Exponential Growth

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Exponential Growth
Exponential Growth
 Discrete Compounding
 Suppose that you were going to
invest $5000 in an IRA earning
interest at an annual rate of 5.5%.
How much interest would you earn
during the 1st year? How much is in
the account after 1 year?
Exponential Growth
 Interest after 1 year:
i1  5000
0.055  $275



P
r
 Account value after 1 year:

F1  5000
0.055   5000 1.055   $5275
  5000



P
i1 $275
 What would happen during the 2nd
year?
Exponential Growth
 Interest made during the 2nd year:
i2  5275
0.055  $290.13


F
1
r
 Value of account after 2nd year:

F2  5275
0.055   5275 1.055   $5,565.13
  5275

F
1
i2 $290.13
 What about for the 3rd year?
Exponential Growth
 Interest made during 3rd year:


i3  $5,565.13
0
.
055

$
306
.
08

 
F2
r
 Value of the account after 3rd year:
F3  5565
.
13  5565 .130.055   5565 .131.055   $5,871.21



F2
i3 $306.08
Exponential Growth
 Summarizing our calculations:
P  5000
F1  50001.055  $5,275
F2  5275
 1.055  50001.055  $5,565.13
2
5000 1.055 
F3  5565
.
131.055  50001.055  $5,871.21


3
5000 1.055 2
Exponential Growth
 From our calculations, a $5,000
investment into an account with an
annual interest rate of 5.5% will have
a value of F after t years according to
the formula:
Ft  50001.055
t
Exponential Growth
 In general, P dollars invested at an annual
rate r, has a value of F dollars after t years
according to:
Ft  P1  r 
t
 Notice that the interest was paid on a
yearly basis, while our money remained in
the account. This is called compounding
annually or one time per year.
Exponential Growth
 What would happen if the interest
was paid more times during the year?
 Suppose interest is collected at the
end of each quarter, (interest is paid
four times each year). What would
happen to our investment?
Exponential Growth
 Since the annual interest rate is 5.5%
this rate needs to be adjusted so that
interest is paid on a quarterly basis.
The quarterly rate is:
r 5.5
r1  
 1.375 %
4
n
4
Exponential Growth
 Interest made during 1st quarter:
 0.055
i  5000
  $68.75

4
P 



1
4
r1
4
 Value of account after 1st quarter:
 0.055
 0.055
F1  5000
 5000
 50001 
 $5,068.75



4
4
4




P



i1
4
Exponential Growth
 Interest made during the 2nd quarter:
 0.055
i  5068.75


 4   $69.70

F


2
4
1
4
r1
4
 Value of account after 2nd quarter:
 0.055
 0.055
F2  5068.75
 5068.75
 5068.751 
 $5,138.45







4
4 
 4

F1

4
i 2 $69.70
4
Exponential Growth
 Interest made during the 3rd quarter:
 0.055
i  5138.45


 4   $70.65

F


3
4
2
4
r1
4
 Account value after 3rd quarter:
 0.055
 0.055
F3  5138.45
 5138.45
 5,138.451 
 $5,209.10







4
4 
 4

F2

4
i 3 $70.65
4
Exponential Growth
 Interest made during the 4th quarter:
 0.055
i  5209.10


 4   $71.63

F


4
4
3
4
r1
4
 Account value after 4th quarter:
 0.055
 0.055
F4  5209.10
 5209.10
 5209.101 
 $5,280.72







4
4 
 4

F3

4
i 4 $71.63
4
Exponential Growth
 Summarizing our results for 1 year:
P  5000
 0.055
F1  50001 
  $5,068.75
4
4 

2
 0.055
 0.055
F2  5068
.
75
1


5000

1 
  $5,138.45



4
4 
4 


F1
4
3
 0.055
 0.055
F3  5138.45
1


5000


1 
  $5,209.10



4
4 
4 


F2
4
4
 0.055
 0.055
F4  5209.10


 1  4   50001  4   $5,280.72
4




F3
4
Exponential Growth
 Notice that the exponent corresponds
to the number of quarters in a year:





