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PERT: Modeling Activity Duration Uncertainty
In the last handout, we developed a network model for a project as shown below.
2
E:3
A:20
1
4
C:5
5
F:6
B:30
D:10
3
Analysis showed that the critical path was B-D-F with a length of 46. We recognized this
as the critical path because it was the longest path through the network. In practice,
activity durations are not known in advance, but instead are estimated. What is the
impact on our management of the project?
Two major impacts: (1) We are actually uncertain about the length of the
critical path, and (2) we do not really know which path is critical (which path
is the longest)!
Uncertain activity times can be handled in two basic ways: (1) we can assume
that the critical path will be the path we have identified, and make probabilistic
arguments about the length of the known critical path, or (2) we can use simulation to
generate the probability distribution associated with the length of the project.
In either case, it is necessary to gather information about the probability
distribution of each activity. This can be accomplished in an easy way. We will assume
that each activity time can be described by a triangular distribution as shown below.
Probability distribution of an activity time, t
a
m
b
where
a = the shortest time needed to complete the activity (most optimistic),
b = the longest time needed to complete the activity (most pessimistic),
m = the most common time needed to complete the activity.
Then the mean and variance of t can be estimated by
( a  m  b)
3
2
(a  b 2  m 2  ma  mb  ab)
Var (t ) 
18
E (t ) 
The computations for our project are shown in the table below.
Activity
A
B
C
D
E
F
a
12
25
2
7
2
2
b
27
35
7
12
4
8
m
21
30
6
11
3
8
Expected
Value
20
30
5
10
3
6
Variance
9.50
4.17
1.17
1.17
0.17
2.00
If the critical (longest) path of our project is B-D-F, and if T is the duration of the project,
then since T  t B  t D  t F we have:
E[T ]  E[t B  t D  t F ]
 E[t B ]  E[t D ]  E[t F ]
 30  10  6
 46
Var (T )   T2  Var (t B  t D  t F )
 Var (t B )  Var (t D )  Var (t F )
(when?)
 4.17  1.17  2.00
 7.33
and
 T  Var (T )

7.33
 2.71
This shows explicitly that the project duration is not fixed at 46 days. Instead, T has
natural variation induced by the variability in the activity times. This does not tell us
what the distribution of T looks like. We only know two pieces of information about that
distribution, namely its mean and standard deviation (or variance).
It is interesting feature of nature that the form of the distribution of T is hinted at
by the fact that T is a sum. Roughly speaking, when a quantity is derived from a sum and
there is not a dominant summand in the sum, then the Central Limit Theorem can apply
and we can deduce that T will have an approximate normal distribution! This is a small
sum involving only three terms; however, the distribution of each activity time has a
well-behaved distribution. In particular, the triangular distributions are unimodal. What
does this mean?
In the worksheet PERT Simulation 1 of the file Dist of T assuming critical
path.xlsx, we have the results of 1000 simulations of the duration of this project assuming
that the critical path is B-D-F. We will consider how to replicate this simulation in a
moment, but for our purposes the real issue is the shape of the distribution of T.
Using the Distribution of T
Knowing that T has an approximate normal distribution provides us with a lot of
information. In particular, a normal distribution is entirely distinguished by its mean and
standard deviation. But we know that
 T  E[T ]  46
 T  2.71
Thus we have specified the entire distribution of T. In the EXCEL worksheet Computing
the dist of T, we have used the EXCEL function
norm.dist(x,mean,stdev,cumulative)
to compute the probability of completing the project by a specified day, starting from day
40 through day 54. This gives the decision maker much more information when
committing to complete the project by a given day.
Alternatively, we have used the EXCEL function
norm.inv(prob, mean, stdev)
to compute the day of completion given the probability.
Next we remove our assumption about which path is critical by exploring how to
simulate the project.