--- title: "CPM network, building project (A to K)" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-network-eleven-act-s73" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-eleven-act-s73 site: BCT Engineering --- # CPM network, building project (A to K) Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2073 Shrawan, 13 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-network-eleven-act-s73 - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Draw a network diagram and find out EST, EFT, LST, LFT, TF, FF, independent float, interfering float and the project completion time of a building project having the following details. What is the significance of critical path in the network analysis? | Activity | Immediate Predecessor | Duration (weeks) | |---|---|---| | A | - | 10 | | B | - | 12 | | C | - | 9 | | D | A | 8 | | E | A | 5 | | F | B | 13 | | G | C | 6 | | H | C | 4 | | I | D | 15 | | J | E, F, G | 7 | | K | H | 9 | ## Short answer The building project completes in 33 weeks. Critical path: A–D–I. The critical path represents the minimum project duration; activities on it have zero total float and any delay extends the project. ## 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 (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only. ```text (1) --A(10)--> (2) (1) --B(12)--> (3) (1) --C(9)--> (4) (2) --D(8)--> (5) (2) --E(5)--> (6) (3) --F(13)--> (6) (4) --G(6)--> (6) (4) --H(4)--> (7) (5) --I(15)--> (8) (6) --J(7)--> (8) (7) --K(9)--> (8) ``` The network has 8 events and no dummy activities. The 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 + 10 = 10 - B: EST = 0; EFT = 0 + 12 = 12 - C: EST = 0; EFT = 0 + 9 = 9 - G: EST = EFT of C = 9; EFT = 9 + 6 = 15 - H: EST = EFT of C = 9; EFT = 9 + 4 = 13 - D: EST = EFT of A = 10; EFT = 10 + 8 = 18 - E: EST = EFT of A = 10; EFT = 10 + 5 = 15 - F: EST = EFT of B = 12; EFT = 12 + 13 = 25 - K: EST = EFT of H = 13; EFT = 13 + 9 = 22 - I: EST = EFT of D = 18; EFT = 18 + 15 = 33 - J: EST = max(E:15, F:25, G:15) = 25; EFT = 25 + 7 = 32 Project duration = largest EFT = **33 weeks**. #### Backward pass (LFT and LST) Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D. - J: LFT = project duration = 33; LST = 33 - 7 = 26 - I: LFT = project duration = 33; LST = 33 - 15 = 18 - K: LFT = project duration = 33; LST = 33 - 9 = 24 - F: LFT = LST of J = 26; LST = 26 - 13 = 13 - E: LFT = LST of J = 26; LST = 26 - 5 = 21 - D: LFT = LST of I = 18; LST = 18 - 8 = 10 - H: LFT = LST of K = 24; LST = 24 - 4 = 20 - G: LFT = LST of J = 26; LST = 26 - 6 = 20 - C: LFT = min(G:20, H:20) = 20; LST = 20 - 9 = 11 - B: LFT = LST of F = 13; LST = 13 - 12 = 1 - A: LFT = min(D:10, E:21) = 10; LST = 10 - 10 = 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 | 10 | 0 | 10 | 0 | 10 | 0 | 0 | 0 | 0 | | B | 12 | 0 | 12 | 1 | 13 | 1 | 0 | 1 | 0 | | C | 9 | 0 | 9 | 11 | 20 | 11 | 0 | 11 | 0 | | D | 8 | 10 | 18 | 10 | 18 | 0 | 0 | 0 | 0 | | E | 5 | 10 | 15 | 21 | 26 | 11 | 10 | 1 | 10 | | F | 13 | 12 | 25 | 13 | 26 | 1 | 0 | 1 | 0 | | G | 6 | 9 | 15 | 20 | 26 | 11 | 10 | 1 | 0 | | H | 4 | 9 | 13 | 20 | 24 | 11 | 0 | 11 | 0 | | I | 15 | 18 | 33 | 18 | 33 | 0 | 0 | 0 | 0 | | J | 7 | 25 | 32 | 26 | 33 | 1 | 1 | 0 | 0 | | K | 9 | 13 | 22 | 24 | 33 | 11 | 11 | 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 | 10 | 10 | 0 | | 3 | 12 | 13 | 1 | | 4 | 9 | 20 | 11 | | 5 | 18 | 18 | 0 | | 6 | 25 | 26 | 1 | | 7 | 13 | 24 | 11 | | 8 | 33 | 33 | 0 | #### Critical path and project duration - Critical activities (TF = 0): A, D, I - Critical path: A - D - I (= 10+8+15 = 33 weeks) **Answer: the building project is completed in 33 weeks; critical path = A - D - I** #### Significance of the critical path - It gives the minimum time in which the project can be completed (33 weeks here). - Critical activities (A, D, I) have zero float; a delay in any of them delays the building by the same time, so they need close supervision and first priority for labour, material and equipment. - To shorten the project (crashing), only critical activities need to be speeded up; shortening non-critical activities wastes money. - Non-critical activities (e.g. B, F, H, K) have float and can be rescheduled to level the resources. - It is the basis for monitoring and updating the schedule; the critical path can change when non-critical activities use up their float. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. J has predecessors E (EFT 15), F (EFT 25) and G (EFT 15). What is the EST of J? A. 15 B. 25 C. 40 D. 32 Answer: B. EST is the largest predecessor EFT = 25 (from F), so EFT of J = 32. (Key point: Forward pass for J) 2. What are the project completion time and critical path? A. 32 weeks, B-F-J B. 33 weeks, A-D-I C. 33 weeks, B-F-J D. 22 weeks, C-H-K Answer: B. A+D+I = 10+8+15 = 33 weeks, the longest path with TF = 0. B-F-J is only 32. (Key point: Project time and critical path) 3. Activity C has EST 0 and LFT 20 with duration 9. What is its total float? A. 0 B. 9 C. 20 D. 11 Answer: D. LST = 20 - 9 = 11, so TF = LST - EST = 11 - 0 = 11 weeks. (Key point: Total float of C) 4. Why is the critical path significant when the project must be shortened? A. Any activity can be crashed with the same effect B. Only critical activities need to be speeded up; shortening non-critical ones wastes money C. Critical activities have float to give away D. The critical path never changes during execution Answer: B. Critical activities have zero float and decide the duration, so only reducing them shortens the project. 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