Transcript Section 3B Putting Numbers in Perspective

Section 4A
The Power of
Compounding
Pages 210-222
4-A
Definitions

The principal in financial formulas ‘initial amount’
upon which interest is paid.

Simple interest is interest paid only on the
original principal, and not on any interest added at
later dates.

Compound interest is interest paid on both the
original principal and on all interest that has been
added to the original principal.
Example
4-A
Simple Interest – 5.0%
Principal
Time
(years)
$1000
0
Interest Paid Total
$0
$1000
$1000
1
$1000×.05=$50
$1050
$1000
2
$50
$1100
$1000
3
$50
$1150
$1000
4
$50
$1200
$1000
5
$50
$1250
$1000
10
$1000+ $50×10 = $1500
Example
4-A
Compound Interest – 5.0%
Principal
Time Interest Paid Total
(years)
$1000
0
$0
$1000
$1000
1
$50
$1050
$1050
2
$52.50
$1102.50
$1102.50
3
$55.13
$1157.63
$1157.63
4
$57.88
$1215.51
$1215.51
5
$60.78
$1276.29
Comparing Compound/Simple
Interest – 5.0%
Principal
Time
(years)
Interest
Paid
Total
Total
Compound Simple
$1000
0
$0
$1000 $1000
$1000
1
$50.00
$1050 $1050
$1050
2
$52.50
$1102.50 $1100
$1102.5
3
$55.13
$1157.63 $1150
$1157.63
4
$57.88
$1215.51 $1200
$1215.51
5
$60.78
$1276.29 $1250
Compound Interest – 7%
Principal
Time
(years)
Interest
Paid
Total
Compound
$1000
0
$0
$1000
$1000
1
$70
$1070
$1070
2
$74.90
$1144.90
$1144.90
3
$80.14
$1225.04
$1225.04
4
$85.75
$1310.79
4-A
General Formual for Compound Interest:
Year 1:
$1000 + $1000(.05) = $1050
= $1000×(1+.05)
Year 2: $1050 + $1050(.05) = $1102.50
= $1050(1+.05)
= $1000(1+.05)(1+.05)
= $1000(1+.05)2
Year 3: $1102.50+ $1102.50(.05) = $1157.63
= $1102.50(1+.05)
= ($1000(1+.05)2)(1+.05)
= $1000(1+.05)3
Amount after year t = $1000(1+.05)t
4-A
General Compound Interest Formula
A = P 1 + i



t



A = accumulated balance after t years
P = starting principal
i = interest rate (written as a decimal)
t = number of years
4-A
Suppose an aunt gave $5000 to a child born
3/8/07. The child’s parents promptly invest it in
a money market account at 4.91% compounded
yearly, and forget about it until the child is 25
years old. How much will the account be worth
then?
A = P 1 + i



t



Amount after year 25 = $5000×(1.0491)25
=$5000×(3.314531691...)
= $16,572.66
4-A
Suppose you are trying to save today for a
$10,000 down payment on a house in ten
years. You’ll save in a money market
account that pays 4.5% compounded
annually (no minimum balance). How much
do you need to put in the account now?
A = P 1 + i



t



$10,000 = $P×(1.045)10
so
$10,000 = $P
(1.045)10
= $6,439.28

Note: 1.04510 = 1.552969422...


$10,000/ 1.5 = $6666.67
$10,000 / 1.6 = $6250
$10,000/ 1.55 = $6451.61
$10,000/ 1.552 = $6443.30
$10,000 / 1.553 = $6439.15
$10,000/1.55297 = $6439.27

$10,000/(1.04510) = $6439.28

Don’t round in the intermediate steps!!!




The Power of Compounding
On July 18, 1461, King Edward IV of England borrows
the equivalent of $384 from New College of Oxford.
The King soon paid back $160, but never repaid the
remaining $224.
This debt was forgotten for 535 years.
In 1996, a New College administrator rediscovered the
debt and asked for repayment of $290,000,000,000
based on an interest rate of 4% per year.
WOW!
Example
4-A
4-A
Compounding Interest
(More than Once a Year)
You deposit $5000 in a bank account that pays an
APR of 4.5% and compounds interest monthly.
How much money will you have after 1 year? 2
years? 5 years?
APR is annual percentage rate
APR of 3% means monthly rate is 4.5%/12 = .375%
4-A
General Compound Interest Formula
A = P 1 + i



A
P
i
t
t



= accumulated balance after t years
= starting principal
= interest rate (as a decimal)
= number of years
4-A
Time
Accumulated Value
0 months
$5000
1 month
1.00375 × $5000
2 months
(1.00375)2 × $5000
3 months
(1.0375)3 × $5000
4 months
(1.00375)4 × $5000
5 months
(1.00375)5 × $5000
6 months
(1.00375)6 × $5000
7 months
(1.00375)7 × $5000
8 months
(1.00375)8 × $5000
9 months
(1.00375)9 × $5000
10 months
(1.00725)10 × $5000
11 months
(1.00725)11 × $5000
1 yr = 12 m
(1.00375)12 × $5000 = $5229.70
2 yr = 24 m
(1.00375)24 × $5000 = $5469.95
5 yr = 60 m
(1.00375)60 × $5000 = $6258.98
4-A
Compound Interest Formula
for Interest Paid n Times per Year





APR
A = P 1 +
n
A
P
APR
n
Y
=
=
=
=
=
(nY )





accumulated balance after Y years
starting principal
annual percentage rate (as a decimal)
number of compounding periods per year
number of years (may be a fraction)
4-A
You deposit $1000 at an APR of 3.5% compounded quarterly.
Determine the accumulated balance after 10 years.





