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).

Increasing Case (Future Value): The future value $FV$ after $n$ periods is given by
$FV = PV\left(1 + \frac{r}{100}\right)^n$
Decreasing Case (Depreciation): If the present value $PV$ decreases by $r\%$ per period, the formula takes the form:
$FV = PV\left(1 - \frac{r}{100}\right)^n$

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, $PV = 2000$, $r = 8$, $n = 10$.
(a) $FV = PV\left(1 + \frac{r}{100}\right)^n = 2000\left(1 + \frac{8}{100}\right)^{10} = 2000(1.08)^{10} \cong \mathbf{4317.85}$
(b) $FV = PV\left(1 - \frac{r}{100}\right)^n = 2000\left(1 - \frac{8}{100}\right)^{10} = 2000(0.92)^{10} \cong \mathbf{868.78}$

⚠️ NOTICE: Using GDC (Financial Mode)

Navigate to MENU $\rightarrow$ Financial $\rightarrow$ Compound Interest.
For Example 1(a): $n=10, \quad I\%=8, \quad PV=-2000$ (use "-" because the money is paid out).
Leave $PMT=0, \quad P/Y=1, \quad C/Y=1$.
Press $FV$ to obtain $\mathbf{4317.85}$. For question (b), set $I\%=-8$ to obtain $\mathbf{868.78}$.

2. Explanation for the Common Ratio $R$

This is in fact a geometric sequence with common ratio $R = 1 + \frac{r}{100}$. Suppose an amount $PV = 2000$ is invested for $r=8\%$ per year:

In our example In general
Present Value 2000 PV
Interest $2000 \times \frac{8}{100}$ $PV \times \frac{r}{100}$
After 1 year $2000 + 2000 \times \frac{8}{100}$
$= 2000\left(1 + \frac{8}{100}\right)$		 $PV + PV\frac{r}{100}$
$= PV\left(1 + \frac{r}{100}\right)$
  • If an amount increases by $r\%$, it is multiplied by $R = 1 + \frac{r}{100}$.
  • After two years, it is multiplied again by $R$, resulting in $FV = PV\left(1 + \frac{r}{100}\right)^2$.
  • Thus, after $n$ years: $FV = PV\left(1 + \frac{r}{100}\right)^n$.

Translations of $r\%$ into $R$:

$r\%$ Increasing ($R = 1 + \frac{r}{100}$) Decreasing ($R = 1 - \frac{r}{100}$)
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 value of year 1 ($u_1$) or value of year 0 ($PV$).

PROBLEM 1
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: $u_n = 1000 \times (1.12)^{n-1}$
PROBLEM 2 (investment)
Rate of increase 12%
Present value: $PV = 1000$
After 1 year: $1000 \times 1.12$
After $n$ years: $FV = 1000 \times (1.12)^n$

(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 10th box?

Solution:

Here $u_1 = 100$ and $r = 1.10$.
Thus, $u_{10} = 100(1.10)^9 \cong \mathbf{235.8}$.
(This is in fact the FV formula, but evaluating that after 9 boxes $FV = 100(1.10)^9$).

EXAMPLE 3 (if the question is about 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: $FV = 2000\left(1 + \frac{8}{100}\right)^n > 5000$

Method A (trial and error): Check several values for $n$ by GDC:
For $n=11$, $FV = 4663.27$.
For $n=12$, $FV = 5036.34$.
Therefore, $n = \mathbf{12}$.
Method B (by using GDC-Financial mode):
Set $I\%=8, PV=-2000, FV=5000$. Keep $PMT=0, P/Y=1, C/Y=1$. Press F1: [n] to obtain $n=11.9$. Since complete years are required, the first integer above 11.9 is accepted, which is $n = \mathbf{12}$.
Method C (by using Solver in the GDC):
Solve the equation $FV=5000$, that is $2000(1.08)^n = 5000$. The solution is $n \cong 11.9$. Thus $n = \mathbf{12}$.
Method D (by using logarithms):
Solve the exponential equation $2000(1.08)^n = 5000$ using logs. The solution is $n = \frac{\log 2.5}{\log 1.08} \cong 11.9$. 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 $FV = 800,000\left(1 + \frac{5.2}{100}\right)^7 \cong \mathbf{1,140,775}$.
(b) The formula works for the past as well.
The population of the city 7 years ago was $FV = 800,000\left(1 + \frac{5.2}{100}\right)^{-7} \cong \mathbf{561,022}$.
[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 $800,000\left(1 + \frac{5.2}{100}\right)^n = 1,600,000$.
Using a GDC finds $n = 13.7$. 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 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:

$FV = PV\left(1 + \frac{r}{100k}\right)^{kn}$

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: $FV = 1000 \times (1 + 0.06)^{2 \times 5} = \mathbf{1791}$
  • c) Quarterly: $FV = 1000 \times (1 + 0.03)^{4 \times 5} = \mathbf{1806}$
  • d) Monthly: $FV = 1000 \times (1 + 0.01)^{12 \times 5} = \mathbf{1817}$

NOTICE for GDC-Financial mode: The number of periods is denoted by C/Y. Check the values of FV for $n=5, I\%=12, PV=-1000, PMT=0, P/Y=1$ while setting $C/Y = 1, 2, 4, 12$ respectively.

