Estimated activity times and precedences are given below:
Activity | Optimistic | Most Likely | Pessimistic | Required Precedence |
A | 6 | 7 | 14 | |
B | 8 | 10 | 12 | |
C | 2 | 3 | 4 | |
D | 6 | 7 | 8 | A |
E | 5 | 5.5 | 9 | B, C |
F | 5 | 7 | 9 | B, C |
G | 4 | 6 | 8 | D, E |
H | 2.7 | 3 | 3.5 | F |
What is the probability that the project will be completed within:
A. 21 Days
B. 22 Days
C. 25 Days
Solution:
Activity | Optimistic time-a | Expected completion time-m | Pessimistic time-b | Expected time= (a+4*m+ b)/6 | Variance, (sigma)^2= (b-a/6)^2 |
A | 6 | 7 | 14 | 8.00 | 1.78 |
B | 8 | 10 | 12 | 10.00 | 0.44 |
C | 2 | 3 | 4 | 3.00 | 0.11 |
D | 6 | 7 | 8 | 7.00 | 0.11 |
E | 5 | 5.5 | 9 | 6.00 | 0.44 |
F | 5 | 7 | 9 | 7.00 | 0.44 |
G | 4 | 6 | 8 | 6.00 | 0.44 |
H | 2.7 | 3 | 3.5 | 3.03 | 0.02 |
Path | Duration |
ADG | 21.00 |
BEG | 23.00 |
BFH | 20.03 |
CEG | 15.00 |
CFH | 13.03 |
Expected project duration is 23 days through BEG
Answer: the probability that the project will be completed within:
A. 21 Days: 0.0416
step 1 | we will find the variance of the tasks which lie on critical path | 1.33 | |||
step 2 | mean project time (u) of critical path is= | 23.00 | |||
step 3 | Required completion time is | 21 | |||
step 4 | standard deviation= sqrt(variance)= | 1.155 | |||
step 5 | because Z= (given completion time- u)/standard deviation | -1.732 | |||
step 6 | P(z)= using NORM.S.DIST(z,true) | 0.0416 |
Answer: the probability that the project will be completed within:
B. 22 Days: 0.1932
step 1 | we will find the variance of the tasks which lie on critical path | 1.33 | |||
step 2 | mean project time (u) of critical path is= | 23.00 | |||
step 3 | Required completion time is | 22 | |||
step 4 | standard deviation= sqrt(variance)= | 1.155 | |||
step 5 | because Z= (given completion time- u)/standard deviation | -0.866 | |||
step 6 | P(z)= using NORM.S.DIST(z,true) | 0.1932 |
Answer: the probability that the project will be completed within:
C. 25 Days: 0.9584
step 1 | we will find the variance of the tasks which lie on critical path | 1.33 | |||
step 2 | mean project time (u) of critical path is= | 23.00 | |||
step 3 | Required completion time is | 25 | |||
step 4 | standard deviation= sqrt(variance)= | 1.155 | |||
step 5 | because Z= (given completion time- u)/standard deviation | 1.732 | |||
step 6 | P(z)= using NORM.S.DIST(z,true) | 0.9584 |
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