Chapter 21: Savings Models
October 16, 2013
Chapter 21: Savings Models
Last time
Arithmetic Growth
Simple Interest
Geometric Growth
Compound Interest
A limit to Compounding
Chapter 21: Savings Models
Problems
Question: I put $1,000 dollars in a savings account with 2%
nominal interest per year. How much money will I have after 10
years? with Simple Interest? Compounded annually? Compounded
quarterly? Compounded daily? Compounded continuously?
Question: I had a CD with National City Bank through 2010 that
paid 4.69% interest compounded daily. What was the APY fro this
rate?
Question: A Paper Series EE Savings Bond is sold at half face
value, to the full face value by 20 years from the issue date. What
is the minimum APY for such a bond
Chapter 21: Savings Models
Question 1 Answer
I put $1,000 dollars in a savings account with 2% nominal interest
per year. How much money will I have after 10 years? with Simple
Interest?
Simple Interest
For a principal P and an annual rate of interest r, the interest
earned in t years is
I = Prt
and the total amount A accumulated in the account is
A = P(1 + rt)
Answer:
A = 1000(1 + 0.02(10)) = 1000(1.2) = 1200
Chapter 21: Savings Models
Question 1 Answer
Compound Interest Formula for an Annual Rate
An initial principal P in an account that pays interest at a a
nominal annual rate r compounded m times per year, grows after t
years to
r mt
A=P 1+
m
Answer: Compounded annually
0.02 10
) = 1000(1.02)10 = 1218.99
1
Answer: Compounded quarterly
A = 1000(1 +
0.02 10(4)
)
= 1000(1.005)40 = 1220.79
4
Answer: Compounded daily
A = 1000(1 +
A = 1000(1 +
0.02 3650
)
= 1221.40
365
Chapter 21: Savings Models
Question 1 Answer
I put $1,000 dollars in a savings account with 2% nominal interest
per year. How much money will I have after 10 years?
Compounded continuously?
Continuous Interest Formula
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
Answer: Compounded continuously?
A = 1000e 0.02(10) = 1000e 0.2 = 1221.4027
Chapter 21: Savings Models
Question 2 Answer
Question: I had a CD with National City Bank through 2010 that
paid 4.69% interest compounded daily. What was the APY fro this
rate?
Formula for APY
APY = (1 +
where
APY
r
m
=
=
=
r m
) −1
m
annual percentage yield (effective annual rate)
nominal interest rate
number of compounding periods per year
Answer:
APY = (1 +
0.0469 365
) − 1 = .04801, 4.8%
365
Chapter 21: Savings Models
Question 3 Answer
Question: A Paper Series EE Savings Bond is sold at half face
value, to the full face value by 20 years from the issue date. What
is the minimum APY for such a bond?
Solve:
P(1 + APY )20 = 2P
(1 + APY )20 = 2
1 + APY = 21/20
APY = 21/20 − 1
APY = 0.0352649
Chapter 21: Savings Models
Problem
If I put a $1,000 a year into a savings account with APY 10%.
How much money will I have at the end of 5 years?
Money is first deposited Jan 1, 2001 and interest is first calculated
Dec 31 2001. How money will there be in the savings account on
Jan 2, 2006? Answer:
A = 1000 + 1000(1 + .10) + 1000(1 + .10)2 + 1000(1 + .10)3
+ 1000(1 + .10)4 + 1000(1 + .10)5
A = 1000(1 + 1.1 + 1.12 + 1.13 + 1.14 + 1.15
A = 6998.95
Chapter 21: Savings Models
This Time
A Model for Saving
Present Value and Inflation
Chapter 21: Savings Models
Geometric Series
Formula for Geometric Series
S = a0 + a0 x + a0 x 2 + a0 x 3 + · · · + a0 x n−1
= a0 1 + x + x 2 + x 3 + · · · + x n
n
x −1
S = a0
, x 6= 1
x −1
Chapter 21: Savings Models
Savings Formula
(1 +
(1 + i)n − 1
A=d
=d
i
where
A=
d=
n = mt
r=
m=
t=
i= r/m
r mt
m)
r
m
−1
amount accumulated
regular deposit of payment at the end of each period
number of periods
nominal annual interest rate
number of compounding periods per year
number of years
periodic rate, the interest rate per compounding period
Chapter 21: Savings Models
Payment Formula
Question: What if I want 5,000 dollars in savings account with
APY 10% at the end of five years, how much money will I need to
deposit at the end of every year?
