---
title: "TCTO numerical: optimal crashed time, Rs 2500/week"
subject: "Construction Management"
chapter: 2
key: "semester_8/construction_management:tcto-2078-chaitra"
source: https://www.bctengineer.com/q/semester_8/construction_management/tcto-2078-chaitra
site: BCT Engineering
---

# TCTO numerical: optimal crashed time, Rs 2500/week

Construction Management · Chapter 2: Construction Planning and Scheduling

- Asked in: 2078 Chaitra, 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-2078-chaitra
- Whole chapter: https://www.bctengineer.com/md/semester_8/construction_management/chapter-2

## Question

Consider the data of a project as shown in table below. If the indirect cost per week is Rs. 2500, find the optimal crashed project completion time and corresponding minimum cost.

| Activity | Immediate Predecessor | Normal Time (Week) | Normal Cost (Rs.) | Crash Duration (Week) | Crash Cost (Rs.) |
|---|---|---|---|---|---|
| P | - | 7 | 2500 | 4 | 2200 |
| Q | - | 9 | 4500 | 7 | 4200 |
| R | P | 5 | 3500 | 4 | 3000 |
| S | P | 8 | 5500 | 7 | 5100 |
| T | Q | 9 | 3000 | 8 | 2500 |
| U | Q | 12 | 5500 | 12 | - |
| V | S, T | 4 | 1500 | 2 | 1800 |

## Short answer

Crash critical activities with smallest cost slope as long as indirect saving (Rs 2500/week) exceeds slope. Normal 22 weeks at Rs 79,000. Optimum: 19 weeks, total cost Rs 71,950. No further crashing reduces total cost.

## Answer

#### Given data and method
Indirect cost = 2,500 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 P, Q, R, S and T the table lists the crash cost below the normal cost, which is impossible for a crash. These columns are taken as interchanged, i.e. the smaller value is the normal cost and the larger is the crash cost. U cannot be crashed (crash duration equals normal 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 |
|---|---|---|---|---|---|
| P | 7 | 4 | 2,200 | 2,500 | 100 |
| Q | 9 | 7 | 4,200 | 4,500 | 150 |
| R | 5 | 4 | 3,000 | 3,500 | 500 |
| S | 8 | 7 | 5,100 | 5,500 | 400 |
| T | 9 | 8 | 2,500 | 3,000 | 500 |
| U | 12 | 12 | 5,500 | 5,500 | cannot crash |
| V | 4 | 2 | 1,500 | 1,800 | 150 |

#### Paths at normal duration
- P-R = 12 weeks
- P-S-V = 19 weeks
- Q-T-V = 22 weeks
- Q-U = 21 weeks

Critical activities (on the longest path or paths): Q, T, V; normal duration = 22 weeks; normal direct cost = 24,000 Rs.

#### Crashing, step by step
| Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total |
|---|---|---|---|---|---|---|
| 0 | - | - | 22 | 24,000 | 55,000 | 79,000 |
| 1 | Q | 150 | 21 | 24,150 | 52,500 | 76,650 |
| 2 | Q | 150 | 20 | 24,300 | 50,000 | 74,300 |
| 3 | V | 150 | 19 | 24,450 | 47,500 | 71,950 |

#### Result
- **Optimum (minimum-cost) schedule:** duration 19 weeks, direct cost 24,450 + indirect 47,500 = **71,950 Rs**.
- Activities crashed at the optimum: Q by 2, V by 1. Durations: P=7, Q=7, R=5, S=8, T=9, U=12, V=3.
- **Minimum possible duration:** 19 weeks, with total cost 71,950 Rs (direct 24,450 + indirect 47,500).

**Answer: optimal crashed project completion time = 19 weeks; minimum total cost = Rs 71,950.**

## Check yourself

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

1. What assumption is made about the table?
   A. Normal and crash cost columns are interchanged for P, Q, R, S, T
   B. Costs are doubled
   C. U is crashed
   D. Indirect cost is ignored
   Answer: A. Crash cost cannot be below normal cost, so the columns are swapped. (Key point: Data assumption)

2. Which is the critical path at normal duration?
   A. Q-T-V, 22 weeks
   B. P-S-V, 19 weeks
   C. Q-U, 21 weeks
   D. P-R, 12 weeks
   Answer: A. Q-T-V = 9+9+4 = 22 weeks. (Key point: Critical path)

3. Why can U not be crashed?
   A. Its slope is highest
   B. Its crash duration equals normal duration
   C. It is not on any path
   D. It is critical
   Answer: B. Crash duration 12 equals normal 12, so no time can be saved. (Key point: Cannot crash)

4. What is the optimal completion time and cost?
   A. 22 weeks, Rs 79,000
   B. 19 weeks, Rs 71,950
   C. 20 weeks, Rs 74,300
   D. 21 weeks, Rs 76,650
   Answer: B. Crashing Q by 2 and V by 1 gives 19 weeks, direct 24,450 + indirect 47,500. (Key point: Optimum)

## Close variants set in other exams

- [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×

## Related in Construction Management

- [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: lowest cost schedule, Rs 400/week](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2075-bhadra) — Construction Management, chapter 2
- [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: seven activities, Rs 225/day](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Atcto-2078-kartik) — Construction Management, chapter 2