MATHEMATICS 3
Operational Analysis
Štefan Berežný
Applied informatics
Košice - 2010
PERT
PERT = Project (or Program)
Evaluation and
Review
Technique:
PERT is a model for project management
designed to analyze and represent the tasks
involved in completing a given project.
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PERT
PERT – especially, the time needed to
complete each task, and identifying the
minimum time needed to complete the total
project.
PERT – is applied to very large-scale, onetime, complex, non-routine infrastructure and
Research and Development projects.
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PERT
The PERT method is similar to CPM method,
but it is applied to stochastic models. Most is
used in projects which are not adequately
tested and there are enough high-quality
estimates of each activity. For each activity it
is necessary to determine the next 3 estimates:
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PERT
Optimistic time (a):
the minimum possible time required to
accomplish a task, assuming everything
proceeds better than is normally expected.
This estimate reflects the shortest possible
duration of the action in the implementation
of activities expected ideal running.
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PERT
Pessimistic time (b):
the maximum possible time required to
accomplish a task, assuming everything goes
wrong. Pessimistic estimate of the duration of
the activity. This rating reflects the longest
possible time for completion of the action in
the implementation of this activity is
calculated with all logically possible
obstacles.
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PERT
Normal time (m):
the best estimate of the time required to
accomplish a task, assuming everything
proceeds as normal.
The estimate of normal activity (Let us denote
by m) that corresponds to the most likely
estimate of the duration of the activity.
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PERT
Expected time ( t ):
Suppose that the duration of action tij is a
random variable that can be described by a
known statistical distribution, which is best
known. distribution of β, which can be
approximated by a normal Gaussian
distribution.
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PERT
The numerical characteristics of this
distribution is:
2
ba
a

4
m

b
b

a
 
t
2 
6
6
36
s
s2
t
- standard deviation
- variance
- average duration of activity tij
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PERT
Example:
Using the PERT method, find the critical path in the project,
which is shown in the chart below. (The figures are in weeks
in order: estimate: optimistic, normal and pessimistic). Also
see:
(1) the likelihood of the project, if extend the
duration of two weeks over a time corresponding to
the
critical
path
method
using
CPM?
(2) what should be the duration of the project that
this time the project executed with a probability of
90%?
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PERT
4, 6, 8
V2
8, 10, 18
V3
V4
2, 2, 2
0, 0, 0
2, 3, 10
8, 9, 16
4, 6, 14
12, 15, 30
V1
5, 6, 7
3, 4, 5
5, 7, 15
4, 5, 6
2, 3, 4
7, 10, 25
V8
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4, 7, 14
V9
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PERT
Solution:
First, we verify whether the graph is acyclic
by ATN (ATN = Algorithm for Topological
Numbering of the vertecies).
If it is, so we can find some of its topological
number.
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PERT
Solution:
In the table is calculated the average duration of
each activity and their variances. Then we get a
new network, which will have a topological
numbering of the verticies and the average
duration of each activity and to the network we
apply the algorithm to find the critical path. The
algorithm is shown in the graph with special
vertices, but the chart table for the network.
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PERT
6
6
2
11
7
2
0
4
10
7
17
1
9
8
4
6
4
8
5
3
12
3
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PERT
Solution:
T = 38 weeks. The project will be completed
with a probability of 50% for 38 weeks. The
critical path is 1 - 4 - 6 - 8 - 9
Hence we get the total variance is equal to the
sum of the variances of sub-activities which form
the critical path, thus we get that:
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PERT
Solution:
   
2
2
1,4
2
4,6

2
6,8

2
8,9
324 100 144 4 572





 15,89
36 36 36 36 36

  15,89  4
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PERT
Solution:
If we to prolong the project for 2 weeks, so
we get that x = 40, m = T = 38 and  = 4.
Hence we get that value
40  38
u
 0,5
4
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PERT
Solution:
The value of the standardized normal
distribution function is F (0,5) = 0.69146.
Based on the calculated values, we found that
the likelihood of the project for 40 weeks is
approximately 70%.
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PERT
Solution:
If we put the (u) = 0,9, both tables we see
that this equality is satisfied for the value
u = 1.29 and receive equal:
x  38
1, 29 
4
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PERT
Solution:
From there to express x = 43.16. The project
will be implemented in approximately 90%
probability for 43 weeks.
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Thank you for your attention
Štefan Berežný
Department Of Mathematics
FEI TU Košice
B. Němcovej 32
040 02 Košice
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