Table 3.9
All activity times for the project are in weeks.
$$\begin{array} { | c | c | c | c | c | }
\hline \text { Activity } & \begin{array} { c }
\text { Optimistic } \
\text { Time }
\end{array} & \begin{array} { c }
\text { Most Likely } \
\text { Time }
\end{array} & \begin{array} { c }
\text { Pessimistic } \
\text { Time }
\end{array} & \begin{array} { c }
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \text { A } & 3 & 4 & 5 & \cdots \
\text { B } & 5 & 7 & 9 & - \
\text { C } & 2 & 3 & 10 & \text { A } \
\text { D } & 2 & 5 & 8 & \text { B } \
\text { E } & 1 & 7 & 7 & \text { B } \
\text { F } & 3 & 5 & 7 & \text { C,D } \
\text { G } & 7 & 8 & 9 & \text { D,E } \
\text { H } & 2 & 6 & 10 & F \
\hline
\end{array}$$
-Using Table 3.9,what is the expected time of completion of the project?
A) less than 22 weeks
B) 22 weeks
C) 23 weeks
D) greater than 23 weeks

Question

Table 3.9
All activity times for the project are in weeks.
 $$\begin{array} { | c | c | c | c | c | } 
\hline \text { Activity } & \begin{array} { c } 
\text { Optimistic } \
\text { Time }
\end{array} & \begin{array} { c } 
\text { Most Likely } \
\text { Time }
\end{array} & \begin{array} { c } 
\text { Pessimistic } \
\text { Time }
\end{array} & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \text { A } & 3 & 4 & 5 & \cdots \
\text { B } & 5 & 7 & 9 & - \
\text { C } & 2 & 3 & 10 & \text { A } \
\text { D } & 2 & 5 & 8 & \text { B } \
\text { E } & 1 & 7 & 7 & \text { B } \
\text { F } & 3 & 5 & 7 & \text { C,D } \
\text { G } & 7 & 8 & 9 & \text { D,E } \
\text { H } & 2 & 6 & 10 & F \
\hline
\end{array}$$
-Using Table 3.9,what is the expected time of completion of the project?
A) less than 22 weeks
B) 22 weeks
C) 23 weeks
D) greater than 23 weeks

Quizplus · Accepted Answer

To calculate the expected time of completion for each activity, we can use the following formula: 
$E(T) = \frac{O+4M+P}{6}$
where O is the optimistic time, M is the most likely time, and P is the pessimistic time.

Using this formula, we can compute the expected time for each activity and the total expected time for the project:

$E(A) = \frac{3+4(4)+5}{6} = 4$
$E(B) = \frac{5+4(7)+9}{6} = 7$
$E(C) = \frac{2+4(3)+10}{6} = 4.17$
$E(D) = \frac{2+4(5)+8}{6} = 5$
$E(E) = \frac{1+4(7)+7}{6} = 5$
$E(F) = \frac{3+4(5)+7}{6} = 4.83$
$E(G) = \frac{7+4(8)+9}{6} = 7.17$
$E(H) = \frac{2+4(6)+10}{6} = 5.33$

The expected time of completion for the project is the sum of the expected times of all activities along the critical path, which is: A-C-F-H. Therefore, the total expected time for the project is:
$E(Total) = E(A)+E(C)+E(F)+E(H) = 4+4.17+4.83+5.33 = 18.33$

Since we rounded each expected time to two decimal places, we can also round the total expected time to the nearest integer, which is 18.

However, we need to account for the fact that there may be some variability in the expected times due to the optimistic and pessimistic estimates. To calculate the standard deviation of the project completion time, we can use the following formula: 
$\sigma = \frac{P-O}{6}$

Using this formula for each activity and the critical path, we can calculate the standard deviation of the project completion time:
$\sigma(A) = \frac{5-3}{6} = 0.33$
$\sigma(B) = \frac{9-5}{6} = 0.67$
$\sigma(C) = \frac{10-2}{6} = 1.33$
$\sigma(D) = \frac{8-2}{6} = 1$
$\sigma(E) = \frac{7-1}{6} = 1$
$\sigma(F) = \frac{7-3}{6} = 0.67$
$\sigma(G) = \frac{9-7}{6} = 0.33$
$\sigma(H) = \frac{10-2}{6} = 1.33$

The variance of the critical path is the sum of the variances of each activity on the path. Therefore:
$\sigma_{Path}^2 = 0.33^2+1.33^2+0.67^2+1.33^2 \approx 3.06$
$\sigma_{Path} \approx 1.75$

Finally, we can use the following formula to calculate the probability of completing the project within a certain time frame: 
$P(T \leq X) = \Phi\left(\frac{X-E(T)}{\sigma}\right)$
where $\Phi(z)$ is the cumulative distribution function of the standard normal distribution, which can be looked up from a table or calculator.

For X = 22 weeks, we have: 
$P(T \leq 22) = \Phi\left(\frac{22-18.33}{1.75}\right) \approx 0.961$
For X = 23 weeks, we have: 
$P(T \leq 23) = \Phi\left(\frac{23-18.33}{1.75}\right) \approx 0.986$

Therefore, the best choice is C) 23 weeks, since there is a higher probability of completing the project within 23 weeks than within 22 weeks.

Table 39 All Activity Times for the Project Are in Weeks

Table 39
All Activity Times for the Project Are in Weeks