INSE 6230
Total Quality Project Management
Lecture 3
Project Time Management II.

Network diagrams
◦ AOA
◦ AON


CPM
Shortening project schedule
◦ Crashing
◦ PERT
2






3
Defining activities: identifying the specific activities that
the project team members and stakeholders must
perform to produce the project deliverables
Sequencing activities: identifying and documenting the
relationships between project activities
Estimating activity resources: estimating how many
resources a project team should use to perform project
activities
Estimating activity durations: estimating the number of
work periods that are needed to complete individual
activities
Developing the schedule: analyzing activity sequences,
activity resource estimates, and activity duration
estimates to create the project schedule
Controlling the schedule: controlling and managing
changes to the project schedule


Network diagramming is a technique for showing the activity
precedence relationships in a project
A network diagram is a schematic display of the logical
relationships among, or sequencing of, project activities
Activity on arrow
network (AOA)
Activity on node
network (AON)
1
A
0
B
C
A
2
D
3
Start
C

D
End
The two networks and methods are both widely used, and
both of them should be learned.
◦
4
B
Historically, CPM uses AON, PERT uses AOA



Also called Arrow Diagramming Method (ADM)
Arrows:
◦ Arrows represent particular activities
◦ Arrows show the activity sequencing
Nodes:
◦ Nodes are the starting and ending points of activities
 AOA network diagram can only show finish-to-start dependencies (FS)
◦ Nodes represent events, milestones, or the completion of one or a
collection of activities
 The first node signifies the start of a project, while the last node
represents its end
5
Activity-on-Arrow (AOA) Network
1. Find all of the activities that start at node 0. Draw their finish
nodes and draw arrows between node 0 and those finish
nodes. Put the activity letter or name and duration estimate on
the associated arrow.
2. Continue drawing the network diagram, working from left to
right. Look for bursts and merges. Bursts occur when a single
node is followed by two or more activities. A merge occurs
when two or more nodes precede a single node.
3. Continue drawing the project network diagram until all
activities that have dependencies are included on the diagram.
4. As a rule of thumb, all arrowheads should face toward the
right, and no arrows should cross on an AOA network diagram.
7
Example 4-3
Suppose a project consists of the five activities A, B, C, D and E that
satisfy the following relationship:
1. Neither A nor B has any immediate predecessors
2. A is an immediate predecessor of C
3. B is an immediate predecessor of D
4. C and D are immediate predecessors of E
A
1
C
E
0
3
B
2
D
4
Example 4-3: revised
A
C
1
E
0
B
1. Neither A nor B has any immediate predecessors
2. A is an immediate predecessor of C
3’. A and B are immediate predecessors of D (revised)
4. C and D are immediate predecessors of E
A
0
C
A
D
2
E
1
2
B
3
3
D
C
1
E
P
0
3
B
2
D
4
4
Dummy Activities in AOA Networks
Dummy activity

◦
No duration and no resources
◦
Graphically represented by a dashed line in a network
Purpose of a dummy activity

◦
Maintains the precedence relationships between activities
◦
Provides unique activity designations (e.g. C starts in 1 and finishes in 2)
Only in Activity-on-Arrow (AOA) networks!

◦
AON networks do not need dummy activities
Dummy Activities in AOA Networks
Y

Consider the activities associated with the fabrication of a steel
component shown in table below. Draw the Activity-on-Arrow (AOA)
network diagram for this project, determine the project duration and
the critical path.
Activity
A
B
C
D
E
F
G
Precedence and durations of activities
Description
Predecessors
Preliminary design
--Evaluation of design
A
Contract negotiation
--Preparation of fabrication
C
plant
B, C
Final design
D, E
Fabrication of product
F
Shipment of product to owner
Duration
6
1
8
5
9
12
3
12
Precedence and durations of activities
Description
Predecessors
Preliminary design
--Evaluation of design
A
Contract negotiation
--Preparation of fabrication plant
C
Final design
B, C
Fabrication of product
D, E
Shipment of product to owner
F
Activity
A
B
C
D
E
F
G
A
0
B
1
3
E
X
C
2
D
Duration
6
1
8
5
9
12
3
F
5
G
6
4
13

