 You should be able to:
1.
Discuss the behavioral aspects of projects in terms of project
personnel and the project manager
2. Explain the nature and importance of a work breakdown structure
in project management
3.
Give a general description of PERT/CPM techniques
4. Construct simple network diagrams
5. List the kinds of information that a PERT or CPM analysis can
provide
6. Analyze networks with deterministic times
7. Analyze networks with probabilistic times
8. Describe activity ‘crashing’ and solve typical problems
17-1
 Projects
 Unique, one-time operations designed to accomplish a
specific set of objectives in a limited time frame
 Examples:
 The Olympic Games
 Producing a movie
 Software development
 Product development
 ERP implementation
17-2
17-3
 WBS
 A hierarchical listing of what must be done during a
project
 Establishes a logical framework for identifying the required
activities for the project
1.
2.
3.
Identify the major elements of the project
Identify the major supporting activities for each of the
major elements
Break down each major supporting activity into a list of
the activities that will be needed to accomplish it
17-4
17-5
17-6
 PERT (program evaluation and review technique)
and CPM (critical path method) are two techniques
used to manage large-scale projects
 By using PERT or CPM Managers can obtain:
1.
2.
3.
4.
A graphical display of project activities
An estimate of how long the project will take
An indication of which activities are most critical to timely
project completion
An indication of how long any activity can be delayed without
delaying the project
17-7
 Network diagram
 Diagram of project activities that shows sequential relationships by
use of arrows and nodes
 Activity on arrow (AOA)
 Network diagram convention in which arrows designate activities
 Activity on node (AON)
 Network convention in which nodes designate activities
 Activities
 Project steps that consume resources and/or time
 Events
 The starting and finishing of activities
17-8
Activity on Activity Meaning
Node (AON)
(a) A
C
B
A comes before B, which comes before C.
A
(b)
C
A and B must both be completed before
C can start.
B
B
(c)
B and C cannot begin until A is completed.
A
C
17-9
Activity on Activity Meaning
Node (AON)
A
C
(d)
B
D
A
C
B
D
(e)
C and D cannot begin until both A and B are
completed.
C cannot begin until both A and B are
completed; D cannot begin until B is
completed. A dummy activity is introduced
in AOA.
17-10
Activity on Activity Meaning
Node (AON)
A
B
(f)
C
D
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.
17-11
Project Activities and Predecessors
Activity
A
Description
Build internal components
Immediate
Predecessors
—
B
Modify roof and floor
—
C
Construct collection stack
A
D
Pour concrete and install frame
A, B
E
Build high-temperature burner
C
F
Install pollution control system
C
G
Install air pollution device
D, E
H
Inspect and test
F, G
17-12
A
Activity A
(Build Internal Components)
B
Activity B
(Modify Roof and Floor)
Start
Start
Activity
17-13
Activity A Precedes Activity C
A
C
B
D
Start
Activities A and B
Precede Activity D
17-14
F
A
C
E
Start
H
B
D
G
Arrows Show Precedence
Relationships
17-15
 Deterministic
 Time estimates that are fairly certain
 Probabilistic
 Time estimates that allow for variation
17-16
Activity
A
B
C
D
E
F
G
H
Description
Time (weeks)
Build internal components
2
Modify roof and floor
3
Construct collection stack
2
Pour concrete and install frame
4
Build high-temperature burner
4
Install pollution control system
3
Install air pollution device
5
Inspect and test
2
Total Time (weeks)
25
17-17
Critical Path
Path
Path duration
A-C-F-H
2+2+3+2=9
A-C-E-G-H
2 + 2 + 4 + 5 + 2 = 15
A-D-G-H
2 + 4 + 5 + 2 = 13
B-D-G-H
3 + 4 + 5 + 2 = 14
Critical path = Longest path A-C-E-G-H
Project duration = 15 weeks
17-18
 Finding ES and EF involves a forward pass
through the network diagram
 Early start (ES)
 The earliest time an activity can start
 Assumes all preceding activities start as early as possible
 For nodes with one entering arrow
 ES = EF of the entering arrow
 For activities leaving nodes with multiple entering arrows
 ES = the largest of the largest entering EF
 Early finish (EF)
 The earliest time an activity can finish
 EF = ES + t
17-19
 Finding LS and LF involves a backward pass through
the network diagram
 Late Start (LS)
 The latest time the activity can start and not delay the project
 The latest starting time for each activity is equal to its latest finishing time
minus its expected duration:
 LS = LF - t
 Late Finish (LF)
 The latest time the activity can finish and not delay the project
 For nodes with one leaving arrow, LF for nodes entering that node equals
the LS of the leaving arrow
 For nodes with multiple leaving arrows, LF for arrows entering node equals
the smallest of the leaving arrows
17-20
 Slack can be computed one of two ways:
 Slack = LS – ES
 Slack = LF – EF
 Critical path
 The critical path is indicated by the activities with zero
