1.5 Percentage Change
1. Exponential Growth and Decay
In problems involving interest rates $r\%$ or population growth $r\%$, an initial quantity (Present Value, $PV$) increases or decreases by $r\%$ per period (yearly, monthly, etc.).
Increasing Case (Future Value): The future value $FV$ after $n$ periods is given by:
$$FV = PV\left(1 + \dfrac{r}{100}\right)^n$$
Decreasing Case (Depreciation): If the present value $PV$ decreases by $r\%$ per period, the formula is:
$$FV = PV\left(1 - \dfrac{r}{100}\right)^n$$
EXAMPLE 1
- (a) An amount of 2000 euros is invested at 8% per year. What is the amount returned after 10 years?
- (b) An amount of 2000 euros is depreciated by 8% every year. What is the amount returned after 10 years?
Solution:
For both problems, we are given: $PV = 2000$, $r = 8$, $n = 10$.
- (a) The future value for growth is:
$$FV = PV\left(1 + \dfrac{r}{100}\right)^n = 2000\left(1 + \dfrac{8}{100}\right)^{10} = 2000(1.08)^{10} \approx 4317.85$$
- (b) The future value for depreciation is:
$$FV = PV\left(1 - \dfrac{r}{100}\right)^n = 2000\left(1 - \dfrac{8}{100}\right)^{10} = 2000(0.92)^{10} \approx 868.78$$
2. Explanation for the Common Ratio $R$
Percentage change functions as a geometric sequence with a common ratio $R = 1 + \dfrac{r}{100}$. Suppose an amount $PV = 2000$ is invested at $r=8\%$ per year:
| Period | In our example | In general |
|---|---|---|
| Present Value | $2000$ | $PV$ |
| Interest | $2000 \times \dfrac{8}{100}$ | $PV \times \dfrac{r}{100}$ |
| After 1 year | $$2000 + 2000 \times \dfrac{8}{100}$$ $$= 2000\left(1 + \dfrac{8}{100}\right)$$ | $$PV + PV\left(\dfrac{r}{100}\right)$$ $$= PV\left(1 + \dfrac{r}{100}\right)$$ |
- If an amount increases by $r\%$, it is multiplied by $R = 1 + \dfrac{r}{100}$.
- After two years, it is multiplied again by $R$, resulting in $FV = PV\left(1 + \dfrac{r}{100}\right)^2$.
- Thus, after $n$ years: $FV = PV\left(1 + \dfrac{r}{100}\right)^n$.
Translations of $r\%$ into $R$:
| $r\%$ | Increasing ($R = 1 + \dfrac{r}{100}$) | Decreasing ($R = 1 - \dfrac{r}{100}$) |
|---|---|---|
| $12\%$ | $R = 1.12$ | $R = 0.88$ |
| $20\%$ | $R = 1.20$ | $R = 0.80$ |
| $5\%$ | $R = 1.05$ | $R = 0.95$ |
| $7.2\%$ | $R = 1.072$ | $R = 0.928$ |
Be careful: The initial amount may be mentioned as the value of year 1 ($u_1$) or the initial present value ($PV$). This changes the exponent slightly.
- Rate of increase: $12\%$
- Amount in year 1: $u_1 = 1000$
- In 2nd year: $u_2 = 1000 \times 1.12$
- In $n$th year: $u_n = 1000(1.12)^{n-1}$
- Rate of increase: $12\%$
- Present value: $PV = 1000$
- After 1 year: $FV_1 = 1000 \times 1.12$
- After $n$ years: $FV_n = 1000(1.12)^n$
Note: Mind that the exponent in Problem 2 is $n$ and not $n-1$. In both cases, the growth is exponential.
EXAMPLE 2
There are ten boxes in a row. The first box contains 100€ and any subsequent box contains 10% more than the previous one. What is the amount in the 10th box?
Solution:
Here, $u_1 = 100$ and the common ratio is $R = 1.10$. Thus, the amount in the 10th box is:
$$u_{10} = 100(1.10)^9 \approx 235.8$$
(This is equivalent to the $FV$ formula, evaluated after 9 successive increases: $FV = 100(1.10)^9$).
