4.3 continuous compound interests perta x

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 period N–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.



Continuous Compound Interest

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?






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






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

For 100 times a year, 1000.08i = = 0.0008,






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

For 100 times a year, 1000.08i = = 0.0008,

N = (20 years)(100 times per years) = 2000






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,







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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,


Hence A = 1000(1 + 0.00008 )20000






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,


Hence A = 1000(1 + 0.00008 )20000 4952.72 $






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,


Hence A = 1000(1 + 0.00008 )20000 4952.72 $

For 10000 times a year, 100000.08i = = 0.000008,






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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,


Hence A = 1000(1 + 0.00008 )20000 4952.72 $

For 10000 times a year, 100000.08i = = 0.000008,







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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,


Hence A = 1000(1 + 0.00008 )20000 4952.72 $

For 10000 times a year, 100000.08i = = 0.000008,


Hence A = 1000(1 + 0.000008 )200000


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

For 100 times a year, 1000.08i = = 0.0008,


Hence A = 1000(1 + 0.0008 )2000 4949.87 $

For 1000 times a year, 10000.08i = = 0.00008,


Hence A = 1000(1 + 0.00008 )20000 4952.72 $

For 10000 times a year, 100000.08i = = 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)


Continuous Compound InterestFormula 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

Formula for Continuously Compounded Return (Perta)

Let P = principal


t = number of years



Example C. We deposited $1000 in an account compounded

continuously.

a. if r = 8%, how much will be there after 20 years?

Continuous Compound InterestFormula for Continuously Compounded Return (Perta)

Let P = principal


t = number of years




continuously.


P = 1000, r = 0.08, t = 20.


Let P = principal


t = number of years




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




continuously.



return is A = 1000*e0.08*20 = 1000*e1.6


Let P = principal


t = number of years




continuously.



return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$


Let P = principal


t = number of years




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




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




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




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?r = 12%, A = 1000*e0.12*20 = 1000e 2.4 11023.18$


Let P = principal


t = number of years




continuously.



return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$


c. If r = 16%, how much will be there after 20 years?


Let P = principal


t = number of years




continuously.



return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$



r = 16%, A = 1000*e0.16*20


Let P = principal


t = number of years




continuously.



return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$



r = 16%, A = 1000*e0.16*20 = 1000*e 3.2


Let P = principal


t = number of years




continuously.



return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$


c. If r = 16%, how much will be there after 20 years?r = 16%, A = 1000*e0.16*20 = 1000*e 3.2 24532.53$


Let P = principal


t = number of years



Continuous Compound InterestAbout the Number e

Just as the number π, the number e 2.71828… occupies a

special place in mathematics.



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.






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…







the number,

( 2.71828…)the same as

( )1,2 1 …( )4,

5 4

( )3,4 3

( )2,3 2

2.71828…which is







the number,

This number emerges often in the calculation of problems in

physical science, natural science, finance and in mathematics.


( )1,2 1 …( )4,

5 4

( )3,4 3

( )2,3 2

2.71828…which is







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.


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


Continuous Compound InterestGrowth 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.




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

Continuous Compound InterestIf 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–xAn Exponential Decay





For example,



In finance, shrinking values is

called “depreciation” or

”devaluation”.






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.






For example,










Hence if P = $1, after 5 years, its purchasing power is

1*e–0.04(5) = $0.82 or 82 cents.






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:


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).
