habitat for humanity international is a nonprofit organization dedicated to eliminating poverty housing…
habitat for humanity international is a nonprofit organization dedicated to eliminating poverty housing worldwide. suppose the following table contains estimates of activity times (in days) involved in the construction of a house that habitat for humanity is building.\n| activity | optimistic | most probable | pessimistic |\n| ---- | ---- | ---- | ---- |\n| a | 3 | 4.0 | 5 |\n| b | 8 | 9.0 | 10 |\n| c | 7 | 7.5 | 10.5 |\n| d | 7 | 9.0 | 10 |\n| e | 6 | 7.0 | 9 |\n| f | 6 | 7.0 | 8 |\n(a) compute the expected activity completion times and the variance for each activity. (round your answers to two decimal places.)\n| activity | expected times | variance |\n| ---- | ---- | ---- |\n| a | | |\n| b | | |\n| c | | |\n| d | | |\n| e | | |\n| f | | |\n(b) an analyst determined that the critical path consists of activities b - d - f. compute the expected project completion time and the variance of this path. (round your answers to two decimal places.)\nexpected project completion time\nvariance of projection completion time
Answer
Explanation:
Step1: Recall expected - time formula
The formula for the expected time $t_e$ of an activity with optimistic time $t_o$, most - probable time $t_m$, and pessimistic time $t_p$ is $t_e=\frac{t_o + 4t_m+t_p}{6}$. The formula for the variance $\sigma^2$ of an activity is $\sigma^2=\left(\frac{t_p - t_o}{6}\right)^2$.
Step2: Calculate expected time and variance for Activity A
For Activity A: $t_o = 3$, $t_m = 4$, $t_p = 5$. $t_e=\frac{3 + 4\times4+5}{6}=\frac{3 + 16+5}{6}=\frac{24}{6}=4.00$ $\sigma^2=\left(\frac{5 - 3}{6}\right)^2=\left(\frac{2}{6}\right)^2=\frac{4}{36}\approx0.11$
Step3: Calculate expected time and variance for Activity B
For Activity B: $t_o = 8$, $t_m = 9$, $t_p = 10$. $t_e=\frac{8 + 4\times9+10}{6}=\frac{8 + 36+10}{6}=\frac{54}{6}=9.00$ $\sigma^2=\left(\frac{10 - 8}{6}\right)^2=\left(\frac{2}{6}\right)^2=\frac{4}{36}\approx0.11$
Step4: Calculate expected time and variance for Activity C
For Activity C: $t_o = 7$, $t_m = 7.5$, $t_p = 10.5$. $t_e=\frac{7+4\times7.5 + 10.5}{6}=\frac{7+30 + 10.5}{6}=\frac{47.5}{6}\approx7.92$ $\sigma^2=\left(\frac{10.5 - 7}{6}\right)^2=\left(\frac{3.5}{6}\right)^2=\frac{12.25}{36}\approx0.34$
Step5: Calculate expected time and variance for Activity D
For Activity D: $t_o = 7$, $t_m = 9$, $t_p = 10$. $t_e=\frac{7 + 4\times9+10}{6}=\frac{7 + 36+10}{6}=\frac{53}{6}\approx8.83$ $\sigma^2=\left(\frac{10 - 7}{6}\right)^2=\left(\frac{3}{6}\right)^2=\frac{9}{36}=0.25$
Step6: Calculate expected time and variance for Activity E
For Activity E: $t_o = 6$, $t_m = 7$, $t_p = 9$. $t_e=\frac{6 + 4\times7+9}{6}=\frac{6 + 28+9}{6}=\frac{43}{6}\approx7.17$ $\sigma^2=\left(\frac{9 - 6}{6}\right)^2=\left(\frac{3}{6}\right)^2=\frac{9}{36}=0.25$
Step7: Calculate expected time and variance for Activity F
For Activity F: $t_o = 6$, $t_m = 7$, $t_p = 8$. $t_e=\frac{6 + 4\times7+8}{6}=\frac{6 + 28+8}{6}=\frac{42}{6}=7.00$ $\sigma^2=\left(\frac{8 - 6}{6}\right)^2=\left(\frac{2}{6}\right)^2=\frac{4}{36}\approx0.11$
Step8: Calculate expected project - completion time for B - D - F
The expected project - completion time for the path B - D - F is the sum of the expected times of activities B, D, and F. $t_{total}=t_{eB}+t_{eD}+t_{eF}=9.00 + 8.83+7.00=24.83$
Step9: Calculate variance of the path B - D - F
The variance of the path B - D - F is the sum of the variances of activities B, D, and F. $\sigma_{total}^2=\sigma_{B}^2+\sigma_{D}^2+\sigma_{F}^2=0.11 + 0.25+0.11=0.47$
Answer:
| Activity | Expected Times | Variance |
|---|---|---|
| A | 4.00 | 0.11 |
| B | 9.00 | 0.11 |
| C | 7.92 | 0.34 |
| D | 8.83 | 0.25 |
| E | 7.17 | 0.25 |
| F | 7.00 | 0.11 |
Expected project completion time: 24.83 Variance of projection completion time: 0.47