--- title: "CPM network, 12 activities (A=8 to L=3)" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-network-twelve-act-b66" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-twelve-act-b66 site: BCT Engineering --- # CPM network, 12 activities (A=8 to L=3) Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2066 Bhadra (old course), 16 marks - Video: [Use forward and backward pass to determine project duration and critical path](https://www.youtube.com/watch?v=4oDLMs11Exs) — Engineer4Free - 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-twelve-act-b66 - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question A project has the following schedule. Construct the network diagram and compute EST, EFT, LST, LFT, TF, FF, IF and interfering floats for each activity and find the critical path. | Activity | Time | Predecessor | |---|---|---| | A | 8 | None | | B | 2 | None | | C | 1 | A | | D | 9 | B | | E | 4 | B | | F | 5 | C, D | | G | 6 | E | | H | 3 | E | | I | 3 | G, H | | J | 5 | H | | K | 2 | I, J | | L | 3 | E, F | ## Short answer Draw the activity-on-arrow network with dummies, then do forward pass for EST/EFT and backward pass for LST/LFT. Floats: TF=LST-EST, FF=min successor EST minus EFT, IntF=TF-FF, IndF=max(0, min succ EST minus max pred LFT minus D). Project duration is 19 time units; critical path B-D-F-L. ## Answer IF is taken as independent float. 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. ```text (1) --A(8)--> (2) (1) --B(2)--> (3) (2) --C(1)--> (4) (3) --D(9)--> (4) (3) --E(4)--> (5) (4) --F(5)--> (7) (5) --H(3)--> (6) (5) - - dummy - -> (7) (5) --G(6)--> (8) (6) - - dummy - -> (8) (6) --J(5)--> (9) (7) --L(3)--> (10) (8) --I(3)--> (9) (9) --K(2)--> (10) ``` The network has 10 events and 2 dummy activities. Start event is (1) and the end event is (10). #### 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 + 8 = 8 - B: EST = 0; EFT = 0 + 2 = 2 - D: EST = EFT of B = 2; EFT = 2 + 9 = 11 - E: EST = EFT of B = 2; EFT = 2 + 4 = 6 - G: EST = EFT of E = 6; EFT = 6 + 6 = 12 - H: EST = EFT of E = 6; EFT = 6 + 3 = 9 - C: EST = EFT of A = 8; EFT = 8 + 1 = 9 - J: EST = EFT of H = 9; EFT = 9 + 5 = 14 - F: EST = max(C:9, D:11) = 11; EFT = 11 + 5 = 16 - I: EST = max(G:12, H:9) = 12; EFT = 12 + 3 = 15 - K: EST = max(I:15, J:14) = 15; EFT = 15 + 2 = 17 - L: EST = max(E:6, F:16) = 16; EFT = 16 + 3 = 19 Project duration = largest EFT = **19 time units**. #### Backward pass (LFT and LST) Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D. - L: LFT = project duration = 19; LST = 19 - 3 = 16 - K: LFT = project duration = 19; LST = 19 - 2 = 17 - I: LFT = LST of K = 17; LST = 17 - 3 = 14 - F: LFT = LST of L = 16; LST = 16 - 5 = 11 - J: LFT = LST of K = 17; LST = 17 - 5 = 12 - C: LFT = LST of F = 11; LST = 11 - 1 = 10 - H: LFT = min(I:14, J:12) = 12; LST = 12 - 3 = 9 - G: LFT = LST of I = 14; LST = 14 - 6 = 8 - E: LFT = min(G:8, H:9, L:16) = 8; LST = 8 - 4 = 4 - D: LFT = LST of F = 11; LST = 11 - 9 = 2 - B: LFT = min(D:2, E:4) = 2; LST = 2 - 2 = 0 - A: LFT = LST of C = 10; LST = 10 - 8 = 2 #### 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 | 8 | 0 | 8 | 2 | 10 | 2 | 0 | 2 | 0 | | B | 2 | 0 | 2 | 0 | 2 | 0 | 0 | 0 | 0 | | C | 1 | 8 | 9 | 10 | 11 | 2 | 2 | 0 | 0 | | D | 9 | 2 | 11 | 2 | 11 | 0 | 0 | 0 | 0 | | E | 4 | 2 | 6 | 4 | 8 | 2 | 0 | 2 | 0 | | F | 5 | 11 | 16 | 11 | 16 | 0 | 0 | 0 | 0 | | G | 6 | 6 | 12 | 8 | 14 | 2 | 0 | 2 | 0 | | H | 3 | 6 | 9 | 9 | 12 | 3 | 0 | 3 | 0 | | I | 3 | 12 | 15 | 14 | 17 | 2 | 0 | 2 | 0 | | J | 5 | 9 | 14 | 12 | 17 | 3 | 1 | 2 | 0 | | K | 2 | 15 | 17 | 17 | 19 | 2 | 2 | 0 | 0 | | L | 3 | 16 | 19 | 16 | 19 | 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 | 8 | 10 | 2 | | 3 | 2 | 2 | 0 | | 4 | 11 | 11 | 0 | | 5 | 6 | 8 | 2 | | 6 | 9 | 12 | 3 | | 7 | 16 | 16 | 0 | | 8 | 12 | 14 | 2 | | 9 | 15 | 17 | 2 | | 10 | 19 | 19 | 0 | #### Critical path and project duration - Critical activities (TF = 0): B, D, F, L - Critical path: B - D - F - L (= 2+9+5+3 = 19 time units) **Answer: project duration = 19 time units; critical path = B - D - F - L** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the critical path and project duration? A. B-D-F-L, 19 B. A-C-F-L, 17 C. B-E-H-J-K, 16 D. B-E-G-I-K, 17 Answer: A. L ends at 19 and B, D, F, L have TF = 0: 2+9+5+3 = 19. (Key point: Critical path) 2. EST of F is 11. Why? A. It is 8 from A B. It is the larger of EFT of C (9) and EFT of D (11) C. It equals EFT of C D. It equals EFT of E Answer: B. F follows C and D, so EST = max(9, 11) = 11. (Key point: EST of F) 3. LFT of E is 8. Where does this come from? A. Project duration minus 11 B. Smallest LST of successors G(8), H(9), L(16) C. Largest LST of successors D. EFT of B Answer: B. LFT is the smallest LST of the successors: min(8, 9, 16) = 8. (Key point: LFT of E) 4. J has EFT 14 and its successor K has EST 15. What is FF of J, and what is TF? A. FF = 3, TF = 1 B. FF = 0, TF = 3 C. FF = 1, TF = 3 D. FF = 1, TF = 1 Answer: C. FF = 15 - 14 = 1; TF = LST - EST = 12 - 9 = 3. 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