Also called Precedence Diagramming Method
(PDM)

Nodes (circles or boxes) represent activities

Arrows represent precedence relationships
More popular than ADM method and used by project
management software

◦ No need to use dummy activities
◦ Can show different task dependency types (FS, SS, FF, SF)
14
Activity-on-Node (AON) Network
New activity nodes
representing the
beginning and the end
have been added

Gantt chart, AOA and AON Networks
Gantt chart
1
A
B
Activity-on-Arrow
C
D
0
2
A
3
B
Activity-on-Node
Start
C
D
End
AOA and AON Networks
Activity-on-Arrow
Activity-on-Node



18
CPM is a network diagramming technique used
to predict total project duration
A critical path for a project is the series of
activities that determines the earliest time by
which the project can be completed
The critical path is the longest path through the
network diagram





19
First develop a good network diagram
Add the duration estimates for all activities on each path
through the network diagram
The longest path is the critical path
If one (or more) of the activities on the critical path takes
longer than planned, the whole project schedule will slip
unless the project manager takes corrective action
It is important to update project schedule information to meet
time goals for a project
◦ The critical path may change as the project progresses
◦ If you know the project completion date will slip, negotiate
with the project sponsor
20

Common misconceptions:
◦ The critical path is not the one with the most critical
activities! Critical path is concerned with the time
dimension of a project
◦ The critical path is not the shortest path through the
diagram!

21
There can be more than one critical path if the
lengths of two or more paths are the same
2
2 + 2 + 6 + 3 = 13 days
22
Activity
A
B
C
D
E
F
G
A
0
Precedence and durations of activities
Description
Predecessors
Preliminary design
--Evaluation of design
A
Contract negotiation
--Preparation of fabrication plant
C
Final design
B, C
Fabrication of product
D, E
Shipment of product to owner
F
B
1
X
C
2
The length of A-B-E-F-G is 31
The length of C-X-E-F-G is 32
The length of C-D-F-G is 28
3
E
D
Duration
6
1
8
5
9
12
3
F
5
G
6
4
The critical path is C – (X) – E – F – G
The project duration is 32 days.
23

Duration D(i)
- estimated time necessary to perform an
activity

Early Start ES(i)
- earliest time an activity can start

Early Finish EF(i)
- earliest time an activity can finish:
A
B
Start
D
C
End
EF(i) = ES(i) + D(i)

Late Finish LF(i)
- latest time an activity can be finished without
delaying the completion date of the project

Late Start LS(i)
- latest time that an activity can be started
without delaying the completion date of the
project:
LS(i) = LF(i) – D(i)
24
ES
EF
Activity Number: ID
LS
D
LF
Scheduling on Critical Path (CP)
• Activities that have different earliest and latest
start times (i.e., ES(i) < LS(i)) can be scheduled to
start anytime between ES(i) and LS(i)
• Schedule Flexibility: The concept of float is to use part
or all of this allowable range to schedule an activity without
delaying the completion of the project
• Activities on a critical path
do not have floats
TF  0
FF  0
• They have to be scheduled at their ES(i)
 no flexibility on CP!!!
ES(i)  LS(i)
EF(i)  LF(i)
CPM with AON Networks
Steps for CPM Algorithm (AON)
1.
Activity Numbering
2.
Forward Pass (ES & EF)
3.
Backward Pass (LF & LS)
4.
Identifying the Critical Path
5.
Float times
ES
CPM with AON: Step 1. Activity Numbering
Number: ID
LS


EF
D
Step 1: Give the starting activity number 0.
Step 2: Give the next number to any unnumbered activity whose
predecessor activities are each already numbered. (e.g., activity 0 is a
predecessor activity for 1 and 2)
Repeat Step 2 until all activities are numbered.
1: A
3: B
10
5
0: Start
4: D
4
0
2: C
8
5: End
0
LF
ES
CPM with AON: Step 2. Forward Pass
Number: ID
LS



