---
title: "TCTO numerical: lowest cost schedule, Rs 400/week"
subject: "Construction Management"
chapter: 2
key: "semester_8/construction_management:tcto-2075-bhadra"
source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2075-bhadra
site: BCT Engineering
---

# TCTO numerical: lowest cost schedule, Rs 400/week

Construction Management · Chapter 2: Construction Planning and Scheduling

- Asked in: 2075 Bhadra, 7 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-2075-bhadra
- Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2

## Question

The details of a project are shown below. If the indirect cost per week is Rs. 400, find the lowest cost schedule of the project.

| Activity | Predecessor(s) | Time (weeks) Normal | Time (weeks) Crash | Cost (Rs) Normal | Cost (Rs) Crash |
|---|---|---|---|---|---|
| A | - | 8 | 5 | 2000 | 2200 |
| B | - | 10 | 8 | 3800 | 4000 |
| C | B | 6 | 5 | 2700 | 2600 |
| D | A | 9 | 6 | 4100 | 4300 |
| E | C | 10 | 9 | 3100 | 3350 |
| F | B | 12 | 12 | 3100 | 3100 |

## Short answer

Time-cost trade-off: crash B by 2, C by 1, and E by 1 week. Lowest-cost schedule is 22 weeks, total cost Rs 28,050. Further crashing E would cost Rs 250/week vs. Rs 400 saved, so 22 weeks is optimal.

## Answer

#### Given data and method
Indirect cost = 400 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.

Assumption: for C the table shows crash cost (2,600) lower than normal cost (2,700). The two values are taken as interchanged, so normal cost = Rs 2,600 and crash cost = Rs 2,700. F cannot be crashed.

$$
\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}
$$

| Activity | Normal dur | Crash dur | Normal cost | Crash cost | Slope |
|---|---|---|---|---|---|
| A | 8 | 5 | 2,000 | 2,200 | 66.67 |
| B | 10 | 8 | 3,800 | 4,000 | 100 |
| C | 6 | 5 | 2,600 | 2,700 | 100 |
| D | 9 | 6 | 4,100 | 4,300 | 66.67 |
| E | 10 | 9 | 3,100 | 3,350 | 250 |
| F | 12 | 12 | 3,100 | 3,100 | cannot crash |

#### Paths at normal duration
- A-D = 17 weeks
- B-C-E = 26 weeks
- B-F = 22 weeks

Critical activities (on the longest path or paths): B, C, E; normal duration = 26 weeks; normal direct cost = 18,700 Rs.

#### Crashing, step by step
| Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total |
|---|---|---|---|---|---|---|
| 0 | - | - | 26 | 18,700 | 10,400 | 29,100 |
| 1 | B | 100 | 25 | 18,800 | 10,000 | 28,800 |
| 2 | B | 100 | 24 | 18,900 | 9,600 | 28,500 |
| 3 | C | 100 | 23 | 19,000 | 9,200 | 28,200 |
| 4 | E | 250 | 22 | 19,250 | 8,800 | 28,050 |

#### Result
- **Optimum (minimum-cost) schedule:** duration 22 weeks, direct cost 19,250 + indirect 8,800 = **28,050 Rs**.
- Activities crashed at the optimum: B by 2, C by 1, E by 1. Durations: A=8, B=8, C=5, D=9, E=9, F=12.
- **Minimum possible duration:** 22 weeks, with total cost 28,050 Rs (direct 19,250 + indirect 8,800).

**Answer: lowest-cost schedule = 22 weeks, total cost Rs 28,050.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which path is critical at normal durations (A-D = 17, B-C-E = 26, B-F = 22 weeks)?
   A. A-D
   B. B-F
   C. B-C-E
   D. A-D and B-F together
   Answer: C. B-C-E is the longest path at 26 weeks, so B, C, E are critical. (Key point: Critical path at normal)

2. Why is F ignored for crashing?
   A. It has the highest slope
   B. Its crash duration equals its normal duration
   C. It is not on any path
   D. It has zero cost
   Answer: B. F is 12 weeks both normal and crash, so it cannot be shortened. (Key point: Activity that cannot be crashed)

3. What is the lowest-cost schedule?
   A. 26 weeks, Rs 29,100
   B. 24 weeks, Rs 28,500
   C. 23 weeks, Rs 28,200
   D. 22 weeks, Rs 28,050
   Answer: D. Crashing B by 2, C by 1, E by 1 gives direct 19,250 + indirect 8,800 = 28,050 Rs at 22 weeks. (Key point: Total cost at optimum)

## Related in Construction Management

- [TCTO numerical: least cost schedule, Rs 2000/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2068-bhadra) — Construction Management, chapter 2
- [TCTO numerical: optimal crashed time, Rs 3000/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2073-magh) — Construction Management, chapter 2, asked 2×
- [TCTO numerical: optimal crashed time, Rs 2500/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2078-chaitra) — Construction Management, chapter 2, asked 2×
- [TCTO numerical: optimal crashed result, Rs 300/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2073-bhadra) — Construction Management, chapter 2
- [TCTO numerical: seven activities, Rs 2000/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2079-shrawan) — Construction Management, chapter 2
- [TCTO numerical: seven activities, Rs 225/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2078-kartik) — Construction Management, chapter 2

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