---
title: "Choose machine under 5% inflation"
subject: "Engineering Economics"
chapter: 8
key: "semester_6/engineering_economics:machine-xy-inflation-5-percent"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/machine-xy-inflation-5-percent
site: BCT Engineering
---

# Choose machine under 5% inflation

Engineering Economics · Chapter 8: Inflation and Its Impact on Project Cashflows

- Asked in: 2072 Asoj, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/machine-xy-inflation-5-percent
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-8

## Question

Choose the best project from the following alternatives.
| Project | Machine X | Machine Y |
|---|---|---|
| First Cost | 15,00000 | 20,00000 |
| Life | 7 years | 7 years |
| Salvage Value | 200000 | 300000 |
| Annual operating and maintenance cost | 300000 | 250000 |
Assume an average inflation of 5% for the next five years and interest rate is 15% / year.

## Short answer

Compute inflation-free rate i' = (15%-5%)/1.05 = 9.524%. Present worth of costs: Machine X Rs 2,877,913, Machine Y Rs 3,077,730. Select X since it has lower cost.

## Answer

**Assumptions:** costs and salvage values are given in today's (constant) dollars; inflation of 5% a year applies throughout the 7-year life (the "5 years" in the question is taken as the whole planning period); 15% is the market interest rate. The machines give the same service, so the one with the lower cost is chosen.

Inflation-free rate: i' = (i − f)/(1 + f) = (0.15 − 0.05)/1.05 = **9.524%**. Evaluating constant-dollar costs at i' gives the same result as converting to actual dollars and discounting at 15%.

Factors at i' = 9.524%: (P/A,i',7) = 4.9457; (P/F,i',7) = 0.5290.

```text
PW(X) = 15,00,000 + 3,00,000(P/A,i',7) − 2,00,000(P/F,i',7)
      = 15,00,000 + 3,00,000(4.9457) − 2,00,000(0.5290)
      = 2,877,913

PW(Y) = 20,00,000 + 2,50,000(P/A,i',7) − 3,00,000(P/F,i',7)
      = 20,00,000 + 2,50,000(4.9457) − 3,00,000(0.5290)
      = 3,077,730
```
| Machine | PW of cost (Rs) | AEC at i' (Rs/yr) |
|---|---|---|
| X | 2,877,913 | 581,902 |
| Y | 3,077,730 | 622,305 |

(Without inflation, discounting at 15% gives PW(X) = 2,672,939 and PW(Y) = 2,927,324, so the ranking is unchanged.)

**Answer: Select Machine X.** It has the lower present worth of cost (Rs 2,877,913 against Rs 3,077,730).

## Check yourself

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

1. With i = 15% and f = 5%, what rate is used for the constant-dollar costs?
   A. 10%
   B. 20.75%
   C. 9.524%
   D. 15%
   Answer: C. i' = (0.15 - 0.05)/1.05 = 9.524%. (Key point: Inflation-free rate)

2. How is the PW of cost of a machine formed?
   A. First cost + O&M x (P/A) + salvage x (P/F)
   B. First cost + O&M x (P/A) - salvage x (P/F)
   C. First cost - O&M x (P/A) - salvage x (P/F)
   D. First cost x (P/A) - O&M
   Answer: B. Salvage is a receipt at year 7 so its present worth is subtracted from the cost. (Key point: PW of cost formula)

3. Which machine is selected and why?
   A. Y, because of lower annual operating cost
   B. Y, because of higher salvage value
   C. X, because it has the lower PW of cost (2,877,913 vs 3,077,730)
   D. Either, since inflation changes the ranking
   Answer: C. Both give the same service, so choose the lower cost; X is cheaper and the ranking is unchanged without inflation. (Key point: Decision)

## Related in Engineering Economics

- [Decision tree for plant improvement alternatives](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adecision-tree-plant-alternatives) — Engineering Economics, chapter 6, asked 2×
- [IRR to select best of A, B, C (15 years)](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-select-abc-15yr) — Engineering Economics, chapter 4
- [IRR choose best of four, 10-year study](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-choose-best-abcd-study-10) — Engineering Economics, chapter 4
- [Repeatability assumption, select best of three projects](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Arepeatability-abc-14) — Engineering Economics, chapter 4, asked 2×
- [Nepal Airlines jet plane choice, 4 years and infinite](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Anepal-airlines-jet-plane-4yr-infinite) — Engineering Economics, chapter 4
- [PW with differential inflation rates](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Apw-constant-dollars-differential-inflation) — Engineering Economics, chapter 8

## Related in other subjects

- [RoR best installed capacity (NRs, 45 years)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-nrs-10-percent) — Hydropower Engineering, chapter 3
- [Choose the best answer (grammar) 2073 Chaitra](https://www.bctengineer.com/question.md?key=semester_5%2Fcommunication_english%3Achoose-best-answer-grammar-set-2073c) — Communication English, chapter 3
- [Regression of maintenance cost on car age](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Acar-age-maintenance-regression) — Probability and Statistics, chapter 5
- [Gumbel design flood, 21 years data, 40-year life](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-21yr-life40-rel85-b) — Engineering Hydrology, chapter 6, asked 3×