EXAMPLE 3 (Finding the number of years $n$)
An amount of 2000 euros is invested at 8% per year. After how many complete years does the amount exceed 5000?
Solution:
We need to solve the inequality:
$$FV = 2000\left(1 + \dfrac{8}{100}\right)^n > 5000$$
Method A (Trial and error): Check values for $n$: For $n=11$, $FV = 4663.27$. For $n=12$, $FV = 5036.34$. Therefore, $n = 12$.
Method B (Using logarithms): Solve the exponential equation $2000(1.08)^n = 5000$:
$$1.08^n = 2.5 \implies n = \dfrac{\log 2.5}{\log 1.08} \approx 11.9$$
Thus, $n = 12$ complete years.
EXAMPLE 4
The current population of a city is 800,000. The population increases by 5.2% every year. Find:
- (a) the population of the city after 7 years;
- (b) the population of the city 7 years ago;
- (c) after how many complete years the population of the city doubles.
Solution:
- (a) Exponential growth with $PV=800,000$ and $r=5.2\%$. The population after 7 years is:
$$FV = 800,000\left(1 + \dfrac{5.2}{100}\right)^7 \approx 1,140,775$$
- (b) The formula works for the past as well. The population 7 years ago was:
$$FV = 800,000\left(1 + \dfrac{5.2}{100}\right)^{-7} \approx 561,022$$
[For the future, multiply by 1.052 every year; for the past, divide by 1.052].
- (c) Solve the equation $FV = 2 \times 800,000$:
$$800,000\left(1 + \dfrac{5.2}{100}\right)^n = 1,600,000$$
Solving this yields $n \approx 13.7$. The population doubles after 14 complete years.
3. Compounded Interest in $k$ Time Periods
Suppose an initial amount $PV=1000$€ is invested with a nominal interest rate of 12% per year. The interest may be compounded in $k$ periods per year:
- Semiannually (half-yearly): $k=2$
- Quarterly: $k=4$
- Monthly: $k=12$
In general, the Future Value formula becomes:
$$FV = PV\left(1 + \dfrac{r}{100k}\right)^{kn}$$
Let's observe how the final amount varies after 5 years for an initial amount of 1000 euros at 12% nominal annual interest:
- (a) Yearly: $FV = 1000 \times (1.12)^5 = 1762.34$
- (b) Half-yearly: $FV = 1000 \times \left(1 + \dfrac{12}{100 \times 2}\right)^{2 \times 5} = 1000 \times (1.06)^{10} = 1790.85$
- (c) Quarterly: $FV = 1000 \times \left(1 + \dfrac{12}{100 \times 4}\right)^{4 \times 5} = 1000 \times (1.03)^{20} = 1806.11$
- (d) Monthly: $FV = 1000 \times \left(1 + \dfrac{12}{100 \times 12}\right)^{12 \times 5} = 1000 \times (1.01)^{60} = 1816.70$
4. Investment With Regular Payments
Suppose $PV$ is invested with an annual interest rate $r\%$, giving a multiplier $R = 1 + \dfrac{r}{100}$. An extra payment $PMT$ is invested at the end of each year.
The value of the investment after $n$ years is given by the formula:
$$FV = \left[ PV \times R^n \right] + PMT \times \left( \dfrac{R^n - 1}{R - 1} \right)$$
This formula can be derived by evaluating each component separately:
- The initial $PV$ is invested for $n$ years: $PV\left(1 + \dfrac{r}{100}\right)^n = PV \times R^n$
- The 1st $PMT$ is invested for $n-1$ years: $PMT \times R^{n-1}$
- The 2nd $PMT$ is invested for $n-2$ years: $PMT \times R^{n-2}$
- ... The last payment made at the very end is simply $PMT$ (invested for 0 years).
The sum of the $n$ payment terms forms a Geometric Series with $u_1 = PMT$ and common ratio $R$. Thus:
$$S_n = PMT \times \left( \dfrac{R^n - 1}{R - 1} \right)$$
(If the initial investment is equal to the regular payment, $PV = PMT$, a Geometric Series of $n+1$ terms is obtained. If the last payment is omitted, subtract one $PMT$ from the final total).
