--- title: "TCTO numerical with time-cost chart (five activities)" subject: "Construction Management" chapter: 2 key: "semester_8/construction_management:tcto-2079-jestha" source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2079-jestha site: BCT Engineering --- # TCTO numerical with time-cost chart (five activities) Construction Management · Chapter 2: Construction Planning and Scheduling - Asked in: 2079 Jestha, 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-jestha - Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2 ## Question Perform time-cost trade-off based on given information regarding a project comprising five different activities. Also, draw the time-cost trade-off chart. Take time dependent cost per week of Rs 1.6 thousand. | Activity | Predecessors | Time (Week) Normal | Time (Week) Crash | Cost (Thousand Rs.) Normal | Cost (Thousand Rs.) Crash | |---|---|---|---|---|---| | A | - | 26 | 18 | 4 | 20 | | B | - | 28 | 22 | 5 | 20 | | C | A | 10 | 5 | 6 | 11 | | D | A | 4 | 2 | 8 | 11 | | E | B, D | 12 | 6 | 3 | 14 | ## Short answer Time-cost trade-off crashes the cheapest critical activity per week while saving Rs 1.6 thousand/week indirect. Normal 42 weeks at Rs 93.2k. Optimum: 40 weeks, total cost Rs 93,000. Minimum duration: 28 weeks at Rs 113.8k. ## Answer #### Given data and method Indirect cost = 1.6 thousand Rs/week. 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. To draw the full trade-off chart, crashing is continued to the minimum possible duration. $$ \text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}} $$ | Activity | Normal dur | Crash dur | Normal cost | Crash cost | Slope | |---|---|---|---|---|---| | A | 26 | 18 | 4 | 20 | 2 | | B | 28 | 22 | 5 | 20 | 2.5 | | C | 10 | 5 | 6 | 11 | 1 | | D | 4 | 2 | 8 | 11 | 1.5 | | E | 12 | 6 | 3 | 14 | 1.83 | #### Paths at normal duration - A-C = 36 weeks - A-D-E = 42 weeks - B-E = 40 weeks Critical activities (on the longest path or paths): A, D, E; normal duration = 42 weeks; normal direct cost = 26 thousand Rs. #### Crashing, step by step (costs in thousand Rs) | Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total | |---|---|---|---|---|---|---| | 0 | - | - | 42 | 26.00 | 67.2 | 93.20 | | 1 | D | 1.5 | 41 | 27.50 | 65.6 | 93.10 | | 2 | D | 1.5 | 40 | 29.00 | 64.0 | 93.00 | | 3 | E | 1.83 | 39 | 30.83 | 62.4 | 93.23 | | 4 | E | 1.83 | 38 | 32.67 | 60.8 | 93.47 | | 5 | E | 1.83 | 37 | 34.50 | 59.2 | 93.70 | | 6 | E | 1.83 | 36 | 36.33 | 57.6 | 93.93 | | 7 | C+E | 2.83 | 35 | 39.17 | 56.0 | 95.17 | | 8 | C+E | 2.83 | 34 | 42.00 | 54.4 | 96.40 | | 9 | A+B | 4.5 | 33 | 46.50 | 52.8 | 99.30 | | 10 | A+B | 4.5 | 32 | 51.00 | 51.2 | 102.20 | | 11 | A+B | 4.5 | 31 | 55.50 | 49.6 | 105.10 | | 12 | A+B | 4.5 | 30 | 60.00 | 48.0 | 108.00 | | 13 | A+B | 4.5 | 29 | 64.50 | 46.4 | 110.90 | | 14 | A+B | 4.5 | 28 | 69.00 | 44.8 | 113.80 | (At 40 weeks A-D-E and B-E are both critical and D is fully crashed, so E, common to both, is crashed. At 36 weeks A-C also becomes critical, so C+E is the cheapest cut of all three paths. At 34 weeks D and E are at their crash limits, so A and B must be crashed together. At 28 weeks B and E are both fully crashed, so B-E cannot be shortened further.) #### Result - **Optimum (minimum-cost) schedule:** duration 40 weeks, direct cost 29 + indirect 64 = **93 thousand Rs**. - Activities crashed at the optimum: D by 2. Durations: A=26, B=28, C=10, D=2, E=12. - Crashing beyond 40 weeks costs more than the Rs 1.6 thousand/week saved (E slope 1.83 > 1.6), so the total cost rises. - **Minimum possible duration:** 28 weeks, with total cost 113.8 thousand Rs (direct 69 + indirect 44.8). #### Time-cost trade-off chart (total cost against duration) ```text Duration Total cost (Rs '000); bar length = cost above 90 (weeks) 42 | ###### 93.20 (normal) 41 | ###### 93.10 40 | ###### 93.00 <-- minimum total cost (optimum) 39 | ###### 93.23 38 | ####### 93.47 37 | ####### 93.70 36 | ######## 93.93 35 | ########## 95.17 34 | ############# 96.40 33 | ################### 99.30 32 | ######################## 102.20 31 | ############################## 105.10 30 | #################################### 108.00 29 | ########################################## 110.90 28 | ################################################ 113.80 (crash) ``` On graph paper, plot direct cost (falling with duration), indirect cost (a straight line rising at 1.6 per week) and their sum; the total cost curve is lowest at 40 weeks. **Answer: optimum duration 40 weeks; minimum total cost Rs 93.0 thousand (direct 29.0 + indirect 64.0). The shortest possible duration is 28 weeks, at a total cost of Rs 113.8 thousand.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which path is critical and how long is it? A. A-C, 36 weeks B. A-D-E, 42 weeks C. B-E, 40 weeks D. A-D-E, 40 weeks Answer: B. A-D-E = 26+4+12 = 42 weeks. (Key point: Critical path) 2. What is D's cost slope? A. 1.5 thousand per week B. 2 thousand C. 1 thousand D. 2.5 thousand Answer: A. (11-8)/(4-2) = 1.5. (Key point: Slope of D) 3. Why is crashing stopped after D by 2 weeks? A. E's slope 1.83 exceeds the 1.6 saved per week B. D is at crash limit and E is cheaper C. A has the lowest slope D. B-E becomes longer Answer: A. Crashing E costs more than the weekly indirect saving; D is also at its limit. (Key point: Stop rule) 4. What is the optimum? A. 42 weeks, 93.2 B. 40 weeks, 93.0 thousand C. 38 weeks, 93.47 D. 28 weeks Answer: B. At 40 weeks, direct 29 + indirect 64 = 93 thousand. 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