2. Chapter 3
Project Management

3. OBJECTIVES Definition of Project Management Work Breakdown Structure
Project Control Charts
Structuring Projects
Critical Path Scheduling 2

4. Project Management Defined
Project is a series of related jobs usually directed toward some major output and requiring a significant period of time to perform
Project Management are the management activities of planning, directing, and controlling resources (people, equipment, material) to meet the technical, cost, and time constraints of a project 3

5. Gantt Chart Activity 1 Activity 2 Activity 3 Activity 4 Activity 5
Vertical Axis: Always Activities or Jobs
Horizontal bars used to denote length of time for each activity or job.
Activity 1
Activity 2
Activity 3
Activity 4
Activity 5
Activity 6
Time
Horizontal Axis: Always Time 6

6. Structuring Projects: Pure Project Advantages
A pure project is where a self-contained team works full-time on the project
Structuring Projects: Pure Project Advantages
The project manager has full authority over the project
Team members report to one boss
Shortened communication lines
Team pride, motivation, and commitment are high 8

7. Structuring Projects: Pure Project Disadvantages
Duplication of resources
Organizational goals and policies are ignored
Lack of technology transfer
Team members have no functional area "home" 8

8. Functional Project
A functional project is housed within a functional division
President
Research and
Development
Engineering
Manufacturing
Project A B C D E F G H I
Example, Project “B” is in the functional area of Research and Development. 9

9. Structuring Projects Functional Project: Advantages
A team member can work on several projects
Technical expertise is maintained within the functional area
The functional area is a “home” after the project is completed
Critical mass of specialized knowledge 10

10. Structuring Projects Functional Project: Disadvantages
Aspects of the project that are not directly related to the functional area get short-changed
Motivation of team members is often weak
Needs of the client are secondary and are responded to slowly 11

11. Matrix Project Organization Structure
President
Research and
Development
Engineering
Manufacturing
Marketing
Manager
Project A
Project B
Project C 12

12. Structuring Projects Matrix: Advantages
Enhanced communications between functional areas
Pinpointed responsibility
Duplication of resources is minimized
Functional “home” for team members
Policies of the parent organization are followed 13

13. Structuring Projects Matrix: Disadvantages
Too many bosses
Depends on project manager’s negotiating skills
Potential for sub-optimization 14

14. Work Breakdown Structure
A work breakdown structure defines the hierarchy of project tasks, subtasks, and work packages
Program
Project 1
Project 2
Task 1.1
Subtask 1.1.1
Work Package
Level 1 2 3 4
Task 1.2
Subtask 1.1.2
Work Package 4

15. Network-Planning Models
A project is made up of a sequence of activities that form a network representing a project
The path taking longest time through this network of activities is called the “critical path”
The critical path provides a wide range of scheduling information useful in managing a project
Critical Path Method (CPM) helps to identify the critical path(s) in the project networks 15

16. Prerequisites for Critical Path Methodology
A project must have:
well-defined jobs or tasks whose completion marks the end of the project;
independent jobs or tasks;
and tasks that follow a given sequence. 16

17. Types of Critical Path Methods
CPM with a Single Time Estimate
Used when activity times are known with certainty
Used to determine timing estimates for the project, each activity in the project, and slack time for activities
CPM with Three Activity Time Estimates
Used when activity times are uncertain
Used to obtain the same information as the Single Time Estimate model and probability information
Time-Cost Models
Used when cost trade-off information is a major consideration in planning
Used to determine the least cost in reducing total project time 15

18. PERT and CPM Network techniques Developed in 1950’s
CPM by DuPont for chemical plants (1957)
PERT by Booz, Allen & Hamilton with the U.S. Navy, for Polaris missile (1958)
Consider precedence relationships and interdependencies
Each uses a different estimate of activity times

19. Six Steps PERT & CPM
Define the project and prepare the work breakdown structure
Develop relationships among the activities - decide which activities must precede and which must follow others
Draw the network connecting all of the activities

20. Six Steps PERT & CPM
Assign time and/or cost estimates to each activity
Compute the longest time path through the network – this is called the critical path
Use the network to help plan, schedule, monitor, and control the project

21. A Comparison of AON and AOA Network Conventions
Activity on Activity Activity on
Node (AON) Meaning Arrow (AOA)
A comes before B, which comes before C
(a) A B C
A and B must both be completed before C can start
(b) A C B
B and C cannot begin until A is completed
(c) B A C
Figure 3.5

