--- title: "CPM precedence diagram, 9 activities" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-network-precedence-k72" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-precedence-k72 site: BCT Engineering --- # CPM precedence diagram, 9 activities Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2072 Kartik - Video: [Introduction to Critical Path Method | Construction of Network Diagrams AOA and AON](https://www.youtube.com/watch?v=sOzBh0f2hO4) — Math Hands-On with Python - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-precedence-k72 - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Draw the CPM network diagram (or precedence diagram) from the following activity relationships. Compute the total minimum project time of completion, critical activities and ES, EF, LS, LF, TF, FF, IntF and IndF. Also mark the critical path in the network diagram. | Activity | Duration | Predecessor | Successor | |---|---|---|---| | A | 3 | - | B, C, D | | B | 5 | A | E | | C | 5 | A | E, F | | D | 6 | A | G | | E | 2 | B, C | G | | F | 3 | C | G | | G | 4 | D, E, F | I | | H | 2 | D | I | | I | 6 | G, H | - | ## Short answer Total minimum project time is 21 time units. Critical activities: A, C, F, G, I (critical path A–C–F–G–I) with zero total float. Other floats (FF, IntF, IndF) for these are also zero; non‑critical activities have positive floats. ## Answer Total minimum project time, critical activities, the eight CPM values (ES, EF, LS, LF, TF, FF, IntF, IndF) and the critical path are found below. Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float. #### Network diagram (activity-on-arrow) Each activity is an arrow with its duration in brackets (time units). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only. Double arrows (==>) mark the critical path. ```text (1) ==A(3)==> (2) (2) ==C(5)==> (3) (2) --B(5)--> (4) (2) --D(6)--> (5) (3) - - dummy - -> (4) (3) ==F(3)==> (6) (4) --E(2)--> (6) (5) - - dummy - -> (6) (5) --H(2)--> (7) (6) ==G(4)==> (7) (7) ==I(6)==> (8) ``` The network has 8 events and 2 dummy activities. The dummy 3-4 makes E depend on both B and C; the dummy 5-6 makes G depend on D while H follows D alone. The start event is (1) and the end event is (8). Critical path on the network: (1) - A - (2) - C - (3) - F - (6) - G - (7) - I - (8). #### Forward pass (EST and EFT) Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D. - A: EST = 0; EFT = 0 + 3 = 3 - B: EST = EFT of A = 3; EFT = 3 + 5 = 8 - C: EST = EFT of A = 3; EFT = 3 + 5 = 8 - D: EST = EFT of A = 3; EFT = 3 + 6 = 9 - E: EST = max(B:8, C:8) = 8; EFT = 8 + 2 = 10 - F: EST = EFT of C = 8; EFT = 8 + 3 = 11 - H: EST = EFT of D = 9; EFT = 9 + 2 = 11 - G: EST = max(D:9, E:10, F:11) = 11; EFT = 11 + 4 = 15 - I: EST = max(G:15, H:11) = 15; EFT = 15 + 6 = 21 Project duration = largest EFT = **21 time units**. #### Backward pass (LFT and LST) Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D. - I: LFT = project duration = 21; LST = 21 - 6 = 15 - G: LFT = LST of I = 15; LST = 15 - 4 = 11 - H: LFT = LST of I = 15; LST = 15 - 2 = 13 - F: LFT = LST of G = 11; LST = 11 - 3 = 8 - E: LFT = LST of G = 11; LST = 11 - 2 = 9 - D: LFT = min(G:11, H:13) = 11; LST = 11 - 6 = 5 - C: LFT = min(E:9, F:8) = 8; LST = 8 - 5 = 3 - B: LFT = LST of E = 9; LST = 9 - 5 = 4 - A: LFT = min(B:4, C:3, D:5) = 3; LST = 3 - 3 = 0 #### Floats - **Total float** $TF = LST - EST = LFT - EFT$ - **Free float** $FF = \min(\text{EST of successors}) - EFT$ - **Interfering float** $IntF = TF - FF$ - **Independent float** $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$ #### Result table | Activity | D | EST | EFT | LST | LFT | TF | FF | Int.F | Ind.F | |---|---|---|---|---|---|---|---|---|---| | A | 3 | 0 | 3 | 0 | 3 | 0 | 0 | 0 | 0 | | B | 5 | 3 | 8 | 4 | 9 | 1 | 0 | 1 | 0 | | C | 5 | 3 | 8 | 3 | 8 | 0 | 0 | 0 | 0 | | D | 6 | 3 | 9 | 5 | 11 | 2 | 0 | 2 | 0 | | E | 2 | 8 | 10 | 9 | 11 | 1 | 1 | 0 | 0 | | F | 3 | 8 | 11 | 8 | 11 | 0 | 0 | 0 | 0 | | G | 4 | 11 | 15 | 11 | 15 | 0 | 0 | 0 | 0 | | H | 2 | 9 | 11 | 13 | 15 | 4 | 4 | 0 | 2 | | I | 6 | 15 | 21 | 15 | 21 | 0 | 0 | 0 | 0 | Event times (TE from the forward pass, TL from the backward pass): | Event | TE (earliest) | TL (latest) | Slack | |---|---|---|---| | 1 | 0 | 0 | 0 | | 2 | 3 | 3 | 0 | | 3 | 8 | 8 | 0 | | 4 | 8 | 9 | 1 | | 5 | 9 | 11 | 2 | | 6 | 11 | 11 | 0 | | 7 | 15 | 15 | 0 | | 8 | 21 | 21 | 0 | #### Critical path and project duration - Critical activities (TF = 0): A, C, F, G, I - Critical path: A - C - F - G - I (= 3+5+3+4+6 = 21 time units) **Answer: total minimum project time = 21 time units; critical path = A - C - F - G - I** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. E depends on B and C, but F depends only on C. How is this shown in the AOA network? A. F is drawn starting at the end of B B. A dummy from C's end event to B's end event, with F starting from C's end event C. E is drawn twice D. Activity C is split into two with no dummy Answer: B. The dummy carries C's dependency to E without making F depend on B. (Key point: Dummy activity) 2. G has predecessors D (EFT 9), E (EFT 10) and F (EFT 11). What is the EST of G? A. 9 B. 10 C. 11 D. 30 Answer: C. EST is the largest predecessor EFT = 11, so G's EFT = 15. (Key point: Forward pass for G) 3. What is the minimum project time and critical path? A. 21, A-C-F-G-I B. 21, A-D-G-I C. 20, A-B-E-G-I D. 15, A-C-F-G Answer: A. A+C+F+G+I = 3+5+3+4+6 = 21, the longest path, with all TF = 0. (Key point: Project time and critical path) 4. H has EST 0, EFT 2, LFT 15 and its successor I has EST 15. What are TF and FF of H? A. TF 13, FF 13 B. TF 13, FF 0 C. TF 15, FF 13 D. TF 2, FF 2 Answer: A. TF = LFT - EFT = 15 - 2 = 13, and FF = EST of I - EFT = 15 - 2 = 13. Interfering float is 0. 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