---
title: "Feasibility of constant-dollar growing series"
subject: "Engineering Economics"
chapter: 8
key: "semester_6/engineering_economics:constant-dollar-series-feasibility-20000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/constant-dollar-series-feasibility-20000
site: BCT Engineering
---

# Feasibility of constant-dollar growing series

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

- Asked in: 2069 Bhadra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/constant-dollar-series-feasibility-20000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-8

## Question

A series of five constant dollar (or real dollar) incomes (beginning with \$5000 at the end of the first year) are increasing at the rate of 7% per year for five years. Inflation free interest rate is 5% and inflation is 8%. Is it feasible investment if investment cost is \$20,000?

## Short answer

Market interest rate i = i' + f + i'f = 5% + 8% + 0.4% = 13.4%. Discounting the growing constant-dollar series at i'=5% yields present worth of incomes $24,734. Since NPW = $24,734 - $20,000 = $4,734 > 0, the investment is feasible.

## Answer

The incomes are in constant dollars, growing 7% a year: A_n = 5,000(1.07)^(n−1). The investment is made now, so it is an actual dollar amount of 20,000.

#### Market interest rate
i = i' + f + i'f = 0.05 + 0.08 + (0.05)(0.08) = **13.4%**

#### Convert to actual dollars and discount at 13.4%
Actual income = constant income × (1.08)^n.
| Year | Constant \$ | Actual \$ = const × 1.08^n | (1.134)^−n | PW |
|---|---|---|---|---|
| 1 | 5,000 | 5,400 | 0.8818 | 4,762 |
| 2 | 5,350 | 6,240 | 0.7776 | 4,853 |
| 3 | 5,724 | 7,211 | 0.6857 | 4,945 |
| 4 | 6,125 | 8,333 | 0.6047 | 5,039 |
| 5 | 6,554 | 9,630 | 0.5333 | 5,135 |

```text
PW of incomes = 24,734
NPW = −20,000 + 24,734 = 4,734
```
**Check (constant dollars at i' = 5%):** a growing series with g = 7% and i' = 5% gives
PW = Σ 5,000(1.07)^(n−1)/(1.05)^n = (5,000/1.05) × [1 − (1.07/1.05)⁵] / [1 − 1.07/1.05] = 24,734, the same value.

**Answer:** PW of income = \$24,734 > \$20,000 (NPW = \$4,734), so the investment is **feasible**.

## Check yourself

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

1. With i' = 5% and f = 8%, what is the market interest rate?
   A. 13.4%
   B. 13%
   C. 3%
   D. 12.6%
   Answer: A. i = 0.05 + 0.08 + 0.05 x 0.08 = 13.4%. (Key point: Market rate)

2. What is the year-2 income in actual dollars?
   A. 5,350
   B. 5,400
   C. 6,240
   D. 6,000
   Answer: C. Constant income 5,000(1.07) = 5,350, then x (1.08)^2 = 6,240. (Key point: Conversion of incomes)

3. What is the conclusion on feasibility?
   A. Not feasible, since PW of income 24,734 > 20,000 is a loss
   B. Feasible: NPW = 24,734 - 20,000 = 4,734 > 0
   C. Not feasible because incomes are only 5,000
   D. Feasible only if discounted at 5%
   Answer: B. The PW of incomes exceeds the investment, so NPW is positive and the investment is feasible. (Key point: Decision)

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