---
title: "Present equivalent of fuel costs, actual dollar"
subject: "Engineering Economics"
chapter: 8
key: "semester_6/engineering_economics:actual-dollar-fuel-cost-waste-plant"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/actual-dollar-fuel-cost-waste-plant
site: BCT Engineering
---

# Present equivalent of fuel costs, actual dollar

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

- Asked in: 2070 Bhadra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/actual-dollar-fuel-cost-waste-plant
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-8

## Question

The annual fuel cost required to operate a small solid waste treatment plant are projected to be Rs. 200000 without considering any future inflation. The best estimate indicates that the annual inflation free interest rate i' will be 6% and the general inflation rate, f, will be 5%. If the plant has the remaining useful life of four years, what is the present equivalent of its fuel costs? Use actual dollar analysis.

## Short answer

Market rate i = 6%+5%+0.3%=11.3%. Inflated annual costs: 210,000, 220,500, 231,525, 243,101. Present worth at 11.3% = Rs 693,021 (equivalent to constant-dollar PW at 6%).

## Answer

Actual-dollar analysis converts the inflation-free (constant-dollar) cash flows into actual dollars and discounts them at the **market interest rate**.

#### Step 1: market interest rate
i = i' + f + i'f = 0.06 + 0.05 + (0.06)(0.05) = **11.3%**

#### Step 2: actual-dollar fuel costs
Fuel cost in year n = 2,00,000(1.05)^n.

#### Step 3: present worth at 11.3%
| Year | Actual-dollar cost | (1.113)^−n | Present worth |
|---|---|---|---|
| 1 | 210,000 | 0.8985 | 188,679 |
| 2 | 220,500 | 0.8073 | 177,999 |
| 3 | 231,525 | 0.7253 | 167,924 |
| 4 | 243,101 | 0.6517 | 158,419 |

```text
PW = 188,679 + 177,999 + 167,924 + 158,419 = Rs 693,021
```
**Check:** the same result comes from discounting constant dollars at i' = 6%: 2,00,000(P/A,6%,4) = 2,00,000(3.4651) = Rs 693,021. (The two agree because 200,000(1.05)^n/(1.113)^n = 200,000/(1.06)^n.)

**Answer:** The present equivalent of the fuel costs is **Rs 693,021** (about Rs 6.93 lakh).

## Check yourself

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

1. With i' = 6% and f = 5%, what is the market interest rate for actual-dollar analysis?
   A. 11.3%
   B. 11%
   C. 1%
   D. 10.5%
   Answer: A. i = i' + f + i'f = 0.06 + 0.05 + 0.003 = 11.3%. (Key point: Market rate)

2. What is the fuel cost in year 2 in actual dollars?
   A. 200,000
   B. 210,000
   C. 220,500
   D. 240,000
   Answer: C. 200,000(1.05)^2 = 220,500. (Key point: Actual dollar costs)

3. Why does discounting constant dollars at 6% give the same PW (Rs 693,021)?
   A. Because 6% equals the market rate
   B. Because inflation is ignored in actual-dollar analysis
   C. Because fuel costs do not inflate
   D. Because 200,000(1.05)^n/(1.113)^n = 200,000/(1.06)^n
   Answer: D. The inflation factor cancels against the market rate, leaving discounting at i' = 6%. (Key point: Check)

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- [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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