The precedence diagram of activities as follows :
1 |
2 |
3 |
|
4 |
5 |
7 |
|
6 |
|||
8 |
The possible paths and their corresponding expected durations as follows :
1-4-6-8 = 6 + 4 + 16 + 2 = 28
1-5-8 = 6 + 15 + 2 = 23
2-8 = 40 + 2 = 42
3-7-8 = 18 + 9 + 2 = 29
Since , 2-8 has the longest duration, it forms the critical path. The expected project duration ( which is same as expected duration of critical path ) will be 42 periods
Variance of the duration of critical path = Variance of activity 2 + Variance of activity 8 = 4 + 1 = 5
Therefore, Standard deviation of duration of activities on critical path = Square root ( Variance ) = Square root ( 5 ) = 2.236
Let z value corresponding to the probability of completing the project within 29 periods = Z1
Accordingly,
Expected project duration + Z1 x standard deviation of expected project duration = 29
Or, 40 + 2.236.Z1 = 29
Or, 2.236.Z1 = 29 – 40 = - 11
Or, Z1 = - 11/2.236 = - 4.919 ( - 4.9 rounded to 2 decimal places )
Corresponding value of probability for Z =- 4.9 as derived from standard normal distribution table is 0.0000005
PROBABILITY OF COMPLETING THE PROJECT IN 29 PERIODS = 0.0000009 |
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