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Project management; Chapter7
Department of Business Administration
FALL 2010-2011
Chapter 7: Project Management
MGMT 405, POM, 2010/11. Lec Notes
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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.
2
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Project Management; Chapter7
Projects
JAN
FEB
MAR
APR
MAY
JUN
Build A
A Done
Build B
B Done
Build C
C Done
Build D
On time!
Ship
Project: Unique, one-time operations designed
to accomplish a specific set of objectives in a
limited time frame.
3
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Project Management; Chapter7
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
4
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Project Management; Chapter7
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
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Key Decisions
 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
6
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Project Management; Chapter7
Project Manager
Responsible for:
Work
Quality
Human Resources
Time
Communications
Costs
7
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Project Management; Chapter7
Ethical Issues
 Temptation to understate costs
 Withhold information
 Misleading status reports
 Falsifying records
 Comprising workers’ safety
 Approving substandard work
8
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Project Management; Chapter7
Project Life Cycle
Concept
Planning
Execution
Management
Feasibility
Termination
Concept: A proposal needed
Feasibility: Cost, benefit and risk analyses
Planning: find out the necessary human resources, time and cost
Execution: control for time, available resource and cost
Termination: It should be reevaluated for the sake of project’s safety
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Work Breakdown Structure
Project X
Level 1
Level 2
Level 3
Level 4
10
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Project Management; Chapter7
Planning and Scheduling
Gantt Chart
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
11
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Project Management; Chapter7
PERT and CPM
 PERT: Program Evaluation and Review Technique
 CPM: Critical Path Method
Both techniques are widely used for planning and
coordinating large-scale projects.
 Using the two techniques, manager are able to obtain:
 Graphically displays project activities
 Estimates how long the project will take
 Indicates most critical activities
 Show where delays will not affect project
12
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Project Management; Chapter7
The Network Diagram
 Network (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.
13
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Project Management; Chapter7
The Network Diagram
 Path
 Sequence of activities that leads from the starting node
to the finishing node
 Critical path
 The longest path; determines expected project duration
 Critical activities
 Activities on the critical path
 Slack
 Allowable slippage for path; the difference the length
of path and the length of critical path
14
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Project Management; Chapter7
Project Network – Activity on Arrow
AOA
Locate
facilities
Order
furniture
4
Furniture
setup
2
Remodel
1
5
6
Move
in
Interview
3
Hire and
train
15
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Project Management; Chapter7
Project Network – Activity on Node
Order
furniture
AON
Locate
facilities
Furniture
setup
2
6
1
Move
in
Remodel
5
S
Interview
3
7
Hire and
train
4
16
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Project Management; Chapter7
Network Conventions
a
c
a
c
b
a
a
c
c
b
b
b
d
Dummy
activity
17
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Project Management; Chapter7
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.
 Deterministic
 Time estimates that are fairly certain
 Probabilistic
 Estimates of times that allow for variation
18
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Project Management; Chapter7
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
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Answer-Bank Network convention
 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
b
7
S
k
h
2
MGMT 405, POM, 2010/11. Lec Notes
17
End
20
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Project Management; Chapter7
Answer-Bank Network convention
 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
S
k
End
h
21
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Project Management; Chapter7
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=30
e-f-m= 3+6+8=17
g-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
22
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Project Management; Chapter7
Example-Bank Network Figure
Bank Network question
8 weeks
6 weeks
4
3 weeks
2
11 weeks
1
5
Move
in
6
1 week
4 weeks
9 weeks
3
23
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Project Management; Chapter7
Example-Bank Network Figure
 Given the information on the bank network:
 Determine
 The length of each path
 The critical path
 The expected length of the project
 The 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.
24
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Project Management; Chapter7
Answer-Bank Network Figure
Critical Path
Path
Length
Slack
(weeks)
1-2-3-4-5-6
1-2-5-6
1-3-5-6
18
20
14
2
0
6
25
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Computing Algorithm
 Network activities
 ES: early start
 EF: early finish-EF=ES+t
 LS: late start-LS=LF-t
 LF: late finish
 Used to determine
ES t
LS
ES t
EF
LF
EF
 Expected project duration
 Slack time-LS-ES or LF-EF
 Critical path
26
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Example-ES-EF-LS-LF-slack
Required: Compute slack time, ES, EF, LS and LF
4
Forward pass
2
ES t
EF
19 1 20
1
5
6
LS
ES t
EF: early finish-EF=ES+t
LS: late start-LS=LF-t
Slack time-LS-ES or LF-EF
3
LF
EF
Backward pass
27
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Project Management; Chapter7
Answer-ES-EF-LS-LF-slack
Forward pass
10
8
2
0
0
0
8
8
8
6
4
10
4
MGMT 405, POM, 2010/11. Lec Notes
3
EF
4
16
14
2
8 11
8
0
EF: early finish-EF=ES+t
LS: late start-LS=LF-t
Slack time-LS-ES or LF-EF
ES t
2
1
6
0
6
16
14
19
19
10
4
6
3
19
17
19 1 20
5
9
19
13
6
19
19
0
1
20
20
ES t
LS
EF
LF
Backward pass
28
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Project Management; Chapter7
Probabilistic Time Estimates
 Optimistic time
 Time required under optimal conditions
 Pessimistic time
 Time required under worst conditions
 Most likely time
 Most probable length of time that will be required
29
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Probabilistic Estimates
Beta Distribution is generally used to describe the inherent variability in time
Estimates. Although there is no real theoretical justification for using the Beta
Distribution, it has certain features that make it attractive in practice.
to
Activity
start
tm
Optimistic
time
te
Most likely
time (mode)
tp
Pessimistic
time
30
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Project Management; Chapter7
Expected Time
te
t
+
4t
+t
o
m
p
=
6
te = expected time
to = optimistic time
tm = most likely time
tp = 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
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Variance
2
2
(t
–
t
)
= p o
36
2 = variance
to = optimistic time
tp = 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.
