#### Chapter 11 Language of Descriptive Statistics

Section 11.2 Frequency Distributions and Percentage Calculation

# 11.2.3 Calculation of Interest

In the calculation of interest, we distinguish between simple interest and compound interest. For simple interest, the interest is paid at the end of the interest period. For compound interest, the interest is also paid on the previously accumulated interest.
##### Info 11.2.7

When simple interest is applied, a quantity $K$ that increases by $p %$ every year will increase after $t$ years ($t\in ℕ$) to

${K}_{t}\mathrm{ }=\mathrm{ }K·\left(1+t·\frac{p}{100}\right).$

Note that $p$ itself can be a decimal, for example, for $p=2.5$, the percent value is $2.5%=0.025$.
##### Exercise 11.2.8
What is the final capital for an initial capital of $K=4,000$ EUR after an interest period of $t=10$ years, when simple interest is applied at a rate of $p=2.5%$ p. a.?
Answer: ${K}_{10}$$=$
EUR.

##### Exercise 11.2.9
What is the initial capital $K$ that had been deposited at the 1$\mathrm{st}$ of January, 2000 to get paid an end capital of ${K}_{12}=10,000$ EUR at the 31$\mathrm{st}$ of December, 2011 when simple interest is applied at a rate of $p=5%$ p. a.?
Answer: $K$$=$
EUR.

While simpleinterest is simply paid after an interest period, compound interest carries over into the next interest period, i.e. the interest will be added to the initial capital or will be capitalised:
##### Example 11.2.10
For a bank account at the end of an interest period, an initial capital of $1,000$ EUR is deposited at an interest rate of $8%$. After one year the deposit (in EUR) in the bank account is
• $1,000+\frac{1,000·8}{100}=1,000·\left(1+\frac{8}{100}\right)=1,000·1.08=1,080$.
• This deposit is invested for an additional year at the same interest rate of $8%$. Then, the deposit (in EUR) after two years is $1,080·1.08=1,000·1.{08}^{2}=1,000·{\left(1+\frac{8}{100}\right)}^{2}$.
• The deposit increases by a factor of $1.08$ per year. Hence, the deposit (in EUR) after $t$ years ($t\in ℕ{}_{0}$) is

$1,000·1.{08}^{t}\mathrm{ }=\mathrm{ }1,000·{\left(1+\frac{8}{100}\right)}^{t}.$

Thus, the compound interest is based on the following formula:
##### Info 11.2.11

A quantity $K$ that increases every year by an amount of $p %$ will have been increased after $t$ years ($t\in ℕ{}_{0}$) to

$K·{\left(1+\frac{p}{100}\right)}^{t} .$

Here, $1+\frac{p}{100}$ is called the growth factor for a growth of $p %$.

In an advert offering deposit accounts or loans, the interest is usually given as a rate per year, even if the actual interest period differs. This interest period is the time between two successive dates at which the interest payments are due. On a deposit account the interest period is one year, though it is becoming more common to offer other interest periods. For example, for short-term loans the interest is paid daily or monthly.
If a bank offers a yearly interest rate of $9 %$ with monthly interest payments, then at the end of each month $\left(\frac{1}{12}\right)·9%=0,75%$ of the capital will be credited.
##### Info 11.2.12

The yearly rate is divided by the number of interest periods to obtain the periodic rate (the interest rate per period).

Suppose an investment of ${S}_{0}$ EUR yields $p %$ interest per interest period. After $t$ periods ($t\in ℕ{}_{0}$) the investment will have been increased to

${S}_{t}\mathrm{ }=\mathrm{ }{S}_{0}·\left(1+r{\right)}^{t}\mathrm{ }\mathrm{ }\text{with}\mathrm{ }\mathrm{ }r\mathrm{ }=\mathrm{ }\frac{p}{100} .$

In every period the investment increases by a factor of $1+r$, and we say "the interest rate equals $p %$" or "the periodic rate equals $r$". Suppose interest is credited at a rate of $\frac{p}{n} %$ to the capital at $n$ different times, more or less evenly distributed over the year. Then, the capital is multiplied by a factor of

${\left(1+\frac{r}{n}\right)}^{n}$

every year. After $t$ years the capital has increased to

${S}_{0}·{\left(1+\frac{r}{n}\right)}^{n·t} .$

##### Example 11.2.13
A capital of $5,000$ EUR is deposited for $t=8$ years in a bank account at a yearly interest rate of $9 %$, where the interest is paid quarterly. The periodic rate $\frac{r}{n}$ here is

$\frac{r}{n}\mathrm{ }=\mathrm{ }\frac{0.09}{4}\mathrm{ }=\mathrm{ }0.0225 ,$

and for the number of periods $n·t$ we have $n·t=4·8=32$. Thus, after $t=8$ years the deposit has increased to

$5000·\left(1+0.0225{\right)}^{32}\mathrm{ }\approx \mathrm{ }10190.52\mathrm{ }\text{EUR} .$

##### Exercise 11.2.14
A capital of ${K}_{0}=8,750$ EUR is deposited for $t=4$ years at an interest rate of $p=3,5%$ p. a., and the interest is capitalised.
1. After one year the amount of capital is ${K}_{1}$$=$ .
2. After two years the amount of capital is ${K}_{2}$$=$ .
3. After three years the amount of capital is ${K}_{3}$$=$ .
4. The final capital is ${K}_{4}$$=$ .
Specify all values rounded mathematically to the second fractional digit. Round only after carrying out the calculations. For these calculations, you are allowed to use a calculator.

A consumer who wants to take out a loan always faces several offers from competing banks. Thus, it is extremely useful to compare the different offers.
##### Example 11.2.15
Let us consider an offer providing an yearly interest rate of $9%$, where the interest is charged at a monthly ($12$ times per year) rate of $0.75%$. If no interest is paid off in the meantime, the initial debt will increase to

${S}_{0}·{\left(1+\frac{0.09}{12}\right)}^{12}\mathrm{ }\approx \mathrm{ }{S}_{0}·1.094$

after one year. The interest to be paid off is approximately

$1.094·{S}_{0}-{S}_{0}\mathrm{ }=\mathrm{ }0.094·{S}_{0} .$

As long as no interest is paid off, the debt will increase at a constant rate which is approximately $9.4%$ per year. This is why we may speak of an "effective" annual interest rate. In the example above the effective annual interest rate is $9.4%$.
##### Info 11.2.16

If interest is paid $n$ times per year at a periodic rate of $\frac{r}{n}$ per period, then the effective annual interest rate $R$ is defined by

$R\mathrm{ }=\mathrm{ }{\left(1+\frac{r}{n}\right)}^{n}-1 .$