--- title: "CPM with mason resource aggregation chart" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:cpm-resource-aggregation-mason" source: https://www.bctengineer.com/q/semester_7/project_engineering/cpm-resource-aggregation-mason site: BCT Engineering --- # CPM with mason resource aggregation chart Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2071 Chaitra, 8+5 marks - Video: [Resource histogram explained for project management](https://www.youtube.com/watch?v=NwyCTXQraLk) — 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-resource-aggregation-mason - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Draw the network diagram of the given project having the following activities. Obtain the project duration, critical path, TF, FF and interfering float. Prepare the resource aggregation chart and allocate the mason using early start schedule. | Activity | Duration (days) | Mason (per day) | |---|---|---| | 1-2 | 3 | 1 | | 2-3 | 3 | 2 | | 2-4 | 4 | 4 | | 2-5 | 2 | 2 | | 3-10 | 3 | 2 | | 4-6 | 2 | 3 | | 4-7 | 4 | 3 | | 5-9 | 4 | 4 | | 6-8 | 2 | 2 | | 7-9 | 4 | 1 | | 8-9 | 3 | 2 | | 9-11 | 3 | 4 | | 10-11 | 2 | 2 | | 11-12 | 2 | 1 | ## Short answer Project duration: 20 days, critical path: 1–2–4–7–9–11–12. With early start schedule, mason requirement peaks at 12 on days 8–9; total mason‑days = 101. ## Answer The activities are given by their event numbers (i-j), so each activity is an arrow from event i to event j. #### Network diagram ```text (1) --1-2(3)--> (2) (2) --2-3(3)--> (3) (2) --2-4(4)--> (4) (2) --2-5(2)--> (5) (3) --3-10(3)--> (10) (4) --4-6(2)--> (6) (4) --4-7(4)--> (7) (5) --5-9(4)--> (9) (6) --6-8(2)--> (8) (7) --7-9(4)--> (9) (8) --8-9(3)--> (9) (9) --9-11(3)--> (11) (10) --10-11(2)--> (11) (11) --11-12(2)--> (12) ``` #### Event times TE from the forward pass, TL from the backward pass (days). | Event | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | |---|---|---|---|---|---|---|---|---|---|---|---|---| | TE | 0 | 3 | 6 | 7 | 5 | 9 | 11 | 11 | 15 | 9 | 18 | 20 | | TL | 0 | 3 | 13 | 7 | 11 | 10 | 11 | 12 | 15 | 16 | 18 | 20 | #### Activity times and floats $TF = TL_j - TE_i - D$, $FF = TE_j - TE_i - D$, $IntF = TF - FF$. | Activity | D | Masons/day | EST | EFT | LST | LFT | TF | FF | Int.F | |---|---|---|---|---|---|---|---|---|---| | 1-2 | 3 | 1 | 0 | 3 | 0 | 3 | 0 | 0 | 0 | | 2-3 | 3 | 2 | 3 | 6 | 10 | 13 | 7 | 0 | 7 | | 2-4 | 4 | 4 | 3 | 7 | 3 | 7 | 0 | 0 | 0 | | 2-5 | 2 | 2 | 3 | 5 | 9 | 11 | 6 | 0 | 6 | | 3-10 | 3 | 2 | 6 | 9 | 13 | 16 | 7 | 0 | 7 | | 4-6 | 2 | 3 | 7 | 9 | 8 | 10 | 1 | 0 | 1 | | 4-7 | 4 | 3 | 7 | 11 | 7 | 11 | 0 | 0 | 0 | | 5-9 | 4 | 4 | 5 | 9 | 11 | 15 | 6 | 6 | 0 | | 6-8 | 2 | 2 | 9 | 11 | 10 | 12 | 1 | 0 | 1 | | 7-9 | 4 | 1 | 11 | 15 | 11 | 15 | 0 | 0 | 0 | | 8-9 | 3 | 2 | 11 | 14 | 12 | 15 | 1 | 1 | 0 | | 9-11 | 3 | 4 | 15 | 18 | 15 | 18 | 0 | 0 | 0 | | 10-11 | 2 | 2 | 9 | 11 | 16 | 18 | 7 | 7 | 0 | | 11-12 | 2 | 1 | 18 | 20 | 18 | 20 | 0 | 0 | 0 | #### Project duration and critical path - Critical activities (TF = 0): 1-2, 2-4, 4-7, 7-9, 9-11, 11-12 - Critical path: 1-2-4-7-9-11-12 (3 + 4 + 4 + 4 + 3 + 2 = 20 days) **Answer: project duration = 20 days; critical path = 1-2-4-7-9-11-12.