks)

2 = 9p

d - = 47 - 44 = 1p 3p


Probability of Meeting a Deadline

P(T ≤ d) P(T ≤ d)

–3.0 0.0014 0 0.50

–2.5 0.0062 0.25 0.60

–2.0 0.023 0.5 0.69

–1.75 0.040 0.75 0.77

–1.5 0.067 1.0 0.84

–1.25 0.11 1.25 0.89

–1.0 0.16 1.5 0.933

–0.75 0.23 1.75 0.960

–0.5 0.31 2.0 0.977

–0.25 0.40 2.5 0.9938

0 0.50 3.0 0.9986

d p

p

d p

p


Spreadsheet for PERT Three-Estimate Approach

3

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B C D E F G H I J KTime Estimates On Mean

Activity o m p Critical Path

A 1 2 3 * 2 0.1111 Mean CriticalB 2 3.5 8 * 4 1 PathC 6 9 18 * 10 4 44D 4 5.5 10 6 1 9E 1 4.5 5 * 4 0.4444F 4 4 10 * 5 1 P(T<=d) = 0.8413G 5 6.5 11 7 1 whereH 5 8 17 9 4 d = 47I 3 7.5 9 7 1J 3 9 9 * 8 1K 4 4 4 4 0L 1 5.5 7 * 5 1M 1 2 3 2 0.1111N 5 5.5 9 * 6 0.4444


Considering Time-Cost Trade-Offs

Question: If extra money is spent to expedite the project, what is the least expensive way of attempting to meet the target completion time (40 weeks)?

CPM Method of Time-Cost Trade-Offs:– Crashing an activity refers to taking special costly measures to reduce the duration

of an activity below its normal value. Special measures might include overtime, hiring additional temporary help, using special time-saving materials, obtaining special equipment, etc.

– Crashing the project refers to crashing a number of activities to reduce the duration of the project below its normal value.


Time-Cost Graph for an Activity

Activity duration

Activity cost

Crash cost

Normal cost Normal

Crash

Crash time Normal time


Time-Cost Trade-Off Data for Reliable’s Project

Time (weeks) Cost

MaximumReduction

in Time (weeks)

Crash Costper Week

SavedActivity Normal Crash Normal Crash

A 2 1 $180,000 $280,000 1 $100,000

B 4 2 320,000 420,000 2 50,000

C 10 7 620,000 860,000 3 80,000

D 6 4 260,000 340,000 2 40,000

E 4 3 410,000 570,000 1 160,000

F 5 3 180,000 260,000 2 40,000

G 7 4 900,000 1,020,000 3 40,000

H 9 6 200,000 380,000 3 60,000

I 7 5 210,000 270,000 2 30,000

J 8 6 430,000 490,000 2 30,000

K 4 3 160,000 200,000 1 40,000

L 5 3 250,000 350,000 2 50,000

M 2 1 100,000 200,000 1 100,000

N 6 3 330,000 510,000 3 60,000


Marginal Cost Analysis for Reliable’s ProjectInitial Table

Length of Path

Activityto

CrashCrashCost

AB

CD

GH

M

AB

CE

HM

AB

CE

FJK

N

AB

CE

FJL

N

AB

CIJK

N

AB

CIJL

N

40 31 43 44 41 42


Marginal Cost Analysis for Reliable’s ProjectTable After Crashing One Week

Length of Path

Activityto

CrashCrashCost

AB

CD

GH

M

AB

CE

HM

AB

CE

FJK

N

AB

CE

FJL

N

AB

CIJK

N

AB

CIJL

N

40 31 43 44 41 42

J $30,000 40 31 42 43 40 41


Marginal Cost Analysis for Reliable’s ProjectTable After Crashing Two Weeks

Length of Path

Activityto

CrashCrashCost

AB

CD

GH

M

AB

CE

HM

AB

CE

FJK

N

AB

CE

FJL

N

AB

CIJK

N

AB

CIJL

N

40 31 43 44 41 42

J $30,000 40 31 42 43 40 41

J $30,000 40 31 41 42 39 40


Marginal Cost Analysis for Reliable’s ProjectTable After Crashing Three Weeks

Length of Path

Activityto

CrashCrashCost

AB

CD

GH

M

AB

CE

HM

AB

CE

FJK

N

AB

CE

FJL

N

AB

CIJK

N

AB

CIJL

N

40 31 43 44 41 42

J $30,000 40 31 42 43 40 41

J $30,000 40 31 41 42 39 40

F $40,000 40 31 40 41 39 40


Marginal Cost Analysis for Reliable’s ProjectFinal Table After Crashing Four Weeks

