Chapter 3
Mathematics of
Finance
Section 2
Compound and
Continuous Interest
Learning Objectives for Section 3.2
Compound and
Continuous Compound Interest
 The student will be able to compute compound and
continuous compound interest.
 The student will be able to compute the growth rate of a
compound interest investment.
 The student will be able to compute the annual
percentage yield of a compound interest investment.
Barnett/Ziegler/Byleen College Mathematics 12e
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Table of Content




Compound Interest
Continuous Compound Interest
Growth and Time
Annual Percentage Yield
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Terms




compound interest
rate per compounding period
continuous compound interest
APY (Annual Percentages Yield)
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Compound Interest
 Unlike simple interest, compound interest on an amount
accumulates at a faster rate than simple interest. The basic
idea is that after the first interest period, the amount of
interest is added to the principal amount and then the
interest is computed on this higher principal. The latest
computed interest is then added to the increased principal
and then interest is calculated again. This process is
completed over a certain number of compounding periods.
The result is a much faster growth of money than simple
interest would yield.
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Example
Suppose a principal of $1,000 was invested in an
account paying 8% annual interest compounded
quarterly. How much would be in the account after one
year?
Compounding quarterly means that the interest is paid
at the end of each 3-month period and the interest as
well as the principal earn interest for the next quarter.
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Solution
 Solution: Using the simple interest formula A = P (1 + rt)
we obtain:
 At the end of the first quarter will have:
A = P (1 + rt)
A = 1000[1 + 0.08(1/4)]
A = 1000(1.02) = $1.020
 Now the principal is $1,020. At the end of the second
quarter will have :
A = 1020[1 + 0.08(1/4)]
A = 1020(1.02) = $1.040.40
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Solution
(cont.)
 At the end of the third quarter will have :
A = 1.040.40[1 + 0.08(1/4)]
A = 1.040.40(1.02) = $1.061.21
 At the end of the fourth quarter will have :
A = 1.061.21[1 + 0.08(1/4)]
A = 1.061.21(1.02) = $1.082.43
• Compare with simple interest:
A = P(1 + rt) = $1,000[1 + 0.08(1)]
A = $1,000[10.08] = $1,080 $2,43 more
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Solution
(cont.)
 Looking over the calculations …
A = 1.000(1.02)
A = [1.000(1.02)](1.02) = 1,000(1.02)2
A = [1.000(1.02)2](1.02) = 1,000(1.02)3
A = [1.000(1.02)3](1.02) = 1,000(1.02)4
• At the end of n quarters:
A = 1.000(1.02)n
or
A = 1.000[1 + 0.08(1/4)]n
A = 1.000[1 + 0.08/4]n
End 1st quarter
End 2nd quarter
End 3rd quarter
End 4th quarter
End nth quarter
Rate per compounding period = annual nominal rate / number of
compounding periods per year
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Solution
 In general
A = P(1 + i)
A = [P(1 + i)](1 + i) = P(1 + i)2
A = [P(1 + i)2](1 + i) = P(1 + i)3
(cont.)
End 1st period
End 2nd period
End 3rd period
….
A = [P(1 + i)(n-1)](1 + i) = P(1 + i)n
End nth period
Then it can be summarized …
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General Formula
 This formula can be generalized to
A  P 1  i 
n
r
i
m
n  mt
r

A  P 1  
 m
mt
where A is the future amount, P is the principal, r is the interest
rate as a decimal, m is the number of compounding periods in
one year, and t is the total number of years. To simplify the
formula,
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Compound Interest General Formula
A  P 1  i 
n
where i = r/m
A = amount (future amount) at the end of n periods
P = principal (present value)
r = annual nominal rate
m = number of compounding periods per year
i = rate per compounding period
t = time (years)
n = total number of compounding periods = mt
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Graphical Illustration of
Compound Interest
Growth of 1.00 compounded monthly at 6% annual interest
over a 15 year period (Arrow indicates an increase in value of
almost 2.5 times the original amount.)
Growth of 1.00 compounded monthly at 6% annual interest
3
over a 15 year period
2.5
2
1.5
1
0.5
0
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Example
 Find the amount to which $1500 will grow if compounded
quarterly at 6.75% interest for 10 years.
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Example
 Find the amount to which $1500 will grow if compounded
quarterly at 6.75% interest for 10 years.
 Solution: Use
n
A  P 1  i 
10(4)
 0.0675 
A  1500 1 

