Chapter 7: Project Management

© Stevenson, McGraw Hill, 2007- Assoc. Prof. Sami Fethi, EMU, All Right Reserved.

Project management; Chapter7

MGMT 405, POM, 2010/11. Lec Notes

Chapter 7: Project Management

Department of Business Administration

FALL 2010-2011

2

MGMT 405, POM, 2010/11. Lec Notes © Stevenson, McGraw Hill, 2007- Assoc. Prof. Sami Fethi, EMU, All Right Reserved.

Project Management; Chapter7

Outline: What You Will Learn . . .

Discuss the behavioral aspects of projects in terms of project personnel and the project manager.

Discuss the nature and importance of a work breakdown structure in project management.

Give a general description of PERT/CPM techniques. Construct simple network diagrams. List the kinds of information that a PERT or CPM analysis can

provide. Analyze networks with deterministic times. Analyze networks with probabilistic times. Describe activity “crashing” and solve typical problems.

3



Project:Project: Unique, one-time operations designed Unique, one-time operations designed to accomplish a specific set of objectives in a to accomplish a specific set of objectives in a limited time frame.limited time frame.

Build A

A Done

Build B

B Done

Build C

C Done

Build D

Ship

JAN FEB MAR APR MAY JUN

On time!

Projects

4



Project Management How is it different?

Limited time frame Narrow focus, specific objectives Less bureaucratic

Why is it used? Special needs Pressures for new or improves products or services

What are the Key Metrics Time Cost Performance objectives

What are the Key Success Factors? Top-down commitment Having a capable project manager Having time to plan Careful tracking and control Good communication

5



Project Management What are the Major Administrative Issues?

Executive responsibilities Project selection Project manager selection Organizational structure

Organizational alternatives Manage within functional unit Assign a coordinator Use a matrix organization with a project leader

What are the tools? Work breakdown structure Network diagram Gantt charts Risk management

6



Deciding which projects to implement Criteria-attractive-cost and benefit-available fund

Selecting a project manager Central person

Selecting a project team Person’s knowledge and skills-relationship with others

Planning and designing the project Goals-timetable-budget-resources

Managing and controlling project resources Personnel-equipment-budget

Deciding if and when a project should be terminated Likelihood of success-costs-resources

Key Decisions

7



Project Manager

Responsible for:

Work Quality

Human Resources Time

Communications Costs

8



Temptation to understate costsWithhold informationMisleading status reportsFalsifying recordsComprising workers’ safetyApproving substandard work

Ethical Issues

9



Project Life Cycle

Concept

FeasibilityFeasibility

PlanningPlanning

ExecutionExecution

TerminationTermination

Man

agem

ent

Concept: A proposal neededFeasibility: Cost, benefit and risk analysesPlanning: find out the necessary human resources, time and costExecution: control for time, available resource and costTermination: It should be reevaluated for the sake of project’s safety

10



Work Breakdown Structure

Project XProject X

Level 1

Level 2

Level 3

Level 4

11



Planning and Scheduling

MAR APR MAY JUN JUL AUG SEP OCT NOV DEC

Locate new facilities

Interview staff

Hire and train staff

Select and order furniture

Remodel and install phones

Move in/startup

Gantt Chart

12



PERT and CPM

PERT: Program Evaluation and Review TechniqueCPM: Critical Path MethodBoth techniques are widely used for planning and

coordinating large-scale projects. Using the two techniques, manager are able to obtain:

Graphically displays project activitiesEstimates how long the project will take Indicates most critical activitiesShow where delays will not affect project

13



The Network DiagramNetwork (precedence) diagram – diagram of

project activities that shows sequential relationships by the use of arrows and nodes.

Activity-on-arrow (AOA) – a network diagram convention in which arrows designate activities.

Activity-on-node (AON) – a network diagram convention in which nodes designate activities.

Activities – steps in the project that consume resources and/or time.

Events – the starting and finishing of activities, designated by nodes in the AOA convention.