Length of Path

Activityto

CrashCrashCost

AB

CD

GH

M

AB

CE

HM

AB

CE

FJK

N

AB

CE

FJL

N

AB

CIJK

N

AB

CIJL

N

40 31 43 44 41 42

J $30,000 40 31 42 43 40 41

J $30,000 40 31 41 42 39 40

F $40,000 40 31 40 41 39 40

F $40,000 40 31 39 40 39 40


Project Network After Crashing

A

START

G

H

M

F

J

FINISH

K L

N

D IE

C

B

2

4

10

746

7

9

3

6

4 5

6

2

S = (0, 0) F = (2, 2)

S = (2, 2) F = (6, 6)

S = (16, 16) F = (22, 22)

S = (16, 16) F = (20, 20)

S = (16, 16) F = (23, 23)

S = (20, 20) F = (23, 23)

S = (22, 22) F = (29, 29)

S = (6, 6) F = (16, 16)

S = (0, 0) F = (0, 0)

S = (23, 23) F = (29, 29)

S = (29, 29) F = (34, 34)

S = (34, 34) F = (40, 40)

S = (29, 30) F = (33, 34)

S = (29, 29) F = (38, 38)

S = (38, 38) F = (40, 40)

S = (40, 40) F = (40, 40)

0

0


Using LP to Make Crashing Decisions

• Restatment of the problem: Consider the total cost of the project, including the extra cost of crashing activities. The problem then is to minimize this total cost, subject to the constraint that project duration must be less than or equal to the time desired by the project manager.

• The decisions to be made are the following:1. The start time of each activity.2. The reduction in the duration of each activity due to crashing.3. The finish time of the project (must not exceed 40 weeks).

• The constraints are:1. Time Reduction ≤ Max Reduction (for each activity).2. Project Finish Time ≤ Desired Finish Time.3. Activity Start Time ≥ Activity Finish Time of all predecessors (for each activity).4. Project Finish Time ≥ Finish Time of all immediate predecessors of finish node.


Spreadsheet Model

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B C D E F G H I J KMaximum Crash Cost

Time Cost Time per Week Start Time FinishActivity Normal Crash Normal Crash Reduction saved Time Reduction Time

A 2 1 $180,000 $280,000 1 $100,000 0 0 2B 4 2 $320,000 $420,000 2 $50,000 2 0 6C 10 7 $620,000 $860,000 3 $80,000 6 0 16D 6 4 $260,000 $340,000 2 $40,000 16 0 22E 4 3 $410,000 $570,000 1 $160,000 16 0 20F 5 3 $180,000 $260,000 2 $40,000 20 2 23G 7 4 $900,000 $1,020,000 3 $40,000 22 0 29H 9 6 $200,000 $380,000 3 $60,000 29 0 38I 7 5 $210,000 $270,000 2 $30,000 16 0 23J 8 6 $430,000 $490,000 2 $30,000 23 2 29K 4 3 $160,000 $200,000 1 $40,000 30 0 34L 5 3 $250,000 $350,000 2 $50,000 29 0 34M 2 1 $100,000 $200,000 1 $100,000 38 0 40N 6 3 $330,000 $510,000 3 $60,000 34 0 40

Max TimeProject Finish Time 40 <= 40

Total Cost $4,690,000


Mr. Perty’s Conclusions

• The plan for crashing the project only provides a 50 percent chance of actually finishing the project within 40 weeks, so the extra cost of the plan ($140,000) is not justified. Therefore, Mr. Perty rejects any crashing at this stage.

• The extra cost of the crashing plan can be justified if it almost certainly would earn the bonus of $150,000 for finishing the project within 40 weeks. Therefore, Mr. Perty will hold the plan in reserve to be implemented if the project is running well ahead of schedule before reaching activity F.

• The extra cost of part or all of the crashing plan can be easily justified if it likely would make the difference in avoiding the penalty of $300,000 for not finishing the project within 47 weeks. Therefore, Mr. Perty will hold the crashing plan in reserve to be partially or wholly implemented if the project is running far behind schedule before reaching activity F or activity J.