EXAMPLE 5
An initial amount of 1000 euros and an extra amount of 1000 euros at the end of each year is invested with an interest rate of 12% compounded yearly ($R=1.12$). Find the value after 7 years.
Solution:
Using the geometric series formula for 8 total payments (initial + 7 end-of-year payments):
$$FV = 1000 \times \left( \dfrac{1.12^8 - 1}{1.12 - 1} \right) = 12299.69$$
Remark: If the last payment is not included (e.g., withdrawn just before the 7th year ends), subtract one payment: $12299.69 - 1000 = 11299.69$.
EXAMPLE 6
An amount of 1000 euros is invested at 12% compounded yearly ($R=1.12$). An extra payment of 300 euros is added at the end of every year. Find the value after 7 years.
Solution:
Combine the compound interest on the $PV$ and the geometric series of the $PMT$:
$$FV = 1000 \times 1.12^7 + 300 \times \left( \dfrac{1.12^7 - 1}{1.12 - 1} \right) = 5237.38$$
NOTICE (Adjusting for Compounding Periods): If the annual interest rate $r\%$ is compounded in $k$ periods per year, but payments are made yearly, then the effective multiplier for 1 full year is $R = \left(1 + \dfrac{r}{100k}\right)^k$. The $FV$ formulas remain valid using this effective $R$.
EXAMPLE 7
An amount of 1000 euros is invested with an interest rate of 12% compounded monthly. An extra payment of 300 euros is added at the end of each year. Find the value after 7 years.
Solution:
First, find the effective annual multiplier $R$:
$$R = \left(1 + \dfrac{0.12}{12}\right)^{12} = 1.01^{12} \approx 1.12682503$$
The value of the investment after 7 years is:
$$FV = 1000 \times R^7 + 300 \times \left( \dfrac{R^7 - 1}{R - 1} \right) \approx 5397.73$$
NOTICE (Payments matching Compounding Periods): If interest is compounded in $k$ periods AND regular payments also occur $k$ times a year, the multiplier per period is $R = \left(1 + \dfrac{r}{100k}\right)$. Now $n$ becomes the total number of periods: $n = k \times (\text{years})$.
EXAMPLE 8
An amount of 1000 euros is invested at 12% compounded monthly. An extra payment of 300 euros is added at the end of every month. Find the value after 7 years.
Solution:
The period is monthly, so $k=12$.
$$R = \left(1 + \dfrac{0.12}{12}\right) = 1.01$$
The total number of periods over 7 years is $n = 7 \times 12 = 84$. The value is:
$$FV = 1000 \times 1.01^{84} + 300 \times \left( \dfrac{1.01^{84} - 1}{1.01 - 1} \right) \approx 41508.41$$
5. Annuity & Amortization
An amount $PV$ is invested, but an amount $PMT$ is regularly withdrawn. The geometric series representing the payments is simply subtracted from the growing present value:
$$FV = \left[ PV \times R^n \right] - PMT \times \left( \dfrac{R^n - 1}{R - 1} \right)$$
EXAMPLE 9
An amount of 1000 euros is invested at 12% compounded monthly. A withdrawal of 150 euros is made at the end of each year.
Solution:
Find the effective annual ratio since withdrawals are yearly but interest is monthly:
$$R = \left(1 + \dfrac{0.12}{12}\right)^{12} = 1.01^{12} \approx 1.12682503$$
The value of the investment after 7 years is:
$$FV = 1000 \times R^7 - 150\left(\dfrac{R^7 - 1}{R - 1}\right) = 761.22$$
Amortization: If the annual withdrawal is greater than the annual interest generated, $FV$ will eventually reach zero. Amortization refers to the time needed for this to occur. To find when the balance reaches zero, solve for $n$ where $FV = 0$:
$$1000 \times R^n = 150\left(\dfrac{R^n - 1}{R - 1}\right)$$
The algebraic solution yields $n \approx 15.64$. Therefore, the last withdrawal will take place after 16 years.