slack
17-21
Perform a Critical Path Analysis
Activity Name
or Symbol
A
Earliest
Start
ES
EF
Latest
Start
LS
LF
2
Earliest
Finish
Latest
Finish
Activity Duration
17-22
Begin at starting event and work forward
Earliest Start Time Rule:
 If an activity has only a single 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}
17-23
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
17-24
ES
EF = ES + Activity time
Start
0
0
0
17-25
EF of A =
ES of A + 2
ES
of A
0
Start
0
A
0
2
0
2
17-26
0
A
2
0
Start
0
0
2
EF of B =
ES of B + 3
ES
of B
B
0
3
3
17-27
0
A
2
2
0
Start
2
C
4
2
0
0
0
B
3
3
17-28
0
A
2
2
0
Start
2
C
4
2
0
= Max (2, 3)
0
D
3
0
B
3
7
3
4
17-29
0
A
2
2
2
0
Start
C
4
2
0
0
0
B
3
3
3
D
7
4
17-30
0
A
2
2
2
0
Start
C
4
4
2
F
7
3
0
4
0
E
8
13
4
0
B
3
3
3
D
4
7
H
15
2
G
8
13
5
17-31
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}
17-32
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
17-33
0
A
2
2
2
0
Start
C
4
4
2
F
7
3
0
4
0
E
8
13
13
4
0
B
3
3
H
2
15
15
LS = LF
D – Activity time
G
3
7
4
8
13
5
LF = EF
of Project
17-34
0
A
2
2
2
0
Start
C
4
4
10
2
F
3
7
13
E
0
8 of
LF =4 Min(LS
following activity)
0
13
13
4
0
B
3
3
3
D
4
7
H
2
15
15
G
8
13
5
17-35
LF = Min(4, 10)
0
A
2
2
2
0
Start
2
C
2
4
4
4
10
0
4
4
0
0
B
3
3
3
D
4
7
E
4
F
3
7
13
8
13
8
13
H
2
15
15
G
8
13
8
13
5
17-36
0
0
0
0
Start
0
A
2
2
2
2
2
C
2
4
4
4
10
0
4
0
4
0
1
B
3
3
3
4
4
D
4
E
4
F
3
7
13
8
13
8
13
H
2
15
15
G
7
8
13
8
8
13
5
17-37
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
17-38
Earliest Earliest
Start
Finish
Activity
ES
EF
A
B
C
D
E
F
G
H
0
0
2
3
4
4
8
13
2
3
4
7
8
7
13
15
Latest
Start
LS
Latest
Finish
LF
Slack
LS – ES
On
Critical
Path
0
1
2
4
4
10
8
13
2
4
4
8
8
13
13
15
0
1
0
1
0
6
0
0
Yes
No
Yes
No
Yes
No
Yes
Yes
17-39
 Knowledge of slack times provides managers with
information for planning allocation of scarce
resources
 Control efforts will be directed toward those activities that might
be most susceptible to delaying the project
 Activity slack times are based on the assumption that all of the
activities on the same path will be started as early as possible and
not exceed their expected time
 If two activities are on the same path and have the same slack, this
will be the total slack available to both
17-40
 The beta distribution is generally used to describe the
inherent variability in time estimates
 The probabilistic approach involves three time
estimates:
 Optimistic time, (to)
 The length of time required under optimal conditions
 Pessimistic time, (tp)
 The length of time required under the worst conditions
 Most likely time, (tm)
 The most probable length of time required
17-41
 The expected time, te ,for an activity is a weighted
average of the three time estimates:
te 
to  4t m  t p
6
 The expected duration of a path is equal to the sum of
the expected times of the activities on that path:
Pathmean  of expectedtimesof activitieson thepath
17-42
z
Specified time - Path mean
Path standard deviation
17-43
 A project is not complete until all project activities are complete
 It is risky to only consider the critical path when assessing the
probability of completing a project within a specified time.
 To determine the probability of completing the project within a particular
time frame
 Calculate the probability that each path in the project will be
completed within the specified time
 Multiply these probabilities
 The result is the probability that the project will be completed
within the specified time
17-44
 The standard deviation of each activity’s time is estimated
as one-sixth of the difference between the pessimistic and
optimistic time estimates. The variance is the square of
the standard deviation:
 t p  to 
 

6


2
2
 Standard deviation of the expected time for the path
 path 
 Variances
of activities on path 
17-45
Optimistic
Activity
to
A
B
C
D
E
F
G
H
1
2
1
2
1
1
3
1
Most
Likely
Pessimistic
Expected
Time
Variance
tm
tp
t = (to + 4 tm + tp)/6
[(tp – to)/6]2
2
3
2
4
4
2
4
2
3
4
3
6
7
9
11
3
2
3
2
4
4
3
5
2
4/36
4/36
4/36
16/36
36/36
64/36
64/36
4/36
17-46
 Knowledge of expected path times and their standard
deviations enables managers to compute probabilistic
estimates about project completion such as:
 The probability that the project will be completed by a
certain time
 The probability that the project will take longer than its
expected completion time
17-47
Path variance is computed by
summing the variances of
activities on the path
p2 = Project variance
= (variances of activities
on the path)
17-48
Variance of critical path
p2 = 4/36 + 4/36 + 36/36 + 64/36 + 4/36
= 112/36 = 3.1111
Standard deviation of critical path
p =
=
Variance of critical path
3.1111 = 1.7638 weeks
17-49
Note:
 Total project completion times follow a
normal probability distribution
 Activity times are statistically
independent
17-50
Standard deviation = 1.7638 weeks
15 Weeks
(Expected Completion Time)
17-51
What is the probability this project can
be completed on or before the 16 week
deadline?
