The director of continuing education at Bluebird University just approved the planning for a sales training seminar. Her administrative assistant identified the various activities that must be done and their relationships to each other as shown below.
Time Estimates |
|||||
Activity |
Description |
Immediate Predecessor(s) |
Optimistic |
Most likely |
Pessimistic |
A |
Prepare course outline |
None |
3 |
4 |
5 |
B |
Identify potential teachers |
None |
4 |
6 |
8 |
C |
Send potential students course outline |
A |
8 |
10 |
12 |
D |
Select teacher |
B |
2 |
3 |
4 |
E |
Register students |
C |
4 |
8 |
9 |
F |
Select classroom |
D, E |
1 |
2 |
6 |
G |
Notify students and teacher of class location |
F |
2 |
4 |
6 |
Te = (4*Tm + To+Tp)/6
Var = [ (Tp-To)/6}2
activity | To | Tm | Tp | Te | (Var)^0.5 | Var |
A | 3 | 4 | 5 | 4 | 0.333333 | 0.111111 |
B | 4 | 6 | 8 | 6 | 0.666667 | 0.444444 |
C | 8 | 10 | 12 | 10 | 0.666667 | 0.444444 |
D | 2 | 3 | 4 | 3 | 0.333333 | 0.111111 |
E | 4 | 8 | 9 | 7.5 | 0.833333 | 0.694444 |
F | 1 | 2 | 6 | 2.5 | 0.833333 | 0.694444 |
G | 2 | 4 | 6 | 4 | 0.666667 | 0.444444 |
a.
There are two paths in the diagram
ACEFG with length = 28 ( this is critical path)
BDFG with length = 15.5
Variance of the critical path = 2.388
SD of the critical path = (Var)1/2
= 1.545
The activities on critical path are A, C, E, F, G and the duration is 28.
Probability that the project will finish before 30 days is given by
z = value in case - mean / SD
= 30-28 / 1.545 =1.294
which gives probability of 0.902
The director of continuing education at Bluebird University just approved the planning for a sales training...
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