Answer:
Path | Duration | Variance |
1-4-6-8 | 36.00 | 16.00 |
1-5-8 | 30.00 | |
2-8 | 32.00 | |
3-7-8 | 25.00 |
1-4-6-8 is a longest and hence critical path, its variance is 16
Activity | Expected time | Variance | Critical |
1 | 15.00 | 3.00 | Yes |
2 | 30.00 | 4.00 | No |
3 | 13.00 | 2.00 | No |
4 | 4.00 | 2.00 | Yes |
5 | 13.00 | 5.00 | No |
6 | 15.00 | 10.00 | Yes |
7 | 10.00 | 2.00 | No |
8 | 2.00 | 1.00 | Yes |
The probability of completing in 32 periods
step 1 | we will find the variance of the tasks which lie on critical path | 16.00 | |||
step 2 | mean project time (u) of critical path is= | 36.00 | |||
step 3 | Required completion time is | 32 | |||
step 4 | standard deviation= sqrt(variance)= | 4.000 | |||
step 5 | because Z= (given completion time- u)/standard deviation | -1.000 | |||
step 6 | P(z)= using NORM.S.DIST(z,true) | 0.159 |
Answer: The probability of completing in 32 periods is 0.159
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