22. A Comparison of AON and AOA Network Conventions
Activity on Activity Activity on
Node (AON) Meaning Arrow (AOA)
C and D cannot begin until A and B have both been completed
(d) A B C D
C cannot begin until both A and B are completed; D cannot begin until B is completed. A dummy activity is introduced in AOA
(e) C A B D
Dummy activity
Figure 3.5

23. A Comparison of AON and AOA Network Conventions
Activity on Activity Activity on
Node (AON) Meaning Arrow (AOA)
B and C cannot begin until A is completed. D cannot begin until both B and C are completed. A dummy activity is again introduced in AOA.
(f) A C D B
Dummy activity
Figure 3.5

24. Immediate Predecessors
AON Example
Milwaukee Paper Manufacturing's Activities and Predecessors
Activity
Description
Immediate Predecessors A
Build internal components — B
Modify roof and floor C
Construct collection stack D
Pour concrete and install frame
A, B E
Build high-temperature burner F
Install pollution control system G
Install air pollution device
D, E H
Inspect and test
F, G
Table 3.1

25. AON Network for Milwaukee Paper
Start B
Activity A
(Build Internal Components)
Start Activity
Activity B
(Modify Roof and Floor)
Figure 3.6

26. AON Network for Milwaukee Paper
Activity A Precedes Activity C A
Start B C D
Activities A and B Precede Activity D
Figure 3.7

27. AON Network for Milwaukee PaperG E F H C A
Start D B
Arrows Show Precedence Relationships
Figure 3.8

28. AOA Network for Milwaukee Paper1 3 2
(Modify Roof/Floor) B
(Build Internal Components) A 5 D
(Pour Concrete/ Install Frame) 4 C
(Construct Stack) 6
(Install Controls) F
(Build Burner) E
(Install Pollution Device) G
Dummy Activity H
(Inspect/ Test) 7
Figure 3.9

29. Determining the Project Schedule
Perform a Critical Path Analysis
The critical path is the longest path through the network
The critical path is the shortest time in which the project can be completed
Any delay in critical path activities delays the project
Critical path activities have no slack time

30. Determining the Project Schedule
Perform a Critical Path Analysis
Activity Description Time (weeks)
A Build internal components 2
B Modify roof and floor 3
C Construct collection stack 2
D Pour concrete and install frame 4
E Build high-temperature burner 4
F Install pollution control system 3
G Install air pollution device 5
H Inspect and test 2
Total Time (weeks) 25
Table 3.2

31. Determining the Project Schedule
Perform a Critical Path Analysis
Activity Description Time (weeks)
A Build internal components 2
B Modify roof and floor 3
C Construct collection stack 2
D Pour concrete and install frame 4
E Build high-temperature burner 4
F Install pollution control system 3
G Install air pollution device 5
H Inspect and test 2
Total Time (weeks) 25
Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completed
Earliest finish (EF) = earliest time at which an activity can be finished
Latest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire project
Latest finish (LF) = latest time by which an activity has to be finished so as to not delay the completion time of the entire project
Table 3.2

32. Determining the Project Schedule
Perform a Critical Path Analysis A
Activity Name or Symbol
Earliest Start ES
Earliest Finish EF
Latest Start LS
Latest Finish LF
Activity Duration 2
Figure 3.10

33. Forward Pass Begin at starting event and work forward
Earliest Start Time Rule:
If an activity has only one immediate predecessor, its ES equals the EF of the predecessor
If an activity has multiple immediate predecessors, its ES is the maximum of all the EF values of its predecessors
ES = Max (EF of all immediate predecessors)

34. Forward Pass Begin at starting event and work forward
Earliest Finish Time Rule:
The earliest finish time (EF) of an activity is the sum of its earliest start time (ES) and its activity time
EF = ES + Activity time

35. ES/EF Network for Milwaukee PaperES
EF = ES + Activity time
Start

36. ES/EF Network for Milwaukee Paper2
EF of A = ES of A + 2
ES of A A 2
Start

37. ES/EF Network for Milwaukee Paper
Start A 2 3
EF of B = ES of B + 3
ES of B B 3

38. ES/EF Network for Milwaukee Paper
Start A 2 C 2 4 B 3

39. ES/EF Network for Milwaukee Paper
Start A 2 C 2 4 D 4 3
= Max (2, 3) 7 B 3

40. ES/EF Network for Milwaukee Paper
Start A 2 C 2 4 D 4 3 7 B 3

41. ES/EF Network for Milwaukee Paper
Start A 2 C 2 4 E 4 F 3 G 5 H 2 8 13 15 7 D 4 3 7 B 3
Figure 3.11

42. Backward Pass Begin with the last event and work backwards
Latest Finish Time Rule:
If an activity is an immediate predecessor for just a single activity, its LF equals the LS of the activity that immediately follows it
If an activity is an immediate predecessor to more than one activity, its LF is the minimum of all LS values of all activities that immediately follow it
LF = Min (LS of all immediate following activities)

