--- title: "TCTO numerical: seven activities, Rs 2000/day" subject: "Construction Management" chapter: 2 key: "semester_8/construction_management:tcto-2079-shrawan" source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2079-shrawan site: BCT Engineering --- # TCTO numerical: seven activities, Rs 2000/day Construction Management · Chapter 2: Construction Planning and Scheduling - Asked in: 2079 Shrawan, 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-2079-shrawan - Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2 ## Question From the given data in table below. Calculate minimum cost and minimum duration schedule using time cost trade off analysis. Indirect cost is Rs. 2000/day. | Activity | Predecessor | Normal Duration (days) | Crash Duration (days) | Normal cost (Rs.) | Crash Cost (Rs.) | |---|---|---|---|---|---| | 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 critical activities one day at a time by cheapest slope while saving Rs 2000/day indirect cost. Normal 21 days at Rs 88,000. Optimum: 13 days, total cost Rs 77,000. Minimum duration: 11 days at Rs 77,800. ## Answer #### Given data and method Indirect cost = 2,000 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 | 42,000 | 88,000 | | 1 | D | 200 | 20 | 46,200 | 40,000 | 86,200 | | 2 | A | 400 | 19 | 46,600 | 38,000 | 84,600 | | 3 | A | 400 | 18 | 47,000 | 36,000 | 83,000 | | 4 | A | 400 | 17 | 47,400 | 34,000 | 81,400 | | 5 | G | 800 | 16 | 48,200 | 32,000 | 80,200 | | 6 | G | 800 | 15 | 49,000 | 30,000 | 79,000 | | 7 | G | 800 | 14 | 49,800 | 28,000 | 77,800 | | 8 | B+D | 1,200 | 13 | 51,000 | 26,000 | 77,000 | | 9 | E+F | 2,400 | 12 | 53,400 | 24,000 | 77,400 | | 10 | E+F | 2,400 | 11 | 55,800 | 22,000 | 77,800 | (Rows with several activities show activities crashed together because two paths are critical.) #### Result - **Optimum (minimum-cost) schedule:** duration 13 days, direct cost 51,000 + indirect 26,000 = **77,000 Rs**. - Activities crashed at the optimum: A by 3, B by 1, D by 2, G by 3. Durations: A=2, B=3, C=3, D=4, E=5, F=4, G=3. - **Minimum possible duration:** 11 days, with total cost 77,800 Rs (direct 55,800 + indirect 22,000). **Answer: minimum-cost schedule = 13 days, total cost Rs 77,000; minimum-duration schedule = 11 days, total cost Rs 77,800. Crashing beyond 13 days costs more (Rs 2,400 per day) than the Rs 2,000/day saved.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the normal critical path and its duration? A. A-B-E-G, 20 days B. A-C-G, 14 days C. A-D-F-G, 21 days D. A-B-E-G, 21 days Answer: C. A-D-F-G = 5+6+4+6 = 21 days, the longest path. (Key point: Critical path) 2. Which critical activity has the lowest slope and what is it? A. A, 400 B. D, 200 C. F, 900 D. G, 800 Answer: B. D: (4400-4000)/2 = 200 per day. (Key point: Cheapest slope) 3. What is the minimum-cost schedule? A. 13 days, Rs 77,000 B. 11 days, Rs 77,800 C. 14 days, Rs 77,800 D. 21 days, Rs 88,000 Answer: A. Total cost is lowest at 13 days, Rs 77,000. (Key point: Optimum) 4. Why does the minimum duration of 11 days cost more? A. Further crashing costs Rs 2,400/day (E+F) against Rs 2,000 saved B. Indirect cost rises C. Slopes are lower D. Normal costs rise Answer: A. Crashing E+F together costs more than the indirect saving. 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