EF
D
Step 1: Let ES(0) = 0
Step 2: For j = 1,2,3,...,n (where n is the End node), let
ES(j) = maximum {EF(i)}
where the maximum is computed over all activities (i) that have j as their
successor.
Step 3: EF(j) = ES(j) + Dj
0
10
10
15
1: A
3: B
10
5
15
0
0: Start
19
4: D
4
0
0
8
2: C
8
19
5: End
0
LF
ES
CPM with AON: Step 3. Backward Pass
Number: ID
LS



EF
D
Step 1: Let LS(n) equal the required completion time of the project.
(Let LS(n) = ES(n))
Step 2: For i = n-1, n-2, ..., 0, let
LF(i) = minimum {LS(j)}
where the minimum is computed over all activities (j) that have i as their
predecessor.
Step 3: LS(i) = LF(i) - Di
0
10
10
1: A
0
10
15
3: B
10
10
5
15
15
0
0: Start
0
19
19
4: D
15
0
0
8
2: C
7
8
15
4
5: End
19
19
0
LF
CPM with AON: Step 4. Critical Path
ES
Number: ID
LS
0
10
10
1: A
0
10
10
10
5
15
0: Start
19
19
4: D
15
0
4
5: End
19
8
2: C
7
8
LF
15
15
0
D
3: B
0
0
EF
15
Path A-B-D: 10+5+4=19
Path C-D: 8+4=12
Condition: ES(i) = LS(i)
19
0
CPM with AON: Step 5. Float Times
ES
EF
Number: ID
LS
D
• Total float (TF) = Total slack
the maximum amount of delay which can be assigned to
any activity without delaying the entire project
TF ( j )  LS( j )  ES ( j )
Free float (FF) = Free slack
the amount of delay which can be assigned to any
activity without delaying subsequent activities
•
FF ( j )  min ES (k ) EF ( j )
k
k : the successor activities of activity j
•
For activities on critical path:
TF ( j )  0
FF ( j )  0
LF
CPM with AON: Step 5 Float Times
ES
Number: ID
LS
FF = 0
TF = 0
0
10
10
10
15
10
10
5
15
15
0: Start
19
19
4: D
15
0
0
8
8
FF = 7
TF = 7
4
5: End
19
19
0
FF = 0
TF = 0
2: C
7
LF
3: B
0
0
D
FF = 0
TF = 0
1: A
0
EF
15
TF ( j )  LS( j )  ES ( j )
FF ( j )  min ES (k ) EF ( j )
k
k : the successor activities of activity j
CPM with AON: Step 5. Float Times
ES
Number: ID
LS
FF = 0
TF = 0
0
10
10
10
3: B
10
10
5
15
15
0: Start
19
19
4: D
15
0
0
8
8
FF = 2
TF = 7
4
FF = 0
TF = 0
2: C
7
LF
15
0
0
D
FF = 0
TF = 0
1: A
0
EF
15
10
14
X
15
4
19
FF = 5
TF = 5
5: End
19
19
0
CPM with AON: Example
AON diagram
4: G
30
9: B
1: A
15
5: E
8: F
45
30
30
11: D
45
0: Start
10: C
0
15
2: I
5
6: H
15
12: End
3: J
5
0
7: K
30
ES
CPM with AON: Example
AON diagram
15
EF
Number: ID
LS
45
D
LF
4: G
60
0
90
90
15
1: A
0
30
15
15
15
60
60
5: E
15
45
60
60
9: B
90
90
8: F
30
120
30
120
120
90
0: Start
0
0
10
5
D
45
105
10: C
5
105 15
2: I
165
11:
90
0
0
120
15
15
120
30
6: H
75
15
90
165
0
5
12:
3: J
55
5
5
60
35
7: K
60
30
90
165
End
0
165
ES
CPM with AON: Example
AON diagram
15
EF
Number: ID
LS
45
D
LF
4: G
60
0
90
90
15
1: A
0
30
15
15
15
60
60
5: E
15
45
60
60
9: B
90
90
8: F
30
120
30
120
120
90
0: Start
0
0
10
5
D
45
105
10: C
5
105 15
2: I
165
11:
90
0
0
120
15
15
120
30
6: H
75
15
90
165
0
5
12:
3: J
55
5
5
60
35
7: K
60
30
90
165
End
0
165
ES
CPM with AON: Example
AON diagram
15
EF
Number: ID
45
LS
D
LF
4: G
60
30
90
TF=45, FF=45
0
1: A
0
90
15
15
15
15
TF=0, FF=0
0
60
5: E
15
45
60
TF=0, FF=0
0
60
9: B
90
90
8: F
30
90
30
0
30
D
45
75
15
120
TF=15, FF=15
90
165
5
12:
3: J
55
5
5
60
TF=55, FF=0
35
7: K
60
30
90
TF=55, FF=55
165
TF=0, FF=0
6: H
TF=60, FF=60
0
105
165
11:
120
TF=0, FF=0
105 15
15
10
5
15
TF=10, FF=10
120
10: C
5
2: I
120
TF=0, FF=0
90
0: Start
0
60
120
165
End
0
Activity E
3
Activity B
8
Activity H
4
Activity F
Activity A
Activity I
4
0
0
Activity C
3
Activity G
5
Activity D
5
38
8
11
Activity E
9
0
3
12
8
Activity B
0
12
8
8
Activity H
12
8
0
0
8
0
4
16
12
16
Activity F
Activity A
16
12
4
Activity I
16
0
0
3
Activity C
3
5
8
5
10
Activity G
11
0
5
5
16
Activity D
6
39
5
11
The critical path is ………A-B-F-H-I……….
The total duration of the project is ……16……..
For activity E, determine ES:……8……….. and LF ………12……
For activity G, determine EF:……10……….. and LS ………11…
For activity C, determine TF:………5……… and FF ………2……
For activity D, determine TF:………6………
For activity F, determine FF ……0…………