Scheduling and Controlling Project Costs

• PERT/Cost is a systematic procedure (normally computerized) to help the project manager plan, schedule, and control costs.

• Assumption: A common assumption when using PERT/Cost is that the costs of performing an activity are incurred at a constant rate throughout its duration.


Budget for Reliable’s Project

ActivityEstimated

Duration (weeks)Estimated

CostCost per Weekof Its Duration

A 2 $180,000 $90,000

B 4 320,000 80,000

C 10 620,000 62,000

D 6 260,000 43,333

E 4 410,000 102,500

F 5 180,000 36,000

G 7 900,000 128,571

H 9 200,000 22,222

I 7 210,000 30,000

J 8 430,000 53,750

K 4 160,000 40,000

L 5 250,000 50,000

M 2 100,000 50,000

N 6 330,000 55,000


PERT/Cost Spreadsheet (Earliest Start Times)

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B C D E F G H I JEstimatedDuration Estimated Start Cost Per Week Week Week Week Week

Activity (weeks) Cost Time of Its Duration 1 2 3 4A 2 $180,000 0 $90,000 $90,000 $90,000 $0 $0B 4 $320,000 2 $80,000 $0 $0 $80,000 $80,000C 10 $620,000 6 $62,000 $0 $0 $0 $0D 6 $260,000 16 $43,333 $0 $0 $0 $0E 4 $410,000 16 $102,500 $0 $0 $0 $0F 5 $180,000 20 $36,000 $0 $0 $0 $0G 7 $900,000 22 $128,571 $0 $0 $0 $0H 9 $200,000 29 $22,222 $0 $0 $0 $0I 7 $210,000 16 $30,000 $0 $0 $0 $0J 8 $430,000 25 $53,750 $0 $0 $0 $0K 4 $160,000 33 $40,000 $0 $0 $0 $0L 5 $250,000 33 $50,000 $0 $0 $0 $0M 2 $100,000 38 $50,000 $0 $0 $0 $0N 6 $330,000 38 $55,000 $0 $0 $0 $0

Weekly Project Cost $90,000 $90,000 $80,000 $80,000Cumulative Project Cost $90,000 $180,000 $260,000 $340,000


PERT/Cost Spreadsheet (Earliest Start Times)

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B E W X Y Z AA AB AC AD AEStart Week Week Week Week Week Week Week Week Week

Activity Time 17 18 19 20 21 22 23 24 25A 0 $0 $0 $0 $0 $0 $0 $0 $0 $0B 2 $0 $0 $0 $0 $0 $0 $0 $0 $0C 6 $0 $0 $0 $0 $0 $0 $0 $0 $0D 16 $43,333 $43,333 $43,333 $43,333 $43,333 $43,333 $0 $0 $0E 16 $102,500 $102,500 $102,500 $102,500 $0 $0 $0 $0 $0F 20 $0 $0 $0 $0 $36,000 $36,000 $36,000 $36,000 $36,000G 22 $0 $0 $0 $0 $0 $0 $128,571 $128,571 $128,571H 29 $0 $0 $0 $0 $0 $0 $0 $0 $0I 16 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $0 $0J 25 $0 $0 $0 $0 $0 $0 $0 $0 $0K 33 $0 $0 $0 $0 $0 $0 $0 $0 $0L 33 $0 $0 $0 $0 $0 $0 $0 $0 $0M 38 $0 $0 $0 $0 $0 $0 $0 $0 $0N 38 $0 $0 $0 $0 $0 $0 $0 $0 $0

$175,833 $175,833 $175,833 $175,833 $109,333 $109,333 $194,571 $164,571 $164,571$1,295,833 $1,471,667 $1,647,500 $1,823,333 $1,932,667 $2,042,000 $2,236,571 $2,401,143 $2,565,714


PERT/Cost Spreadsheet (Latest Start Times)

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B C D E F G H I JEstimatedDuration Estimated Start Cost Per Week Week Week Week Week

