Compound Interest

Tags: 
simple interest
compound interest
present amount
future amount
nominal rate
interest rate
continuous compounding

In compound interest, the interest earned by the principal at the end of each interest period (compounding period) is added to the principal. The sum (principal + interest) will earn another interest in the next compounding period.
 

Consider $1000 invested in an account of 10% per year for 3 years. The figures below shows the contrast between simple interest and compound interest.
 

At 10% simple interest, the $1000 investment amounted to $1300 after 3 years. Only the principal earns interest which is $100 per year.
 

000-simple-interest.gif

 

At 10% compounded yearly, the $1000 initial investment amounted to $1331 after 3 years. The interest also earns an interest.
 

000-compound-interest.gif

 

Elements of Compound Interest
$ P $ = principal, present amount
$ F $ = future amount, compound amount
$ i $ = interest rate per compounding period
$ r $ = nominal annual interest rate
$ n $ = total number of compounding in t years
$ t $ = number of years
$ m $ = number of compounding per year

$ i = \dfrac{r}{m} $   and   $ n = mt $

 

Future amount,

$ F = P(1 + i)^n $   or   $ F = P\left(1 + \dfrac{r}{m} \right)^{mt} $

The factor   $ (1 + i)^n $   is called single-payment compound-amount factor and is denoted by   $ (F/P, \, i, \, n) $.
 

Present amount,

$ P = \dfrac{F}{(1 + i)^n} $

The factor   $ \dfrac{1}{(1 + i)^n} $   is called single-payment present-worth factor and is denoted by   $ (P/F, \, i, \, n) $.
 

Number of compounding periods,

$ n = \dfrac{\ln (F / P)}{\ln (1 + i)} $

 

Interest rate per compounding period,

$ i = \sqrt[n]{\dfrac{F}{P}} - 1 $

 

Values of   $ i $   and   $ n $
In most problems, the number of years   $ t $   and the number of compounding periods per year   $ m $   are given. The example below shows the value of   $ i $   and   $ n $.

Example
Number of years,   $ t = 5 \text{ years} $
Nominal rate,   $ r = 18\% $

  • Compounded annually ($ m = 1 $)

    $ n = 1(5) = 5 $

    $ i = 0.18 / 1 = 0.18 $

  • Compounded semi-annually ($ m = 2 $)

    $ n = 2(5) = 10 $

    $ i = 0.18 / 2 = 0.09 $

  • Compounded quarterly ($ m = 4 $)

    $ n = 4(5) = 20 $

    $ i = 0.18 / 4 = 0.045 $

  • Compounded semi-quarterly ($ m = 8 $)

    $ n = 8(5) = 40 $

    $ i = 0.18 / 4 = 0.0225 $

  • Compounded monthly ($ m = 12 $)

    $ n = 12(5) = 60 $

    $ i = 0.18 / 12 = 0.015 $

  • Compounded bi-monthly ($ m = 6 $)

    $ n = 6(5) = 30 $

    $ i = 0.18 / 6 = 0.03 $

  • Compounded daily ($ m = 360 $)

    $ n = 360(5) = 1800 $

    $ i = 0.18 / 360 = 0.0005 $

 

Continuous Compounding (m → ∞)
In continuous compounding, the number of interest periods per year approaches infinity. From the equation
$ F = \left( 1 + \dfrac{r}{m} \right)^{mt} $
 

when   $ m \to \infty $,   $ mt = \infty $,   and   $ \dfrac{r}{m} \to 0 $.   Hence,
$ \displaystyle F = P \lim_{m \to \infty}\left( 1 + \dfrac{r}{m} \right)^{mt} $
 

Let   $ x = \dfrac{r}{m} $.   When   $ m \to \infty $,   $ x \to 0 $,   and   $ m = \dfrac{r}{x} $.

$ \displaystyle F = P \lim_{x \to 0}(1 + x)^{\frac{r}{x}t} $

$ \displaystyle F = P \lim_{x \to 0}(1 + x)^{\frac{1}{x}rt} $
 

From Calculus,   $ \displaystyle \lim_{x \to \infty}(1 + x)^{1/x} = e $,   thus,

$ F = Pe^{rt} $

 

Submitted by Romel Verterra on August 27, 2011 - 7:32am
  • 10148 reads
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