Three main techniques for shortening schedules
◦ Reduce the duration of activities on the critical path
by adding more resources or changing their scope
◦ Fast tracking activities by doing them in parallel or
overlapping them
 Can end up lengthening the schedule since starting tasks
too soon may increase project risk and result in rework
◦ Crashing activities by obtaining the greatest amount of
schedule compression for the least incremental cost
 Crash time
 an amount of time an activity can be reduced
 Crash cost
 cost of reducing activity time
40



Find critical path (CP) by CPM
Determine Crash Cost per period (week, month) for each
activity Crash cost per period =
Find an activity on the CP with the minimum crash cost per
period and reduce its duration as much as possible
◦ Reduction should fall within its allowable range
◦ Reduction can be carried out only to the point where another path
becomes critical!
 If the reduction goes beyond the point where another path becomes
critical, the costs may be incurred unnecessarily
 If 2 paths become CP simultaneously, activities on both CPs must be
reduced by the same amount

Continue until you reach the desired project completion time
◦ When reducing it is necessary to keep up with all the network paths
41

In a house building project you are asked to deliver the house in
30 weeks, but you cannot deliver it before 36 weeks based on
your original schedule and budget. How much extra cost would
need to be incurred to complete the house by this time?
House building network:
2
8
Start
3
4
Activity ID
Activity duration (in weeks)
4
12
1
7
4
12
3
4
5
4
6
4
2
8
Start
B:
C:
D:
12
1
7
4
12
3
4
A:
4
1-2-4-7
12 + 8 + 12 + 4 = 36 months
1-2-5-6-7
12 + 8 + 4 + 4 + 4 = 32 months
1-3-4-7
12 + 4 + 12 + 4 = 32 months
1-3-5-6-7
12 + 4 + 4 + 4 + 4 = 28 months
5
4
6
4
 Critical path