14



The Network DiagramPath

Sequence of activities that leads from the starting node to the finishing node

Critical pathThe longest path; determines expected project duration

Critical activitiesActivities on the critical path

SlackAllowable slippage for path; the difference the length

of path and the length of critical path

15



Project Network – Activity on Arrow

1

2

3

4

5 6

Locatefacilities

Orderfurniture

Furnituresetup

InterviewHire andtrain

Remodel

Move in

AOA

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Project Network – Activity on Node

1

2

3

5

6

Locatefacilities

Orderfurniture

Furnituresetup

Interview

RemodelMove in

4

Hire andtrain

7S

AON

17



Network Conventions

a

b

c ab

c

a

b

c

d

a

b

c

Dummyactivity

18



Time Estimates

The main determinant of the way PERT and CPM networks are analysed and interpreted is whether activity time estimates are probabilistic or deterministic.

DeterministicTime estimates that are fairly certain

ProbabilisticEstimates of times that allow for variation

19



Example-Bank Network convention

The following table contains information related to the major activities of a research project. Use the information to do the following:(a) Draw a precedence diagram using AOA and AON(b) Find the critical path based AOA.(c) Determine the expected length of the project.

Activity Immediate Predecessor Expected Time (days)

a - 5

c a 8

d c 2

b a 7

e - 3

f e 6

i b, d 10

m f,i 8

g - 1

h g 2

k h 17

end k,m

20



Answer-Bank Network convention

c

a

f

g

h

d

Activities with no predecessors are at the beginning (life side) of the network. Activities with multiple predecessors are located at path intersections. Use first AOA

5

8 2b

7

e

36

i

10

m

8

1

2

k

17

S

End

21



Answer-Bank Network convention

c

a

f

g h

Activities with no predecessors are at the beginning (life side) of the network. Activities with multiple predecessors are located at path intersections. Use Second AON

b

e

mS

End

b

d

i

f

k

22



Example-Bank Network convention

(b)Find the critical path based AOA.

a-c-d-i-m*=5+8+2+10+8=33#

a-b-i-m=5+7+10+8=30e-f-m= 3+6+8=17g-h-k=1+2+17=20

a-c-d-i-m*-Critical path

(c) Determine the expected length of the project.

33 # -Expected project duration

23



Example-Bank Network Figure

1

2

3

4

5 6

8 weeks

6 weeks

3 weeks

4 weeks9 weeks

11 weeks

1 week

Locate

facilities

Order

furniture Fu

rnitu

re

setup

InterviewHire

and train

Remodel Move in

Bank Network questionBank Network question

24



Example-Bank Network Figure

Given the information on the bank network:

DetermineThe length of each pathThe critical pathThe expected length of the projectThe amount of slack time for each path Knowledge of slack times provides managers with

information for planning allocation of scarce resources and for directing control efforts toward those activities that may be most susceptible to delaying the project.

25



Answer-Bank Network Figure

P a t h L e n g t h( w e e k s )

S l a c k

1 - 2 - 3 - 4 - 5 - 61 - 2 - 5 - 61 - 3 - 5 - 6

1 82 01 4

206

Critical PathCritical Path

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Network activitiesES: early startEF: early finish-EF=ES+tLS: late start-LS=LF-tLF: late finish

Used to determineExpected project durationSlack time-LS-ES or LF-EFCritical path

Computing Algorithm

ES t EF

LS LFES t EF

27



Example-ES-EF-LS-LF-slack

1

2

3

4

5 60

8 8

8 6 1

4

14 3 17

0 4 44

9

13

8 11 1919 1 20

LS LFES t EF

ES t EF

EF: early finish-EF=ES+tLS: late start-LS=LF-tSlack time-LS-ES or LF-EF

Required: Compute slack time, ES, EF, LS and LF

Forward pass

Backward pass

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Answer-ES-EF-LS-LF-slack

1

2

3

4

5 60

8 8

8 6 1

4

14 3 17

0 4 44

9

13

8 11 1919 1 20

ES t EFLS LF

ES t EF

19 1 2019 200

EF: early finish-EF=ES+tLS: late start-LS=LF-tSlack time-LS-ES or LF-EF

10 9 194 136

6 4 10 0 46

8 11 19 8 190

16 3 1914 172

10 6 168 14 2

0 8 80 80

Forward pass

Backward pass

29



Probabilistic Time Estimates

Optimistic timeTime required under optimal conditions

Pessimistic timeTime required under worst conditions

Most likely timeMost probable length of time that will be required

30



Probabilistic Estimates

Activitystart

Optimistictime

Most likelytime (mode)

