1. **State the problem:** We have a project with multiple jobs and their time estimates: optimistic ($t_0$), most likely ($t_m$), and pessimistic ($t_p$). We need to draw the project network and find the probability of completing the project in 40 days. 2. **Calculate the expected time ($t_e$) and variance ($\sigma^2$) for each job:** The expected time is given by $$t_e = \frac{t_0 + 4t_m + t_p}{6}$$ The variance is $$\sigma^2 = \left(\frac{t_p - t_0}{6}\right)^2$$ 3. **List the jobs and their times:** | Job | $t_0$ | $t_m$ | $t_p$ | |-----|-------|-------|-------| | 1-2 | 1 | 7 | 13 | | 1-6 | 4 | 5 | 5 | | 2-3 | 2 | 5 | 14 | | 2-4 | 2 | 14 | 8 | | 3-5 | 2 | 5 | 3 | | 6-7 | 7 | 10 | 17 | | 5-8 | 6 | 5 | 26 | | 7-8 | 5 | 3 | 8 | 4. **Calculate $t_e$ and $\sigma^2$ for each job:** - Job 1-2: $t_e = \frac{1 + 4(7) + 13}{6} = \frac{1 + 28 + 13}{6} = \frac{42}{6} = 7$; $\sigma^2 = \left(\frac{13 - 1}{6}\right)^2 = \left(\frac{12}{6}\right)^2 = 2^2 = 4$ - Job 1-6: $t_e = \frac{4 + 4(5) + 5}{6} = \frac{4 + 20 + 5}{6} = \frac{29}{6} \approx 4.83$; $\sigma^2 = \left(\frac{5 - 4}{6}\right)^2 = \left(\frac{1}{6}\right)^2 = 0.0278$ - Job 2-3: $t_e = \frac{2 + 4(5) + 14}{6} = \frac{2 + 20 + 14}{6} = \frac{36}{6} = 6$; $\sigma^2 = \left(\frac{14 - 2}{6}\right)^2 = 2^2 = 4$ - Job 2-4: $t_e = \frac{2 + 4(14) + 8}{6} = \frac{2 + 56 + 8}{6} = \frac{66}{6} = 11$; $\sigma^2 = \left(\frac{8 - 2}{6}\right)^2 = 1^2 = 1$ - Job 3-5: $t_e = \frac{2 + 4(5) + 3}{6} = \frac{2 + 20 + 3}{6} = \frac{25}{6} \approx 4.17$; $\sigma^2 = \left(\frac{3 - 2}{6}\right)^2 = \left(\frac{1}{6}\right)^2 = 0.0278$ - Job 6-7: $t_e = \frac{7 + 4(10) + 17}{6} = \frac{7 + 40 + 17}{6} = \frac{64}{6} \approx 10.67$; $\sigma^2 = \left(\frac{17 - 7}{6}\right)^2 = \left(\frac{10}{6}\right)^2 = 2.78$ - Job 5-8: $t_e = \frac{6 + 4(5) + 26}{6} = \frac{6 + 20 + 26}{6} = \frac{52}{6} \approx 8.67$; $\sigma^2 = \left(\frac{26 - 6}{6}\right)^2 = \left(\frac{20}{6}\right)^2 = 11.11$ - Job 7-8: $t_e = \frac{5 + 4(3) + 8}{6} = \frac{5 + 12 + 8}{6} = \frac{25}{6} \approx 4.17$; $\sigma^2 = \left(\frac{8 - 5}{6}\right)^2 = \left(\frac{3}{6}\right)^2 = 0.25$ 5. **Draw the project network:** The network nodes and jobs are: - Start at 1 - Jobs: 1-2, 1-6 - From 2: jobs 2-3, 2-4 - From 3: job 3-5 - From 6: job 6-7 - From 5: job 5-8 - From 7: job 7-8 - End at 8 6. **Find all paths from start (1) to end (8) and their expected durations:** - Path 1: 1-2-3-5-8 $$7 + 6 + 4.17 + 8.67 = 25.84$$ - Path 2: 1-2-4 (ends at 4, no path to 8, discard) - Path 3: 1-6-7-8 $$4.83 + 10.67 + 4.17 = 19.67$$ 7. **Identify the critical path:** The longest expected duration path is 1-2-3-5-8 with $25.84$ days. 8. **Calculate the variance of the critical path:** Sum variances of jobs on critical path: $$4 + 4 + 0.0278 + 11.11 = 19.14$$ 9. **Calculate the standard deviation of the critical path:** $$\sigma = \sqrt{19.14} \approx 4.38$$ 10. **Calculate the Z-score for completing in 40 days:** $$Z = \frac{40 - 25.84}{4.38} = \frac{14.16}{4.38} \approx 3.23$$ 11. **Find the probability:** Using standard normal distribution tables, $P(Z \leq 3.23) \approx 0.9993$. **Final answer:** The probability of completing the project in 40 days is approximately **99.93%**.