Minimum project completion
time at this point is 36 weeks.
Can we complete it earlier?
Activity
Normal
Time
(Weeks)
Crash
Time
(Weeks)
1
2
3
4
5
6
7
12
8
4
12
4
4
4
7
5
3
9
1
1
3
Normal
Cost
Crash
Cost
$3,000
2,000
4,000
50,000
500
500
15,000
$5,000
3,500
7,000
71,000
1,100
1,100
22,000
$75,000
$110,700
Total
Allowable
Crash Time
(Weeks)
Crash Cost – Normal Cost
Normal Time – Crash Time
44
5
3
1
3
3
3
1
Crash
Cost per
Week
$400
500
3,000
7,000
200
200
7,000
CP
$500
$7000
4
2
8
$7000
12
7
4
1
Project Duration:
36 weeks
12
$400
3
4
$3000
5
4
6
4
$200
$200
Path durations:
1-2-4-7: 36
1-2-5-6-7: 32
1-3-4-7: 32
1-3-5-6-7: 28
•The cheapest activity to crash is Activity 1
•Max allowable time it can be reduced is 5 weeks
•Can we really reduce it by 5 weeks?
•Remember, you can reduce only until your CP reaches
another path(s), at which point you need to check whether
another path(s) did not become critical
•Your maximum crashing time is thus 4 weeks, because at 32
weeks you reach other 2 paths and you need to recalculate CP
45
$500
Crash 1 by 4 weeks:
$7000
4
2
8
$7000
12
7
4
1
Project Duration:
32 weeks
8
$400
3
4
$3000
5
4
6
4
$200
$200
Path durations:
1-2-4-7: 32
1-2-5-6-7: 28
1-3-4-7: 28
1-3-5-6-7: 24
•Project duration is now 32 weeks
•Additional cost: 4 weeks per $400 = $1,600
•CP is still the same 1-2-4-7
•We can still continue crashing Activity 1 (still it is cheapest)
•Max allowable time it can still be reduced is 1 week
46
$500
Crash 1 by 1 week:
$7000
4
2
8
$7000
12
7
4
1
Project Duration:
31 weeks
7
$400
3
4
$3000
5
4
6
4
$200
$200
•Project duration is now 31 weeks
•Additional cost: 1 week per $400 = $400
•Total crashing cost: $1,600 + $400 = $2,000
•CP is still the same 1-2-4-7
Path durations:
1-2-4-7: 31
1-2-5-6-7: 27
1-3-4-7: 27
1-3-5-6-7: 23
• Which activity to crash now?
•Activity 1 cannot be crashed anymore, we reached allowable max
•Activity 2 is cheapest one to crash now
•Max allowable time it can be reduced is 3 weeks
•But we need to crash it by 1 week only (from 31 to 30 weeks) in
47
order to reach the desirable project duration
$500
Crash 2 by 1 week:
$7000
4
2
7
$7000
12
7
4
1
Project Duration:
30 weeks
7
$400
3
4
$3000
5
4
6
4
$200
$200
•Project duration is now 30 weeks
•Additional cost: 1 week per $500 = $500
•Total crashing cost: $1,600 + $400 + $500 = $2,500
•CP is still the same 1-2-4-7
Path durations:
1-2-4-7: 30
1-2-5-6-7: 26
1-3-4-7: 27
1-3-5-6-7: 23
•The project duration has now been reduced to the desirable
30 weeks for a total crashing cost (additional cost) of $2,500
•Following this procedure the network can be crashed to 24 weeks
at a total additional cost of $31,500.
48