32
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Example-Probabilistic Time Estimates
Given the following diagram:
Optimistic
Compute
time
The expected time
The expected duration
Identify the critical path
The variance
The std. deviation
3-4-5
d
Most likely
time
Pessimistic
time
2-4-6
b
3-5-7
e
5-7-9
f
4-6-8
h
33
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Project Management; Chapter7
Answer-Probabilistic Time Estimates
34
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Project Management; Chapter7
Answer-Probabilistic Time Estimates
Tabc = 10.0
Tdef = 16.0
Tghi = 13.50
4.00
b
4.00
d
5.0
e
7.0
f
6.0
h
35
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Project Management; Chapter7
Path Probabilities
Z =
Specified time – Path mean
Path 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.
36
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Project Management; Chapter7
Example-The Path probability
 Given the information on the example of probabilistic
time estimates (the previous example):
 Determine
 The 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.
37
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Project Management; Chapter7
Answer-The Path probability
 Determine
 The probability that the project can be completed within
17 weeks of its start. Path: a-b-c
17 – 10
Z =
0.97
=7.22
 Determine
Prob.comp in 17 week=1.00
Appendix B, Table B, p.p 884/5
 The probability that the project will be completed within
17 weeks of its start. Path: d-e-f
17 – 16 =1
Z =
1
Prob.comp in 17 week=0.8413
Appendix B, Table B, p.p 885
38
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Project Management; Chapter7
Answer-The Path probability
 Determine
 The probability that the project will be completed within
17 weeks of its start. Path: g-h-i
Z = 17 – 13.5 =3.27
1.07
Prob.comp in 17 week=1.00
Appendix B, Table B, p.p 884/5
Prob finish in 17 week=1.00 X 0.8413 X 1.00= 0.8413
39
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Answer-The Path probability
 Determine
 The probability that the project can be completed within
15 weeks of its start. Path: a-b-c
15 – 10
Z =
0.97
 Determine
=5.15
Prob.comp in 15 week=1.00
Appendix B, Table B, p.p 884/5
 The probability that the project will be completed within
15 weeks of its start. Path: d-e-f
15 – 16 =-1.00 Prob.comp in 15 week=0.1587
Z =
1
Appendix B, Table B, p.p 885
40
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Answer-The Path probability
 Determine
 The probability that the project will be completed within
15 weeks of its start. Path: g-h-i
Z = 15 – 13.5 =1.40
1.07
Prob.comp in 15 week=0.9192
Appendix B, Table B, p.p 884/5
Prob finish in 15 week=1.00 X 0.1587 X 0.9192= 0.1459
 The probability that the project will not be completed within 15 weeks of
its start: 1- 0.1459=0.8541
41
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Project Management; Chapter7
Answer-The Path probability-Graphically
17
Weeks
1.00
a-b-c
Weeks
10.0
0.8413
d-e-f
16.0
Weeks
1.00
g-h-i
13.5
Weeks
42
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
Answer-The Path probability-Graphically
15
Weeks
1.00
a-b-c
10.0
Weeks
0.1587
d-e-f
16.0
Weeks
0.9192
g-h-i
13.5
Weeks
43
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Project Management; Chapter7
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
44
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Project Management; Chapter7
Time-Cost Trade-Offs: Crashing
Total
cost
Expected indirect costs
Shorten
CRASH
Cumulative
(direct)
cost of
crashing
Shorten
Optimum
45
MGMT 405, POM, 2010/11. Lec Notes
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Project Management; Chapter7
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.
Activity
Immediate predecessor
Normal 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
46
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Project Management; Chapter7
 (a) Determine the critical path.
2
f
4
d
47
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Project Management; Chapter7
Answer-Crashing
(a) Determine which activities are on the critical path, its length, and the
length of the other path
Path
length
a-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 days
c
$ 300
1
e
600
2
d
700
3
f
800
1
48
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Project Management; Chapter7
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
49
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Project Management; Chapter7
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
activity crashed
c
e
cost
$300
600
17
f
800
50
MGMT 405, POM, 2010/11. Lec Notes
© Stevenson, McGraw Hill, 2007- Assoc. Prof. Sami Fethi, EMU, All Right Reserved.
Project Management; Chapter7
Advantages of PERT
 Forces managers to organize
 Provides graphic display of activities
 Identifies
 Critical activities
 Slack activities
4
2
1
5
6
3
51
MGMT 405, POM, 2010/11. Lec Notes
© Stevenson, McGraw Hill, 2007- Assoc. Prof. Sami Fethi, EMU, All Right Reserved.
Project Management; Chapter7
Limitations of PERT
 Important activities may be omitted
 Precedence relationships may not be correct
 Estimates may include
a fudge factor
4
 May focus solely
on critical path
2
1
5
6
142 weeks
3
52
MGMT 405, POM, 2010/11. Lec Notes
© Stevenson, McGraw Hill, 2007- Assoc. Prof. Sami Fethi, EMU, All Right Reserved.
Project Management; Chapter7
Thanks
53
MGMT 405, POM, 2010/11. Lec Notes
© Stevenson, McGraw Hill, 2007- Assoc. Prof. Sami Fethi, EMU, All Right Reserved.
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