--- title: "CPM network, 13 activities (A=2 to M=7)" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-network-thirteen-act-b81" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-thirteen-act-b81 site: BCT Engineering --- # CPM network, 13 activities (A=2 to M=7) Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2081 Bhadra, 12 marks - Video: [Project Management: Finding the Critical Path, duration and Project Duration | Critical Path Method](https://www.youtube.com/watch?v=tOJ8IuIqBpw) — Civil Engineering Exam - 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-thirteen-act-b81 - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Draw the CPM network diagram and compute EST, EFT, LST, LFT, TF, FF, Int.F and Ind.F from the information given below. Also compute the project duration and mark the critical path. | Activity | Duration (days) | Predecessor | |---|---|---| | A | 2 | - | | B | 4 | - | | C | 3 | A | | D | 5 | A, B | | E | 5 | B | | F | 1 | C | | G | 4 | D | | H | 5 | D | | I | 7 | E | | J | 2 | D, F | | K | 3 | H, I | | L | 6 | K | | M | 7 | G | ## Short answer Forward/backward pass yields project duration 25 days. Critical path is B-E-I-K-L; activities B, E, I, K, L have zero total float. ## Answer 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 (days). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only. ```text (1) --A(2)--> (2) (1) --B(4)--> (3) (2) --C(3)--> (4) (2) - - dummy - -> (6) (3) --E(5)--> (5) (3) - - dummy - -> (6) (4) --F(1)--> (10) (5) --I(7)--> (9) (6) --D(5)--> (7) (7) --G(4)--> (8) (7) --H(5)--> (9) (7) - - dummy - -> (10) (8) --M(7)--> (12) (9) --K(3)--> (11) (10) --J(2)--> (12) (11) --L(6)--> (12) ``` The network has 12 events and 3 dummy activities. Start event is (1) and the end event is (12). #### 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 + 2 = 2 - B: EST = 0; EFT = 0 + 4 = 4 - C: EST = EFT of A = 2; EFT = 2 + 3 = 5 - D: EST = max(A:2, B:4) = 4; EFT = 4 + 5 = 9 - E: EST = EFT of B = 4; EFT = 4 + 5 = 9 - F: EST = EFT of C = 5; EFT = 5 + 1 = 6 - G: EST = EFT of D = 9; EFT = 9 + 4 = 13 - H: EST = EFT of D = 9; EFT = 9 + 5 = 14 - I: EST = EFT of E = 9; EFT = 9 + 7 = 16 - J: EST = max(D:9, F:6) = 9; EFT = 9 + 2 = 11 - M: EST = EFT of G = 13; EFT = 13 + 7 = 20 - K: EST = max(H:14, I:16) = 16; EFT = 16 + 3 = 19 - L: EST = EFT of K = 19; EFT = 19 + 6 = 25 Project duration = largest EFT = **25 days**. #### 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 = 25; LST = 25 - 6 = 19 - K: LFT = LST of L = 19; LST = 19 - 3 = 16 - M: LFT = project duration = 25; LST = 25 - 7 = 18 - J: LFT = project duration = 25; LST = 25 - 2 = 23 - I: LFT = LST of K = 16; LST = 16 - 7 = 9 - H: LFT = LST of K = 16; LST = 16 - 5 = 11 - G: LFT = LST of M = 18; LST = 18 - 4 = 14 - F: LFT = LST of J = 23; LST = 23 - 1 = 22 - E: LFT = LST of I = 9; LST = 9 - 5 = 4 - D: LFT = min(G:14, H:11, J:23) = 11; LST = 11 - 5 = 6 - C: LFT = LST of F = 22; LST = 22 - 3 = 19 - B: LFT = min(D:6, E:4) = 4; LST = 4 - 4 = 0 - A: LFT = min(C:19, D:6) = 6; LST = 6 - 2 = 4 #### 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 | 2 | 0 | 2 | 4 | 6 | 4 | 0 | 4 | 0 | | B | 4 | 0 | 4 | 0 | 4 | 0 | 0 | 0 | 0 | | C | 3 | 2 | 5 | 19 | 22 | 17 | 0 | 17 | 0 | | D | 5 | 4 | 9 | 6 | 11 | 2 | 0 | 2 | 0 | | E | 5 | 4 | 9 | 4 | 9 | 0 | 0 | 0 | 0 | | F | 1 | 5 | 6 | 22 | 23 | 17 | 3 | 14 | 0 | | G | 4 | 9 | 13 | 14 | 18 | 5 | 0 | 5 | 0 | | H | 5 | 9 | 14 | 11 | 16 | 2 | 2 | 0 | 0 | | I | 7 | 9 | 16 | 9 | 16 | 0 | 0 | 0 | 0 | | J | 2 | 9 | 11 | 23 | 25 | 14 | 14 | 0 | 0 | | K | 3 | 16 | 19 | 16 | 19 | 0 | 0 | 0 | 0 | | L | 6 | 19 | 25 | 19 | 25 | 0 | 0 | 0 | 0 | | M | 7 | 13 | 20 | 18 | 25 | 5 | 5 | 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 | 2 | 6 | 4 | | 3 | 4 | 4 | 0 | | 4 | 5 | 22 | 17 | | 5 | 9 | 9 | 0 | | 6 | 4 | 6 | 2 | | 7 | 9 | 11 | 2 | | 8 | 13 | 18 | 5 | | 9 | 16 | 16 | 0 | | 10 | 9 | 23 | 14 | | 11 | 19 | 19 | 0 | | 12 | 25 | 25 | 0 | #### Critical path and project duration - Critical activities (TF = 0): B, E, I, K, L - Critical path: B - E - I - K - L (= 4+5+7+3+6 = 25 days) **Answer: project duration = 25 days; critical path = B - E - I - K - L** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Activity D has predecessors A (EFT 2) and B (EFT 4). What is the EST of D? A. 4 B. 2 C. 6 D. 9 Answer: A. EST = max(2, 4) = 4, giving EFT = 9. (Key point: Forward pass rule) 2. Which is the critical path? A. A - D - H - K - L B. B - E - I - K - L C. A - C - F - J D. B - D - G - M Answer: B. B 4 + E 5 + I 7 + K 3 + L 6 = 25 days, with zero total float. (Key point: Critical path) 3. Activity J has EST 9 and LST 23. What is its total float? A. 9 B. 23 C. 14 D. 2 Answer: C. TF = LST - EST = 23 - 9 = 14 days. (Key point: Total float) 4. Activity D has successors G (LST 14), H (LST 11) and J (LST 23). What is the LFT of D? A. 14 B. 23 C. 6 D. 11 Answer: D. LFT is the smallest LST of the successors: min(14, 11, 23) = 11, so LST = 11 - 5 = 6. (Key point: Backward pass rule) ## Close variants set in other exams - [CPM network, 13 activities (A to M)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-thirteen-act-k82) — Project Engineering, chapter 3, asked 5× - [CPM network, 9 activities (A=3 to I=1)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-b82) — Project Engineering, chapter 3, asked 5× - [CPM network, 9 activities (A=5 to I=2)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-a75) — Project Engineering, chapter 3, asked 5× ## Related in Project Engineering - [CPM network, road project (days)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-road-project-b79) — Project Engineering, chapter 3 - [CPM network, 12 activities (A=10 to L=8)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-twelve-act-as67) — Project Engineering, chapter 3 - [CPM network, 10 activities (A=4 to J=9)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-as70) — Project Engineering, chapter 3 - [CPM network, 10 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-a74) — Project Engineering, chapter 3 - [CPM network, 10 activities (A=10 to J=4)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-ten-act-k79) — Project Engineering, chapter 3 - [CPM network, building project (A to K)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-eleven-act-s73) — Project Engineering, chapter 3 ## Related in other subjects - [CPM schedule, critical path and float](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_engineering_and_management%3Acpm-network-critical-path-float) — Industrial Engineering and Management, chapter 3