--- title: "Time cost trade-off concept with example" subject: "Construction Management" chapter: 2 key: "semester_8/construction_management:tcto-concept" source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-concept site: BCT Engineering --- # Time cost trade-off concept with example Construction Management · Chapter 2: Construction Planning and Scheduling - Asked in: 2074 Bhadra, 6 marks; 2071 Bhadra, 6 marks; 2072 Asoj; 2065 Baisakh (old course); 2064 Poush (old course) (asked 5 times) - Video: [CE 404 - CEPM | Crashing & Time Cost Trade Off | Needs & Methods of Crashing | Mod 2 - Lec 1 | KTU](https://www.youtube.com/watch?v=UgMHsI5t7dY) — Alex -E Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/construction_management/tcto-concept - Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2 ## Question Explain the concept of time cost trade-off (and its use in determining the duration of a project) with an example. ## Short answer Time-cost trade-off balances direct and indirect costs to find the minimum total cost. Crashing critical activities with the lowest cost slope first, while the slope is less than indirect cost saved per day. For the example, crashing B (Rs 300/day) then A (Rs 500/day) yields an optimum duration of 10 days and total cost of Rs 9,300. ## Answer **Time-cost trade-off (TCTO)** is the technique of shortening a project's duration by crashing (speeding up) selected activities while watching the total cost. Direct cost (labour, materials, machines) rises as an activity is crashed, while indirect cost (overheads, supervision, penalties) falls as the project becomes shorter. The sum of the two is lowest at an optimum duration. ```text Cost ^ |T T T = total cost | T T D = direct cost | T T I = indirect cost | T T | T * T * = minimum total cost |D | D I | D I | D I | D I | I D | I D | I D D +-------------------------------------> Duration crash optimum normal ``` #### Use in determining project duration - Cost slope of an activity = (crash cost - normal cost) / (normal time - crash time). - Only critical activities are crashed, cheapest slope first, until the slope exceeds the indirect cost saved per unit time or the crash limit is reached. - The duration with the minimum total cost is the optimum duration; the shortest feasible duration is the crash duration. #### Example Two activities in series: A (normal 6 days, Rs 3,000; crash 4 days, Rs 4,000) and B (normal 5 days, Rs 2,000; crash 4 days, Rs 2,300). Indirect cost Rs 400/day. Slope of A = (4000 - 3000)/(6 - 4) = Rs 500/day; slope of B = (2300 - 2000)/(5 - 4) = Rs 300/day. | Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total | |---|---|---|---|---|---|---| | 0 | - | - | 11 | 5,000 | 4,400 | 9,400 | | 1 | B | 300 | 10 | 5,300 | 4,000 | 9,300 | | 2 | A | 500 | 9 | 5,800 | 3,600 | 9,400 | | 3 | A | 500 | 8 | 6,300 | 3,200 | 9,500 | Crashing B costs Rs 300 and saves Rs 400, so it is worthwhile. Crashing A (Rs 500) costs more than the Rs 400 saved. **The optimum duration is 10 days with a total cost of Rs 9,300** (against Rs 9,400 at the normal duration). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When an activity is crashed, what happens to direct and indirect cost? A. Both rise B. Both fall C. Direct rises, indirect falls as the project shortens D. Direct falls, indirect rises Answer: C. Crashing needs extra resources so direct cost rises, while shorter duration cuts overheads. (Key point: Cost behaviour) 2. Where is the optimum duration? A. At minimum total cost B. At normal duration always C. At minimum direct cost D. At zero indirect cost Answer: A. The lowest point of the total cost curve is the optimum. (Key point: Optimum) 3. How is the cost slope calculated? A. (normal cost - crash cost)/(crash time) B. (crash cost - normal cost)/(normal time - crash time) C. (normal time - crash time)/(crash cost - normal cost) D. Crash cost/normal time Answer: B. Slope is extra cost per unit time saved. (Key point: Cost slope) 4. Which activities are crashed first? A. Non-critical with highest slope B. Any activity C. Critical activities with the cheapest slope D. Longest activities Answer: C. Only critical activities shorten the project, cheapest first. (Key point: Which to crash) 5. In the example (indirect Rs 400/day), what is the optimum? A. 11 days, Rs 9,400 B. 10 days, Rs 9,300 C. 9 days, Rs 9,400 D. 8 days, Rs 9,500 Answer: B. Crashing B (300) saves 400, A (500) would cost more than the saving, so 10 days at Rs 9,300. (Key point: Example result) ## Related in Construction Management - [Time cost trade-off and steps in planning](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-steps-planning) — Construction Management, chapter 2 - [TCTO, crashing and reducing project duration](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-crashing-reduce-schedule) — 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 ## Related in other subjects - [Concept of project finance](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aproject-finance-concept) — Project Engineering, chapter 6 - [Cut-off and time-current characteristics of fuse](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Afuse-cutoff-time-current) — Switchgear and Protection, chapter 3 - [Prospective current, cut-off current, time delay fuse](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aprospective-cutoff-time-delay-fuse) — Switchgear and Protection, chapter 3 - [Project proposal; cost benefit analysis for road](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aproposal-cba-road-project) — Project Engineering, chapter 2