1. **Problem Statement:** We are given a list of activities with their precedence relationships and expected completion times. We need to find the critical path and the project duration. 2. **Key Concepts:** - The **critical path** is the longest path through the network diagram, determining the shortest possible project duration. - The **project duration** is the total time taken along the critical path. - We use the **Critical Path Method (CPM)** which involves calculating earliest start (ES), earliest finish (EF), latest start (LS), and latest finish (LF) times for each activity. 3. **Step 1: List activities and times:** - A: 7 days - B: 9 days (after A) - C: 5 days (after A) - D: 9 days (after A) - E: 8 days (after B) - F: 10 days (after B) - G: 12 days (after D) - H: 5 days (after C, E, G) - I: 6 days (after D) - J: 4 days (after I) - K: 5 days (after F, H, J) - L: 8 days (after I) 4. **Step 2: Calculate earliest start (ES) and earliest finish (EF):** - ES(A) = 0, EF(A) = 0 + 7 = 7 - ES(B) = EF(A) = 7, EF(B) = 7 + 9 = 16 - ES(C) = EF(A) = 7, EF(C) = 7 + 5 = 12 - ES(D) = EF(A) = 7, EF(D) = 7 + 9 = 16 - ES(E) = EF(B) = 16, EF(E) = 16 + 8 = 24 - ES(F) = EF(B) = 16, EF(F) = 16 + 10 = 26 - ES(G) = EF(D) = 16, EF(G) = 16 + 12 = 28 - ES(H) = max(EF(C), EF(E), EF(G)) = max(12, 24, 28) = 28, EF(H) = 28 + 5 = 33 - ES(I) = EF(D) = 16, EF(I) = 16 + 6 = 22 - ES(J) = EF(I) = 22, EF(J) = 22 + 4 = 26 - ES(K) = max(EF(F), EF(H), EF(J)) = max(26, 33, 26) = 33, EF(K) = 33 + 5 = 38 - ES(L) = EF(I) = 22, EF(L) = 22 + 8 = 30 5. **Step 3: Calculate latest finish (LF) and latest start (LS):** - Project duration = max(EF(K), EF(L)) = max(38, 30) = 38 - LF(K) = 38, LS(K) = 38 - 5 = 33 - LF(L) = 38 (project end), LS(L) = 38 - 8 = 30 (adjusted to project end) - LF(J) = LS(K) = 33, LS(J) = 33 - 4 = 29 - LF(I) = min(LS(J), LS(L)) = min(29, 30) = 29, LS(I) = 29 - 6 = 23 - LF(H) = LS(K) = 33, LS(H) = 33 - 5 = 28 - LF(F) = LS(K) = 33, LS(F) = 33 - 10 = 23 - LF(E) = LS(H) = 28, LS(E) = 28 - 8 = 20 - LF(G) = LS(H) = 28, LS(G) = 28 - 12 = 16 - LF(D) = min(LS(G), LS(I)) = min(16, 23) = 16, LS(D) = 16 - 9 = 7 - LF(C) = LS(H) = 28, LS(C) = 28 - 5 = 23 - LF(B) = min(LS(E), LS(F)) = min(20, 23) = 20, LS(B) = 20 - 9 = 11 - LF(A) = min(LS(B), LS(C), LS(D)) = min(11, 23, 7) = 7, LS(A) = 7 - 7 = 0 6. **Step 4: Identify critical path:** Activities with zero slack (ES = LS and EF = LF) are critical. - A: ES=0, LS=0 (critical) - D: ES=7, LS=7 (critical) - G: ES=16, LS=16 (critical) - H: ES=28, LS=28 (critical) - K: ES=33, LS=33 (critical) Critical path: A → D → G → H → K 7. **Step 5: Project duration:** Sum of durations on critical path = 7 + 9 + 12 + 5 + 5 = 38 days **Final answer:** The critical path is **A-D-G-H-K** with a project duration of **38 days**.