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Program Evaluation

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variance of path duration = s2 = S Vt. X = T Zspath ... expected duration of a project is 40 days and the variance of the critical path ... – PowerPoint PPT presentation

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Title: Program Evaluation


1
Program Evaluation Review Technique(PERT)
  • 3 duration time estimates
  • optimistic (to), most likely (tm), pessimistic
    (tp)
  • Activity duration
  • mean te (to 4tm tp) / 6
  • variance Vt (tp to) / 62
  • Path duration
  • mean of path duration T S te
  • variance of path duration s2 S Vt

2
  • X T Zspath
  • Z is number of standard deviations that X is from
    the mean.
  • Example If the mean duration of the critical
    path is 55 days and the variance of this path is
    16, what is the longest the project should take
    using a 95 confidence level?

T 55
for 95, Z 1.64 from Z-table
X T Zscp
55 1.64(4)
61.56 days
3
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4
probability of being late .05
Zscp
1.64(4)
actual project duration
T 55
X
61.56
.05 in tail ? Z 1.64
5
PERT Example
scp 3
  • If the expected duration of a project is 40 days
    and the variance of the critical path is 9 days,
    what is the probability that the project will
    complete in less than 45 days?
  • in more than 35 days?
  • in less than 35 days?
  • in between 35 and 45 days?

X T Zscp
45 40 Z(3)
Z 1.667
prob .952
95.2
Z 1.667
prob .952
95.2
35 40 Z(3)
1 - .952 .048
4.8
prob(Xgt45) 1 - .952 .048
prob(35ltXlt45)
1 prob(Xlt35) prob(Xgt45)
.904
or 90.4
1 - .048 - .048
6
1 - .048 - .048 .904
.048
.048
Zscp
Zscp
Z 1.667
Z 1.667
actual project duration
T 40
35
45
7
PERT Example
  • The expected duration of a project is 200 days,
    and the standard deviation of the critical path
    is 10 days. Predict a completion time that you
    are 90 sure you can meet.

90 ? Z 1.28
X T Zscp
200 1.28(10)
212.8 days
8
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