1. Chapter 5: Project Management
Department of Business Administration
SPRING
Chapter 5: Project Management

2. 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. Projects
Build A
A Done
Build B
B Done
Build C
C Done
Build D
Ship
JAN
FEB
MAR
APR
MAY
JUN
On time!
Project: Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame.

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. 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

7. Project Manager Responsible for: Work Quality Human Resources Time
Communications Costs

8. Technical Information about projects
U.S.A spend $2.3 for projects every year.
Cross the world, almost $10 billion is spent for projects.
A project manager earns averagely $82,000.
Project management Institute (PMI) was established in In 1990, The institute had 17,500 members. In 2001, there were 86,000 members. Nowadays, this number reached to 100,000 members.

9. Ethical Issues Temptation to understate costs Withhold information
Misleading status reports
Falsifying records
Comprising workers’ safety
Approving substandard work

10. Concept Project Life Cycle Feasibility Planning Execution Termination
Management
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

11. Work Breakdown Structure
Project X
Level 1
Level 2
Level 3
Level 4

12. 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

13. 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

14. 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.

15. 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
Slack is the length of the time where an activity can be delayed without interfering with the project completion.
Dummy
Using this variable does not cost any burden for the company. This exists solely for the purpose of establishing precedence relationships for the sake of simplicity and is not asssigned any time.

16. Project Network – Activity on Arrow1 2 3 4 5 6
Locate facilities
Order furniture
Furniture setup
Interview
Hire and train
Remodel
Move in
AOA

17. Project Network – Activity on Node1 2 3 5 6
Locate facilities
Order furniture
Furniture setup
Interview
Remodel
Move in 4
Hire and train 7 S
AON

18. Network Conventions a c b Dummy activity d a and b must be completed
before c can start
a must be completed
before b or c can start
a must be completed
before c can start//
b and dummy must be completed
before c can start
a and b must be completed
before b or c can start

19. Computing Algorithm Network activities Used to determine
ES: early start
EF: early finish-EF=ES+t
LS: late start-LS=LF-t
LF: late finish
Used to determine
Expected project duration
Slack time-LS-ES or LF-EF
Critical path-longest period
ES t EF
LS LF
ES t EF

20. Example 1-Bank Network convention
The following table contains information related to the major activities of a research project. Use the information to do the following:
Draw a precedence diagram using AOA and AON
Find the critical path based AOA.
Determine the expected length of the project.
The amount of slack time for each path
Activity
Immediate Predecessor
Expected Time (days) a - 5 c 8 d 2 b 7 e 3 f 6 i
b, d 10 m
f,i g 1 h k 17
end
k,m

21. 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.
(a) Use first AOA; the precedence diagram using AOA is constructed as follows: c a f g h d 8 2 b 7 i S 5 10 e 6 m 3 8 1 k
End 17 2

22. Example-Bank Network convention
(b)Find the critical path based AOA.
a-c-d-i-m*= =33#
a-b-i-m= =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

23. Answer-Bank Network convention
(d) calculate the amount of slack time for each path
LS LF
ES t EF 13 13 c a f g h d 5 13 13 15 5 8 15 5 2 15 15 8 5 b 12 15 25 5 7 i 25 19 19 25 10 25 S 5 9 16 3 25 3 e 6 m 33 13 3 33 8 33 14 16 20 1 14 1 16 k 3 3 1
End 17 2

24. Example 2-Bank Network convention
The following table contains information related to the major activities of a research project. Use the information to do the following:
Draw a precedence diagram using AOA
Find the critical path based AOA.
Determine the expected length of the project.
The amount of slack time for each path
Activity
Immediate Predecessor
Expected Time (days) a - 8 b 6 d 3 c 11 e 4 f 9 g
c, d, f 1
end

25. Example 2-Bank Network Figure1 2 3 4 5 6
8 weeks
6 weeks
3 weeks
4 weeks
9 weeks
11 weeks
1 week
Locate facilities
Order furniture
Furniture setup
Interview
Hire and train
Remodel
Move in
(a) Bank Network question

26. Answer-Bank Network Figure
(b) and (c) Critical 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.

27. Example-ES-EF-LS-LF-slack
(d) Required: Compute slack time, ES, EF, LS and LF 1 2 3 4 5 6
Forward pass
ES t EF
LS LF
ES t EF
EF: early finish-EF=ES+t
LS: late start-LS=LF-t
Slack time-LS-ES or LF-EF
Backward pass

28. Answer (d- cont...) -ES-EF-LS-LF-slack
Forward pass 2 1 2 3 4 5 6
ES t EF
14 17 2
0 8
19 20 6
4 13 6
LS LF
ES t EF
EF: early finish-EF=ES+t
LS: late start-LS=LF-t
Slack time-LS-ES or LF-EF
Backward pass

29. PERT model-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

30. 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

31. 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.
Activity
start
Optimistic time
Most likely
time (mode)
Pessimistic
time to tp tm te

32. te = to + 4tm +tp 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

33. (tp – to)2 2 = 36 Variance 2 = variance to = optimistic time
2 =
(tp – to)2 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.