4 

A  2929.50
 Helpful hint: Be sure to do the arithmetic using the rules for
order of operations.
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Same Problem Using
Simple Interest
Using the simple interest formula, the amount to which $1500
will grow at an interest of 6.75% for 10 years is given by
A = P (1 + rt)
= 1500(1+0.0675(10)) = 2512.50
which is more than $400 less than the amount earned using
the compound interest formula.
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Changing the number of
compounding periods per year
To what amount will $1500 grow if compounded daily at
6.75% interest for 10 years?
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Changing the number of
compounding periods per year
To what amount will $1500 grow if compounded daily at
6.75% interest for 10 years?
10(365)
0.0675 

A  1500 1 

365


Solution:
= 2945.87
This is about $15.00 more than compounding $1500 quarterly
at 6.75% interest.
Since there are 365 days in year (leap years excluded), the
number of compounding periods is now 365. We divide the
annual rate of interest by 365. Notice, too, that the number of
compounding periods in 10 years is 10(365)= 3650.
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Changing the number of
compounding periods per year
Example 1
If $1,000 is invested at 8% compound:
A.
B.
C.
D.
Annually
Semiannual
Quarterly
Monthly
What is the amount after 5 years?
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Changing the number of
compounding periods per year
Example 1 - Solution
If $1,000 is invested at 8% compound, what is the amount after
5 years?
A. Annually
B. Semiannual
n = 5 ; I = r = 0.08
A = P(1 + i)n
= 1,000(1 + 0.08)5
= 1,000(1.469328)
= $1,469.33
n = (2)5 = 10
I = r/n = 0.08/2 = 0.04
A = P(1 + i)n
= 1,000(1 + 0.04)10
= 1,000(1.480244)
= $1,480.24
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Changing the number of
compounding periods per year
Example 1 - Solution
If $1,000 is invested at 8% compound, what is the amount after
5 years?
D. Monthly
C. Quarterly
n = (4)5 = 20
I = r/n = 0.08/4 = 0.02
A = P(1 + i)n
= 1,000(1 + 0.02)20
= 1,000(1.485947)
= $1,485.95
Barnett/Ziegler/Byleen College Mathematics 12e
n = (12)5 = 60
I = r/n = 0.08/12 = 0.006666
A = P(1 + i)n
= 1,000(1 + 0.00666)60
= 1,000(1.489846)
= $1,489.85
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Effect of Increasing the
Number of Compounding Periods
 If the number of compounding periods per year is
increased while the principal, annual rate of interest and
total number of years remain the same, the future amount
of money will increase slightly.
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Continuous Compound Interest
Previously we indicated that increasing the number of
compounding periods while keeping the interest rate,
principal, and time constant resulted in a somewhat higher
compounded amount. What would happen to the amount if
interest were compounded daily, or every minute, or every
second?
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Answer to Continuous Compound
Interest Question
As the number m of compounding periods per year increases
without bound, the compounded amount approaches a limiting
value. This value is given by the following formula:
r mt
rt
lim m P(1  )  Pe
m
rt
A  Pe
Here A is the compounded amount.
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Example of
Continuously Compounded Interest
What amount will an account have after 10 years if $1,500
is invested at an annual rate of 6.75% compounded
continuously?
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Example of
Continuously Compounded Interest
What amount will an account have after 10 years if $1,500
is invested at an annual rate of 6.75% compounded
continuously?
Solution: Use the continuous compound interest formula
A = Pert with P = 1500, r = 0.0675, and t = 10:
A = 1500e(0.0675)(10) = $2,946.05
That is only 18 cents more than the amount you receive by
daily compounding.
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Changing the number of
compounding periods per year
To what amount will $1500 grow if compounded daily at
6.75% interest for 10 years?
10(365)
0.0675 