The network and durations given below show the original schedule for a
project. Based on the information in the table you can decrease the
durations of activities at an additional expense. The owner wants you to you
to finish the project in 120 days at the minimum possible cost.
Determine the project duration and its cost before and after crashing and
show the details about the activities which should be crashed. What is (are)
A
the critical path(s)?
120
1
B
20
2
C
3
40
D
30
4
E
50
5
F
60
49
Activity Normal duration
(days)
A
120
B
20
C
40
D
30
E
50
F
60
Crash Duration
(days)
100
15
30
20
35
40
Normal
Cost
Crash
Cost
12000
1800
16000
1400
3600
13500
14000
2800
17000
2000
5400
14500
Activity
Normal
duration (days)
A
B
C
D
E
F
•
Crash
duration
(days)
100
15
30
20
35
40
120
20
40
30
50
60
Normal
Cost
Crash
Cost
Crash cost
per day
12000
1800
16000
1400
3600
3500
14000
2800
17000
2000
5400
14500
200
100
60
120
∑ 38300
At this point:
•
The critical path is B-C-D-E
•
The cost of the project is $ 38 300
•
The project duration is 140 days.
Path durations:
A: 120
B-C-D-E: 140
B-F-E: 130
A
120
1
B
20
2
C
3
40
F
60
50
D
30
4
E
50
5
Activity
Normal
duration (days)
A
B
C
D
E
F
Crash
duration
(days)
100
15
30
20
35
40
120
20
40
30
50
60
Normal
Cost
Crash
Cost
Crash cost
per day
12000
1800
16000
1400
3600
3500
14000
2800
17000
2000
5400
14500
200
100
60
120
550
∑ 38300
•
The first activity to crash is activity D. It will be crashed by 10 days for an additional
cost of $600.
We have 2 critical paths now! B-C-D-E and B-F-E (130 days)
Both should be reduced!
A
Path durations:
A: 120
B-C-D-E: 130
B-F-E: 130
120
1
B
20
2
C
3
40
F
60
51
D
30
20
4
E
50
5
Activity
A
B
C
D
E
F
Normal
duration (days)
Crash
duration
(days)
100
15
30
20
35
40
120
20
40
30
50
60
Normal
Cost
Crash
Cost
Crash cost
per day
12000
1800
16000
1400
3600
3500
14000
2800
17000
2000
5400
14500
200
100
60
120
550
∑ 38300
•
What activities can we reduce in order to reduce the duration of BOTH critical paths?
•
•
•
•
Only B: $200 per day
C and F: $100 + $550 = $650 per day
D and F: $60 + $550 = $610 per day
Only E: $120 per day
1
B
20
2
A
120
C
3
40
F
60
52
Path durations:
A: 120
B-C-D-E: 130
B-F-E: 130
D
30
20
4
E
50
5
Activity
Normal
duration (days)
A
B
C
D
E
F
120
20
40
30
50
60
Crash
duration
(days)
100
15
30
20
35
40
Normal
Cost
Crash
Cost
Crash cost
per day
12000
1800
16000
1400
3600
3500
14000
2800
17000
2000
5400
14500
100
200
100
60
120
550
•
∑ 38300
The second activity to crash is activity E. It will be crashed by 10 days for an
additional cost of $ 1 200.
•
After crashing, the project cost is $ 40 100 and the project duration is 120 days.
•
The final critical path(s) are B-C-D-E, B-F-E and A.
Path durations:
A: 120
B-C-D-E: 120
B-F-E: 120
A
120
1
B
20
2
C
3
40
F
53
60
D
2030
4
E
50
40
5
Dealing with Uncertainty
• There may be a significant amount of uncertainty associated with the
actual task durations
• Ignore the uncertainty, and schedule the project using the expected or
most likely duration for each activity.
• Drawbacks:
• Typically results in overly optimistic schedules
• The use of single activity durations often produces a rigid, inflexible
mindset on the part of schedulers and the loss of confidence in the
realism of a schedule
• Include a contingency allowance in the estimate of activity durations.
• E.g., an activity with an expected duration of 2 days might be
•
scheduled for a period of 2.2 days, including a 10% contingency.
Systematic use of contingency factors can result in more accurate
schedules
• Use more elaborate techniques to deal with uncertainty
• PERT
• Monte Carlo simulation
PERT (Project Evaluation and Review Technique)
PERT is a network analysis technique used to estimate
project duration when there is a high degree of
uncertainty about the activity durations
• A commonly used formal method for dealing with
uncertainty in project scheduling.
• Applies the CPM to a weighted duration estimate
Procedure:
• Calculate the means of durations for each activity based
•
•
on three point estimates
Determine critical path using the means
Find the expected project duration and the variance
PERT: Three point estimates
The mean and variance for each activity duration are typically computed from the
three point estimates:
• optimistic (ai,j)
• most likely (mi,j)
• pessimistic (bi,j)
Mean:
Variance:
Project duration measures:
• The expected project duration E(T) is equal to the sum of the
expected durations of the activities along the critical path.
• The variance σ 2(T) in the duration of CP is calculated as the sum
of the variances along the critical path.
• Assuming that activity durations are independent random variables
PERT: Example 1
PERT: Example 1
k
Calculate CP without considering uncertainty:
ABDI: 21 weeks
ACDI: 19 weeks
ACEGI: 25 weeks
The critical path A – C – E – G – I
ACEHI: 23 weeks
ACFHI: 21 weeks
Completion time without considering uncertainty is 25 weeks
PERT: Example 1
If we can obtain three point estimates we can incorporate
uncertainty into the project duration calculations
PERT: Example 1
k
Calculate CP while considering uncertainty:
ABDI: 23.5 weeks
ACDI: 20.84 weeks
ACEGI: 26 weeks
The critical path A – C – E – G – I
ACEHI: 24.5 weeks
ACFHI: 23.17 weeks
Completion time with considering uncertainty,
i.e. expected project duration, is 26 weeks
PERT: Example 1
In order to determine the variance in duration of the critical path you can
calculate variances only for activities on the critical path.
The critical path: A – C – E – G – I
Variance:
Standard deviation:
PERTquestions
Probabilistic network analysis:
T
T ~ N ( ,  2 )
T