43. Backward Pass Begin with the last event and work backwards
Latest Start Time Rule:
The latest start time (LS) of an activity is the difference of its latest finish time (LF) and its activity time
LS = LF – Activity time

44. LS/LF Times for Milwaukee Paper4 F 3 G 5 H 2 8 13 15 7 D C B
Start A 13
LS = LF – Activity time
LF = EF of Project 15
Figure 3.12

45. LS/LF Times for Milwaukee Paper4 F 3 G 5 H 2 8 13 15 7 D C B
Start A
LF = Min(LS of following activity) 10 13
Figure 3.12

46. LS/LF Times for Milwaukee Paper
LF = Min(4, 10) 4 2 E 4 F 3 G 5 H 2 8 13 15 7 10 D C B
Start A
Figure 3.12

47. LS/LF Times for Milwaukee Paper4 F 3 G 5 H 2 8 13 15 7 10 D C B
Start A 1
Figure 3.12

48. Computing Slack Time
After computing the ES, EF, LS, and LF times for all activities, compute the slack or free time for each activity
Slack is the length of time an activity can be delayed without delaying the entire project
Slack = LS – ES or Slack = LF – EF

49. Computing Slack Time A 0 2 0 2 0 Yes B 0 3 1 4 1 No C 2 4 2 4 0 Yes
Earliest Earliest Latest Latest On Start Finish Start Finish Slack Critical Activity ES EF LS LF LS – ES Path
A Yes
B No
C Yes
D No
E Yes
F No
G Yes
H Yes
Table 3.3

50. Critical Path for Milwaukee Paper4 F 3 G 5 H 2 8 13 15 7 10 D C B
Start A 1
Figure 3.13

51. ES – EF Gantt Chart for Milwaukee Paper
A Build internal components
B Modify roof and floor
C Construct collection stack
D Pour concrete and install frame
E Build high-temperature burner
F Install pollution control system
G Install air pollution device
H Inspect and test

52. LS – LF Gantt Chart for Milwaukee Paper
A Build internal components
B Modify roof and floor
C Construct collection stack
D Pour concrete and install frame
E Build high-temperature burner
F Install pollution control system
G Install air pollution device
H Inspect and test

53. Variability in Activity Times
CPM assumes we know a fixed time estimate for each activity and there is no variability in activity times
PERT uses a probability distribution for activity times to allow for variability

54. Variability in Activity Times
Three time estimates are required
Optimistic time (a) – if everything goes according to plan
Most–likely time (m) – most realistic estimate
Pessimistic time (b) – assuming very unfavorable conditions

55. Variability in Activity Times
Estimate follows beta distribution
Expected time:
Variance of times:
t = (a + 4m + b)/6
v = [(b – a)/6]2

56. Variability in Activity Times
Estimate follows beta distribution
Expected time:
Variance of times:
t = (a + 4m + b)/6
v = [(b − a)/6]2
Probability
Optimistic Time (a)
Most Likely Time (m)
Pessimistic Time (b)
Activity Time
Probability of 1 in 100 of > b occurring
Probability of 1 in 100 of < a occurring

57. Computing Variance A 1 2 3 2 .11 B 2 3 4 3 .11 C 1 2 3 2 .11
Most Expected Optimistic Likely Pessimistic Time Variance Activity a m b t = (a + 4m + b)/6 [(b – a)/6]2 A B C D E F G H
Table 3.4

58. Probability of Project Completion
Project variance is computed by summing the variances of critical activities
s2 = Project variance
= (variances of activities on critical path) p

59. Probability of Project Completion
Project variance is computed by summing the variances of critical activities
Project variance
s2 = = 3.11
Project standard deviation
sp = Project variance
= = 1.76 weeks p

60. Probability of Project Completion
PERT makes two more assumptions:
Total project completion times follow a normal probability distribution
Activity times are statistically independent

61. Probability of Project Completion
Standard deviation = 1.76 weeks
15 Weeks
(Expected Completion Time)
Figure 3.15

62. Probability of Project Completion
What is the probability this project can be completed on or before the 16 week deadline?
Z = – /sp
= (16 wks – 15 wks)/1.76
= 0.57
due expected date date of completion
Where Z is the number of standard deviations the due date lies from the mean

63. Probability of Project Completion
From Appendix I
What is the probability this project can be completed on or before the 16 week deadline?
Z = − /sp
= (16 wks − 15 wks)/1.76
= 0.57
due expected date date of completion
Where Z is the number of standard deviations the due date lies from the mean