Activity (weeks) Cost Time of Its Duration 1 2 3 4A 2 $180,000 0 $90,000 $90,000 $90,000 $0 $0B 4 $320,000 2 $80,000 $0 $0 $80,000 $80,000C 10 $620,000 6 $62,000 $0 $0 $0 $0D 6 $260,000 20 $43,333 $0 $0 $0 $0E 4 $410,000 16 $102,500 $0 $0 $0 $0F 5 $180,000 20 $36,000 $0 $0 $0 $0G 7 $900,000 26 $128,571 $0 $0 $0 $0H 9 $200,000 33 $22,222 $0 $0 $0 $0I 7 $210,000 18 $30,000 $0 $0 $0 $0J 8 $430,000 25 $53,750 $0 $0 $0 $0K 4 $160,000 34 $40,000 $0 $0 $0 $0L 5 $250,000 33 $50,000 $0 $0 $0 $0M 2 $100,000 42 $50,000 $0 $0 $0 $0N 6 $330,000 38 $55,000 $0 $0 $0 $0

Weekly Project Cost $90,000 $90,000 $80,000 $80,000Cumulative Project Cost $90,000 $180,000 $260,000 $340,000


PERT/Cost Spreadsheet (Latest Start Times)

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10111213141516171819202122

B E W X Y Z AA AB AC AD AEStart Week Week Week Week Week Week Week Week Week

Activity Time 17 18 19 20 21 22 23 24 25A 0 $0 $0 $0 $0 $0 $0 $0 $0 $0B 2 $0 $0 $0 $0 $0 $0 $0 $0 $0C 6 $0 $0 $0 $0 $0 $0 $0 $0 $0D 20 $0 $0 $0 $0 $43,333 $43,333 $43,333 $43,333 $43,333E 16 $102,500 $102,500 $102,500 $102,500 $0 $0 $0 $0 $0F 20 $0 $0 $0 $0 $36,000 $36,000 $36,000 $36,000 $36,000G 26 $0 $0 $0 $0 $0 $0 $0 $0 $0H 33 $0 $0 $0 $0 $0 $0 $0 $0 $0I 18 $0 $0 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000J 25 $0 $0 $0 $0 $0 $0 $0 $0 $0K 34 $0 $0 $0 $0 $0 $0 $0 $0 $0L 33 $0 $0 $0 $0 $0 $0 $0 $0 $0M 42 $0 $0 $0 $0 $0 $0 $0 $0 $0N 38 $0 $0 $0 $0 $0 $0 $0 $0 $0

$102,500 $102,500 $132,500 $132,500 $109,333 $109,333 $109,333 $109,333 $109,333$1,222,500 $1,325,000 $1,457,500 $1,590,000 $1,699,333 $1,808,667 $1,918,000 $2,027,333 $2,136,667


Cumulative Project Costs

8 16 24 32 40

$1 million

$2 million

$3 million

$4 million

$5 million

Earliest start time project cost schedule

Feasible region for project costs

Latest start time project cost schedule

Project cost schedule for both earliest and latest start times

0Week

Cumulative Project Cost


PERT/Cost Report after Week 22

ActivityBudgeted

CostPercent

CompletedValue

CompletedActual Cost

to DateCost Overrun

to Date

A $180,000 100% $180,000 $200,000 $20,000

B 320,000 100 320,000 330,000 10,000

C 620,000 100 620,000 600,000 –20,000

D 260,000 75 195,000 200,000 5,000

E 410,000 100 410,000 400,000 –10,000

F 180,000 25 45,000 60,000 15,000

I 210,000 50 105,000 130,000 25,000

Total $2,180,000 $1,875,000 $1,920,000 $45,000


Project Management

• Characteristics of Projects– Unique, one-time operations

– Involve a large number of activities that must be planned and coordinated

– Long time-horizon

– Goals of meeting completion deadlines and budgets

• Examples– Building a house

– Planning a meeting

– Introducing a new product

• PERT—Project Evaluation and Review TechniqueCPM—Critical Path Method

– A graphical or network approach for planning and coordinating large-scale projects.


Example: Building a House

Activity Time (Days)ImmediatePredecessor

Foundation 4 —

Framing 10 Foundation

Plumbing 9 Framing

Electrical 6 Framing

Wall Board 8 Plumbing, Electrical

Siding 16 Framing

Paint Interior 5 Wall Board

Paint Exterior 9 Siding

Fixtures 6 Int. Paint, Ext. Paint


Gantt Chart

Foundation

Framing

Plumbing

Electrical

Siding

Wall Board

Paint Interior

Paint Exterior

Fixtures

Activity Start 5 10 15 20 25 30 35 40 45 50Days After Start

Days After StartStart 5 10 15 20 25 30 35 40 45 50


PERT and CPM

• Procedure1. Determine the sequence of activities.

