Question 17.6: Computing the Probability That a Project Will and Will Not B...
Computing the Probability That a Project Will and Will Not Be Completed by a Specified Time
Using the information from Example 5, answer the following questions:
a. Can the paths be considered independent? Why?
b. What is the probability that the project can be completed within 17 weeks of its start?
c. What is the probability that the project will be completed within 15 weeks of its start?
d. What is the probability that the project will not be completed within 15 weeks of its start?
a. Yes, the paths can be considered independent, because no activity is on more than one path and you have no information suggesting that any activity times are interrelated.
b. To answer questions of this nature, you must take into account the degree to which the path distributions “overlap” the specified completion time. This overlap concept is illustrated in the accompanying figure, which shows the three path distributions, each centered on that path’s expected duration, and the specified completion time of 17 weeks. The shaded portion of each distribution corresponds to the probability that the part will be completed within the specified time. Observe that paths a–b–c and g–h–i are well enough to the left of the specified time, so that it is highly likely that both will be finished by week 17, but the critical path overlaps the specified completion time. In such cases, you need consider only the distribution of path d–e–f in assessing the probability of completion by week 17.
To find the probability for a path, you must first compute the value of z using Formula 17–8 for the path. For example, for path d–e–f, we have:
(17.6.b)
Turning to Appendix B, Table B, with z = +1.00, you will find that the area under the curve to the left of z is .8413. The computations are summarized in the following table. Note: If the value of z exceeds +3.00, treat the probability of completion as being equal to 1.000.
| Path | z=\frac{17 − Expected path duration}{Path standard deviation} | Probability of Completion in 17 Weeks |
| a–b–c | \frac{17-10}{0.97}=+7.22 | 1.00 |
| d–e–f | \frac{17-16}{1.00}=+1.00 | 0.8413 |
| g–h–i | \frac{17-13.5}{1.07}=+3.27 | 1.00 |
P ( Finish by week 17 ) = P(Path a–b–c finish) × P ( Path d–e–f finish ) × P ( Path g–h–i finish)
= 1.00 × .8413 × 1.00 = .8413
c. For a specified time of 15 weeks, the z values are
| Path | z=\frac{15 − Expected path duration}{Path standard deviation} | Probability of Completion in 15 Weeks |
| a–b–c | \frac{15-10.00}{0.97}=+5.15 | 1.00 |
| d–e–f | \frac{15-16.00}{1.00}=-1.00 | .1587 |
| g–h–i | \frac{15-13.50}{1.07}=+1.40 | .9192 |
Paths d–e–f and g–h–i have z values that are less than +3.00.
From Appendix B, Table B, the area to the left of z = –1.00 is .1587, and the area to the left of z = +1.40 is .9192. The path distributions are illustrated in the figure. The joint probability of all finishing before week 15 is the product of their probabilities:
1.00(.1587)(.9192) = .1459.
d. The probability of not finishing before week 15 is the complement of the probability obtained in part c: 1 − .1459 = .8541.
(17.6.d)
