Question 17.5: Computing Expected Activity Times and Variances, and the Exp...
Computing Expected Activity Times and Variances, and the Expected Duration and Standard Deviation of Each Path
The network diagram for a project is shown in the accompanying figure, with three time estimates for each activity. Activity times are in weeks. Do the following:
a. Compute the expected time for each activity and the expected duration for each path.
b. Identify the critical path.
c. Compute the variance of each activity and the variance and standard deviation of each path.
a.
| TIMES | ||||||
| Path | Activity | t_o | t_m | t_p | t_0=\frac{t_o +4t_m +t_p}{6} | Path Total |
| a–b–c | a | 1 | 3 | 4 |
\left. \begin{matrix} 2.83 \\ 4.00 \\ 3.17 \end{matrix} \right\} |
10.00 |
| b | 2 | 4 | 6 | |||
| c | 3 | 3 | 5 | |||
| d–e–f | d | 3 | 4 | 5 |
\left. \begin{matrix} 4.00 \\ 5.00\\ 7.00 \end{matrix} \right\} |
16.00 |
| e | 3 | 5 | 7 | |||
| f | 5 | 7 | 9 | |||
| g–h–i | g | 2 | 3 | 6 |
\left. \begin{matrix} 3.33\\ 6.00 \\ 4.17 \end{matrix} \right\} |
13.50 |
| h | 4 | 6 | 8 | |||
| i | 3 | 4 | 6 | |||
b. The path that has the longest expected duration is the critical path. Because path d–e–f has the largest path total, it is the critical path.
c.
| TIMES | |||||||
| Path | Activity | t_o | t_m | t_p | \sigma ^2_{oct}=\frac{(t_p-t_o)^2}{36} | σ^2_{Path} | σ_{path} |
| a–b–c | a | 1 | 3 | 4 |
\left. \begin{matrix} (4 − 1)^2/36 = 9/36 \\ (6 − 2)^2/36 = 16/36 \\ (5 − 2)^2/36 = 9/36 \end{matrix}\right\} |
34/36 = 0.944 |
0.97 |
| b | 2 | 4 | 6 | ||||
| c | 2 | 3 | 5 | ||||
| d–e–f | d | 3 | 4 | 5 |
\left.\begin{matrix} (5 − 3)^2/36 = 4/36 \\(7 − 3)^2/36 = 16/36 \\(9 − 5)^2/36 = 16/36 \end{matrix}\right\} |
36/36 = 1.00 |
1.00 |
| e | 3 | 5 | 7 | ||||
| f | 5 | 7 | 9 | ||||
| g–h–i | g | 2 | 3 | 6 |
\left. \begin{matrix} (6 − 2)^2/36 = 16/36 \\(8 − 4)^2/36 = 16/36\\(6 − 3)^2/36 = 9/36\end{matrix} \right\} |
41/36 = 1.139 |
1.07 |
| h | 4 | 6 | 8 | ||||
| i | 3 | 4 | 6 | ||||