34. Example 3-Probabilistic Time Estimates
The following table contains information related to the major activities of a research project. Use the information to do the following:
Draw its precedence diagram.
Find the critical Path.
Determine the expected time of the project.
Calculate variance or standart deviation of the project
Compute the probability that the project can be completed within 15 and 17 weeks of its start.
Activity
Immediate Predecessor
O-M-P a -
1-3-4 b
2-4-6 c
2-3-5 d
3-4-5 e
3-5-7 f
5-7-9 g
2-3-6 h
4-6-8 i
3-4-6
end
c, f, i

35. Answer-Probabilistic Time Estimates
(a) Draw its precedence diagram
1-3-4 a
3-4-5 d
3-5-7 e
5-7-9 f
2-4-6 b
4-6-8 h
2-3-6 g
3-4-6 i
2-3-5 c
Optimistic
time
Most likely
Pessimistic

36. Answer b,c and d-Probabilistic Time Estimates

37. Answer-Probabilistic Time Estimates
Tabc = 10.0 Tdef = 16.0 (critical path) Tghi = 13.50
2.83 a
4.00 d
5.0 e
7.0 f b
6.0 h
3.33 g
4.17 i
3.17 c

38. Specified time – Path mean Path standard deviation
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 is almost equal to

39. Example-The Path probability
Given the information on the example of probabilistic time estimates (part (e) in the previous example):
e) 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.

40. Answer-The Path probability
Determine (e)
The probability that the project can be completed within 17 weeks of its start. Path: a-b-c
17 – 10
0.97
Z =
=7.22
Prob.comp in 17 week=1.00
Appendix B, Table B, p.p 884/5
Determine
The probability that the project will be completed within 17 weeks of its start. Path: d-e-f
17 – 16 1 =1
Prob.comp in 17 week=0.8413
Appendix B, Table B, p.p 885
Z =

41. Answer-The Path probability
Determine
The probability that the project will be completed within 17 weeks of its start. Path: g-h-i
17 – 13.5
1.07
Prob.comp in 17 week=1.00
Appendix B, Table B, p.p 884/5
Z =
=3.27
Prob finish in 17 week=1.00 X X 1.00=

42. 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
0.97
Z =
=5.15
Prob.comp in 15 week=1.00
Appendix B, Table B, p.p 884/5
Determine
The probability that the project will be completed within 15 weeks of its start. Path: d-e-f
15 – 16 1
=-1.00
Prob.comp in 15 week=0.1587
Appendix B, Table B, p.p 885
Z =

43. Answer-The Path probability
Determine
The probability that the project will be completed within 15 weeks of its start. Path: g-h-i
15 – 13.5
1.07
Prob.comp in 15 week=0.9192
Appendix B, Table B, p.p 884/5
Z =
=1.40
Prob finish in 15 week=1.00 X X =
The probability that the project will not be completed within 15 weeks of its start: =0.8541

44. Answer-The Path probability-Graphically17
Weeks
1.00
a-b-c
Weeks
10.0
0.8413
d-e-f
Weeks
16.0
1.00
g-h-i
Weeks
13.5

45. Answer-The Path probability-Graphically15
Weeks
1.00
a-b-c
Weeks
10.0
0.1587
d-e-f
Weeks
16.0
0.9192
g-h-i
Weeks
13.5

46. Table 12.3, page 569

47. Time-cost Trade-offs: Crashing Excluded from the exam topics
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

48. Time-Cost Trade-Offs: Crashing
Excluded from the exam topics
Total
cost
Shorten
Cumulative (direct)
cost of crashing
Expected indirect costs
Optimum
CRASH

49. Example-Crashing
Using the following information, develop the optimal time cost solution. Indirect costs are $ 1000 per day.
Determine which activities are on the critical path, its length, and the length of the other path
Rank the critical activities in order of lowest crashing cost, and determine the number of days each can be crashed.
Determine the critical path after each reduction by shortening the project.
Excluded from the exam topics
Activity
Normal time
Crash time
Cost per day to crash a 6 c 10 8
$500 d 5 4
300 b 1
700 e 9 7
600 f 2
800

50. 10 6 b a 2 f 5 9 c e 4 d Determine the expected length of the project.
Excluded from the exam topics 6 a 4 d 5 c 10 b 9 e 2 f

51. Answer-Crashing
Excluded from the exam topics
Determine which activities are on the critical path, its length, and the length of the other path
Path length
a-b-f
c-d-e-f (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 $ e d f

52. 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
c-d-e-f c no further reduction possible d $ 700 e f
Excluded from the exam topics

53. 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 $ 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=
a-b-f
c-d-e-f
activity crashed c e f
cost $
Excluded from the exam topics

54. Thanks