Transcript Problem 13-28

Problem 13-28
The estimated times (in weeks) and immediate predecessors for the
activities in a project are given in the following table. Assume that the
activity times are independent.
Immediate
Activity Predecessor
a
m
b
A
-
9
10
11
B
-
4
10
16
C
A
9
10
11
D
B
5
8
11
Kelly Devilbiss - Problem 13-28
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(a)
Calculate the expected time and variance for each
activity.
Activity
a
m
b
Expected time
t= [(a+4m+b)/6]
A
9
10
11
10
4/36
B
4
10
16
10
144/36
C
9
10
11
10
4/36
D
5
8
11
8
36/36
Kelly Devilbiss - Problem 13-28
Variance
[(b-a)/6]2
2
(b)
What is the expected completion time of the critical
path? What is the expected completion time of the
other patient in the network?
Critical Path AC takes 20 weeks
Critical Path BD takes 18 weeks
Kelly Devilbiss - Problem 13-28
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(c)
What is the variance of the critical path?
What is the variance of the other path in the
network?
Critical Path AC:
σ2CP = σ2A + σ2C
= 4/36 + 4/36 = 8/36
= 0.222
Critical Path BD:
σ2CP = σ2B + σ2D
= 144/36 + 36/36 = 180/36
=5
Kelly Devilbiss - Problem 13-28
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(d)
If the time to complete path A-C is normally
distributed, what is the probability that this path will
be finished in 22 weeks or less?
t = 20
variance (σ2) = 0.222
standard deviation (√0.222) = 0.47
P(x ≤ 22) = P[(22 – 20)/0.47] = 4.26
Using Appendix A: 4.26 = 1.00
Kelly Devilbiss - Problem 13-28
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(e)
If the time to complete path B-D is normally
distributed, what is the probability that this path will
be finished in 22 weeks or less?
t = 18
variance (σ2) = 5
standard deviation (√5) = 2.24
P(x ≤ 22) = P[(22 – 18)/2.24] = 1.79
Using Appendix A: 1.79 = 0.96327
Kelly Devilbiss - Problem 13-28
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(f)
Explain why the probability that the critical path will
be finished in 22 weeks or less is not necessarily the
probability that the project will be finished in 22
weeks or less.
Path B-D has more variability and has a higher
probability of exceeding 22 weeks.
Kelly Devilbiss - Problem 13-28
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