Consider the following network for a small maintenance project (all times are in days). [7 Marks]
|
Activity |
Immediate Predecessor [s] |
Optimistic Time |
Most Likely Time |
Pessimistic Time |
|
A |
- |
2 |
3 |
4 |
|
B |
- |
5 |
6 |
7 |
|
C |
- |
5 |
6 |
7 |
|
D |
A |
3 |
4 |
5 |
|
E |
A |
2 |
3 |
4 |
|
F |
C |
3 |
4 |
5 |
|
G |
C |
8 |
10 |
16 |
|
H |
B, E, F |
5 |
6 |
7 |
|
I |
B, E, F |
7 |
11 |
15 |
|
J |
B, E, F |
2 |
3 |
4 |
|
K |
G, J |
3 |
4 |
5 |
|
L |
D, H |
7 |
11 |
15 |
Using an example, briefly explain how PERT can be used in project management.
a.
The diagram is shown below

b.
The expected completion time is 27 days. The critical path is C-F-H-L
c.
F is on the critical path and thus the float is 0. For the activity G the float will be 27-20.667 = 6.333
d.
The variance of critical path is 2.11. The standard deviation is sqrt(2.11) = 1.45
The z value for 25 days is z = (25-27)/1.45 = -1.37
The corresponding probability P(z=-1.37) = 0.0838
e.
The z value for 30 days is z = (30-27)/1.45 = 2.06
The corresponding probability P(z = 2.06) = 0.9807 for completing below 30 days.
The probability of completing in more than 30 days is 1-0.9807 = 0.0193
Consider the following network for a small maintenance project (all times are in days). [7 Marks]...
If the contracted duration for a project with the following
network is 120 days, what is the probability (in percent) of
meeting the schedule based on PERT analysis? Then, assume that the
owner delays the start by 5 days while keeping the end date as
originally scheduled. What is the probability (in percent) that the
project will still finish at the originally scheduled completion
date? The requested probabilities should be determined based on the
entire network and not just one...
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