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Report
PERT
Program Evaluation and Review Technique
Tong Wang
511D ERB
PERT Example
Consider a small project that involves the following
activities.
Precedin
g
Activit Activity
y
Completion Times (days)
Optimistic Most
Likely
Pessimistic
a
-
5
6
7
b
-
4
5
18
c
a
4
15
20
d
b,c
3
4
5
e
a
5
16
18
PERT Example (cont’d)
• (a) Determine the expected value and the variance of the
completion time for each activity.
• (b) Use the expected times from (a) to find the critical
path.
• (c) Assuming that the normal distribution applies,
determine the probability that the critical path will take
between 18 and 26 days to complete.
• (d) How much time must be allowed to achieve a 90%
probability of timely completion?
• (e)By using modified probability of completion method,
what is the probability that all paths will take before 18
weeks?
Exercise Solution
(a)
0
6
0
6
6
20.5
9.5
24
END
E
14.5
START
B
7
0
7
13
20
D
4
C
14
A
6
6
20
6
20
20
24
20
24
Activit Mean Variance
y
a
6
4/36
b
7
196/36
c
14
256/36
d
4
4/36
e
14.5
169/36
Exercise Solution (cont’d)
(b) By using the expected time (mean) of each activity, we
find that the critical path is A-C-D.
Remark: For this simple project, we can find the longest
path (A-C-D) has the largest expected time, which is the
critical path.
The mean critical path duration is
μ= 6 + 14 + 4 = 24.
The variance of the critical path duration is the sum of the
variances along the path:
σ2cp = (4+256+4) / 36 = 264/36
so that the standard deviation is readily computed as
σcp = 2.708.
Exercise Solution (cont’d)
(c)The interval probability may be computed as the
difference between two cumulative probabilities as
follows:
P(18 ≤ t ≤ 26) = P(t ≤ 26) - P(t ≤ 18).
Two separate z computations are required. First at 26
we have
z26= (26-24) / 2.708=0.739
Then by looking up the normal table with z26=0.739,
we have one result that
P(t ≤ 26) = 0.770
Exercise Solution (cont’d)
Secondly, at 18 we have
z18= (18-24) / 2.708= -2.216
and
P(t ≤ 18) = 0.013
Combining these two results yields the desired probability
as
P(18 ≤ t ≤ 26) = 0.770 - 0.013= 0.757.
Exercise Solution (cont’d)
(d) For 90% probability, we must pick a z value
corresponding to 90% of the area under the normal
curve, 50% left of mean and 40% right of mean, so
z = 1.282.
Then solving for t we have
t = 24 +1.282*2.708 = 27.47 days.
Exercise Solution (cont’d)
(e)There are three paths in total. They are A-C-D, AE, and B-D.
mean
mean
standard deviation
standard
deviation
A-C-D
A-C-D
24
24
2.708
2.708
A-E
A-E
18.5
20.5
2.192
2.192
11
2.356
B-D
B-D
11
2.356
The probabilities for the three paths to be completed 18 weeks
are given below.
P(X1 ≤ 18) = P( z ≤ (18-24) / 2.708) = 0.013
P(X2 ≤ 18) = P( z ≤ (18-20.5) / 2.192) = 0.127
P(X3 ≤ 18) = P( z ≤ (18-11) / 2.356) = 0.999
Then the probability of completing all the paths in 18 weeks is
P(X ≤18) = P(X1 ≤ 18) P(X2 ≤ 18) P(X3 ≤ 18) = 0.0016.
Questions?

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