1. **State the problem:** We have a project with activities A to J, each with durations and dependencies. We need to find the overall project duration, earliest and latest start/finish times for certain activities, slack time for activity C, and the effect of increasing activity G's duration by 5 days. 2. **Key concepts:** - The **critical path** is the longest path through the project network, determining the minimum project duration. - **Earliest start (ES)** and **earliest finish (EF)** times are calculated forward from the start. - **Latest start (LS)** and **latest finish (LF)** times are calculated backward from the project end. - **Slack time** = LS - ES (or LF - EF), the amount of delay allowed without delaying the project. 3. **Calculate earliest start and finish times:** - Activities with no predecessors start at day 0. - EF = ES + duration. | Activity | Duration | Predecessors | ES | EF | |---|---|---|---|---| | A | 6 | - | 0 | 6 | | B | 11 | - | 0 | 11 | | C | 6 | - | 0 | 6 | | D | 4 | A | 6 | 10 | | E | 2 | C | 6 | 8 | | F | 10 | B,D,E | max(EF of B,D,E) = max(11,10,8) = 11 | 21 | | G | 15 | D,E | max(10,8) = 10 | 25 | | H | 8 | F | 21 | 29 | | I | 5 | G,H | max(25,29) = 29 | 34 | | J | 17 | I | 34 | 51 | 4. **Overall project duration:** The EF of the last activity J is 51 days. 5. **Earliest start and finish of activity I:** - ES(I) = 29 days - EF(I) = 34 days 6. **Calculate latest finish and start times (backward pass):** - LF(J) = EF(J) = 51 - LS(J) = LF(J) - duration = 51 - 17 = 34 - LF(I) = LS(J) = 34 - LS(I) = LF(I) - 5 = 29 - LF(G) = LS(I) = 29 - LF(H) = LS(I) = 29 - LS(G) = 29 - 15 = 14 - LS(H) = 29 - 8 = 21 - LF(F) = min(LS(H)) = 21 - LS(F) = 21 - 10 = 11 - LF(B) = LS(F) = 11 - LF(D) = min(LS(F), LS(G)) = min(11,14) = 11 - LF(E) = min(LS(F), LS(G)) = min(11,14) = 11 - LS(B) = 11 - 11 = 0 - LS(D) = 11 - 4 = 7 - LS(E) = 11 - 2 = 9 - LF(A) = LS(D) = 7 - LS(A) = 7 - 6 = 1 - LF(C) = LS(E) = 9 - LS(C) = 9 - 6 = 3 7. **Latest start of D:** 7 days 8. **Latest finish of B:** LF(B) = 11 days 9. **Slack time for C:** Slack = LS - ES = 3 - 0 = 3 days 10. **If G increases by 5 days (from 15 to 20):** - New EF(G) = ES(G) + 20 = 10 + 20 = 30 - EF(I) = max(EF(G), EF(H)) = max(30,29) = 30 - EF(J) = EF(I) + 17 = 30 + 17 = 47 Recalculate backward times for G and I: - LF(J) = 47 - LS(J) = 30 - LF(I) = 30 - LS(I) = 25 - LF(G) = LS(I) = 25 - LS(G) = 25 - 20 = 5 Since LS(G) decreased, check if this affects D and E: - LF(D) = min(LS(F), LS(G)) = min(11,5) = 5 - LS(D) = 5 - 4 = 1 - LF(E) = min(LS(F), LS(G)) = 5 - LS(E) = 5 - 2 = 3 - LF(A) = LS(D) = 1 - LS(A) = 1 - 6 = -5 (negative means A must start earlier) - LF(C) = LS(E) = 3 - LS(C) = 3 - 6 = -3 - LF(B) = LS(F) = 11 - LS(B) = 0 Overall project duration is now EF(J) = 52 days (since EF(I) = 30, EF(J) = 47, but original EF(J) was 51, so re-checking the critical path shows the new duration is 52 days). **Final answers:** - Overall duration: 51 days - Earliest start of I: 29 days - Earliest finish of I: 34 days - Latest start of D: 7 days - Latest finish of B: 11 days - Slack time for C: 3 days - New overall duration if G increases by 5 days: 52 days