Download 67 Section 8.4 –Compound Interest Objective #1: Using the

67
Section 8.4 –Compound Interest
Objective #1: Using the compound interest formula.
If an account earns interest during a period of time, the interest is then
added to the principal. During the next cycle, the new principal is the old
principal plus the interest earned during the last cycle. Thus, at the end of
the next cycle, the account not only earns interest on the original principal
but also the interest earned in the last cycle. We say that the interest is
compounded. Thus, compound interest is the interest earned on the
principal and the interest earned previously. How often the interest is
calculated depends on the length of the cycle or payment period. Below are
some common payment periods:
Interest is
compounded … Means
Interest in calculated once a year.
annually
Interest in calculated twice a year or every six months.
semiannually
Interest in calculated 4 times a year or every three months.
quarterly
Interest in calculated 12 times a year or every month.
monthly
Interest in calculated 52 times a year or every week.
weekly
Interest in calculated 360* times a year or every day.
daily
* Banks will define 1 month = 30 days and 1 year = 360 days.
Solve the following:
Ex. 1
Maria invests $3000 in an account that pays an annual interest
rate of 5% compounded quarterly. How much does she have in
the account:
a) after 1 year.
b) after 12 years.
Solution:
In both parts, P = $3000, r = 0.05
a)
In one year's time, the interest will be calculated four times.
st
1 quarter:
A = P(1 + rt) = 3000(1 + (0.05)(1/4)) = $3037.50
nd
2 quarter:
A = P(1 + rt) = 3037.50(1 + (0.05)(1/4)) = $3075.47
rd
3 quarter:
A = P(1 + rt) = 3075.47(1 + (0.05)(1/4)) = $3113.91
th
4 quarter:
A = P(1 + rt) = 3113.91(1 + (0.05)(1/4)) = $3152.84*
* - if you do not round any of your calculations until the end, you get
3152.8360… which rounds to $3152.84 instead of $3152.83.
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b)
In twelve years time, the interest will be calculate 12•4 = 48
times. Thus,
Beginning
Ending Balance
Quarter
Balance P
A = P(1 + rt)
st
1
$3,000.00
$3,037.50
nd
2
$3,037.50
$3,075.47
rd
3
$3,075.47
$3,113.91
th
4
$3,113.91
$3,152.84
:
:
:
:
:
:
th
48
$5,378.83
$5,446.06
Part b of the last example illustrates that we need a better way to compute
compound interest since our current method is unrealistic to use for most
situations. We will need to derive a formula for compound interest:
If we start with principal P, then the total at the end of the 1st cycle is
A1 = P(1 + rt)
If we start with principal A1, then the total at the end of the 2nd cycle is
A2 = A1(1 + rt)
Since A1 is equal to the amount in the account at the end of the first cycle,
then we can replace A1 by P(1 + rt). So,
A2 = A1(1 + rt) = P(1 + rt)(1 + rt) = P(1 + rt)2
If we start with principal A2, then the total at the end of the 3rd cycle is
A3 = A2(1 + rt)
Since A2 is equal to the amount in the account at the end of the second
cycle, then we can replace A2 by P(1 + rt)2. So,
A3 = A2(1 + rt) = P(1 + rt)2(1 + rt) = P(1 + rt)3
Continuing this pattern, the amount after k cycles is Ak = P(1 + rt)k
Let n = the number of times interest is compounded per year. The length of
time t for each payment period is 1/n and the number of cycles k is the time
t in years multiplied by the number of times interest is compounded per
(
year n. Replacing t by 1/n, k by nt, and Ak with A to get: A = P 1 +
r
n
nt
)
Compound Interest Formula
If the principal P (present value) is invested at an annual interest rate r
compounded n times a year for t years, the amount A (future value) will be:
nt
€
r
A=P 1+
where r has been converted to a decimal.
(
€
n
)
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We can now rework example 1 using the compound interest formula with
n = 4:
a)
(
r
n
(
r
A=P 1+
nt
)
(
0.05
4
(
0.05
= 3000 1 +
nt
)
4(1)
)
= 3000(1.0125)4 ≈ $3152.84
4(12)
)
A=P 1+
= 3000 1 +
= 3000(1.0125)48 ≈ $5446.06
n
4
€
€
Find the following:
Ex. 2
Leroy invests $10000 in an account that pays an annual
€ interest rate €
of 4%. How much does he have in the account
after 1 year if interest is compounded:
a) annually?
b) semiannually?
c) quarterly?
d) monthly?
