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Chapter 21: Savings Models
Lesson Plan
 Arithmetic Growth and Simple Interest
 Geometric Growth and Compound
Interest
 A Limit to Compounding
 A Model for Saving
 Present Value and Inflation
© 2009, W.H. Freeman and Company
For All Practical
Purposes
Mathematical Literacy in
Today’s World, 8th ed.
Chapter 21: Savings Models
Arithmetic Growth and Simple Interest
 Principal – The initial balance of the savings account.
 Interest – Money earned on a savings account or a loan.
 Example: The amount of interest on 10% of the principal of $1000 is
10% × $1000 = 0.10 × $1000 = $100
 Simple Interest
 The method of paying interest only on the initial balance in an
account, not on any accrued interest.
 Example: The following shows the simple interest of a savings account
with a principal of $1000 and a 10% interest rate:
 End of first year, you receive $100 interest.
 The account total at the start of the second year is $1100.
 End of second year, you receive again only $100, which is the
interest from the original balance of $1000.
 Account total at the beginning of the third year is $1200.
 At the end of each year you receive just $100 in interest.
Chapter 21: Savings Models
Arithmetic Growth and Simple Interest
 Bonds
 An obligation to repay a specified
amount of money at the end of a
fixed term, with simple interest
usually paid annually.
 Interest Rate Formula: I = Prt
 The total amount accumulated:
A = P(1 + rt )
Example: Say you bought a 10-year T-note
(U.S. Treasure note) today. What would be
the total amount accumulated in 10 years at
4.0% simple interest?
Answer:
P = $10,000, r = 4.0% = 0.40, and t = 10 yrs.
Interest, I = Prt = (10,000)(0.04)(10) = $4000
Total A= P(1 + rt ) = 10,000(1 + (0.04)(10))
= 10,000(1.40) = $14,000
Interest Rate Formula –
I = Prt
Where:
I = Simple interest earned
P = Principal amount
r = Annual rate of interest
t = Time in years
Total Amount Accumulated
A = P (1 + rt )
Arithmetic Growth – (linear
growth) A = P (1 + rt )
Chapter 21: Savings Models
Geometric Growth and Compound Interest
 Geometric Growth
 Geometric growth is the growth proportional to the amount present
(also called exponential growth).
 Compound Interest
 Interest paid on both the principal and on the accumulated interest.
Rate per Compounding
Period – For a
nominal annual rate
of interest r
compounded n times
per year, the rate per
compounding period
is:
i = r/ n
Compound Interest Formula –
A = P (1 + i ) nt
Where:
A = Amount earned after interest is made
P = Principal amount
i = Interest rate per compounding period,
which is computed as i = r /n
n = Number of compounding periods
t = Time of the loan in years
Chapter 21: Savings Models
Geometric Growth and Compound Interest
 Compounding Period
 The amount of time elapsing before interest is paid.
 For the examples below (annual, quarterly, and monthly compounding),
the amount earned increases when interest is paid more frequently.
Example: Suppose the initial balance is $1000 (P = $1000) and the interest
rate is 10% (r = 0.10). What is the amount earned in 10 years (t = 10) for the
following compounding periods, n?
To answer this problem you need to use the following equations:
Rate per compounding period, i = r / n
Compound Interest Formula, A = P (1 + i ) nt
Annual compounding: i = 0.10, and nt = (1)10 years
A = $1000(1 + 0.10) 10 = $1000(1.10) 10 = $2593.74
Quarterly compounding: i = 0.10/4 = 0.025, and nt = (4)(10) = 40 quarters
A = $1000(1 + 0.025) 40 = $1000(1.025) 40 = $2685.06
Monthly compounding: i = 0.10/12 = 0.008333, and nt = (12)(10) = 120mo.
A = $1000(1 + 0.10/12) 120 = $2707.04
Chapter 21: Savings Models
Geometric Growth and Compound Interest
 Present Value
 The present value P of an amount A to be paid in the future,
after earning compound interest for n compounding periods at
rate i per period is the future amount paid, A divided by (1 + i)nt.
 The equation is
Present Value Formula –
obtained by
manipulating the
A
P
=
formula for finding
(1 + i )nt
Where:
the amount to be
paid in the future
P = Present value (principal amount)
after compound
A = Amount to be paid in the future after
interest is earned.
compounding interest is earned
 With the compound i = Interest rate per compounding period
interest formula
which is computed as i = r /n
A = P (1 + i )nt , we
n = Number of compounding periods in a year
can solve for the
t = The years of the loan
present value, P.
Chapter 21: Savings Models
Geometric Growth and Compound Interest
 Effective Rate
 The effective rate of interest is:
n
 effective rate  1  i   1 where i=r/m and n=mt.
 Annual Percentage Yield (APY)
 The amount of interest earned in 1 year with a principal of $1.
 The annual effective rate of interest.
m
r