The network diagram below represents a project consisting of 9 activities,
the durations of which are uncertain. The activity most likely, optimistic
and pessimistic estimates are indicated in the table below
Determine the completion time of the project without considering
uncertainty, the expected project duration based on the three point
estimates, the corresponding variance and the critical path.
Activity
A
B
C
D
E
F
G
H
I
63
Optimistic Most Likely Pessimistic
Duration
Duration
Duration
(weeks)
(weeks)
(weeks)
3
4
5
2
3
5
6
8
10
5
7
8
6
9
14
10
12
14
2
2
4
4
5
8
4
6
8
Activity
A
B
C
D
E
F
G
H
I
Optimistic Most Likely Pessimistic
Duration
Duration
Duration
(weeks)
(weeks)
(weeks)
3
4
5
2
3
5
6
8
10
5
7
8
6
9
14
10
12
14
2
2
4
4
5
8
4
6
8
Calculate CP without considering uncertainty
For the calculations consider only most likely durations
BFI: 21 weeks
BEGI: 20 weeks
BEH: 17 weeks
ACFI: 30 weeks
ACEH: 26 weeks
ACEGI: 29 weeks
ADGI: 19 weeks
ADH: 16 weeks
64
Critical path is A-C-F-I
Project duration without considering
uncertainty is 30 weeks
Optimistic Most Likely Pessimistic
Duration Duration
Duration
Activity (weeks)
(weeks)
(weeks)
A
3
4
5
B
2
3
5
C
6
8
10
D
5
7
8
E
6
9
14
F
10
12
14
G
2
2
4
H
4
5
8
I
4
6
8

4
3.16
8
6.83
9.33
12
2.33
5.33
6
Calculate CP while considering uncertainty
First compute , i.e. the means for the activity durations
BFI: 21.16 weeks
BEGI: 20.8 weeks
BEH: 17.82 weeks
ACFI: 30 weeks
ACEH: 26.66 weeks
ACEGI: 29.66 weeks
ADGI: 19.16 weeks
ADH: 16.16 weeks
65
Critical path is A-C-F-I
Project duration while considering
uncertainty is 30 weeks
Optimistic Most Likely Pessimistic
Duration Duration
Duration
Activity (weeks)
(weeks)
(weeks)
A
3
4
5
B
2
3
5
C
6
8
10
D
5
7
8
E
6
9
14
F
10
12
14
G
2
2
4
H
4
5
8
I
4
6
8

2
4
3.16
8
6.83
9.33
12
2.33
5.33
6
0111
0.444
0.444
0.444
The critical path is A – C – F – I
In order to determine the variance in duration of the critical path
calculate variances for activities on the critical path.
Variance:  2 = 0.111+0.444+0.444+0.444 = 1.444
Expected project duration based on the three point estimates is 30
weeks and the corresponding variance is 1.444
66
Problems with using PERT method
 Three point estimations involve more work than CPM
 Subjective time estimates
 It is assumed that the activity durations are independent
random variables.
 In practice, the durations are often correlated with each other

68
Project Cost Management