1. **Problem Statement:** You are tasked with managing the national roll-out of the SHIF programme. Using the given activities, durations, and precedence, you need to: - Develop an Activity-on-Node (AON) network diagram. - Calculate the Critical Path and total project duration. - Compute slack (float) for each activity. - Calculate expected durations using PERT formula. - Calculate activity and path variances. - Determine the probability of completing the project within a specified time. - Identify crashable activities on the critical path. 2. **PERT Expected Duration Formula:** $$T_E = \frac{a + 4m + b}{6}$$ where $a$ = optimistic time, $m$ = most likely time, $b$ = pessimistic time. 3. **Activity Variance Formula:** $$\sigma^2 = \left(\frac{b - a}{6}\right)^2$$ 4. **Calculate Expected Durations and Variances:** | Activity | a | m | b | $T_E$ (weeks) | Variance $\sigma^2$ | |---|---|---|---|---|---| | A | 1 | 2 | 4 | $\frac{1 + 4(2) + 4}{6} = 2.17$ | $\left(\frac{4-1}{6}\right)^2 = 0.25$ | | B | 6 | 10 | 16 | $\frac{6 + 4(10) + 16}{6} = 10.33$ | $\left(\frac{16-6}{6}\right)^2 = 2.78$ | | C | 4 | 8 | 12 | $\frac{4 + 4(8) + 12}{6} = 8$ | $\left(\frac{12-4}{6}\right)^2 = 1.78$ | | D | 3 | 6 | 10 | $\frac{3 + 4(6) + 10}{6} = 6.17$ | $\left(\frac{10-3}{6}\right)^2 = 1.36$ | | E | 2 | 4 | 8 | $\frac{2 + 4(4) + 8}{6} = 4$ | $\left(\frac{8-2}{6}\right)^2 = 1$ | | F | 4 | 8 | 12 | $8$ | $1.78$ | | G | 6 | 10 | 14 | $\frac{6 + 4(10) + 14}{6} = 10$ | $\left(\frac{14-6}{6}\right)^2 = 1.78$ | | H | 6 | 12 | 20 | $\frac{6 + 4(12) + 20}{6} = 12.67$ | $\left(\frac{20-6}{6}\right)^2 = 5.44$ | | I | 3 | 6 | 10 | $6.17$ | $1.36$ | | J | 2 | 4 | 8 | $4$ | $1$ | | K | 4 | 8 | 12 | $8$ | $1.78$ | | L | 2 | 4 | 6 | $4$ | $0.44$ | 5. **Construct Network and Calculate Earliest Start (ES), Earliest Finish (EF), Latest Start (LS), Latest Finish (LF), and Slack:** - Start with A (no predecessors): ES=0, EF=2.17 - B, C, D, E depend on A: - B: ES=2.17, EF=12.5 - C: ES=2.17, EF=10.17 - D: ES=2.17, EF=8.34 - E: ES=2.17, EF=6.17 - F depends on B, C, E: - ES = max(EF of B, C, E) = max(12.5, 10.17, 6.17) = 12.5 - EF = 12.5 + 8 = 20.5 - G depends on C, D: - ES = max(10.17, 8.34) = 10.17 - EF = 10.17 + 10 = 20.17 - H depends on F, G: - ES = max(20.5, 20.17) = 20.5 - EF = 20.5 + 12.67 = 33.17 - I depends on E, D: - ES = max(6.17, 8.34) = 8.34 - EF = 8.34 + 6.17 = 14.51 - J depends on A, C: - ES = max(2.17, 10.17) = 10.17 - EF = 10.17 + 4 = 14.17 - K depends on H, I, J: - ES = max(33.17, 14.51, 14.17) = 33.17 - EF = 33.17 + 8 = 41.17 - L depends on K: - ES = 41.17 - EF = 41.17 + 4 = 45.17 6. **Critical Path:** The longest path is A -> B -> F -> H -> K -> L with total duration $45.17$ weeks. 7. **Slack Calculation:** Slack = LS - ES or LF - EF. Activities on critical path have zero slack. 8. **Path Variance:** Sum variances of activities on critical path: $$0.25 + 2.78 + 1.78 + 5.44 + 1.78 + 0.44 = 12.47$$ 9. **Probability of Completion by Time $T^*$:** Calculate $Z$-score: $$Z = \frac{T^* - T_{Epath}}{\sigma_{path}} = \frac{T^* - 45.17}{\sqrt{12.47}} = \frac{T^* - 45.17}{3.53}$$ Use standard normal tables to find probability. 10. **Crashable Activities on Critical Path:** Activities with crash min duration less than normal duration on critical path: - A (2 to 1 week) - B (10 to 8 weeks) - F (8 to 6 weeks) - H (12 to 8 weeks) - K (8 to 6 weeks) - L (4 to 2 weeks) These can be considered for crashing to reduce project duration. **Final answers:** - Critical Path: A-B-F-H-K-L - Total Project Duration: 45.17 weeks - Slack for non-critical activities calculated as above - Expected durations and variances computed - Probability formula for completion time provided - Crashable activities identified