Payment Formula
i
r /m
d =A
=A
(1 + i)n − 1
(1 + mr )mt − 1
Answer:
50000(
.1
) = 818.99
(1.1)5 − 1
Chapter 21: Savings Models
Savings Plans
Sinking Fund
A sinking fund is a savings plan to accumulate a fixed sum by a
particular date, usually through equal periodic deposits.
Annuity
An annuity is a specified number of equal periodic payment.
Chapter 21: Savings Models
Present Value and Inflation
Present Value
The present value of an amount to be paid or received at a
specified time in the future is what that future payment would be
worth today, as determined from a given interest rate and
compounding period.
P=
A
A
=
n
(1 + i)
(1 + r /m)mt
Question: What is the present value of $20,000, 5 years from now,
at an APY of 5%?
Chapter 21: Savings Models
Present Value and Inflation
Present Value
The present value of an amount to be paid or received at a
specified time in the future is what that future payment would be
worth today, as determined from a given interest rate and
compounding period.
P=
A
A
=
n
(1 + i)
(1 + r /m)mt
Question: What is the present value of $20,000, 5 years from now,
at an APY of 5%?
Answer:
P=
20000
= 15670.50
(1 + 0.05)5
Chapter 21: Savings Models
Inflation
Inflation
Inflation is a rise in prices from a set base year.
Annual Rate of Inflation
The annual rate of inflation, a (= 100a%), is the additional
proportionate cost of goods one year later. Goods that cost $ i in
the base year will then cost $ (1+a).
Relative Purchasing Power of a Dollar a Year from Now with
Inflation Rate a
$1
$a
= $1 −
1+a
1+a
Chapter 21: Savings Models
Example
Suppose that there is constant 3% annual inflation from mid-2009
through mid-2013. What will be the projected price in mid-2013 of
an item that cost $100 in mid-2009?
Chapter 21: Savings Models
Example
Suppose that there is constant 3% annual inflation from mid-2009
through mid-2013. What will be the projected price in mid-2013 of
an item that cost $100 in mid-2009?
Answer:
A = 100(1 + 0.03)4 = 112.55
Chapter 21: Savings Models
Depreciation
Question: How much is a mid-2009 dollar worth in terms of a mid
-2013 dollar, with 3% inflation?
a
The lost in purchasing power due to an inflation rate of a is 1+a
The relative purchasing power of P dollars t years from now as
t
a
t
A = P(1 + i) = P 1 −
1+a
Answer:
A = (1 −
0.03 4
) = 0.888487
1.03
Chapter 21: Savings Models
Problems
Question 1: Suppose that you want to save up $2000 for a
semester abroad two years from now. How much do you have to
put away each month in a savings account that earns 2 % interest
compounded monthly?
Question 2: A colleague feels that he will need $1 million in
savings to afford to retire at age 65 and still maintain his current
standard of living. Younger colleague, age 30, decides to begin
savings for retirement based on that advice. How much does the
younger colleague need to save per month to have $ 1 million at
retirement if the fund earns a steady 3% annual interest
compounded monthly?
Question 3: Suppose you start saving for retirement at age 45.
How much do you have to save per month, with a steady return of
6% compounded monthly, to accumulate $250,000 by age 65?
Chapter 21: Savings Models
Problems
Question 4: What is the present value of $10,000, 4 years from
now, at an APY of 5%?
Question 5: What is the present value of $15,000, 10 years from
now, at an APY of 3%?
Question 6: Suppose that inflation proceeds at a constant rate of
2% per year from mid- 2012 through mid 2015.
a) Find the cost in mid-2015 of a basket of goods that cost $1
in mid-2012.
b) What will be the value of a dollar in mid-2015 in constant
mid-2012 dollars?
Chapter 21: Savings Models
Next Time
Real rate of Growth
The real annual rate of growth of an investment at annual interest
rate r with annual inflation rate a is
g=
r −a
1+a
Chapter 21: Savings Models