2. Construct the network or precedence diagram.

3. Starting from the left, compute the Early Start (ES) and Early Finish (EF) time for each activity.

4. Starting from the right, compute the Late Finish (LF) and Late Start (LS) time for each activity.

5. Find the slack for each activity.

6. Identify the Critical Path.


Notation

t Duration of an activity

ES The earliest time an activity can start

EF The earliest time an activity can finish (EF = ES + t)

LS The latest time an activity can start and not delay the project

LF The latest time an activity can finish and not delay the project

Slack The extra time that could be made available to an activity without delaying the project (Slack = LS – ES)

Critical Path The sequence(s) of activities with no slack


PERT/CPM Project Network

START FINISHd

c

f

e

h

g

iba0 0

Foundation Framing

Siding PaintExterior

Plumbing

WallBoard Paint

Interior FixturesElectrical

4 10

9

6

16

8

5

9

6


Calculation of ES, EF, LF, LS, and Slack

• ES = Maximum of EF’s for all predecessors

• EF = ES + t

• LF = Minimum of ES for all successors

• LS = LF – t

• Slack = LS – ES = LF – EF


Building a House: ES, EF, LS, LF, Slack

Activity ES EF LS LF Slack

(a) Foundation 0 4 0 4 0

(b) Framing 4 14 4 14 0

(c) Plumbing 14 23 17 26 3

(d) Electrical 14 20 20 26 6

(e) Wall Board 23 31 26 34 3

(f) Siding 14 30 14 30 0

(g) Paint Interior 31 36 34 39 3

(h) Paint Exterior 30 39 30 39 0

(i) Fixtures 39 45 39 45 0


PERT/CPM Project Network

START0

FINISH0

a

b c

d

e

f

h

g i

j

4

4

4 5 5

533

7

8


Example #2: ES, EF, LS, LF, Slack

Activity ES EF LS LF Slack

a 0 4 0 4 0

b 0 4 1 5 1

c 4 7 5 8 1

d 4 8 4 8 0

e 4 12 5 13 1

f 4 11 6 13 2

g 8 13 8 13 0

h 8 11 10 13 2

i 13 18 13 18 0

j 11 16 13 18 2


PERT with Uncertain Activity Durations

• If the activity times are not known with certainty, PERT/CPM can be used to calculate the probability that the project will complete by time t.

• For each activity, make three time estimates:– Optimistic time: o

– Pessimistic time: p

– Most-likely time: m


Beta Distribution

Assumption: The variability of the time estimates follows the beta distribution.

Elapsed time

p0

Beta distribution

mo


PERT with Uncertain Activity Durations

Goal: Calculate the probability that the project is completed by time t.

Procedure:

1. Calculate the expected duration and variance for each activity.

2. Calculate the expected length of each path. Determine which path is the mean critical path.

3. Calculate the standard deviation of the mean critical path.

4. Find the probability that the mean critical path completes by time t.


Expected Duration and Variance for Activities (Step #1)

• The expected duration of each activity can be approximated as follows:

• The variance of the duration for each activity can be approximated as follows:

2

p o6

2

o 4m p6


Expected Length of Each Path (Step #2)

• The expected length of each path is equal to the sum of the expected durations of all the activities on each path.

• The mean critical path is the path with the longest expected length.


Standard Deviation of Mean Critical Path (Step #3)

• The variance of the length of the path is the sum of the variances of all the activities on the path.

2path = ∑ all activities on path 2

• The standard deviation of the length of the path is the square root of the variance.

path path2


Probability Mean-Critical Path Completes by t (Step #4)

• What is the probability that the mean critical path (with expected length tpath and standard deviation path) has duration ≤ t?

• Use Normal Tables (Appendix A)

z

t (tpath)

path

+ +2 +3Ğ2Ğ3

Path duration

ProbabilityDensityFunction

tpath

t


Example

START0

FINISH

0a

b

c

d

e

2 - 3 - 4

2 - 4 - 5 3 - 4 - 6

1 - 3 - 7 2 - 3 - 8

Question: What is the probability that the project will be finished by day 12?


Expected Duration and Variance of Activities (Step #1)

Activity o m p

a 2 3 4 3.00 1/9

b 2 4 5 3.83 1/4

c 1 3 7 3.33 1

d 3 4 6 4.17 1/4

e 2 3 8 3.67 1

o 4m p6

2 p o6

2


Expected Length of Each Path (Step #2)

Path Expected Length of Path

a - b - d 3.00 + 3.83 + 4.17 = 11

a - c - e 3.00 + 3.33 + 3.67 = 10

The mean-critical path is a - b - d.


