Download Project Duration

4. PERT (PROGRAM EVALUATION AND
REVIEW TECHNIQUE)
Objective:
To understand how to apply the PERT
method to planning construction projects that
are subject to uncertainty:
Summary:
4.1 Introduction
4.2 PERT Procedure
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4.1 INTRODUCTION
PERT is usually used to assess risk in very
risky projects, for example, where:
– some activities have a high degree of
uncertainty about their duration (such as in
design-build);
– the cost of not meeting a deadline is very high
(such as very high penalty payments);
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• Basic assumptions of PERT:
– uncertainty is only important along the
deterministically derived critical path;
– there is no correlation between the durations of
different activities (basic Monte Carlo CPM
assumes this as well);
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activity
activity
‘X’
6
‘Y’
5
9
3
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• the durations of ‘X’ and ‘Y’ are assumed to be independent;
• yet, if they were performed by the same efficient crew, then
if ‘X’ completed quickly, maybe ‘Y’ would be as well;
– the duration of an activity is a random variable that
can be described by the Beta distribution;
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probability density
a  4m  b
the mean t 
6
shaded area
= 1.0
ba
the variance v  

 6 
2
activity duration
a = optimistic
duration
m = most likely
duration
b = pessimistic
duration
Fig. 4-1: Simplified Beta Distribution
• the optimistic duration is your estimate of the shortest the activity could take
to complete (say 1 in 100 observations);
• the pessimistic duration is your estimate of the longest the activity could
take to complete (say 1 in 100);
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• the most likely duration is your estimate of how long it is most likely to take.
4.1 PERT PROCEDURE
The basis of the PERT approach is:
– expected project duration = deterministic project
duration;
– variance on project duration (the square of the standard
deviation) is equal to the sum of each activity’s
variance along the critical path;
– by the Central Limit Theorem, it is assumed that the
distribution for the project duration is Normal;
– with the above parameters, it is now possible to answer
questions such as: what is the probability of completing
the project on time?
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Step 1: Determine the layout of the activity network and
assign durations (a, b, and m) to each activity:
KEY:
ES TF
FF activity
LS ‘X’
a
m
4
EF
IF
LF
b
activity
activity
activity
‘B’
6
‘D’
5
‘F’
5
14
2
14
3
7
activity
3
‘A’
4
end
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5
activity
activity
activity
‘C’
10
‘E’
11
‘G’
6
15
4
12
4
14
Fig. 4-2: Example PERT Network
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Step 2: Find the deterministic critical path:
1: compute the mean durations: t 
a  4m  b
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2: perform a deterministic forward pass to find the deterministic ES and EF
3: perform a deterministic backward pass to find the deterministic LS and LF
4: find the deterministic critical path(s)
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5
12
activity
0
5
8
4
activity
0
3
‘A’
4
t=5
5
11
‘B’
6
t=7
15
14
activity
5
5
‘C’
10
t = 10
25
activity
15
5
18
21
2
‘D’
5
t=6
15
activity
27
14
25
27
3
‘E’
15
4
11
t = 10
‘F’
5
t=5
25
activity
15
15
30
32
7
32
end
32
32
activity
25
12
25
4
‘G’
6
t=7
32
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Fig. 4-3: Deterministic Critical Path
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Step 3: Determine the expected (mean) duration
and variance for the project:
1: the PERT expected project duration is the deterministic critical path: T = 32 days
2: compute the variance for each duration (2 decimal places):
ba
v

 6 
2
3: the PERT variance for the project duration is the sum of the variances along
the critical path (if multiple paths, use the largest for safety): V = 9.12 days
12
5
12
activity
0
5
8
4
activity
0
3
‘A’
5
4
11
t = 5 v = 1.78
21
2
15
15
activity
5
5
25
activity
‘B’
15
6
14
t = 7 v = 2.78
5
18
‘C’
15
10
15
t = 10 v = 2.78
30
activity
‘D’
27
5
14
t = 6 v = 4.00
25
27
3
25
activity
‘E’ 25
15
4
11
12
t = 10 v = 1.78
Fig. 4-4: Project Duration: Expected and Variance
‘F’
32
5
7
t = 5 v = 0.44
32
end
32
32
activity
25
4
‘G’
32
6
14
t = 7 v = 2.78
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Step 4: Evaluate the Project:
Question 1: For the above project, what is the probability of completing
within 35 days?
1: Assume the Central Limit Theorem applies: the combined distributions along
the critical path are assumed to approximate a Normal distribution.
t  T  35  32
2: calculate the ‘z’ value: z  V  9.12  0.99
3: use the z-value in the standard normal variable table to find the probability:
probability of completing within t = 35 days: p = 83.9%
probability density
expected (mean) project
duration T = 32 days
shaded area
equals probability
of completing
within t = 35 days
required project completion
t = 35 days
increasing the variance increases
the width of the distribution
V = 9.12 days
Fig. 4-5: Normal Distribution
Project duration
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Question 2: For the above project, what is the probability of completing
within 25 days?
t  T  25  32
1: calculate the ‘z’ value: z  V  9.12  2.32
2: use the z-value in the standard normal variable table to find the probability:
probability of completing within 25 days = 1.02%
probability density
expected (mean) project
duration T = 32 days
shaded area
equals probability
of completing
within t = 25 days
required project completion
t = 25 days
V = 9.12 days
Fig. 4-6: Normal Distribution
Project duration
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Question 3: For the above project, what project duration has a 90%
probability of being satisficed? (reverse of previous questions)
1: calculate the ‘z’ value from the standard normal variable tables: z = 1.28
t  T 
t  32
2: solve for t in the z-value formula: z  V  1.28  9.12
therefore t = 35.87 days.
probability density
expected (mean) project
duration T = 32 days
shaded area equals
90% probability
of completing
within t = ? days
required project completion
t = ? days
V = 9.12 days
Fig. 4-7: Normal Distribution
Project duration
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Question 4: For the above project, what project duration has a 99%
probability of being satisficed?
1: calculate the ‘z’ value from the standard normal variable tables: z = 2.33
t  T 
t  32
2: solve for t in the z-value formula: z  V  2.33  9.12
therefore t = 39.0 days.
probability density
expected (mean) project
duration T = 32 days
shaded area equals
99% probability
of completing
within t = ? days
required project completion
t = ? days
V =9.12 days
Fig. 4-7: Normal Distribution
Project duration
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