--- title: "TCTO numerical: seven activities, Rs 1000/day" subject: "Construction Management" chapter: 2 key: "semester_8/construction_management:tcto-2070-magh" source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2070-magh site: BCT Engineering --- # TCTO numerical: seven activities, Rs 1000/day Construction Management · Chapter 2: Construction Planning and Scheduling - Asked in: 2070 Magh, 6 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-2070-magh - Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2 ## Question Calculate minimum cost and minimum duration schedule using time cost trade-off analysis of the project with following detail. Indirect cost is Rs. 1000 per day. | Activity | Predecessor | Normal Duration | Crash Duration | Normal Cost | Crash Cost | |---|---|---|---|---|---| | A | - | 5 | 2 | 5000 | 6200 | | B | A | 4 | 3 | 6000 | 7000 | | C | A | 3 | 2 | 5000 | 6500 | | D | A | 6 | 4 | 4000 | 4400 | | E | B | 5 | 3 | 15000 | 18000 | | F | D | 4 | 2 | 3000 | 4800 | | G | C, E, F | 6 | 3 | 8000 | 10400 | ## Short answer Compute cost slopes. Crash cheapest critical activities one day at a time while saving Rs 1000/day indirect cost. Normal 21 days at Rs 67,000. Optimum: 14 days, total cost Rs 63,800. Minimum duration: 11 days at Rs 66,800. ## Answer #### Given data and method Indirect cost = 1,000 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. $$ \text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}} $$ | Activity | Normal dur | Crash dur | Normal cost | Crash cost | Slope | |---|---|---|---|---|---| | A | 5 | 2 | 5,000 | 6,200 | 400 | | B | 4 | 3 | 6,000 | 7,000 | 1,000 | | C | 3 | 2 | 5,000 | 6,500 | 1,500 | | D | 6 | 4 | 4,000 | 4,400 | 200 | | E | 5 | 3 | 15,000 | 18,000 | 1,500 | | F | 4 | 2 | 3,000 | 4,800 | 900 | | G | 6 | 3 | 8,000 | 10,400 | 800 | #### Paths at normal duration - A-B-E-G = 20 days - A-C-G = 14 days - A-D-F-G = 21 days Critical activities (on the longest path or paths): A, D, F, G; normal duration = 21 days; normal direct cost = 46,000 Rs. #### Crashing, step by step | Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total | |---|---|---|---|---|---|---| | 0 | - | - | 21 | 46,000 | 21,000 | 67,000 | | 1 | D | 200 | 20 | 46,200 | 20,000 | 66,200 | | 2 | A | 400 | 19 | 46,600 | 19,000 | 65,600 | | 3 | A | 400 | 18 | 47,000 | 18,000 | 65,000 | | 4 | A | 400 | 17 | 47,400 | 17,000 | 64,400 | | 5 | G | 800 | 16 | 48,200 | 16,000 | 64,200 | | 6 | G | 800 | 15 | 49,000 | 15,000 | 64,000 | | 7 | G | 800 | 14 | 49,800 | 14,000 | 63,800 | | 8 | B+D | 1,200 | 13 | 51,000 | 13,000 | 64,000 | | 9 | E+F | 2,400 | 12 | 53,400 | 12,000 | 65,400 | | 10 | E+F | 2,400 | 11 | 55,800 | 11,000 | 66,800 | (Rows with several activities show activities crashed together because two paths are critical.) #### Result - **Optimum (minimum-cost) schedule:** duration 14 days, direct cost 49,800 + indirect 14,000 = **63,800 Rs**. - Activities crashed at the optimum: A by 3, D by 1, G by 3. Durations: A=2, B=4, C=3, D=5, E=5, F=4, G=3. - **Minimum possible duration:** 11 days, with total cost 66,800 Rs (direct 55,800 + indirect 11,000). **Answer: minimum-cost schedule = 14 days, total cost Rs 63,800; minimum-duration schedule = 11 days, total cost Rs 66,800.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With indirect Rs 1,000/day, why does the optimum shift to 14 days? A. G's slope 800 is below 1,000 but B+D at 1,200 exceeds it B. All slopes are below 1,000 C. D's slope is above 1,000 D. Indirect cost is ignored Answer: A. Crashing continues while slope is below 1,000; B+D (1,200) costs more than it saves. (Key point: Indirect cost effect) 2. What are the minimum-cost duration and cost? A. 13 days, Rs 64,000 B. 14 days, Rs 63,800 C. 11 days, Rs 66,800 D. 21 days, Rs 67,000 Answer: B. At 14 days, 49,800 + 14,000 = 63,800. (Key point: Optimum) 3. Why are B and D crashed together at the 13-day step? A. Two paths (A-B-E-G and A-D-F-G) become critical, so both must be shortened B. B has the lowest slope C. D is not crashable D. To avoid cost Answer: A. Both paths must be reduced to shorten the project. (Key point: Two critical paths) 4. What is the minimum duration schedule and cost? A. 11 days, Rs 66,800 B. 11 days, Rs 55,800 C. 13 days, Rs 64,000 D. 12 days, Rs 65,400 Answer: A. Direct 55,800 + indirect 11,000 = 66,800. 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