--- title: "CPM network, hydropower project (months)" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-network-hydropower-k80" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-network-hydropower-k80 site: BCT Engineering --- # CPM network, hydropower project (months) Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2080 Baisakh, 12 marks - 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-hydropower-k80 - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Draw a network diagram for a hydropower project having the information given below. Find out ES, EF, LS, LF, TF, FF, IntF, IndF and then analyse the situation of the project stating the critical path, project completion time and critical activities. | Activity | Duration (month) | Predecessor | |---|---|---| | A | 3 | - | | B | 5 | - | | C | 8 | - | | D | 4 | A | | E | 7 | A | | F | 2 | B | | G | 1 | C | | H | 7 | A | | I | 4 | E, F, G | | J | 9 | C | ## Short answer CPM analysis gives project duration 17 months. Critical path is C-J, with only activities C and J having 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 (months). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only. ```text (1) --A(3)--> (2) (1) --B(5)--> (3) (1) --C(8)--> (4) (2) --H(7)--> (5) (2) --E(7)--> (6) (2) --D(4)--> (7) (3) --F(2)--> (6) (4) --G(1)--> (6) (4) --J(9)--> (7) (5) - - dummy - -> (7) (6) --I(4)--> (7) ``` The network has 7 events and 1 dummy activity. Start event is (1) and the end event is (7). #### 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 + 5 = 5 - C: EST = 0; EFT = 0 + 8 = 8 - D: EST = EFT of A = 3; EFT = 3 + 4 = 7 - E: EST = EFT of A = 3; EFT = 3 + 7 = 10 - H: EST = EFT of A = 3; EFT = 3 + 7 = 10 - F: EST = EFT of B = 5; EFT = 5 + 2 = 7 - G: EST = EFT of C = 8; EFT = 8 + 1 = 9 - J: EST = EFT of C = 8; EFT = 8 + 9 = 17 - I: EST = max(E:10, F:7, G:9) = 10; EFT = 10 + 4 = 14 Project duration = largest EFT = **17 months**. #### 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 = 17; LST = 17 - 4 = 13 - J: LFT = project duration = 17; LST = 17 - 9 = 8 - G: LFT = LST of I = 13; LST = 13 - 1 = 12 - F: LFT = LST of I = 13; LST = 13 - 2 = 11 - H: LFT = project duration = 17; LST = 17 - 7 = 10 - E: LFT = LST of I = 13; LST = 13 - 7 = 6 - D: LFT = project duration = 17; LST = 17 - 4 = 13 - C: LFT = min(G:12, J:8) = 8; LST = 8 - 8 = 0 - B: LFT = LST of F = 11; LST = 11 - 5 = 6 - A: LFT = min(D:13, E:6, H:10) = 6; LST = 6 - 3 = 3 #### 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 | 3 | 6 | 3 | 0 | 3 | 0 | | B | 5 | 0 | 5 | 6 | 11 | 6 | 0 | 6 | 0 | | C | 8 | 0 | 8 | 0 | 8 | 0 | 0 | 0 | 0 | | D | 4 | 3 | 7 | 13 | 17 | 10 | 10 | 0 | 7 | | E | 7 | 3 | 10 | 6 | 13 | 3 | 0 | 3 | 0 | | F | 2 | 5 | 7 | 11 | 13 | 6 | 3 | 3 | 0 | | G | 1 | 8 | 9 | 12 | 13 | 4 | 1 | 3 | 1 | | H | 7 | 3 | 10 | 10 | 17 | 7 | 7 | 0 | 4 | | I | 4 | 10 | 14 | 13 | 17 | 3 | 3 | 0 | 0 | | J | 9 | 8 | 17 | 8 | 17 | 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 | 6 | 3 | | 3 | 5 | 11 | 6 | | 4 | 8 | 8 | 0 | | 5 | 10 | 17 | 7 | | 6 | 10 | 13 | 3 | | 7 | 17 | 17 | 0 | #### Critical path and project duration - Critical activities (TF = 0): C, J - Critical path: C - J (= 8+9 = 17 months) **Answer: project duration = 17 months; critical path = C - J** #### Analysis of the project situation - Project completion time = **17 months**. - Critical path = **C - J** (8 + 9 = 17 months); critical activities = C and J (zero float). - All other activities (A, B, D, E, F, G, H, I) are non-critical; they can be delayed by their floats (see table) without delaying the project, but a delay of C or J delays the hydropower project month for month. - Management should give priority in resources, funds and supervision to C and J, and use the float of the other activities for levelling the resources. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Activity I follows E (EFT 10), F (EFT 7) and G (EFT 9). What is its EST? A. 10 B. 7 C. 9 D. 26 Answer: A. EST = max(10, 7, 9) = 10, so EFT = 14. (Key point: Forward pass rule) 2. What is the project completion time? A. 14 months B. 17 months C. 20 months D. 25 months Answer: B. The largest EFT is that of J: 8 + 9 = 17 months. (Key point: Project completion time) 3. Which is the critical path and why? A. A - E - I, because it has the most activities B. B - F - I, because it starts earliest C. C - J, because its total float is zero (8 + 9 = 17) D. A - H, because it has the longest float Answer: C. C and J are the only activities with TF = 0 and they sum to the project duration of 17 months. (Key point: Critical path) 4. How should management treat the activities of this project? A. Use the float of C and J for levelling B. Give equal priority to all activities C. Delay C and J since they have float D. Give priority in resources and supervision to C and J, and use the float of others for levelling Answer: D. A delay in C or J delays the project month for month, while others have float that can be used for levelling. 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