Pessimistictime

to tptm te

Beta Distribution is generally used to describe the inherent variability in timeEstimates. Although there is no real theoretical justification for using the BetaDistribution, it has certain features that make it attractive in practice.

31



Expected Time

te = to + 4tm +tp

6

te = expected timeto = optimistic timetm = most likely timetp = pessimistic time

The knowledge of the expected path times and their std. Deviation enables a manager to compute probabilistic estimates of the project completion time as such specific time and scheduled time

32



Variance

(tp – to)2

36

= varianceto = optimistic timetp = pessimistic time

The size of Variance reflects the degree of uncertainty associated with an activity’s time: The large the variance, the greater the uncertainty.

33



ExampleExample-Probabilistic Time Estimates-Probabilistic Time Estimates

1-3-4a

3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

2-3-6g 3-4-6

i

2-3-5cOptimistictime

Most likelytime

Pessimistictime

Given the following diagram: Compute The expected timeThe expected durationIdentify the critical pathThe varianceThe std. deviation

34



Answer-Probabilistic Time EstimatesProbabilistic Time Estimates

35



Answer-Probabilistic Time EstimatesProbabilistic Time Estimates

2.83a

4.00d

5.0e

7.0f

4.00b

6.0h

3.33g 4.17

i

3.17c

Tabc = 10.0Tdef = 16.0Tghi = 13.50

36



Path Probabilities

Z = Specified time – Path meanPath standard deviation

Z indicates how many standard deviations of the path distribution the specified time is beyond the expected path duration. The more positive the value, the better. A negative value of z indicates that the specified time is earlier than the expected path duration. Z=+3.00-probability 100%- From the relevant table +3.00 is almost equal to 0.9987.

37



Example-The Path probability

Given the information on the exampleexample of probabilistic of probabilistic time estimates (the previous example):time estimates (the previous example):

DetermineThe probability that the project can be completed within 17

weeks of its start. The probability that the project will be completed within 15

weeks of its start.The probability that the project will not be completed within

15 weeks of its start.

38



Answer-The Path probabilityDetermine

The probability that the project can be completed within 17 weeks of its start. Path: a-b-c

Z = 17 – 100.97

=7.22Determine

The probability that the project will be completed within 17 weeks of its start. Path: d-e-f

Z = 17 – 161

=1

Prob.comp in 17 week=1.00Appendix B, Table B, p.p 884/5

Prob.comp in 17 week=0.8413Appendix B, Table B, p.p 885

39




The probability that the project will be completed within 17 weeks of its start. Path: g-h-i

Z = 17 – 13.51.07

=3.27 Prob.comp in 17 week=1.00Appendix B, Table B, p.p 884/5

Prob finish in 17 week=1.00 X 0.8413 X 1.00= 0.8413

40




The probability that the project can be completed within 15 weeks of its start. Path: a-b-c

Z = 15 – 100.97

=5.15Determine

The probability that the project will be completed within 15 weeks of its start. Path: d-e-f

Z = 15 – 161

=-1.00

Prob.comp in 15 week=1.00Appendix B, Table B, p.p 884/5

Prob.comp in 15 week=0.1587Appendix B, Table B, p.p 885

41



Answer-The Path probability

DetermineThe probability that the project will be completed within

15 weeks of its start. Path: g-h-i

Z = 15 – 13.51.07

=1.40 Prob.comp in 15 week=0.9192Appendix B, Table B, p.p 884/5

The probability that the project will not be completed within 15 weeks of its start: 1- 0.1459=0.8541