A  1500 1 

365


Solution:
= 2945.87
That is only 18 cents more than the amount you receive by
daily compounding.
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Continuously Compounded Interest
Example 2
What amount will an account have after 2 years if $5,000 is
invested at an annual rate of 8% compoundly
A. daily?
A = P(1 + r/m)mt
P = 5,000; r = 0.08; m = 365
t=2
A = 5,000(1+ 0.08/365)(365)(2)
= $5,867.55
Barnett/Ziegler/Byleen College Mathematics 12e
A. continuously?
A = Pert
P = 5,000; r = 0.08; t = 2
A = 5,000e(0.08)(2)
= $5,867.55
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Growth & Time
 How much at a future date?
 What annual rate has been earned?
 How long take to double the investment?
The formulas for the compound interest and the
continuous interest can be used to answer such
questions.
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Growth & Time
 How long will it take for $5,000
to grow to $15,000 if the money
is invested at 8.5%
compounded quarterly?
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Growth & Time
Solution
 How long will it take for $5,000
to grow to $15,000 if the money
is invested at 8.5%
compounded quarterly?
1. Substitute values in the
compound interest formula.
2. divide both sides by 5,000
3. Take the natural logarithm of
both sides.
4. Use the exponent property of
logarithms
5. Solve for t.
Barnett/Ziegler/Byleen College Mathematics 12e
 Solution:
0.085 4t
15000  5000(1 
)
4
3 = 1.021254t
ln(3) = ln1.021254t
ln(3) = 4t*ln1.02125
ln(3)
t
 13.0617
4ln(1.02125)
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Growth & Time
Example 3
Finding Present Value
How much will you invest now at 10% to have $8,000 toward
the purchase of a car in 5 years if interest is
A. compounded quarterly?
P = ? A = 8,000 i = 0.10/4
n = 4(5) = 20
A = P(1 + i)n
8,000 = P(1 + 0.025)20
P = 8,000 / (1 + 0.025)20
P = 8,000 / (1.638616)
= $4,882.17
Barnett/Ziegler/Byleen College Mathematics 12e
B. compounded continuously?
P = ? A = 8,000 r = 0.10
t=5
A = Pert
8,000 = Pe(0.10)(5)
P = 8,000 /e(0.10)(5)
P = $4,852.25
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Growth & Time
Example 4
Computing Growth Rate
An investment of $10,000 in a growth-oriented fund over a 10years period would have grown in $126,000. What annual
nominal rate would produce the same growth if interest was:
C. compounded quarterly?
A = P(1 + i)n
126,000 = 10,000 (1+ r)10
12.6 = (1+ r)10
Raiz10(12.6) = 1+ r
r = Raiz10(12.6) – 1
r = 0.28836 ó 28.836%
Barnett/Ziegler/Byleen College Mathematics 12e
D. compounded continuously?
A = Pert
126,000 = 10,000e10r
12.6 = e10r
ln 12.6 = 10r
r = ln 12.6/10
r = 0.25337 ó 25.337%
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Growth & Time
Example 5
Computing Growth Time
How long will it take $10,000 to grow to $12,000 if it is invested
at 9% compounded monthly?
Using logarithms & calculator:
A = P(1 + i)n
12,000 = 10,000 (1+ 0.09/12)n
1.2 = 1.0075n
Solve for n by taking logarithms
of both sides:
Barnett/Ziegler/Byleen College Mathematics 12e
ln 1.2 = ln 1.0075n
ln 1.2 = nln 1.0075
n = ln 1.2 / ln 1.0075
n = 24.40 ó 25 months
or 2yr 1mth
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Annual Percentage Yield
 The simple interest rate that will produce the same amount as a
given compound interest rate in 1 year is called the annual
percentage yield (APY). To find the APY, proceed as follows:
Amount at simple interest APY after one year
= Amount at compound interest after one year
m
r

