Simple Interest $(SI)$
Whenever we borrow a certain sum of money, we pay back the original amount accompanied with some additional amount. The borrowed (or lent) sum is called
principal and the additional amount that we pay is called the interest.
The additional amount that we pay is in a way charge or fees for borrowing the money.
Based on the way we calculate interest, it can be either Simple interest or Compound interest.
The formula to calculate Simple interest is $$SI = \frac{P × N × R}{100}$$
where
P is Principal (Borrowed amount)
N is Time period for which the money is borrowed
R is Rate of Interest
The total amount that is repaid is calculated as
$$A = P + SI = P + \frac{P × N × R}{100}$$
E.g. A sum of ₹ 25,000 is borrowed for a period of 3 years at the rate of interest of 9%. Find the simple interest that needs
to be paid by the borrower to the lender.
Solution:
As we have seen above simple interest is calculated as $$SI = \frac{P × N × R}{100}
= \frac{25000 × 3 × 9}{100} = 6750$$
The amount to be repaid is $$A = 25000 + \frac{25000 × 3 × 9}{100} = 31750$$
The interest can be paid out at the end of the 3 years or it can be paid at fixed interval like each year or each quarter or even every month.
This depends on the agreement between the borrower and the lender.
In the above example ₹ 6750 is the interest for 3 years.
If the interest is to be paid every year, borrower will pay ₹ $\frac{6750}{3} = 2250$ to the lender each one year for 3 years.
If the interest is to be paid every month, ₹ $\frac{6750}{36} = 187.5$ is paid as interest by borrower to lender each month for 36 months.
Simple Interest Solved Examples
Compound Interest $(CI)$
Compound interest is the addition of interest to the principal of a loan or deposit, i.e. it involves calculating interest on interest.
Instead of paying interest every period, it is reinvested and added to the principal. This results in increased principal for the next period.
so that interest in the next period is then earned on the principal sum plus previouslyaccumulated interest.
In the above example we calculated simple interest for a sum of ₹ 25,000 for a period of 3 years at the rate of interest of 9%.
Let us now calculate compound interest for the same principal and rate with compounding period/frequency as annual.
For the first year, Interest for first year $$I1 = \frac{25000 × 1 × 9}{100} = 2250$$
For the second year interest calculation, the interest calculated for the first year is added back to the principal.
So the new principal $= 25000 + 2250 = 27250$
Interest for second year $$ I2 = \frac{27250 × 1 × 9}{100} = 2452.5$$
For the third year interest calculation, the interest calculated for the second year is added back to the principal.
So the new principal $= 27250 + 2452.5 = 29702.5$
Interest for third year $$I3 = \frac{29702.5 × 1 × 9}{100} = 2673.225$$
Total interest for 3 years = $I1 + I2 + I3 = 2250 + 2452.5 + 2673.225 = 7375.725$
Notice that this interest is much more than Simple interest calculated.
The direct formula to calculate compound interest is $$ CI = P × (1 + \frac{r}{n})^{nt}  P = 25000 × (1 + \frac{9}{100})^{1 × 3}  25000 = 7375.725$$
where
P is Principal (Amount borrowed or lent)
n is the number of times interest is compounded per year (n = 1 in this example)
t is the number of years the money is borrowed or lent (t = 3 in this example)
r is annual rate of interest
If the compounding of interest happens only once a year (annually), the above formula simplifies to
$$ CI = P × (1 + \frac{r}{100})^{t}  P$$
If the compounding of interest happens twice a year (biannually),
$$ CI = P × (1 + \frac{r}{2 × 100})^{2 × t}  P$$
If the compounding of interest happens four times a year (quarterly),
$$ CI = P × (1 + \frac{r}{4 ×100})^{4 × t}  P$$
In some cases, if there is increase in the first year and then decrease in the second year (both at rate $r$), the above formula can be
rewritten as
$$ CI = \left \{ P × (1 + \frac{r}{100}) × (1  \frac{r}{100}) \right \}  P $$
Note that the rate can be different in the first year and subsequent years.
Compound Interest Solved Examples
Continous Compounding
Typically, as we know, compounding of interest happens on a finite number of periods e.g. yearly, monthly, quarterly etc. However, continous compounding is an
extreme case where your principal constantly earns interest and the interest keeps earning on the interest earned.
Hypothetically, with continuous compounding, interest is calculated and added to the principal every infinitesimally small instant.
While this is not possible in practice, the concept of continuously compounded interest is important in finance.
Suppose you put ₹ 1 in a bank. Assume that the bank pays 100% interest a year, and this is credited to your account at the end of a year.
The interest is calculated as
$$ FV = PV (1 + r)^n = 1 (1+ \frac{100}{100})^1 = 1 (1+1)^1 = 2 $$
where
FV = Future Value
PV = Present Value or Principal
r = Rate of Interest
n = Frequency of Interest calculation
Notice that the frequency is 1 in this case as the interest if given out only once in a year.
Now, assume that the Bank agrees to pay interest twice a year. So instead of 100% at the end of the year, you are going to get 50% every half year.
$$ FV = PV (1 + r)^n = 1 (1+\frac{50}{100})^2 = 1 (1+\frac{1}{2})^2 = 1.5^2 = 2.25 $$
The frequency is 2 in this case as the interest if given out twice a year.
If the Bank now agrees to pay interest every month.
$$ FV = PV (1 + r)^n = 1 (1+1/12)^{12} = 2.61303529022468 $$
If the Bank calculates interest every day,
$$ FV = PV (1 + r)^n = 1 (1+1/365)^{365} = 2.71456748202201 $$
What we are doing here, is to increase the frequency of interest calculation  from yearly to half yearly to monthly to daily etc.
It is obvious that compounding more frequently results in more money in the bank. If we keep on increasing the frequency of interest calculation
like this, the frequency becomes very large say that it tends to infinity. This type of compounding is called as Continous compounding.
Continous compounding results in increased returns (FV) but the rate of increase gradually slows down and eventually hits a limit
which is nothing but $e$.
Frequency

Value of Frequency

Future Value

Annually

$1$

$2.000000000000000$

Semi annually

$2$

$2.250000000000000$

Monthly

$12$

$2.613035290224680$

Daily

$365$

$2.714567482022010$

Hourly

$365 × 24$

$2.718126691617420$

Every minute

$365 × 24 × 60$

$2.718279242588710$

Every second

$365 × 24 × 60 × 60$

$2.718281781302460$


The x axis represents the frequency and y axis represents the FV. See how the graph climbs fast initially but then the
increase gradually slows down and almost becomes flat. It would settle down finally at $e$.

The FV does not go beyond $e$.
Why? To answer this question, we need to evaluate
$$ \sum_{n=0}^\infty \frac{1}{n!}$$
This above summation can also be written as follows
$$ = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + \frac{1}{8!} +... $$
This series is convergent and leads to number $e$.
It was the great mathematician Leonhard Euler who discovered the number e and calculated its value to 23 decimal places.
$e$ is a number given by the structure of nature. It tells the rate of increase (growth) or decrease (decay) of a particular thing
with respect to itself.