---
title: "Constant vs actual dollar; inflation-free rate on loan"
subject: "Engineering Economics"
chapter: 8
key: "semester_6/engineering_economics:constant-actual-dollar-bike-loan"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/constant-actual-dollar-bike-loan
site: BCT Engineering
---

# Constant vs actual dollar; inflation-free rate on loan

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

- Asked in: 2076 Shrawan, 2+4 marks; 2071 Bhadra, 2+4 marks (asked 2 times)
- Video: [✅2080 to 2069 | Inflation full chapter| 5+ question solved| engineering economics](https://www.youtube.com/watch?v=564K0T_Ws-k) — Revise with Roshan 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/constant-actual-dollar-bike-loan
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-8

## Question

Define constant dollar amount and actual dollar amount. Suppose you borrowed Rs 100000 from a bank to buy a bike and you have promised to pay Rs 5500 per month for two years. What is the inflation-free interest rate you are supposed to pay if average inflation rate is 0.75% per month?

## Short answer

Constant dollars represent purchasing power, actual dollars are nominal amounts. The loan’s monthly market rate is 2.35% (from 100,000 = 5,500(P/A,i,24)). With 0.75% monthly inflation, the real monthly rate is (0.02352-0.0075)/1.0075 ≈ 1.59%.

## Answer

#### Constant dollar and actual dollar
- **Actual dollars (current dollars)** are the amounts of money actually paid or received at a given date, including the effect of inflation. A bank instalment of Rs 5,500 is an actual-dollar amount.
- **Constant dollars (real dollars)** are amounts expressed in the purchasing power of a chosen base year (usually today). Actual = constant × (1 + f)^n, where f is the general inflation rate and n the number of periods from the base.

Interest rates: the **market (combined) rate** i applies to actual dollars; the **inflation-free (real) rate** i' applies to constant dollars. They are related by
i = i' + f + i'f, so i' = (i − f)/(1 + f).

#### Loan problem (monthly)
Actual-dollar loan: Rs 1,00,000 repaid by 24 payments of Rs 5,500, so the monthly market rate i satisfies
```text
1,00,000 = 5,500 (P/A, i, 24)   →   (P/A, i, 24) = 18.1818
```
Trial values: i = 2% gives (P/A,2%,24) = 18.9139; i = 3% gives 16.9355. Interpolation: i = 2% + 1% × (18.9139 − 18.1818)/(18.9139 − 16.9355) = 2.37%. Exact solution: **i = 2.352% per month** (nominal 28.22% per year; effective 32.2% per year).

Inflation-free monthly rate, with f = 0.75% per month:
```text
i' = (i − f)/(1 + f) = (2.352% − 0.75%)/1.0075 = 1.590% per month
```
**Answer:** The inflation-free interest rate is about **1.59% per month** (about 20.8% effective per year); the market rate on the loan is 2.35% per month.

## Check yourself

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

1. What distinguishes actual dollars from constant dollars?
   A. Actual dollars exclude inflation; constant dollars include it
   B. Actual dollars are amounts paid at the date including inflation; constant dollars are in the purchasing power of a base year
   C. They are the same unless interest is charged
   D. Actual dollars are used only for loans, constant only for savings
   Answer: B. Actual = constant x (1+f)^n; constant dollars are measured in base-year purchasing power. (Key point: Actual vs constant dollars)

2. How is the monthly market interest rate on the bike loan found?
   A. Solve 1,00,000 = 5,500 (P/A,i,24) for i
   B. Divide 5,500 by 1,00,000
   C. Solve 1,00,000 = 5,500 (F/A,i,24)
   D. Subtract 0.75% from 5,500/1,00,000
   Answer: A. The payments are an annuity of 24 actual-dollar amounts, giving (P/A,i,24) = 18.1818 and i of about 2.35% per month. (Key point: Market rate from loan)

3. Which formula gives the inflation-free rate i' from market rate i and inflation f?
   A. i' = i - f
   B. i' = (i - f)/(1 + f)
   C. i' = (i + f)/(1 + f)
   D. i' = i/(1 + f)
   Answer: B. From i = i' + f + i'f, i' = (i - f)/(1 + f). (Key point: Real rate formula)

4. With i = 2.352% and f = 0.75% per month, what is the inflation-free monthly rate?
   A. 3.10%
   B. 1.60%, found by simple subtraction only
   C. 1.59%
   D. 2.35%
   Answer: C. i' = (2.352 - 0.75)/1.0075 = 1.59% per month. (Key point: Result)

## Related in Engineering Economics

- [Short note: market vs inflation-free interest rate](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Ashort-note-market-vs-inflation-free) — Engineering Economics, chapter 8
- [Present equivalent of fuel costs, actual dollar](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aactual-dollar-fuel-cost-waste-plant) — Engineering Economics, chapter 8
- [Loan 120,000: number of installments and rate](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aloan-120000-installments-rate) — Engineering Economics, chapter 2
- [Inflation, effects, inflation-free rate](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adefine-inflation-effects-implied-rate) — Engineering Economics, chapter 8
- [Car loan interest rate and lump sum at year 3](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acar-loan-installments-lump-sum-year3) — Engineering Economics, chapter 2
- [Define inflation; NPW or IRR in constant dollars](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Ainflation-define-npw-irr) — Engineering Economics, chapter 8

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