2.3 continuous compound interests

Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
Recall the period-compound PINA formula.

Let P = principal
A = accumulation
then P(1 + i) N = A

Let P = principal
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
0 1 2 3 Nth periodN–1

Let P = principal
A = accumulation
then P(1 + i) N = A
P
Rule: Multiply (1 + i) each period forward

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i)

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2 P(1 + i) 3

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
P(1 + i) N

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1 P(1 + i) N = A
Example A. $1,000 is in an account that has a monthly interest
rate of 1%. How much will be there after 60 years?
P(1 + i) N

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
We have P = $1,000, i = 1% = 0.01, N = 60 *12 = 720 months
so by PINA, there will be 1000(1 + 0.01) 720
P(1 + i) N = AP(1 + i) N

Let P = principal
A = accumulation
then P(1 + i) N = A
P
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
We have P = $1,000, i = 1% = 0.01, N = 60 *12 = 720 months
so by PINA, there will be 1000(1 + 0.01) 720 = $1,292,376.71
after 60 years.
P(1 + i) N = AP(1 + i) N

The graphs shown here are the different returns with r = 20%
with different compounding frequencies.
Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia)

with different compounding frequencies. We observe that
I. the more frequently we compound, the bigger the return

with different compounding frequencies. We observe that
I. the more frequently we compound, the bigger the return
II. but the returns do not go above the blue-line
the continuous compound return, which is the next topic.

Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
Continuous Compound Interest

P = 1000, r = 0.08, T = 20,

P = 1000, r = 0.08, T = 20,
For 100 times a year, 100
0.08
i = = 0.0008,

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
N = (20 years)(100 times per years) = 2000

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $
0.08
i = = 0.00008,

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $
0.08
i = = 0.00008,
Hence A = 1000(1 + 0.00008 )20000

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $
0.08
i = = 0.00008,
Hence A = 1000(1 + 0.00008 )20000  4952.72 $

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $
0.08
i = = 0.00008,
Hence A = 1000(1 + 0.00008 )20000  4952.72 $
0.08
i = = 0.000008,

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $
0.08
i = = 0.00008,
Hence A = 1000(1 + 0.00008 )20000  4952.72 $
0.08
i = = 0.000008,
Hence A = 1000(1 + 0.000008 )200000

P = 1000, r = 0.08, T = 20,
0.08
i = = 0.0008,
Hence A = 1000(1 + 0.0008 )2000  4949.87 $
0.08
i = = 0.00008,
Hence A = 1000(1 + 0.00008 )20000  4952.72 $
0.08
i = = 0.000008,
Hence A = 1000(1 + 0.000008 )200000  4953.00 $

We list the results below as the number compounded per year
f gets larger and larger.

4 times a year 4875.44 $

100 times a year 4949.87 $

10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $

10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $


10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
 4953.03 $

10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
 4953.03 $
We call this amount the continuously compounded return.

10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
 4953.03 $
This way of compounding is called compounded continuously.

10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
 4953.03 $
This way of compounding is called compounded continuously.
The reason we want to compute interest this way is because
the formula for computing continously compound return is
easy to manipulate mathematically.

Formula for Continuously Compounded Return (Perta)

Formula for Continuously Compounded Return (Perta)Formula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e  2.71828…

There is no “f” because
it’s compounded continuously
Let P = principal
t = number of years
Per*t = A where e  2.71828…

Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
P = 1000, r = 0.08, t = 20.
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
b. If r = 12%, how much will be there after 20 years?
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20 = 1000e 2.4
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20 = 1000e 2.4  11023.18$
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20 = 1000e 2.4  11023.18$
c. If r = 16%, how much will be there after 20 years?
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20 = 1000e 2.4  11023.18$
r = 16%, A = 1000*e0.16*20
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20 = 1000e 2.4  11023.18$
r = 16%, A = 1000*e0.16*20 = 1000*e 3.2
Let P = principal
t = number of years
Per*t = A where e  2.71828…

continuously.
return is A = 1000*e0.08*20 = 1000*e1.6  4953.03$
r = 12%, A = 1000*e0.12*20 = 1000e 2.4  11023.18$
r = 16%, A = 1000*e0.16*20 = 1000*e 3.2  24532.53$
Let P = principal
t = number of years
Per*t = A where e  2.71828…

About the Number e

Just as the number π, the number e  2.71828… occupies a
special place in mathematics.
About the Number e

special place in mathematics. Where as π  3.14156… is a
geometric constant–the ratio of the circumference to the
diameter of a circle, e is derived from calculations.
About the Number e

For example, the following sequence of numbers zoom–in on
the number,
( )1,2
1 …( )4,
5
4
( )3,4
3
( )2,3
2  2.71828…
About the Number e

the number,
( 2.71828…)the same as
( )1,2
1 …( )4,
5
4
( )3,4
3
( )2,3
2  2.71828…which is
About the Number e

the number,
This number emerges often in the calculation of problems in
physical science, natural science, finance and in mathematics.
( 2.71828…)the same as
( )1,2
1 …( )4,
5
4
( )3,4
3
( )2,3
2  2.71828…which is
About the Number e

the number,
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
This number emerges often in the calculation of problems in
physical science, natural science, finance and in mathematics.
Because of its importance, the irrational number 2.71828…
is named as “e” and it’s called the “natural” base number.
( 2.71828…)the same as
http://www.ndt-ed.org/EducationResources/Math/Math-e.htm
( )1,2
1 …( )4,
5
4
( )3,4
3
( )2,3
2  2.71828…which is
About the Number e

Growth and Decay

Growth and Decay
In all the interest examples we have positive interest rate r,
hence the return A = Perx grows larger as time x gets larger.

