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We borrow money from banks, or invest into saving accounts.

In all cases, there is some interest.

One always returns more than one borrows.

How much more? --By a certain percentage of the principal. That is what interest rate is all about.Assume the interest rate to be 5%.

That means you invest \( \$ 1,000 \), and a year later you have your \(\ $ 1,000\) plus 5% of your \(\$1,000\)

That is \( 1000 + 0.05 \times 1000 = 1050 \) dollars

You may continue that way with extra \(\$ 50 \) every year:

1000 +50=1050 after one year,

1050 +50=1100 after two years,

1100 +50=1150 after three years ...

That does not seem right, however!

After first year you better take your \(\$1050\), and deposit

it to another bank with the same 5% interest rate.

Then, after the second year, instead of \(\$ 1100 \), you have

your \( \$ 1050 \) plus 5% of this \( \$ 1050 \), not just of \( \$ 1000 \)!

That is \( 1050 + 0.05 \times 1050 = 1102.5 \),

which is \( \$ 2.5 \) more.

Do you really need to go to another bank? No, that same will work perfectly. That is compound interest.

In fact, compound interest makes more sense than simple:

During the second year your \( \$ 1,050 \) are in bank,

and the bank owes you 5% out of that sum, not merely of \(\$ 1,000 \).

The difference of \(\$2.5 \) is small, but that grows with years drastically.

It is easy to calculate with simple interest, but how to calculate with compound interest?

\(A = P \times (1+APR)^Y \)

where

\(A\) = accumulated balance

\(P\) = starting principal

\(APR\) = annual percentage rate (in decimal)

\(Y\) = number of years

\(A = P \times (1+APR)^Y \)

Invest \(\$ 1,000 \) into an account with an interest rate of 5%

How much do you have 20 years later?

Solution. \(A = 1000 \times (1+0.05)^{20} = \$ 2653.30 \)

How much is that better than simple interest?

You get extra \(\$50\) every year, and thus 20 years later \( 1000+20 \times 50 = \$ 2000 \)

The difference has become substantial.

Why they pay interest once a year?

Assume you deposit \( \$ 1,000 \), and the APR is 5%

A year later you have \( \$ 1,050 \).

May they do 2.5% after half a year, and again 2.5% after another half a year?

After half a year you have \(1000 + 0.025 \times 1000 = 1025 \),

and after another half a year \(1025 + 0.025 \times 1025 = 1050.63 \).

The difference is 63 cents only.

This difference may be bigger if it is more than twice a year.

\(A = P \times (1+\frac{APR}{n})^{(nY)} \)

where

\(A\) = accumulated balance

\(P\) = starting principal

\(APR\) = annual percentage rate (in decimal)

\(n\) = number of compounding periods per year

\(Y\) = number of years

\(A = P \times (1+\frac{APR}{n})^{(nY)} \)

Invest \(\$ 1,000 \) into an account with an interest rate of 5% compounded monthly

How much do you have 20 years later?

Solution. \(A = 1000 \times (1+\frac{0.05}{12})^{12 \times 20} = \$ 2712.64 \)

Thus we get \(\$ 2712.64 \) with monthly compounding instead of \(\$ 2653.30 \) we had with yearly compounding

One may try to compound even more frequently:

Every week, every day, every hour, every minute ...

The formula still works, but becomes inconvenient.

There is another formula instead, which assumes as frequent compounding as one wants.

This new formula is even simpler.

\(A = P\times e^{(APR \times Y)} \)

where

\(A\) = accumulated balance

\(P\) = starting principal

\(APR\) = annual percentage rate (in decimal)

\(Y\) = number of years

\(e \approx 2.71828\) is a special number, and the calculators know that number

\(A = P \times e^{(APR \times Y)} \)

Invest \(\$ 1,000 \) into an account with an interest rate of 5% with continuous compounding.

How much do you have 20 years later?

Solution. \(A = 1000 \times e^{0.05 \times 20} = \$ 2718.28 \)

Type of interest | Accumulated balance |

Simple interest | 2000 |

Compounded yearly | 2653.30 |

Compounded monthly | 2712.14 |

Compounded continuously | 2718.28 |

APY is the actual percentage by which the balance increases during a year

APY is equal to APR if the interest is compounded yearly, but bigger than APR otherwise

Example. For continuous compounding with APR=20%, what is the APY?

Solution. A \(\$ 1,000 \) investment over a year will become

\(A = 1000 \times e^{0.2 \times 1} = 1221.4\)

The relative difference is \(\frac{1221.4-1000}{1000}=0.221 \), and we see that APY is 22.1%.

How much money should I deposit now in order to have a certain amount in the future?

That is about solving the formulas for the principal, P

\(A = P \times (1+APR)^Y \) implies \(P= \frac{A}{(1+APR)^Y} \)

\(A = P \times (1+\frac{APR}{n})^{(nY)} \) implies \(P= \frac{A}{(1+\frac{APR}{n})^{(nY)} }\)

\(A = P \times e^{(APR \times Y)} \) implies \(P= \frac{A}{e^{(APR \times Y)}} \)