Standard Deviation of Mean-Critical Path (Step #3)

• The variance of the length of the path is the sum of the variances of all the activities on the path.

2path = ∑ all activities on path 2 = 1/9 + 1/4 + 1/4 = 0.61

• The standard deviation of the length of the path is the square root of the variance.

path path2 0.610.78


Probability Mean-Critical Path Completes by t=12 (Step #4)

• The probability that the mean critical path (with expected length 11 and standard deviation has duration ≤ 12?

• Then, from Normal Table: Prob(Project ≤ 12) = Prob(z ≤ 1.41) = 0.92

z

t (tpath)

path

12 110.71

1.41


Time-Cost Tradeoffs

• It is often possible to expedite activities at extra cost (called “crashing”).

• Benefits of completing a project earlier:– Monetary incentives for timely or early completion of project

– Beating the competition to the market

– Reduce indirect cost

• Procedure:1. Obtain estimates of regular and crash times and costs for each activity.

2. Determine the lengths of all paths using regular time for each activity.

3. Identify the critical path(s)

4. Crash activities on the critical path(s) in order of increasing cost as long as crashing costs do not exceed benefits.


Time-Cost Trade-Off Data for House Project

Time (days) Cost

MaximumReduction

in Time (days)

Crash Costper DaySavedActivity Normal Crash Normal Crash

(a) Foundation 4 4 $8,000 $8,000 0 —

(b) Framing 10 8 9,000 9,800 2 $400

(c) Plumbing 9 7 5,600 6,000 2 200

(d) Electrical 6 4 8,400 8,800 2 200

(e) Wall Board 8 6 11,500 12,700 2 600

(f) Siding 16 12 13,400 15,600 4 300

(g) Paint Interior 5 2 3,000 5,700 3 900

(h) Paint Exterior 9 7 5,000 6,400 2 700

(i) Fixtures 7 5 6,500 8,900 2 1,200


Building a House Project Network

Question: Suppose $1,000 in indirect costs can be saved for each day the project is shortened. Which activities should be crashed?

START FINISHd

c

f

e

h

g

iba0 0

Foundation Framing

Siding PaintExterior

Plumbing

WallBoard Paint

Interior FixturesElectrical

4 10

9

6

16

8

5

9

6


Marginal Cost Analysis for House ProjectInitial Table

Length of Path

Activityto Crash

CrashCost abcegi abdegi abfhi

42 39 45


Marginal Cost Analysis for House ProjectAfter Crashing 1 Day

Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44


Marginal Cost Analysis for House ProjectAfter Crashing 2 Days

Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44

(f) Siding $300 42 39 43



Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44

(f) Siding $300 42 39 43

(f) Siding $300 42 39 42



Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44

(f) Siding $300 42 39 43

(f) Siding $300 42 39 42

(b) Framing $400 41 38 41



Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44

(f) Siding $300 42 39 43

(f) Siding $300 42 39 42

(b) Framing $400 41 38 41

(b) Framing $400 40 37 40



Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44

(f) Siding $300 42 39 43

(f) Siding $300 42 39 42

(b) Framing $400 41 38 41

(b) Framing $400 40 37 40

(c&f) Plumbing & Siding $500 39 37 39


Marginal Cost Analysis for House ProjectFinal Table (After Crashing 7 Days)

Length of Path

Activityto Crash


42 39 45

(f) Siding $300 42 39 44

(f) Siding $300 42 39 43

(f) Siding $300 42 39 42

(b) Framing $400 41 38 41

(b) Framing $400 40 37 40

(c&f) Plumbing & Siding $500 39 37 39

(c&h) Plumbing & Paint Ext. $900 38 37 38


Marginal Cost Analysis for House Project

• Seven days have been saved off the length of the project (38 days rather than 45).

– $7,000 saved in indirect costs.

• The total crashing costs were– $300 + $300 + $300 + $400 + $500 + $900 = $2,700

• Total savings = $7,000 – $2,700 = $5,300.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003 8.1 Table of Contents Chapter 8 (PERT/CPM...

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Transcript of McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003 8.1 Table of Contents Chapter 8 (PERT/CPM...