APR
A = P 1 +
n
A
P
APR
n
Y
=
=
=
=
=
(nY )





accumulated balance after 1 year
$1000
3.50% (as a decimal) = .035
4
(410)
10


.035 
A = $10001 +


4 

 $1, 416.91
4-A
Suppose you are trying to save today for a
$10,000 down payment on a house in ten
years. You’ll save in a money market
account with an APR of 4.5% compounded
monthly. How much do you need to put in
the account now?
nY
A = P1 +
$10,000  P  (1 
.045 1210
12
)
APR 
n 
 P  (1.00375)
$10,000
 P  $6,381.65
120
(1.00375)
120
$1000 invested for 1 year at 3.5%
Compounded
Annually
(yearly)
quarterly
monthly
daily
Formula
(1)
.035 
1000 1 +

1 





(4)
.035 
1000 1 +

4 





Total
$1035
$1035.46
(12)
.035 
1000 1 +

12 
$1035.57
(365)
.035 
1000 1 +

365 
$1035.62










$1000 invested for 20 years at 3.5%
Compounded
Annually
(yearly)
quarterly
monthly
daily
Formula
(20)
.035 
1000 1 +

1 





(4 20)
.035 
1000 1 +

4 





(12 20)
.035 
1000 1 +

12 





(36520)
.035 
1000 1 +

365 





Total
$1989.79
$2007.63
$2011.70
$2013.69
$1000 invested for 1 year at 3.5%
Compounded
annually
Total
$1035
Annual Percentage Yield
$1035  $1000
100%  3.5%
$1000
quarterly $1035.46
$1035.46  $1000
100%  3.546%
$1000
monthly $1035.57
$1035.57  $1000
100%  3.557%
$1000
daily $1035.62
$1035.62  $1000
100%  3.562%
$1000
APY = annual percentage yield
APY = relative increase over a year
 amount after 1 year - principalamount 

 100%
principal amount


Ex: Compound daily for a year:
$1035.62  $1000
100%
$1000
= .03562 × 100%
= 3.562%
APR vs APY




APR = annual percentage rate (nominal
rate)
APY = annual percentage yield
(effective yield)
When compounding annually APR = APY
When compounding more frequently,
APY > APR
$1000 invested for 1 year at 3.5%
Compounded
Total
Annual Percentage Yield
annually
$1035
3.5%
quarterly $1035.46
3.546%
monthly $1035.57
3.557%
daily $1035.62
3.562%
$1000 invested for 1 year at 3.5%
Compounded
Total
annually
$1035
quarterly
$1035.46
monthly
$1035.57
daily
$1035.617971
Twice daily
$1035.61884
continuously $1035.619709
4-A
Euler’s Constant e
Investing $1 at a 100% APR for one year, the following table of
amounts — based on number of compounding periods — shows
us the evolution from discrete compounding to continuous
compounding.
(n 1)





1.0 
A = $1 1 +

n 
n = number of compoundings
1 = year
4 = quarters
12 = months
365 = days
365•24 = hours
365•24•60 = minutes
365•24•60•60 = seconds
infinite number of compoundings
A = accumulation
2.0
2.44140625
2.6130352902236
2.7145674820245
2.7181266906312
2.7182792153963
2.7182824725426
e  2.71828182846
Leonhard Euler
(1707-1783)
Compound Interest Formula
for Continuous Compounding
A = P  e ( APR  Y )
A = accumulated balance after Y years
P = principal
APR = annual percentage rate (as a decimal)
Y = number of years (may be a fraction)
e = the special number called Euler’s constant or
the natural number and is an irrational number
approximately equal to 2.71828…
4-A
Example
4-A
4-A
Suppose you have $2000 in an account with an
APR of 5.38% compounded continuously.
Determine the accumulated balance after 1, 5
and 20 years. Then find the APY for this
account.
After 1 year:
(.0538

1)
A(1) = $2000  e
 $2,110.55
4-A
Suppose you have $2000 in an account
with an APR of 5.38% compounded
continuously. Determine the accumulated
balance after 1, 5 and 20 years.
After 5 years: A(5) = $2,000  e (.0538  5)
 $2,617.31
After 20 years:
A(20) = $2,000  e (.0538  20)
 $5,865,85
4-A
Suppose you have $2000 in an account
with an APR of 5.38% compounded
continuously. Then find the APY for this
account.
APY
$2110.55-$2000
=
 100%
$2000
110.55
=
100%
$2000
= 5.5275%
4-A
The Power of Compounding
Simple Interest
A  P  i  P  t 
Compound Interest
Once a year
A  P  (1  APR)
t
“n” times a year
A = P (1+
Continuously
APR (n  Y)
n
)
A = Pe ( APR  Y )
Homework for Wednesday:
Pages 225-226
# 36, 42, 48, 50, 52, 56, 60, 62, 75