Prob finish in 15 week=1.00 X 0.1587 X 0.9192= 0.1459

42



17Weeks

Weeks

Weeks

Weeks

10.0

16.0

13.5

1.00

1.00

a-b-c

d-e-f

g-h-i

Answer-The Path probability-Graphically

0.8413

43



15Weeks

Weeks

Weeks

Weeks

10.0

16.0

13.5

1.00

0.9192

a-b-c

d-e-f

g-h-i

Answer-The Path probability-Graphically

0.1587

44



Time-cost Trade-offs: Crashing In many projects, it is possible to reduce the length of a

project by injecting additional resources. The impetus to shorten projects may reflect efforts to avoid late penalties, or/ to take advantage of monetary incentives for timely completion of a project, or/ to free resources for use on other projects. This is called crashing.

Crash – briefly, shortening activity duration Procedure for crashing

Crash the project one period at a time Only an activity on the critical path Crash the least expensive activity Multiple critical paths: find the sum of crashing the least

expensive activity on each critical path

45



Time-Cost Trade-Offs: Crashing

TotalcostTotalcost

ShortenShorten

ShortenShorten

Cumulative (direct)cost of crashing

Cumulative (direct)cost of crashing

Expected indirect costsExpected indirect costs

Optimum

CRASHCRASH

46



Example-Crashing

Using the following information, develop the optimal time cost solution. Indirect costs are $ 1000 per day.(a) Determine which activities are on the critical path, its length, and the

length of the other path(b) Rank the critical activities in order of lowest crashing cost, and

datermine the number of days each can be crashed.(c) Determine the critical path after each reduction by shortening the

project.

ActivityImmediate

predecessorNormal time Crash time

Cost per day to crash

a - 6 6

c - 10 8 $500

d c 5 4 300

b a 4 1 700

e d 9 7 600

f b,e 2 1 800

47



6a

4d

5c

10b

9 e

2f

(a) Determine the critical path.

48



Answer-Crashing

(a) Determine which activities are on the critical path, its length, and the length of the other path

Path lengtha-b-f 18

c-d-e-f 20 (critical path)

(b) Rank the critical activities in order of lowest crashing cost, and datermine the number of days each can be crashed.

Activity Cost per day to crash Available daysc $ 300 1e 600 2d 700 3f 800 1

49



Answer-Crashing

(c) Determine the critical path after each reduction by shortening the project.(1) Shorten activity c one day at a cost of $ 300. The length of the critical path becomes 19 days.(2) Activity c cannot be shorten any more. Shorten activity e one day at cost of $ 600. The length of the critical path c-d-e-f becomes 18 days which is the same as length of path a-b-f.(3) The path are now both critical, further improvement will necesitate shortening both paths.

Path Activity Cost per day to crash a-b-f a no reduction possible

b $ 500

f 800

c-d-e-f c no further reduction possible d

$ 700 e 600

f 800

50



Answer-Crashing

At the first glance, it would seem that crashing f would not be advantageous, because it has the highest crashing cost. However, f is on both paths, so shortening f by one day would shorten both paths by one day for a cost of $ 800. The option of shortening the least expensive activity on each path would cost $ 500 for b and $ 600 for e or $ 1100. Thus shorten f by one day. The project duration is now 17 days.(4) At this point, no additional improvement is feasible. The cost to crash b is $ 500 and the cost to crash e is $ 600, for a total of $ 1100 and that would exceed the indirect costs of $ 100 per day.(5) The crashing sequence is summarized below:

Length after crashing n days Path n=0 1 2 3

a-b-f 18 18 18 17

c-d-e-f 20 19 18 17

activity crashed c e f cost $300 600 800

51



Advantages of PERT

Forces managers to organizeProvides graphic display of activities Identifies

Critical activitiesSlack activities

1

2

3

4

5 6

52



Limitations of PERT

Important activities may be omitted

Precedence relationships may not be correct

Estimates may include a fudge factor

May focus solelyon critical path

1

2

3

4

5 6

142 weeks

53



Thanks

Documents

Chapter 7: Project Management