COMPOUND INTEREST

by Steven Johnson

Simple Versus Compound Interest

With simple interest, interest is paid on the principal only, not on any past interest that has been collected. Many investments or deposits, however, do earn interest on not only the principal but also on the interest that has been previously earned by the account. This type of interest is called compound interest.
	To see how compound interest is calculated we will use the simple interest formula and an example investment of $1,000 invested at 10% interest calculated at the end of each year.  Let us find the amount of compound interest at the end of three years.  Using the formula for simple interest, I = PRT

Interest = 1000 * .10 * 1

At the end of the first year, the account contains

Principal + Interest = $1,000 + $100 = $1,100

The interest for the second year is paid on the original deposit plus the interest paid on the first year.

Interest = 1100 * 0.10 * 1

The amount earned at the end of the second year.  

Principal + Interest = $1,100 + $110 = $1,210

The interest for the third year is paid on the amount paid at the end of the second year.

Interest = 1210 * 0.10 * 1

The total amount on hand at the end of the third year is 

Amount at the end of second year + Interest = $1,210 + $121 = $1,331

So the total amount on deposit at the end of the third year is $1,331, this is called the compound amount or future amount.  We use the letter, M to symbolize this amount, some times called the maturity value.

	The interest earned during the time of the investment is found by the following formula.
	Interest = Compound amount - Original principal

Interest = 1331 - 1000

The simple interest earned on $1,000 at 10% for three years is $300.  Therefore, the compound interest is $31 more than the simple interest. The advantage of compound interest is that it increases greatly as the number of years increases as is shown in the following graph.

Clear[t]
lissimp1=ListPlot[Table[{t,1000 + (1000 * 0.10 * t)},{t,1,20, 1}],PlotStyle->{PointSize[0.03],RGBColor[0,0,1]},PlotRange->{0,6000},AxesLabel->{"Number of Years","Maturity Value In Thousands of Dollars"},Ticks->{{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20},{0,1000,2000,3000,4000,5000,6000}},DisplayFunction->Identity];

Clear[t]
lissimp2=ListPlot[Table[{t,1000 + (1000 * 0.10 * t)},{t,1,20,1}],PlotStyle->{PointSize[0.03]},PlotRange->{0,6000},AxesLabel->{"Number of Years", "Maturity Value In Thousands of Dollar"}, Ticks->{{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20},{0,1000,2000,3000,4000,5000,6000}},PlotJoined->True,DisplayFunction->Identity];

Clear[n]
liscomp1=ListPlot[Table[{n,1000(1+0.10)^n},{n,1,20,1}],PlotStyle->{PointSize[0.03],RGBColor[0,1,0]},DisplayFunction->Identity];

Clear[n]
liscomp2=ListPlot[Table[{n,1000(1+0.10)^n},{n,1,20,1}],PlotStyle->{PointSize[0.03],RGBColor[0.56,0.14,0.42]},PlotJoined->True,DisplayFunction->Identity];

Show[lissimp1,lissimp2,liscomp1,liscomp2,DisplayFunction->$DisplayFunction];

Computing Compound Interest

The method for finding compound interest can be simplified by multiplying the principal by the expression
		(1+rate of interest)
where rate of interest is expressed as a decimal value of the interest rate per year. Then for the example, we can multiply by the principal at the beginning by 1 + 0.10=1.10.  So 
(((1000*1.10)*1.10)*1.10) = the compounded amount for three years compounded yearly

M = 1000 * 1.10
M = M * 1.10
M = M * 1.10

Or the compounded interest could be calculated as

	M = 1000 * 1.10 * 1.10 * 1.10 = 1000 * (((1.10))^3)  
	
This illustrates the more general form of the compound interest formula

	M = P(1 + ((i))^n)   and  I = M - P
	
where M is the maturity  value, P is the beginning principal,  i is the interest rate per period compounded, and n is the number of periods compounded. For example, if the interest rate per year is 6% and the loan is compounded semiannually, then in the formula i = .06/2 = .03.
Also if the amount is compounded semiannually for 5 years, then n = 5 * 2 = 10 or the amount is compounded 10 times.
But if it is compounded quarterly, then n = 5 * 4 = 20.

Solve the following problem to find the compound interest:
	Bill Jones inherits $20,000 which he deposits in a retirement account that pays interest compounded monthly (1) at 6% and then (2) at 8%.  How much will he have after 25 years in both cases?

Solutions:

P = 20000
n = 25 * 12
i = .06 / 12
M = P(1 + i)^n

So the maturity value M = $89,299.40.  The amount of interest, Int = M - P
then

M = 89299.40
P = 20000
Int = M - P

The interest collected on the investment is $69,299.40.

Continuous Compounding

Is there an advantage to compounding more and more frequently?  We can see the answer to this question in the following table that compares the amount of interest collected as the number of periods of compounding increases.  We will use an investment of $1,000 at 12% per year for 10 years and we will vary the frequency of compounding from annually to every minute then the table compares the interest.

({{\(Frequency\ of

            	Compounding\), \(Number\ of

            	Periods\), Interest}, {Annually, 10, $2, 105.85}, {Semiannually, 20, $2, 207.14}, {Quarterly, 40, $2, 262.04}, {Monthly, 120, $2, 300.39}, {Daily, 3, 650, $2, 319.46}, {Hourly, 87, 600, $2, 320.09}, {Every Minute, 5, 256, 000, $2, 320.11}})

As you can see, there really is not much advantage to increase the frequency of compounding after a desired number.  But since we do sometimes need to compound almost every instant we find the following formula will do just that, compound a every instant or continuously. 
The formula for Continuously Compounding is   M =  P (e^(ni )), where M is the maturity value, P is the principal, n is the number of years, and i is the interest rate per year. The number e is approximately equal to 2.7182818.  Continuous Compounding is used by some government agencies and financial institutions.  For the above investment, we can use this formula to calculate the continuously compounded maturity value as follows:

Clear[P,n,i]
P = 1000
i = 0.12
n = 10
M = P* E^(n*i)

Interest collected would be I = M - P or 3320.12 -1000 or $2,320.12 which is 1 penny more than compounding every minute (from the above table).

Interest = M - P

Therefore, it is easier to compute the interest continuously compounded.

oneplot=Plot[1000*E^(.12x),{x,1,10},PlotStyle->{RGBColor[0,0,1],Thickness[0.01]},PlotRange->{0,3500},AxesLabel->{"Number of Years","Maturity Value"}];

twoplot=Plot[1000(1 + 0.12/n)^10n,{n,1,8760},PlotStyle->{Red,Thickness[0.01]},AxesLabel->{"Frequency","Maturity Value"}];