Growth and Decay
We call an expansion that may be modeled by A = Perx
with r > 0 as “ an exponential growth with growth rate r”.

y = e1x has the growth rate of
r = 1 or 100%.
Growth and Decay
For example,

r = 1 or 100%.
Exponential growth are rapid
expansions compared to other
expansions as shown here
by their graphs.
y = x3
y = 100x
y = ex
Growth and Decay
For example,
An Exponential Growth

r = 1 or 100%.
Exponential growth are rapid
expansions compared to other
expansions as shown here
by their graphs.
y = x3
y = 100x
y = ex
The world population may be
modeled with an exponential
growth with r ≈ 1.1 % or 0.011
or that A ≈ 6.5e0.011t in billions,
with 2011as t = 0.
Growth and Decay
For example,
An Exponential Growth

If the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.

We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.

For example,
y = e–1x has the decay or
contraction rate of I r I = 1 or 100%.
y = e–x
An Exponential Decay

For example,
In finance, shrinking values is
called “depreciation” or
”devaluation”.
y = e–x

For example,
”devaluation”. For example,
a currency that is depreciating
at a rate of 4% annually may be
modeled by A = Pe –0.04x
where x is the number of yearss elapsed.
y = e–x

For example,
Hence if P = $1, after 5 years, its purchasing power is
1*e–0.04(5) = $0.82 or 82 cents.
y = e–x

For example,
Hence if P = $1, after 5 years, its purchasing power is
1*e–0.04(5) = $0.82 or 82 cents. For more information:
y = e–x
http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm

Doubling Time and the 72-Rule
Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.

Solving for the doubling time we have
Perx = 2P or that
erx= 2

Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2)

Perx = 2P or that
rx = In(2) and that the doubling time is
x = In(2)/r

Perx = 2P or that
x = In(2)/r ≈ 0.72/r

Perx = 2P or that
x = In(2)/r ≈ 0.72/r
The last formula for the doubling time x is the 72-Rule, i.e.
x ≈ 0.72/r

Perx = 2P or that
x = In(2)/r ≈ 0.72/r
x ≈ 0.72/r
This 72- Rule gives an easy estimation for the doubling time.

Perx = 2P or that
x = In(2)/r ≈ 0.72/r
x ≈ 0.72/r
For example, if r = 8% then the doubling time is ≈ 72/8 = 9 (yrs).

Perx = 2P or that
x = In(2)/r ≈ 0.72/r
x ≈ 0.72/r
For example, if r = 8% then the doubling time is ≈ 72/8 = 9 (yrs).
if r = 12% then the doubling time is ≈ 72/12 = 6 (yrs).
if r = 18% then the doubling time is ≈ 72/18 = 4 (yrs).

Compound Interest
B. Given the continuous compound annual rate r find the
principal needed to obtain $1,000 after the given amount of time.
1. r = 1%, time = 60 months.
Exercise A.
Given the continuous compound annual interest rate r,
and the time, find the return with a principal of $1,000.
2. r = 1%, time = 60 years.
3. r = 3 %, time = 60 years 4. r = 3½ %, time = 60 months.
5. r = 2¼ %, time = 6 months. 6. r = 1¼ %, time = 5½ years.
7. r = 3 3/8%, time = 52 months 8. r = 3/8%, time = 27 months.
1. r = 1%, time = 60 months. 2. r = 1%, time = 60 years.
3. r = 3 %, time = 60 years 4. r = 3½ %, time = 60 months.
5. r = 2¼ %, time = 6 months. 6. r = 1¼ %, time = 5½ years.
7. r = 3 3/8%, time = 52 months 8. r = 3/8%, time = 27 months.

C. 1. For continuous growth with growth rate r > 0,
the formula ≈ 3/2 x 0.72/r ≈ 1.08/r estimates the time that will
take to triple the original amount. Given annual rate r = 6%,
a. estimate how long it will take to have a return
that’s 3 times the original amount?
b. estimate how long it will take to have a return
c. estimate how long it would take a have a return
a. estimate how long it will take to have a return that’s 4 times
the original amount. How about 8 times the original amount?
b. estimate how long it will take to have a return that’s 12 times
the original amount.
2. By the 72-formula, it will take ≈ 2 x 0.72/r = 1.44/r years to
quadruple the original amount. Given r = 12%,
3. How much off are the estimations in 10. a and b?

4. Similar to C1–C3, given the continuous decay rate –r ( r > 0),
by the half life 72-formula, it will take
≈ 3/2 x 0.72/r ≈ 1.08/r to decay to 1/3 of the initial amount,
≈ 2 x 0.72/r = 1.44/r to decay to ¼ of the original size.
Given r = 4%, estimate how long it will take to decay to 1/6
of the initial amount? 1/9 of the initial amount?
1/12 of the initial amount?
5. Given r = 8%, estimate how long it will take to decay to
1/6 of the initial amount? 1/9 of the initial amount?
1/12 of the initial amount?
14. How much off are the estimations in 14?

2.3 continuous compound interests

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2.3 continuous compound interests