APY  1    1
 m
 Example: With a nominal rate of 6% compounded monthly, what is
the APY?
 Solution:
12
 .06 
APY  1 
  1  .0617  6.17%
 12 
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Chapter 21: Savings Models
Geometric Growth and Compound Interest
 Compound Interest Compared to Simple Interest
 The graph compares the
growth of $1000 with
compound interest and with
simple interest.
 The straight line explains why
growth simple interest is also
known as linear growth.
 Example of geometric and
arithmetic growth:
Thomas Robert Malthus (1766–
1843), an English demographer
and economist, claimed that
human population grows
geometrically but food supplies
grow arithmetically—which he
attributed to future problems.
Chapter 21: Savings Models
A Limit to Compounding
 A Limit to Compound Interest
 The following table shows a trend: More frequent compounding
yields more interest.
 As the frequency of compounding increases, the interest tends
to reach a limiting amount (shown in the right columns).
Comparing Compound Interest
The Value of $1000 at 10% Annual Interest, for Different Compounding Periods
Compounded Compounded Compounded Compounded Compounded
Years
Yearly
Quarterly
Monthly
Daily
Continuously
1
1100.00
1103.81
1104.71
1105.16
1105.17
5
1610.51
1638.62
1645.31
1648.61
1648.72
10
2593.74
2685.06
2707.04
2717.91
2718.28
Chapter 21: Savings Models
A Limit to Compounding
 Continuous Compounding
 As n gets very large, (1+ 1/n)n
approaches the constant e ≈ 2.71828.
 For a principal P deposited in an
account at a nominal annual rate r,
compounded continuously, the
balance after t years is:
A = Pe rt
Yield of $1 at 100% Interest (i = 1)
Compounded n Times per Year
n
(1+ 1/n)n
1
2.0000000
5
2.4883200
10
2.5937424
50
2.6915880
100
2.7048138
1,000
2.7169239
Example: For $1000 at an annual rate of 10%,
compounded n times in the course of a
10,000
2.7181459
single year, what is the balance at the end
100,000
2.7182682
of the year?
1,000,000
2.7182818
 As the quantity gets closer and closer to
10,000,000
2.7182818
$1000( e 0.1) = $1105.17.
It approaches e ≈ 2.71828
 No matter how frequently interest is
(which is the base of the
compounded, the original $1000 at the end
natural logarithms).
of one year cannot grow beyond $1105.17.
Chapter 21: Savings Models
A Model for Saving
 A Savings Plan
 To have a specified amount of money in an account at a particular
time in the future, you need to determine what size deposit you
need to make regularly into an account with a fixed rate of interest.
 Savings Formula
The amount A
accumulated after
a certain period of
time can be
calculated by
stating a uniform
deposit of d per
compounding
period (deposited
at the end of the
period) and using a
certain interest rate
i per compounding
period.
Savings Formula
 1  i n  1 
Ad

