Compound Interest
By
Ms. Karen Overman
using Tan’s 5th edition Applied Calculus for the
managerial , life, and social sciences text
Simple Interest
You may recall from previous math courses the formula
for simple interest, I = Prt, where I is the interest earned
P is the principal or the amount invested, r is the interest
rate and t is the time in years.
The amount after the interest in added, the accumulated
amount is the sum of the principal and the interest or
A = P+I = P+Prt = P(1+rt).
Getting to Compound Interest
Suppose we have an account where the simple interest is
added in each year and then that money also earns interest.
This is called compound interest.
Let’s look at the pattern for several years.
1st year: A(1) = P(1+rt)
The A(1) denotes the amount
after 1 year, etc…
2nd year: A(2) = A(1) (1+rt)
But A(1) = P(1+rt), so …
A(2) = P(1+rt) (1+rt)
So, A(2) = P(1+rt)²
3rd year: A(3) = A(2) (1+rt)
A(3) = P(1+rt)² (1+rt)
A(3) = P(1+rt)³
Are you starting to see a pattern?
A(1) = P(1+rt)
A(2) = P(1+rt)²
A(3) = P(1+rt)³
So, what would be the accumulated amount after n years?
An   P1  rt 
n
It is also possible to have the interest compound more than
once a year, say m times a year. If the interest rate for the
year is r and the interest is compounded m times a year, then
we replace r with i, where i = r/m. And for n to keep the
meaning of how many times it was compounded, n = mt.
This leads us to …
Compound Interest Formula
r
A  P1  i  , where i 
and n  mt
m
n
A = the accumulated amount
P = the principal (amount invested)
r = the yearly interest rate
m = the number of times it is compounded in a year
t = the time in years
Let’s look at an example:
Find the amount in an account after 4 years if $5000 is invested
at 6% per year and is compounded a.) quarterly and b.) monthly.
It helps to identify what is given and what you need to find.
P = 5000, r = 0.06, t = 4 and in a.) m = 4 in b.) m = 12
We are trying to find A in each case.
Now, just substitute into the compound interest formula.
a.)
r 

A  P1  
 m
mt
 0.06 
 50001 

4 

A  50001.01516
A  6344.93
44
Example continued…
 0.06 
A

5000
1 

b.)
12 

124
 50001.00548  6352.45
So after 4 years, if the $5000 is compounded quarterly
(every 3 months) or 4 times a year the accumulated amount
is $6344.93. If the $5000 is compounded monthly or 12
times a year the accumulated amount is $6352.45.
You can see from this example if the yearly interest rate is
constant, the accumulated amount is larger for larger
numbers of compoundings in a single year.
You can see from the previous example that the amount of
money earned depends on the number of times it is
compounded. To serve as a method of comparing interest
rates, the effective rate is the simple interest rate that
would give you the same accumulated amount.
Recall the accumulated amount of simple interest is A=P(1+rt).
Thus the accumulated amount with the effective rate , r e
in 1 year would be A  P1  re  (assuming t = 1).
The accumulated amount if it were compound interest over
1 year would be A  P1  r 
m

m
(assuming t = 1).
Set the accumulated amount with the effective rate (or
simple interest) equal to the accumulated amount from
compounded interest.
r 

P1  re   P1  
m

m
Now, solve for the effective rate.
r 

P1  re   P1  
m

r 

1  re  1  
m

r 

re  1  
m

m
Divide both sides by P .
m
m
1
Subtract1 from both sides.
The Effective Rate is given by
r 

re  1  
m

m
1
where r is the yearly interest rate and m is the number of
compoundings per year.
Let’s look at an example to make this a little more clear.
Let’s find the effective rates for the compound interest
problem we looked at before: $5000 invested for 4 years at
6% per year compounded a.)quarterly and b.)monthly.
r 

re  1  
m

a.) r = 0.06, m = 4
m
1
4
0.06 

4
re  1 
  1  1.015  1  0.613
4 

So, having an interest rate of 6% yearly and compounding it
quarterly is equivalent to having a simple interest rate of
6.13%.
This is what the effective rate does. It serves as a method
of comparing rates to one standard.
b.) r = 0.06, m = 12
0.06 

re  1 

12 

12
 1  1.00512  1  0.617
Let’s look at another problem.
Suppose you want to invest enough now so that you’ll have
$60,000 for your child’s education in 18 years. What would
you need to invest if the interest rate is 4 ½ % compounded
monthly?
Think about the compound interest formula. What parts do
you know and what are you trying to find?
You know: A = $60,000 – that’s how much you want to have
in 18 years, the accumulated amount
r = 4 ½% = 0.045
t = 18 years
m = 12 – compounded monthly
You want to find: P – the amount to invest now
Example continued…
Now simply substitute into the compound interest formula
and solve for P.
r 

A  P1  
 m
mt
 0.045
60,000 P1 

12 

1218
60,000 P1.00375216
60,000
1.00375
216
P
P  26,731.95
Wow, you would need to
invest about $26,732 now!
Fortunately most of us will
invest a little each year to
reach the same goal instead
of all at one time.
Present Value
The previous problem was an example of present value. The
equation for present value is simply the compound interest
formula solved for P.
A  P1  i n Divide both sides by 1  i n .
A
n


P
or
P

A
1

i
1  i n
So the equation for Present Value is
P  A1  i 
n
r
, wherei 
and n  mt.
m
P is called the present value and A is called the future value.
Continuously Compounded Interest
There is also a formula for interest compounded continuously.
To show the derivation of this formula requires a limit we have
not studied so the derivation is omitted here.
Continuously Compounded Interest
A  P ert , where P  P rincipal
r  interest rate
t  time in years
Example
Let’s consider that first problem again only this time we’ll
calculate the accumulated amount after investing $5000
for 4 years at a yearly interest rate of 6% compounded
continuously.
P = 5000, r = 0.06 and t = 4
A  P ert  5000e0.064  5000e0.24  6356.25
So, the accumulated amount will be $6356.25. You should
note this amount is more than both of our previous
calculations where it was compounded quarterly and monthly.
Another example
Using those same numbers, let’s see how long it would take
to double our investment of $5000.
Think about what you know and what you are trying to find!
You know P = 5000, r = 0.06 and if you know you are trying
to double your investment then A = 10,000.
You are trying to find the time to double or t.
Substitute the values into the continuously compounded
interest formula and solve for t.
Example continued…
Substituting into the formula.
A  P ert
10000 5000e0.06t
2  e 0.06t
ln 2  lne0.06t
ln2  0.06 t
ln2
t
0.06
t
ln2
 11.55
0.06
Divide both sides by 5000.
Take the ln of both sides and
solve for t.
So it would take just over
11 ½ years to double the
initial investment.