P(1  APY )  P 1   
 m
m
r

1  APY  1   
 m
m
r

APY  1    1
 m
Barnett/Ziegler/Byleen College Mathematics 12e
This is also called
the effective rate, as
well as APR.
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Annual Percentage Yield
Example
What is the annual percentage yield for money that is
invested at 6% compounded monthly?
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Annual Percentage Yield
Example
What is the annual percentage yield for money that is
invested at 6% compounded monthly?
General formula:
Substitute values:
m
r

APY  1    1
 m
12
 0.06 
APY  1 
  1  0.06168
12 

Effective rate is 0.06168 = 6.168
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Computing the Annual Nominal
Rate Given the APY
 What is the annual nominal rate
compounded monthly for a CD
that has an annual percentage
yield of 5.9%?
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Computing the Annual Nominal
Rate Given the APY
m
 What is the annual nominal rate
compounded monthly for a CD
that has an annual percentage
yield of 5.9%?
 1. Use the general formula for
APY.
 2. Substitute value of APY and 12
for m (number of compounding
periods per year).
 3. Add one to both sides
 4. Take the twelfth root of both
sides of equation.
 5. Isolate r (subtract 1 and then
multiply both sides of equation by
12.
Barnett/Ziegler/Byleen College Mathematics 12e
r

APY  1    1
 m 12
r 

0.059  1    1
 12 
12
r 

1.059  1  
 12 
r 

1.059  1  
 12 
r
12
1.059  1 
12
12
12

12

1.059  1  r
0.057  r
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Annual Percentage Yield
EXAMPLE 6 - Using APY to Compare Investments.
Find the APY for each of the bank in the table and compare
these CD’s.
Certificates of Deposits (CDs)
Bank
Rate
Compounded
Advanta
4.93%
monthly
DeepGreen
4.95%
daily
Charter One
4.97%
quarterly
Liberty
4.94%
continuously
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Annual Percentage Yield
EXAMPLE 6 - Using APY to Compare Investments (contd.)
Find the APY for each of the bank in the table and compare these
CD’s.
= (1 + r/m)m – 1]
Bank
Calculations [APY
Advanta
= (1 + 0.0493/12)^12 – 1 = 0.05043
5.043%
DeepGreen
= (1 + 0.0495/365)^365 – 1 = 0.05074
5.074%
Charter One
= (1 + 0.0497/4)^4 – 1 = 0.05063
5.063%
Liberty
= e^(0.0494) – 1 = 0.05064
5.064%
Barnett/Ziegler/Byleen College Mathematics 12e
APY
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Example of
Annual Percentage Yield
EXAMPLE 7 - Computing the Annual Nominal Rate given
the APY
A savings & loans wants to offer a CD with a monthly
compounding rate that has an APY of 7.5%. What annual
nominal rate compounded monthly should they use?.
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Example of
Annual Percentage Yield
EXAMPLE 7 - Computing the Annual Nominal Rate given the
APY (continued)
APY = (1 + r/m)m - 1
0.075 = (1+ r/12)12 - 1
1.075 = (1+ r/12)12
Raiz12(1.075) = 1+ r/12
Raiz12(1.075) – 1 = r/12
r = 12(Raiz12(1.075) – 1)
Barnett/Ziegler/Byleen College Mathematics 12e
Using the calculator:
r = 0.072539
r = 7.254%
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Chapter 3
Mathematics of
Finance
Section 2
Compound and
Continuous Interest
END
Last Update: Marzo 19/2013