i


Where:
A = Amount accumulated in the future after
compounding interest is earned
d = Uniform deposits (or payments made)
i = Interest rate per compounding period which
is computed as i = r /m
n = Number of compounding, n=mt
t = The years of the savings plan (or loan)
Chapter 21: Savings Models
A Model for Saving
 Payment Formula:
 Solving for d in the Savings Formula so we can calculate how
much our periodic payment should be in order to have $A in the
future yields:


i
d  A

n
 1  i   1 
An annuity is a
specified number of
(usually equal)
periodic payments.
A sinking fund is a savings plan
to accumulate a fixed sum by a
particular date, usually through
equal periodic payments.
Note: The periodic payment for a sinking fund may be calculated using
the Payment Formula.
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Chapter 21: Savings Models
Present Value and Inflation
 Exponential Decay
 Exponential decay is geometric growth with a negative rate of
growth.
 Present Value of a Dollar a Year from Now with Inflation Rate a.
$1
$a
 $1 
1 a
1 a
The quantity i=–a/(1+a) behaves like a
negative interest rate so we can use the
compound interest formula to find the
present value of P dollars t years from
now.
 Example: Suppose a 25% annual inflation rate from mid-2009
through mid-2013. What will be the value of a dollar in mid-2013 in
constant mid-2009 dollars?
Answer: a=0.25, so i=-a/(1+a)=-0.25/(1+.25)=-0.25/1.25=-0.20 and
P 1  i   $11  0.20    0.80   $0.41
t
4
4
Chapter 21: Savings Models
Present Value and Inflation
 Depreciation Example
Suppose you bought a car at the beginning of 2009 for $12,000
and its value in current dollars depreciates steadily at a rate of 15%
per year. What will be its value at the beginning of 2012 in current
dollars?
Answer: Using the compound interest formula, P = $12,000, i = −0.15, and
n = 3. The projected price is A = P (1 + r)n = $12,000 (1 - 0.15)3 =
$7369.50
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Chapter 21: Savings Models
Present Value and Inflation
 Consumer Price Index
 The official measure of inflation is the Consumer Price Index
(CPI), prepared by the Bureau of Labor Statistics.
 This index represents all urban consumers (CPI-U) and covers about
80% of the U.S. population.
 This is the index of inflation that is referred to on television news
broadcasts, in newspapers, and magazine articles.
 Each month, the Bureau of Labor Statistics determines the average
cost of a “market basket” of goods, including food, housing,
transportation, clothing, and other items.
 The base period used to construct the CPI-U is from 1982–1984 and
is set to 100. CPI for other year
cost of market basket in other year
100
= cost of market basket in base year
 From this proportion calculation, you can also compute the cost of an
item in dollars for one year to what it would cost in dollars in a different
year.
cost in year A
CPI for year A
cost in year B
=
CPI for year B
Chapter 21: Savings Models
Present Value and Inflation
 Real Growth
 If your investments are growing at say 6% a year and inflation is
growing at 3% a year, you cannot simply subtract the two to find
the purchasing power of your investments. It’s not that simple.
 At the beginning of the year, your
Where P is the initial
investment would buy the quantity:
investment principle and
qold = P/m
m is the price of goods.
 At the end of the year, the investment
would be:
P(1 + r )
qnew = m(1 + a)
The investment rate is r
and the inflation is a.
Here, you see the two influences on the investment (investment
growth rate r and inflation a) have directly opposite effects.
Fisher’s effect – Named after the economist Irving Fisher (1867–1947).
To
understand why you cannot simply find the difference between interest and
inflation, you must realize that the gain itself is not in original dollars but in
deflated dollars.
Chapter 21: Savings Models
Present Value and Inflation
 Real Rate of Growth
 The real (effective) annual rate of growth of an investment at
annual interest rate r with annual inflation rate a is:
ra
g
1 a
 Example: Suppose you have an investment earning 8% per year
and the annual inflation rate is 3% per year. What is your real rate
of growth?
Answer: r=.008 and a=0.03, so
g=(0.08-0.03)/(1+0.03)=0.05/1.03=0.0485=4.85%