64. Probability of Project Completion
0.57 Standard deviations
Probability (T ≤ 16 weeks) is 71.57%
Weeks Weeks
Time
Figure 3.16

65. Determining Project Completion Time
Probability of 0.99 Z
Probability of 0.01
2.33 Standard deviations
2.33
From Appendix I
Figure 3.17

66. Variability of Completion Time for Noncritical Paths
Variability of times for activities on noncritical paths must be considered when finding the probability of finishing in a specified time
Variation in noncritical activity may cause change in critical path

67. What Project Management Has Provided So Far
The project’s expected completion time is 15 weeks
There is a 71.57% chance the equipment will be in place by the 16 week deadline
Five activities (A, C, E, G, and H) are on the critical path
Three activities (B, D, F) have slack time and are not on the critical path
A detailed schedule is available

68. Trade-Offs And Project Crashing
It is not uncommon to face the following situations:
The project is behind schedule
The completion time has been moved forward
Shortening the duration of the project is called project crashing

69. Factors to Consider When Crashing A Project
The amount by which an activity is crashed is, in fact, permissible
Taken together, the shortened activity durations will enable us to finish the project by the due date
The total cost of crashing is as small as possible

70. Steps in Project Crashing
Compute the crash cost per time period. If crash costs are linear over time:
Crash cost
per period =
(Crash cost – Normal cost)
(Normal time – Crash time)
Using current activity times, find the critical path and identify the critical activities

71. Steps in Project Crashing
If there is only one critical path, then select the activity on this critical path that (a) can still be crashed, and (b) has the smallest crash cost per period. If there is more than one critical path, then select one activity from each critical path such that (a) each selected activity can still be crashed, and (b) the total crash cost of all selected activities is the smallest. Note that a single activity may be common to more than one critical path.

72. Steps in Project Crashing
Update all activity times. If the desired due date has been reached, stop. If not, return to Step 2.

73. Crashing The Project A 2 1 22,000 22,750 750 Yes
Time (Wks) Cost ($) Crash Cost Critical Activity Normal Crash Normal Crash Per Wk ($) Path?
A ,000 22, Yes
B ,000 34,000 2,000 No
C ,000 27,000 1,000 Yes
D ,000 49,000 1,000 No
E ,000 58,000 1,000 Yes
F ,000 30, No
G ,000 84,500 1,500 Yes
H ,000 19,000 3,000 Yes
Table 3.5

74. Crash and Normal Times and Costs for Activity B
| | |
1 2 3 Time (Weeks)
$34,000 —
$33,000 —
$32,000 —
$31,000 —
$30,000 — —
Activity Cost
Crash
Normal
Crash Cost/Wk =
Crash Cost – Normal Cost
Normal Time – Crash Time =
$34,000 – $30,000
3 – 1
= = $2,000/Wk
$4,000
2 Wks
Crash Time
Normal Time
Crash Cost
Normal Cost
Figure 3.18

75. Critical Path And Slack Times For Milwaukee Paper4 F 3 G 5 H 2 8 13 15 7 10 D C B
Start A 1
Slack = 1
Slack = 0
Slack = 6
Figure 3.19

76. Steps in the CPM with Single Time Estimate
1. Activity Identification
2. Activity Sequencing and Network Construction
3. Determine the critical path
From the critical path all of the project and activity timing information can be obtained 15

77. CPM with Single Time Estimate
Consider the following consulting project:
Activity
Designation
Immed. Pred.
Time (Weeks)
Assess customer's needs A
None 2
Write and submit proposal B 1
Obtain approval C
Develop service vision and goals D
Train employees E 5
Quality improvement pilot groups F
D, E
Write assessment report G
Develop a critical path diagram and determine the duration of the critical path and slack times for all activities.

78. First draw the network D(2) E(5) F(5) A(2) B(1) C(1) G(1)
Act. Imed. Pred. Time
A None 2
B A 1
C B 1
D C 2
D(2)
E(5)
E C 5
F D,E 5
G F 1
F(5)
A(2)
B(1)
C(1)
G(1) 18

79. Determine early starts and early finish times
EF=6
D(2)
ES=0
EF=2
ES=2
EF=3
ES=3
EF=4
ES=9
EF=14
ES=14
EF=15
A(2)
B(1)
C(1)
F(5)
G(1)
ES=4
EF=9
Hint: Start with ES=0 and go forward in the network from A to G.
E(5) 21

