--- title: "TCTO numerical: crashing critical path, Rs 30/day" subject: "Construction Management" chapter: 2 key: "semester_8/construction_management:tcto-2069-bhadra" source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2069-bhadra site: BCT Engineering --- # TCTO numerical: crashing critical path, Rs 30/day Construction Management · Chapter 2: Construction Planning and Scheduling - Asked in: 2069 Bhadra (old course), 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2069-bhadra - Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2 ## Question For the activities lying on the critical path of a network, the normal duration and the crash duration along with their respective direct cost are given in the table. | Activity | Normal duration | Normal cost (Rs) | Crash duration | Crash costs (Rs) | |---|---|---|---|---| | 1-2 | 2 days | 50 | 1 day | 75 | | 2-3 | 4 days | 160 | 3 days | 225 | | 3-5 | 9 days | 270 | 6 days | 350 | | 5-6 | 5 days | 100 | 4 days | 250 | The overhead indirect cost is Rs.30/- per day. Crashing is possible for 1 day for each of the activities 1-2, 2-3, and 5-6 and 3 days for activity 3-5. Find the lowest cost schedule by crashing step by step assuming no fresh critical path is developed on crashing a project. ## Short answer Lowest cost schedule: crash 1-2 by 1 day (slope Rs 25/day) and 3-5 by 3 days (slope Rs 26.67/day). Resulting duration 16 days, total cost Rs1,165 (direct Rs685 + overhead Rs480). No other crashing is economical against Rs30/day indirect saving. ## Answer #### Given data and method Indirect cost = 30 Rs/day. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost. The four activities lie in series on the critical path (normal duration 2 + 4 + 9 + 5 = 20 days). Crashing limits: 1-2, 2-3 and 5-6 by 1 day each; 3-5 by 3 days. No new critical path appears. $$ \text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}} $$ | Activity | Normal dur | Crash dur | Normal cost | Crash cost | Slope | |---|---|---|---|---|---| | 1-2 | 2 | 1 | 50 | 75 | 25 | | 2-3 | 4 | 3 | 160 | 225 | 65 | | 3-5 | 9 | 6 | 270 | 350 | 26.67 | | 5-6 | 5 | 4 | 100 | 250 | 150 | #### Paths at normal duration - 1-2 + 2-3 + 3-5 + 5-6 = 20 days Critical activities (on the longest path or paths): 1-2, 2-3, 3-5, 5-6; normal duration = 20 days; normal direct cost = 580 Rs. #### Crashing, step by step | Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total | |---|---|---|---|---|---|---| | 0 | - | - | 20 | 580 | 600 | 1,180 | | 1 | 1-2 | 25 | 19 | 605 | 570 | 1,175 | | 2 | 3-5 | 26.67 | 18 | 631.67 | 540 | 1,171.67 | | 3 | 3-5 | 26.67 | 17 | 658.33 | 510 | 1,168.33 | | 4 | 3-5 | 26.67 | 16 | 685 | 480 | 1,165 | | 5 | 2-3 | 65 | 15 | 750 | 450 | 1,200 | | 6 | 5-6 | 150 | 14 | 900 | 420 | 1,320 | #### Result - **Optimum (minimum-cost) schedule:** duration 16 days, direct cost 685 + indirect 480 = **1,165 Rs**. - Activities crashed at the optimum: 1-2 by 1, 3-5 by 3. Durations: 1-2=1, 2-3=4, 3-5=6, 5-6=5. - **Minimum possible duration:** 14 days, with total cost 1,320 Rs (direct 900 + indirect 420). **Answer: the lowest-cost schedule is obtained by crashing 1-2 by 1 day and 3-5 by 3 days: duration 16 days, total cost Rs 1,165 (direct Rs 685 + overhead Rs 480), against Rs 1,180 at the normal duration. Crashing 2-3 (Rs 65/day) or 5-6 (Rs 150/day) costs more than the Rs 30/day saved.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With slopes 1-2: 25, 2-3: 65, 3-5: 26.67, 5-6: 150 (Rs/day), which is crashed first? A. 3-5 B. 2-3 C. 5-6 D. 1-2 Answer: D. 1-2 has the lowest slope, Rs 25/day, so it is crashed first (only 1 day allowed). (Key point: Cheapest slope first) 2. Overhead is Rs 30/day. Which activities are worth crashing? A. All four B. 1-2 and 3-5 only C. 2-3 and 5-6 only D. None Answer: B. Slopes 25 and 26.67 are below 30; 65 and 150 are above 30 so they raise total cost. (Key point: Comparison with overhead) 3. What is the lowest-cost schedule? A. 20 days, Rs 1,180 B. 14 days, Rs 1,320 C. 16 days, Rs 1,165 D. 15 days, Rs 1,200 Answer: C. Crashing 1-2 by 1 and 3-5 by 3 gives 16 days with direct 685 + overhead 480 = Rs 1,165. (Key point: Result) ## Related in Construction Management - [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 - [TCTO numerical: seven activities, Rs 2000/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2079-shrawan) — Construction Management, chapter 2 - [TCTO numerical: seven activities, Rs 1000/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2070-magh) — Construction Management, chapter 2 - [TCTO numerical: least cost schedule, Rs 2000/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2068-bhadra) — Construction Management, chapter 2 - [TCTO numerical with time-cost chart (five activities)](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2079-jestha) — Construction Management, chapter 2 - [TCTO numerical: lowest cost schedule, Rs 400/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2075-bhadra) — Construction Management, chapter 2 ## Related in other subjects - [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 - [Dummy activity; use of critical path in CPM](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Adummy-critical-path-use) — Project Engineering, chapter 3 - [Critical path and dummy activity significance](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acritical-path-dummy-significance) — Project Engineering, chapter 3 - [EVA; 10 days, 50 labour days at Rs 1000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeva-numerical-labour-days-1000) — Project Engineering, chapter 4, asked 2×