1.5 Percentage Change - Financial Applications
1. Exponential Growth and Decay
In problems with interest rates $r\%$ or population growth $r\%$, some initial quantity (present value $PV$) increases by $r\%$ (per year, per month, or per any period).
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:
2. Explanation for the Common Ratio $R$
This is in fact a geometric sequence with a common ratio $R = 1 + \dfrac{r}{100}$. Suppose an amount $PV = 2000$ is invested for $r=8\%$ per year:
- 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 of the following slight difference: the initial amount may be mentioned as the value of year 1 ($u_1$) or the value of year 0 ($PV$).
- Rate of increase: $12\%$
- Amount in year 1: $u_1 = 1000$
- In $2^{nd}$ year: $u_2 = 1000 \times 1.12$
- In $n^{th}$ year: $u_n = 1000 \times (1.12)^{n-1}$
- Rate of increase: $12\%$
- Present value: $PV = 1000$
- After 1 year: $FV_1 = 1000 \times 1.12$
- After $n$ years: $FV_n = 1000 \times (1.12)^n$
(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 $10^{\text{th}}$ box?
Solution:
EXAMPLE 3
(If the question is about 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:
-
Method A:
(Trial and error): Check several values for $n$:
For $n=11$, $FV = 4663.27$.
For $n=12$, $FV = 5036.34$.
Therefore, $n = \mathbf{12}$. -
Method B:
(Using logarithms):
Solve the exponential equation $2000(1.08)^n = 5000$ using logs. $$1.08^n = 2.5 \implies n = \dfrac{\log 2.5}{\log 1.08} \cong 11.9$$ Thus $n = \mathbf{12}$.
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)
This is exponential growth with $PV=800,000$ and $r=5.2\%$.
The population of the city after $7$ years is: $$FV = 800,000\left(1 + \dfrac{5.2}{100}\right)^7 \cong \mathbf{1,140,775}$$ -
(b)
The formula works for the past as well.
The population of the city $7$ years ago was: $$FV = 800,000\left(1 + \dfrac{5.2}{100}\right)^{-7} \cong \mathbf{561,022}$$ [Short explanation: for the future, multiply by $1.052$ every year; for the past, divide by $1.052$ every year, or otherwise multiply by $1.052^{-1}$ every year]. - (c) Solve the equation $FV = 2 \times 800,000$, which is: $$800,000\left(1 + \dfrac{5.2}{100}\right)^n = 1,600,000$$ Solving this yields $n \cong 13.7$. Therefore, the population doubles after $\mathbf{14}$ complete years.
3. Compounded Interest Compounded in $k$ Time Periods
Suppose that an initial amount $PV=1000$€ is invested with an 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 $FV$ formula takes the form:
It is interesting to see how the final amount varies after $5$ years, for the initial amount of $1000$ euros:
- a) Yearly: $FV = 1000 \times (1.12)^5 = \mathbf{1762}$
- b) Half-yearly: $FV = 1000 \times \left(1 + \dfrac{12}{100 \times 2}\right)^{2 \times 5} = 1000 \times (1 + 0.06)^{10} = \mathbf{1791}$
- c) Quarterly: $FV = 1000 \times \left(1 + \dfrac{12}{100 \times 4}\right)^{4 \times 5} = 1000 \times (1 + 0.03)^{20} = \mathbf{1806}$
- d) Monthly: $FV = 1000 \times \left(1 + \dfrac{12}{100 \times 12}\right)^{12 \times 5} = 1000 \times (1 + 0.01)^{60} = \mathbf{1817}$
4. Investment With Regular Payments
$PV$ is invested with an annual interest rate $r\%$, giving a ratio $R = 1 + \dfrac{r}{100}$. An extra payment $PMT$ is invested at the end of each year.
The value of the investment after $n$ years is given by:
Indeed, 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 $1^{\text{st}}$ $PMT$ is invested for $n-1$ years: $PMT \times R^{n-1}$
- The $2^{\text{nd}}$ $PMT$ is invested for $n-2$ years: $PMT \times R^{n-2}$
- $\dots$ 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 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: $FV = PMT \times \left( \dfrac{R^{n+1} - 1}{R - 1} \right)$. If the last payment is not included, one $PMT$ is simply subtracted from the total).
EXAMPLE 5
An initial amount of $1000$ euros and then an extra amount of $1000$ euros at the end of each year is invested with an interest rate of $12\%$ compounded yearly (so $R=1.12$). Find the value after 7 years.
Solution:
EXAMPLE 6
An amount of $1000$ euros is invested with an interest rate of $12\%$ compounded yearly (so $R=1.12$). An extra payment of $300$ euros is added at the end of every year. Find the value after 7 years.
Solution:
EXAMPLE 7 (Adjusting for Compounding Periods)
NOTICE: If the annual interest rate $r\%$ is compounded in $k$ periods per year, but payments are made yearly, then the effective ratio for 1 full year is $R = \left(1 + \dfrac{r}{100k}\right)^k$. The formulas for $FV$ above are still valid using this effective $R$.
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:
EXAMPLE 8 (Payments matching Compounding Periods)
NOTICE: If the annual interest rate $r\%$ is compounded in $k$ periods and the regular payments also take place $k$ times a year, then the multiplier per period is $R = \left(1 + \dfrac{r}{100k}\right)$. Now $n$ becomes the total number of periods, i.e. $n = k \times (\text{years})$.
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 every month. Find the value after $7$ years.
Solution:
5. Annuity - Amortization
An amount $PV$ is invested, but an amount $PMT$ is regularly withdrawn. The only difference is that now the geometric series representing the payments is subtracted from the growing present value:
EXAMPLE 9
An amount of $1000$ euros is invested with an interest rate of $12\%$ compounded monthly. A withdrawal of $150$ euros is made at the end of each year.
Solution:
In the example above, solve for $n$ the equation $FV = 0$: $$1000 \times R^n = 150\left(\dfrac{R^n - 1}{R - 1}\right)$$ The algebraic solution is $n \cong 15.64$, so the last withdrawal will take place after $16$ years.