Z=
T - TM
p
TM = Path mean completion time
T = Due date
Z = (16 – 15)/1.7638
Where Z is the
number of
standard
deviations the
due date or target
date lies from the
mean or expected
date
= 0.57
17-52
Probability
(T ≤ 16 weeks)
is 0.7157
0.57 Standard deviations
Probability
(T > 16 weeks)
is 1 – 0.7157 =
0.2843
15
Weeks
16
Weeks
Time
17-53
Determine a due date for the project
completion time that will have a 99%
probability of meeting.
Due date = TE + Z p
17-54
Probability
of 0.99
Probability
of 0.01
2.325 Standard
deviations
From Appendix I
0
Z
2.325
17-55
Due date (T) = TM + z p
Probability
of 0.99
Probability
of 0.01
2.325 Standard
deviations
From Appendix I
0
Z
2.325
17-56
Due date (T) = 15 + 2.325 (1.7638)
Probability
of 0.99
Probability
of 0.01
2.325 Standard
deviations
From Appendix I
0
Z
2.325
17-57
Due date (T) = 19.1 weeks
Probability
of 0.99
Probability
of 0.01
2.325 Standard
deviations
From Appendix I
0
Z
2.325
17-58
 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
17-59
1.
2.
3.
4.
5.
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) are not on the
critical path and have slack time
A detailed schedule is available
17-60
 Budget control is an important aspect of project
management
 Costs can exceed budget
 Overly optimistic time estimates
 Unforeseen events
 Unless corrective action is taken, serious cost overruns
can occur
17-61
 Activity time estimates are made for some given level
of resources
 It may be possible to reduce the duration of a project
by injecting additional resources
 Motivations:
 To avoid late penalties
 Monetary incentives
 Free resources for use on other projects
17-62
 Crashing
 Shortening activity durations
 Typically, involves the use of additional funds to support additional
personnel or more efficient equipment, and the relaxing of some work
specifications
 The project duration may be shortened by increasing direct
expenses, thereby realizing savings in
indirect project costs
17-63
 To make decisions concerning crashing requires
information about:
1.
Regular time and crash time estimates for each activity
2.
Regular cost and crash cost estimates for each activity
3.
A list of activities that are on the critical path
 Critical path activities are potential candidates for crashing
 Crashing non-critical path activities would not have an impact on
overall project duration
17-64
 General procedure:
1.
Crash the project one period at a time
2.
Crash the least expensive activity that is on the critical path
3.
When there are multiple critical paths, find the sum of crashing
the least expensive activity on each critical path

If two or more critical paths share common activities,
compare the least expensive cost of crashing a common
activity shared by critical paths with the sum for the separate
critical paths
17-65
17-66
 Among the most useful features of PERT:
1. It forces the manager to organize and quantify available
information and to identify where additional
information is needed
2. It provides the a graphic display of the project and its
major activities
3. It identifies
a. Activities that should be closely watched
b. Activities that have slack time
17-67
 Potential sources of error:
1.
The project network may be incomplete
2. Precedence relationships may not be correctly expressed
3. Time estimates may be inaccurate
4. There may be a tendency to focus on critical path activities to the
exclusion of other important project activities
5. Major risk events may not be on the critical path
17-68
 Technology has benefited project management
 CAD
 To produce updated prototypes on construction and productdevelopment projects
 Communication software
 Helps to keep project members in close contact
 Facilitates remote viewing of projects
 Project management software
 Specialized software used to help manage projects




Assign resources
Compare project plan versions
Evaluate changes
Track performance
17-69
 Advantages include:
 Imposes a methodology and common project management
terminology
 Provides a logical planning structure
 May enhance communication among team members
 Can flag the occurrence of constraint violations
 Automatically formats reports
 Can generate multiple levels of summary and detail reports
 Enables “what if” scenarios
 Can generate a variety of chart types
17-70
 Risks are an inherent part of project management
 Risks relate to occurrence of events that have undesirable
consequences such as
 Delays
 Increased costs
 Inability to meet technical specifications
 Good risk management involves
 Identifying as many risks as possible
 Analyzing and assessing those risks
 Working to minimize the probability of their occurrence
 Establishing contingency plans and budgets for dealing with any
that do occur
17-71
 Projects present both strategic opportunities and risks
 It is critical to devote sufficient resources and attention to projects
 Projects are often employed in situations that are characterized by
significant uncertainties that demand
 Careful planning
 Wise selection of project manager and team
 Monitoring of the project
 Project software can facilitate successful project completion
 Be careful to not focus on critical path activities
to the exclusion of other activities that may
become critical
 It is not uncommon for projects to fail
 When that happens, it can be beneficial to examine the probable reasons
for failure
17-72