The following table shows estimates of activity times (weeks) for a project:
Activity |
Optimistic Time |
Most Probable Time |
Pessimistic Time |
A | 3 | 4.5 | 9 |
B | 1.5 | 2 | 2.5 |
C | 5 | 6 | 7 |
D | 4 | 4.5 | 8 |
E | 5 | 6 | 7 |
F | 1 | 2.5 | 7 |
G | 7 | 9 | 11 |
Suppose that the critical path is A-B-D-F-G. Use Appendix A to answer the questions. Do not round intermediate calculations. Round your answers to three decimal places.
What is the probability that the project will be completed within 25 weeks?
Answer: the probability that the project will be completed within 25 weeks is 0.7208 or 72.08%
Explanation:
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 | 3 | 4.5 | 9 | 5.00 | 1.00 |
B | 1.5 | 2 | 2.5 | 2.00 | 0.03 |
C | 5 | 6 | 7 | 6.00 | 0.11 |
D | 4 | 4.5 | 8 | 5.00 | 0.44 |
E | 5 | 6 | 7 | 6.00 | 0.11 |
F | 1 | 2.5 | 7 | 3.00 | 1.00 |
G | 7 | 9 | 11 | 9.00 | 0.44 |
The duration of critical path ABDFG is 24 weeks,
step 1 | we will find the variance of the tasks which lie on critical path | 2.92 | |||
step 2 | mean project time (u) of critical path is= | 24.00 | |||
step 3 | Required completion time is | 25 | |||
step 4 | standard deviation= sqrt(variance)= | 1.709 | |||
step 5 | because Z= (given completion time- u)/standard deviation | 0.585 | |||
step 6 | P(z)= using NORM.S.DIST(z,true) | 0.7208 |
the probability that the project will be completed within 25 weeks is 0.7208
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