So
So
So
So
So
for
for
for
for
for
1 year there are 4 quarters
2 years there are 8 quarters
3 years there are 12 quarters
4 years there are 16 quarters
t years there are 4t quarters
Exponential Growth
 So the value of a $5,000 investment
with an annual interest rate of 5.5%
compounded quarterly after t years is
given by:
 0.055
Ft  50001 

4 

4t
Exponential Growth
 In general, P dollars invested at an
annual rate r, compounded n times
per year, has a value of F dollars after
t years according to:
 r
F  P 1  
 n
nt
Exponential Growth
 From the last slide, we can also say:
 r
P  F 1  
 n
 nt
 In other words, we can find the
present value (P) by knowing the
future value (F).
Exponential Growth
 Notice for each of the 3 years the
account that is compounded quarterly
is worth more than the one
compounded annually
t
1
2
3
n=1
n=4
F
F
$5,275.00 $5,280.72
$5,565.13 $5,577.21
$5,871.21 $5,890.34
Exponential Growth
 It would seem the larger n is the
more an investment is worth, but
consider:
t
1
2
3
n=52
n=365
F
F
$5,282.55 $5,282.68
$5,581.07 $5,581.34
$5,896.45 $5,896.89
Exponential Growth
 Notice value of the investment is
leveling off when P, r, and t are fixed,
but n is allowed to get really big.
n
 r
 This suggests that 1  n  is leveling
off to some special number
Exponential Growth
 There is a clever technique that
allows us to find this value. We let m
= n/r, so that n = mr. For any value
of r, m gets larger as n increases.
We rewrite the expression:
r 

F  P1 

 rm 
mrt

1
 P 1  
 m 
m rt



Exponential Growth
m
 As m gets big, 1  1   2.7182818284
5905 e

m
10
100
1000
10000
100000
1000000
10000000
100000000
m
1

1  
 m
m
2.59374246
2.704813829
2.716923932
2.718145927
2.718268237
2.718280469
2.718281694
2.718281786
Exponential Growth
 So as m gets large,
r 

F  P1 

 rm 
mrt

1
 P 1  
 m 
 Pe
rt
 This is for continuous compounding
 In Excel, use the function EXP(x)
m rt



Exponential Growth
 So P dollars will grow to F dollars
after t years compounded
continuously at r % by the equation:
F  Pe
rt
 We can also find P by knowing F as
follows:
 rt
P  Fe
Exponential Growth
 How do we compare investments with
different interest rates and different
frequencies of compounding?
 Look at the values of P dollars at the end
of one year
 Compute annual rates that would
produce these amounts without
compounding.
 Annual rates represent the effective
annual yield
Exponential Growth
 In our current example when we
compounded quarterly, after one year
41
we had:
0.055

F1  50001 
  $5280.72
4 

 Notice we gained $280.72 on interest
after a year. That interest represents
a gain of 5.61% on $5000:
Effective
Annual
280 .72
y
 0.0561
Yield (y)
5000
Exponential Growth
 Effective annual yield (Discrete):
 find the difference between our money
after one year and our initial investment
and divide by the initial investment.
 Therefore, interest at an annual rate
r, compounded n times per year has
n
yield y:
r


P 1    P
n
n
 r

y
 1    1
P
 n
Exponential Growth
 You may need to find the annual rate
that would produce a given yield.
n
r

y  1    1
 Need to solve for r :

This tells you the annual
interest rate r that will
produce a given yield when
compounding n times a year.
Note: This is only for
Discrete Compounding
n
r

y  1  1  
 n
n
 y  11/ n  1  r
n
r
1/ n
 y  1  1 
n
1/ n
n  y  1  1  r


Exponential Growth
 Effective Annual Yield (Continuous):
Pe  P
r
y
 e 1
P
r
 Annual interest rate:
r  ln y  1
Exponential Growth
 Ex. Find the final amount if $10,000 is
invested with interest calculated
monthly at 4.7% for 6 years.
 Soln. F  P1 

0.047 126
 10,0001  12 
r nt
n
 $13,250.49
Exponential Growth
 Ex. Find the annual yield on an
investment that computes interest at
4.7% compounded monthly.
r n
 Soln. y  1  n   1
 1 

