--- title: "TCTO numerical: seven activities, Rs 225/day" subject: "Construction Management" chapter: 2 key: "semester_8/construction_management:tcto-2078-kartik" source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2078-kartik site: BCT Engineering --- # TCTO numerical: seven activities, Rs 225/day Construction Management · Chapter 2: Construction Planning and Scheduling - Asked in: 2078 Kartik, 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-2078-kartik - Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2 ## Question Calculate the minimum cost, optimum duration using time cost trade off analysis for the following project schedule consisting of seven activities. Also find the possible minimum time to complete the project. Indirect cost of the project is Rs. 225 per day. | S.N. | Project Activity | Immediate Predecessor | Normal Duration (days) | Crash Duration (days) | Normal Cost (NPR) | Crash Cost (NPR) | |---|---|---|---|---|---|---| | 1 | A | - | 6 | 4 | 1400 | 1900 | | 2 | B | - | 8 | 5 | 2000 | 2800 | | 3 | C | A | 4 | 2 | 1100 | 1500 | | 4 | D | A | 3 | 2 | 800 | 1400 | | 5 | E | B, C | 6 | 3 | 900 | 1600 | | 6 | F | D | 10 | 6 | 2500 | 3500 | | 7 | G | E | 3 | 2 | 500 | 800 | ## Short answer With indirect cost Rs 225/day, cost slopes (cheapest Rs 200/day) exceed the saving. Optimum is normal 19 days, total cost Rs 13,475. Minimum possible time is 12 days (total Rs 15,466.67). ## Answer #### Given data and method Indirect cost = 225 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 | 6 | 4 | 1,400 | 1,900 | 250 | | B | 8 | 5 | 2,000 | 2,800 | 266.67 | | C | 4 | 2 | 1,100 | 1,500 | 200 | | D | 3 | 2 | 800 | 1,400 | 600 | | E | 6 | 3 | 900 | 1,600 | 233.33 | | F | 10 | 6 | 2,500 | 3,500 | 250 | | G | 3 | 2 | 500 | 800 | 300 | #### Paths at normal duration - A-C-E-G = 19 days - A-D-F = 19 days - B-E-G = 17 days Critical activities (on the longest path or paths): A, C, D, E, F, G; normal duration = 19 days; normal direct cost = 9,200 Rs. #### Crashing, step by step | Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total | |---|---|---|---|---|---|---| | 0 | - | - | 19 | 9,200 | 4,275 | 13,475 | | 1 | A | 250 | 18 | 9,450 | 4,050 | 13,500 | | 2 | A | 250 | 17 | 9,700 | 3,825 | 13,525 | | 3 | E+F | 483.33 | 16 | 10,183.33 | 3,600 | 13,783.33 | | 4 | E+F | 483.33 | 15 | 10,666.67 | 3,375 | 14,041.67 | | 5 | E+F | 483.33 | 14 | 11,150 | 3,150 | 14,300 | | 6 | F+G | 550 | 13 | 11,700 | 2,925 | 14,625 | | 7 | B+C+D | 1,066.67 | 12 | 12,766.67 | 2,700 | 15,466.67 | (Rows with several activities show activities crashed together because two or more paths are critical. After step 2, all three paths, including B-E-G, are 17 days long, so a cut must shorten all three: E+F (Rs 483.33) is the cheapest such combination; C+F would leave B-E-G at 17 days.) #### Result - **Optimum (minimum-cost) schedule:** duration 19 days, direct cost 9,200 + indirect 4,275 = **13,475 Rs**. - Activities crashed at the optimum: none. Durations: A=6, B=8, C=4, D=3, E=6, F=10, G=3. - **Minimum possible duration:** 12 days, with total cost 15,466.67 Rs (direct 12,766.67 + indirect 2,700). **Answer: minimum cost Rs 13,475 at the optimum (normal) duration of 19 days, since the cheapest way to shorten both critical paths (A, Rs 250/day) costs more than the Rs 225/day indirect cost saved; the minimum possible time to complete the project is 12 days (total cost Rs 15,466.67).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the cost slope of an activity calculated in time-cost trade-off? A. (Crash cost - Normal cost) / (Normal duration - Crash duration) B. (Normal cost - Crash cost) / (Normal duration + Crash duration) C. Crash cost / Crash duration D. (Crash duration - Normal duration) / (Crash cost - Normal cost) Answer: A. Slope is the extra direct cost per unit of time saved: cost increase divided by days shortened. (Key point: Cost slope formula) 2. Which activities are crashed first in TCTO? A. Any activity with the lowest slope in the network B. Critical activities with the lowest cost slope C. Non-critical activities with the highest slope D. Activities with the longest normal duration Answer: B. Only critical-path activities shorten the project, and the cheapest slope is chosen first. (Key point: Which activities to crash) 3. At normal duration, A-C-E-G and A-D-F both equal 19 days. Why must several activities be crashed together later? A. Because crashing is always done in pairs B. Because both critical paths must be shortened to reduce project duration C. Because slopes of two activities are always added for any crash D. Because indirect cost doubles Answer: B. If only one of the parallel critical paths is shortened, the project duration is unchanged. 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