--- title: "CPM components, 9 activities with successors" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-components-nine-act-c70" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-components-nine-act-c70 site: BCT Engineering --- # CPM components, 9 activities with successors Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2070 Chaitra, 12 marks - Video: [Finding the Critical Path, duration and Project Duration, Critical Path Method, float, EST, EFT, LST](https://www.youtube.com/watch?v=OKKcjeJt4cI) — 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-components-nine-act-c70 - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Find all the components of CPM from the following information. | S.N | Activity | Duration | Predecessor | Successor | |---|---|---|---|---| | 1 | A | 3 | - | D | | 2 | B | 6 | - | E, G, I | | 3 | C | 2 | - | F | | 4 | D | 2 | A | G | | 5 | E | 1 | B | H | | 6 | F | 3 | C | I | | 7 | G | 7 | B, D | - | | 8 | H | 3 | E | - | | 9 | I | 4 | B, F | - | ## Short answer Project duration is 13 time units; critical path B–G. All CPM components (event times, activity times and floats) are computed; only B and G have zero total float and are critical. ## Answer The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method. 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(3)--> (2) (1) --B(6)--> (3) (1) --C(2)--> (4) (2) --D(2)--> (6) (3) --E(1)--> (5) (3) - - dummy - -> (6) (3) - - dummy - -> (7) (4) --F(3)--> (7) (5) --H(3)--> (8) (6) --G(7)--> (8) (7) --I(4)--> (8) ``` The network has 8 events and 2 dummy activities. Start event is (1) and the end event is (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 = 0; EFT = 0 + 6 = 6 - C: EST = 0; EFT = 0 + 2 = 2 - F: EST = EFT of C = 2; EFT = 2 + 3 = 5 - D: EST = EFT of A = 3; EFT = 3 + 2 = 5 - E: EST = EFT of B = 6; EFT = 6 + 1 = 7 - G: EST = max(B:6, D:5) = 6; EFT = 6 + 7 = 13 - I: EST = max(B:6, F:5) = 6; EFT = 6 + 4 = 10 - H: EST = EFT of E = 7; EFT = 7 + 3 = 10 Project duration = largest EFT = **13 time units**. #### Backward pass (LFT and LST) Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D. - H: LFT = project duration = 13; LST = 13 - 3 = 10 - I: LFT = project duration = 13; LST = 13 - 4 = 9 - G: LFT = project duration = 13; LST = 13 - 7 = 6 - E: LFT = LST of H = 10; LST = 10 - 1 = 9 - D: LFT = LST of G = 6; LST = 6 - 2 = 4 - F: LFT = LST of I = 9; LST = 9 - 3 = 6 - C: LFT = LST of F = 6; LST = 6 - 2 = 4 - B: LFT = min(E:9, G:6, I:9) = 6; LST = 6 - 6 = 0 - A: LFT = LST of D = 4; LST = 4 - 3 = 1 #### 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 | 1 | 4 | 1 | 0 | 1 | 0 | | B | 6 | 0 | 6 | 0 | 6 | 0 | 0 | 0 | 0 | | C | 2 | 0 | 2 | 4 | 6 | 4 | 0 | 4 | 0 | | D | 2 | 3 | 5 | 4 | 6 | 1 | 1 | 0 | 0 | | E | 1 | 6 | 7 | 9 | 10 | 3 | 0 | 3 | 0 | | F | 3 | 2 | 5 | 6 | 9 | 4 | 1 | 3 | 0 | | G | 7 | 6 | 13 | 6 | 13 | 0 | 0 | 0 | 0 | | H | 3 | 7 | 10 | 10 | 13 | 3 | 3 | 0 | 0 | | I | 4 | 6 | 10 | 9 | 13 | 3 | 3 | 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 | 4 | 1 | | 3 | 6 | 6 | 0 | | 4 | 2 | 6 | 4 | | 5 | 7 | 10 | 3 | | 6 | 6 | 6 | 0 | | 7 | 6 | 9 | 3 | | 8 | 13 | 13 | 0 | #### Critical path and project duration - Critical activities (TF = 0): B, G - Critical path: B - G (= 6+7 = 13 time units) **Answer: project duration = 13 time units; critical path = B - G** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the forward pass, how is the EST of an activity found? A. EFT minus duration of its successors B. LST of its predecessor C. Smallest EFT of its predecessors D. Largest EFT of its predecessors Answer: D. EST is the largest EFT among the preceding activities (0 for start activities); EFT = EST + D. (Key point: Forward pass rule) 2. What are the project duration and critical path? A. 13 units, A-D-G B. 13 units, B-G C. 10 units, B-I D. 11 units, C-F-I Answer: B. Largest EFT is 13 (G: 6 + 7), and B and G have zero total float, so the critical path is B-G. (Key point: Project duration) 3. Why is LFT of B equal to 6? A. It is the largest of those LST values B. It equals the project duration C. It is the smallest of the LST values of its successors E(9), G(6), I(9) D. It equals EFT of A Answer: C. LFT is the smallest LST of the successors: min(9, 6, 9) = 6, so LST of B = 6 - 6 = 0. (Key point: Backward pass for B) 4. For activity D (EST 3, EFT 5, LST 4, LFT 6, successor G with EST 6), what are TF and FF? A. TF = 1, FF = 0 B. TF = 0, FF = 1 C. TF = 2, FF = 1 D. TF = 1, FF = 1 Answer: D. TF = LST - EST = 1. FF = EST of successor G - EFT of D = 6 - 5 = 1. (Key point: Total and free float) ## Close variants set in other exams - [CPM components (AOA), 10 activities](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-ten-act-b80) — Project Engineering, chapter 3, asked 5× - [CPM components (AOA), 13 activities (months)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-thirteen-act-c75) — Project Engineering, chapter 3, asked 5× ## Related in Project Engineering - [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, 9 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-s65) — Project Engineering, chapter 3, asked 2× - [CPM network, 9 activities with successors](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-nine-act-a76) — Project Engineering, chapter 3 - [CPM components (AOA), 9 activities](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-nine-act-b78) — Project Engineering, chapter 3, asked 4× - [CPM components (AOA), activities A to I](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-aoa-nine-act-c76) — Project Engineering, chapter 3, asked 4× - [CPM precedence diagram, 9 activities](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-precedence-k72) — Project Engineering, chapter 3 ## Related in other subjects - [Essential components of an activity diagram](https://www.bctengineer.com/question.md?key=semester_6%2Fobject_oriented_analysis_and_design%3Aactivity-diagram-components) — Object Oriented Analysis and Design, chapter 2 - [TCTO numerical: seven activities, Rs 225/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2078-kartik) — Construction Management, chapter 2