4. Investment With Regular Payments

$PV$ is invested with an annual interest rate $r\%$: $R = 1 + \frac{r}{100}$. An extra $PMT$ is invested at the end of each year.

The value of the investment after $n$ years is given by:

$FV = \left[ PV \times R^n \right] + PMT \times \left( \frac{R^n - 1}{R - 1} \right)$

Indeed, this can be derived by evaluating each component:

  • $PV$ is invested for $n$ years: $PV(1 + \frac{r}{100})^n = PV \times R^n$
  • $1^{st}$ $PMT$ is invested for $n-1$ years: $PMT \times R^{n-1}$
  • $2^{nd}$ $PMT$ is invested for $n-2$ years: $PMT \times R^{n-2}$
  • The last payment is $PMT$

The sum of the last $n$ terms is a G.S. with $u_1 = PMT$, ratio $= R$. Thus $S_n = PMT \times \left( \frac{R^n - 1}{R - 1} \right)$.

(If each payment is equal to the present value ($PMT = PV$), a G.S. of $n+1$ terms is obtained: $FV = PMT \times \left( \frac{R^{n+1} - 1}{R - 1} \right)$). If the last payment is not included, one $PMT$ is simply subtracted.

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 12% compounded yearly (so $R=1.12$).

The value of the investment after 7 years is $FV = 1000 \times \left( \frac{1.12^8 - 1}{1.12 - 1} \right) = \mathbf{12299.69}$.
By using GDC-Financial mode: $n=7, I\%=12, PV=-1000, PMT=-1000, P/Y=1, C/Y=1$. FV gives 12299.69.
Remark: if the last payment is not included we subtract one PMT: $Value = 12299.69 - 1000 = 11299.69$.

EXAMPLE 6

An amount of 1000 euros is invested with an interest rate 12% compounded yearly (so $R=1.12$). An extra payment of 300 euros is added at the end of every year.

The value of the investment after 7 years is $FV = 1000 \times 1.12^7 + 300 \times \left( \frac{1.12^7 - 1}{1.12 - 1} \right) = \mathbf{5237.38}$.
By using GDC-Financial mode: $n=7, I\%=12, PV=-1000, PMT=-300, P/Y=1, C/Y=1$. FV gives 5237.38.

EXAMPLE 7 & 8 (Adjusting for Compounding Periods)

NOTICE: If the annual interest rate $r\%$ is compounded in $k$ periods then the ratio for 1 year is $R = \left(1 + \frac{r}{100k}\right)^k$. The formulas for FV above are still valid.

Example 7: An amount of 1000 euros is invested with an interest rate 12% compounded monthly. An extra payment of 300 euros is added at the end of each year.
$R = \left(1 + \frac{0.12}{12}\right)^{12} = 1.01^{12} = 1.12682503$.
The value of the investment after 7 years is $FV = 1000 \times R^7 + 300 \times \left( \frac{R^7 - 1}{R - 1} \right) = \mathbf{5397.73}$.
Example 8: NOTICE: If the annual interest rate $r\%$ is compounded in $k$ periods and the regular payments also take place in $k$ periods then $R = \left(1 + \frac{r}{100k}\right)$. Now $n$ is the total number of periods, i.e. $n = k \times (years)$.
An amount of 1000 euros is invested with an interest rate 12% compounded monthly. An extra payment of 300 euros is added at the end of every month.
$R = \left(1 + \frac{0.12}{12}\right) = 1.01$.
The value of the investment after 7 years (so $n = 7 \times 12 = 84$) is $FV = 1000 \times 1.01^{84} + 300 \times \left( \frac{1.01^{84} - 1}{1.01 - 1} \right) = \mathbf{41508.41}$.

5. Annuity - Amortization

An amount $PV$ is invested but an amount $PMT$ is regularly withdrawn. The only difference is that now the part of the payments is subtracted:

$FV = \left[ PV \times R^n \right] - PMT \times \left( \frac{R^n - 1}{R - 1} \right)$

EXAMPLE 9

An amount of 1000 euros is invested with an interest rate 12% compounded monthly. A withdrawal of 150 euros is made at the end of each year.

Solution:

$R = \left(1 + \frac{0.12}{12}\right)^{12} = 1.01^{12} = 1.12682503$.
The value of the investment after 7 years is $FV = 1000 \times R^7 - 150\left(\frac{R^7 - 1}{R - 1}\right) = \mathbf{761.22}$.
Amortization: If (annual withdrawal) > (the annual interest), $FV$ will be zeroed out after a certain time. Amortization refers to the time needed for this to occur.
In the example above, solve for $n$ the equation $FV = 0$, i.e. $1000 \times R^n = 150\left(\frac{R^n - 1}{R - 1}\right)$.
The solution is 15.64, so the last withdrawal will take place after 16 years. By using GDC-Financial mode, set $FV = 0$ and press $n$ to obtain $n=15.64$, so $n=16$.