0.047 12
12
1
 1.048025794 1
 0.0480
 About 4.80%
Exponential Growth
• Ex. Find the rate, compounded weekly,
that has a yield of 9.1%


 521  0.091  1
• Soln. r  n 1  y 1 / n  1
1 / 52
 0.087167685
About 8.72%
Exponential Growth
 Examples that use the word
continuous to describe compounding
period mean you use:
F  Pert
 Ex. How much would you have after 3
years if an investment of $15,000 was
placed into an account that earned
10.3% interest compounded
continuously?
Exponential Growth
 Soln.
F  Pe rt
 15,000e
0.103 3 
 $20,430.94
Exponential Growth
 Ex. Find the annual rate of an investment
that has an annual yield of 9% when
compounded continuously.
 Soln. r  ln y  1
 ln0.09  1
 0.0862
 Approx 8.62%
Exponential Growth
 Where else can compound interest be
used?
 Financing a home
 Financing a car
 Anything where you make monthly
payments (with interest) on money
borrowed
Exponential Growth
 The average cost of a home in Tucson
is roughly around $225,000.
Suppose you were planning to put
down $25,000 now and finance the
rest on a 30 year mortgage at 7%
compounded monthly. How much
would your monthly payments be?
Exponential Growth
 For a 30 year mortgage, you’ll be making
360 monthly payments.
 At the end of the 360 months we want the
present value (P) of all the monthly
payments to add up to the amount you
plan to finance, e.g. $200,000
 The $200,000 is called the principal
Exponential Growth
 Let’s say that Pk represents the
present monthly value k months ago.
 Then after 360 months, we want:
200,000  P1  P2  P3    P360
Exponential Growth
 Since we’re borrowing money here,
each Pk can be expressed as
 r
Pk  F 1  
 n
k
 But where F represents the future
value for Pk. In other words, F is your
monthly payment.
Exponential Growth
 Remember we want:
200,000  P1  P2  P3    P360
 So if we insert:
 r
Pk  F 1  
 n
k
 We have instead:
1
2
 r
 r
 r
200,000  F 1    F 1      F 1  
 n
 n
 n
360
Exponential Growth
 Now for a little algebra (factor out F):
1
2
360

r
r
r 


200,000  F 1    1      1   
 n
 n  
 n 
 Divide both sides by the stuff in [ ]
200,000
 r 
 r
 r
 1     1       1  
 n
 n
 n 
1
2
360



F
Exponential Growth
 The last result will tell us our monthly
payment F:
200,000
 r 
 r
 r
 1     1       1  
 n
 n
 n 
1
2
360



F
 Notice that all we need to is figure
out how to add up the numbers in the
bottom. This is where we use Excel.
Exponential Growth
 Since we’re compounding monthly at
7%, r = 0.07 and n = 12
 So:
200,000
 0.07 
 0.07 
 0.07 
  1 
    1 

 1 
12 
12 
12 



1
2
360



F
Exponential Growth
 We’ll do the rest of our calculation in
Excel
 So our monthly payments F:
F
200,000
360
 0.07  1  0.07  2
 0.07  
  1 
    1 
 
1 
12 
12 
12  



200,000

 1330.60
150.3076
Exponential Growth
 Now that we know what F is we can
figure out what each Pk is.
 Again, each Pk will tell us what F
dollars was worth k months ago
 We’ll again use Excel to answer this
question.
Exponential Growth
 In Excel:
End
This number tells us that our
monthly payment of $1330.60 was
worth $1322.89 one month ago.
Notice that as we descend down
the table the values get smaller
because we’re going farther back in
time.
This number tells us how much of the
monthly payment is for interest. Notice
that as we descend the table the interest
goes up. This tells us that in the beginning
of a payment plan a lot of the monthly
payment is toward interest and only a small
portion is going toward principal while the
reverse is true at the end.
Start
Exponential Growth
 What your outstanding balance looks
like with each monthly payment?
Balance $
Balance
250000
200000
150000
100000
50000
0
-50000 0
Balance
100
200
Months
300
400
Exponential Growth
 Things to notice:
 After 360 months of payments of
1330.61, you’re really paying
$479,019.60 on $200,000 borrowed.
 The mortgage company has made
139% profit on your borrowing
$200,000.
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