80. Determine late starts and late finish times
Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A.
ES=4
EF=6
D(2)
E(5)
ES=0
EF=2
ES=2
EF=3
ES=3
EF=4
ES=9
EF=14
ES=14
EF=15
LS=7
LF=9
F(5)
A(2)
B(1)
C(1)
G(1)
ES=4
EF=9
LS=3
LF=4
LS=2
LF=3
LS=0
LF=2
LS=9
LF=14
LS=14
LF=15
LS=4
LF=9 22

81. Critical Path & Slack ES=4 EF=6 D(2) ES=0 EF=2 ES=2 EF=3 ES=3 EF=4
Slack=(7-4)=(9-6)= 3 Wks
D(2)
ES=0
EF=2
ES=2
EF=3
ES=3
EF=4
ES=9
EF=14
ES=14
EF=15
LS=7
LF=9
F(5)
A(2)
B(1)
C(1)
G(1)
ES=4
EF=9
LS=3
LF=4
LS=2
LF=3
LS=0
LF=2
LS=9
LF=14
LS=14
LF=15
E(5)
LS=4
LF=9
Duration=15 weeks 23

82. Example 2. CPM with Three Activity Time Estimates

83. Example 2. Expected Time Calculations
ET(A)= 3+4(6)+15 6
ET(A)=42/6=7

84. Ex. 2. Expected Time Calculations
ET(B)= 2+4(4)+14 6
ET(B)=32/6=5.333

85. Ex 2. Expected Time Calculations
ET(C)= 6+4(12)+30 6
ET(C)=84/6=14

86. Example 2. Network A(7) B C(14) D(5) E(11) F(7) H(4) G(11) I(18)
(5.333)
C(14)
D(5)
E(11)
F(7)
H(4)
G(11)
I(18)
Duration = 54 Days

87. Example 2. Probability Exercise
What is the probability of finishing this project in
less than 53 days?
D=53
p(t < D) t
TE = 54

88. (Sum the variance along the critical path.)

89. p(t < D) t
D=53
TE = 54
p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156))
There is a 43.8% probability that this project will be completed in less than 53 weeks.

90. Ex 2. Additional Probability Exercise
What is the probability that the project duration will exceed 56 weeks? 30

91. Example 2. Additional Exercise Solution
TE = 54
p(t < D)
D=56
p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312)) 31

92. Time-Cost Models
Basic Assumption: Relationship between activity completion time and project cost
Time Cost Models: Determine the optimum point in time-cost tradeoffs
Activity direct costs
Project indirect costs
Activity completion times 32

93. CPM Assumptions/Limitations
Project activities can be identified as entities (There is a clear beginning and ending point for each activity.)
Project activity sequence relationships can be specified and networked
Project control should focus on the critical path
The activity times follow the beta distribution, with the variance of the project assumed to equal the sum of the variances along the critical path 15

94. Answer: d. All of the above
Question Bowl
Which of the following are examples of Graphic Project Charts?
Gantt
Bar
Milestone
All of the above
None of the above
Answer: d. All of the above 7

95. Answer: d. All of the above
Question Bowl
Which of the following are one of the three organizational structures of projects?
Pure
Functional
Matrix
All of the above
None of the above
Answer: d. All of the above 7

96. Question Bowl Answer: a. SOW (or Statement of Work)
A project starts with a written description of the objectives to be achieved, with a brief statement of the work to be done and a proposed schedule all contained in which of the following?
SOW (statement of work)
WBS (work breakdown structure)
Early Start Schedule
Late Start Schedule
None of the above
Answer: a. SOW (or Statement of Work) 7

97. Question Bowl Answer: c. Slack time
For some activities in a project there may be some leeway from when an activity can start and when it must finish. What is this period of time called when using the Critical Path Method?
Early start time
Late start time
Slack time
All of the above
None of the above
Answer: c. Slack time 7

98. Question Bowl
How much “slack time” is permitted in the “critical path” activity times?
Only one unit of time per activity
No slack time is permitted
As much as the maximum activity time in the network
As much as is necessary to add up to the total time of the project
None of the above
Answer: b. No slack time is permitted (All critical path activities must have zero slack time, otherwise they would not be critical to the project completion time.) 7

99. Question Bowl
When looking at the Time-Cost Trade Offs in the Minimum-Cost Scheduling time-cost model, we seek to reduce the total time of a project by doing what to the least-cost activity choices?
Crashing them
Adding slack time
Subtracting slack time
Adding project time
None of the above
Answer: a. Crashing them (We “crash” the least-cost activity times to seek a reduced total time for the entire project and we do it step-wise as inexpensively as possible.) 7

100. End of Chapter 3