** #### Resource aggregation chart (early start schedule) Each activity starts at its EST and needs the given number of masons every day until its EFT. The masons of all activities working on a given day are added. | Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |---|---|---|---|---|---|---|---|---|---|---| | Masons | 1 | 1 | 1 | 8 | 8 | 10 | 10 | 12 | 12 | 7 | | Day | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | |---|---|---|---|---|---|---|---|---|---|---| | Masons | 7 | 3 | 3 | 3 | 1 | 4 | 4 | 4 | 1 | 1 | ```text 12 | # # 11 | # # 10 | # # # # 9 | # # # # 8 | # # # # # # 7 | # # # # # # # # 6 | # # # # # # # # 5 | # # # # # # # # 4 | # # # # # # # # # # # 3 | # # # # # # # # # # # # # # 2 | # # # # # # # # # # # # # # 1 |# # # # # # # # # # # # # # # # # # # # +---------------------------------------- 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 day (last digit) ``` - Peak requirement = **12 masons** (days 8 and 9). - Total mason-days = 101 (check: the sum of duration x masons over all activities = 101). - Average = 101/20 = 5.05, i.e. about 5 masons per day. #### Allocation of masons The early start allocation needs only 1 mason on days 1-3, then 8 to 12 masons on days 4-9, and then 7, 3, 1, 4 and 1 masons on the later days. The demand is very uneven. Because the non-critical activities 2-3, 2-5, 3-10, 5-9 and 10-11 have a total float of 6-7 days, they can be delayed to cut the peak (for example, by starting 5-9, which needs 4 masons, later) without extending the 20 days. This is resource smoothing. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this event-numbered network, how is total float of activity i-j found? A. TF = TE_j - TE_i - D B. TF = TL_j - TE_i - D C. TF = TL_j - TL_i + D D. TF = TL_i - TE_j - D Answer: B. Total float is the latest finish event time minus the earliest start event time minus the duration: TF = TL_j - TE_i - D. (Key point: Total float formula (event-based)) 2. Activity 2-3 has TF = 7 and FF = 0. What is its interfering float? A. 7 B. 14 C. 3 D. 0 Answer: A. Interfering float = TF - FF = 7 - 0 = 7 days. (Key point: Interfering float) 3. Which activities form the critical path and what is the project duration? A. 1-2-5-9-11-12, 16 days B. 1-2-4-7-9-11-12, 20 days C. 1-2-4-6-8-9-11-12, 19 days D. 1-2-3-10-11-12, 13 days Answer: B. The activities with TF = 0 are 1-2, 2-4, 4-7, 7-9, 9-11, 11-12, and 3+4+4+4+3+2 = 20 days. (Key point: Critical path and duration) 4. How is the daily mason requirement on the aggregation chart obtained? A. Maximum masons of any single activity that day B. Average of all activities' masons C. Masons of the critical activities only D. Sum of masons of all activities working that day when each starts at its EST Answer: D. Each activity starts at its EST and the masons of all activities running on a day are added to give that day's demand. (Key point: Resource aggregation on early start) 5. The early start chart peaks at 12 masons on days 8-9. How can the peak be reduced without extending the 20 days? A. Delay critical activities such as 4-7 B. Delay non-critical activities such as 5-9 using their float C. Shorten activity 1-2 D. Start all activities at their LST including critical ones, ignoring float Answer: B. Non-critical activities with 6-7 days of total float can be shifted later without delaying completion; this is resource smoothing. 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