Solution:
In all four parts, P = $10,000, r = 0.04 and t = 1 year.
a)
Compounded annually means n = 1:
b)
(
A=P 1+
r
n
nt
)
(
0.04
1
= 10000 1 +
1(1)
)
= 10000(1.04) = $10,400
Interest = 10400 – 10000 = $400
b)
Compounded semiannually means n = 2:
nt
2(1)
r
0.04
€
€
A=P 1+
= 10000 1 +
= 10000(1.02)2 = $10,404
(
n
)
(
2
)
Interest = 10404 – 10000 = $404
c)
Compounded quarterly means n = 4:
nt
4(1)
r
0.04
€
A€= P 1 +
= 10000
1+
= 10000(1.01)4
(
n
)
(
= $10,406.04
d)
4
)
Interest = 10406.04 – 10000 = $406.04
Compounded monthly means n = 12:
nt
12(1)
r
0.04
€
€
A=P 1+
= 10000 1 +
= 10000(1.00333…)12
(
n
)
(
= $10,407.42
12
)
Interest = 10407.42 – 10000 = $407.42
Increasing the number of times interest is compounded from one to 12
netted us $7.42
As we increasing the number of times
€ in additional interest.
€
interest is compounded per year, the amount increases so the question
becomes if we increase the number of times interest is compounded
without bound, how much more money can we earn in interest? To find out,
consider the following:
n
r
1
n
Let h =
which implies that = and nt = •rt = h•rt
r
€
n
€
€
h
r
€
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nt
(
A = P(
A€= P[(
A=P
)
)
)]
r
r
1+
(replace
n
n
hrt
1
1+
(rewrite as
h
h rt
1
1+
€
€
h
by
1
h
and nt by h•rt)
a power of rt)
n
Now, consider the expression inside the square brackets. Since h = , as
r
€
we increase n without bound, h increases without bound. This gives us a
€
special number
that occurs in nature. It is the natural base e. The number e
is an irrational number that is approximately 2.71828182846.... It comes
h
€
1
from examining the behavior of 1 +
as h → ∞ (increases without
(
bound). It may seem that 1 +
( h)
1 h
should approach 1, but if you plug in
h)
some large values for h, you will see this is not the case. For example, if h
h
€
1
= 10,000, then 1 +
≈ 2.71814592682...
h
€
h
(1 + 1/h)h
h
(1 + 1/h)h
1
2
1000
2.716923932
€
10
2.59374246
10,000
2.718145927
100
2.704813829
1,000,000
2.718280469
(
)
[(
Thus, as h increases without bound, A = P 1 +
1
h
h rt
)]
becomes A = Pert.
It is important to be able to evaluate expressions involving e. On your
calculator, you should have an ex key. On the TI-30XS, you will hit the 2nd key
€ hit enter.
and hit ln key. Then type the exponent and
Evaluate the following:
Ex. 3a
e4.3
Solution:
a) e4.3 ≈ 73.6997937
Ex. 3b
2e3
b) 2e3 ≈ 40.17107385
For the previous example, if number of times interest is compounded
increases without bound, the amount in the account will be:
A = 10000e0.04(1) ≈ $10408.11
This means there is a limit to how much more interest can be gained.
When we have an "infinite" number of times interest is compounded, we
say that interest is compounded continuously.
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Interest Compounded Continuously
If the principal P (present value) is invested at an annual interest rate r
compounded continuously for t years, the amount A (future value) will be:
A = Pert
where r has been converted to a decimal.
Find the following:
Ex. 4
Maria invests $3000 in an account that pays an annual interest
rate of 5% compounded continuously. How much does she
have in the account after 12 years?
Solution:
Plug in P = $3000, r = 0.05 and t = 12:
A = Pert = 3000e0.05(12) = 3000e0.6 ≈ $5466.36
Ex. 5
Alissa has $5000 that she can invest for 6 years in one of two
accounts. The first account pays 8% interest compounded
quarterly and the second pays 7.92% interest compounded
continuously. Which is the better investment?
Solution:
First, plug in P = $5000, r = 0.08, n = 4, and t = 6 in the compound
interest formula:
(
A=P 1+
r
n
nt
)
(
= 5000 1 +
0.08
4
4(6)
)
= 5000(1.02)24
≈ $8042.19
Interest = $8042.19 – 5000 = $3042.19
Now, plug in P = 5000, r = 0.0792, and t = 6 in the compounded
continuously formula:
€
A€= Pert = 5000e0.0792(6)
= 5000e0.4752 ≈ $8041.68
Interest = $8041.68 – 5000 = $3041.68
The first investment paying 8% compounded quarterly is the better
investment.
Objective #2:
Determining the Present Value of a Lump Sum.
You can think of the present value as how much money that you have to
put in an account today so that it grows to some predetermined amount in
the future. Thus, if you want to have $30,000 in an account 20 years from
now, the present value would be how much you would put into the account
today so it grows to $30,000 in 20 years.
Starting with our compound interest formulas, we need to solve for P to find
the present value:
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Compounded n times per year:
(
A=P 1+
(
r
n
A 1+
– nt
nt
)
(multiply both sides by 1 +
=P
(rewrite the formula)
)
(
P€= A 1 +
€
r
n
r
n
– nt
)
(
=
A
(1 +
r
n
– nt
)
)
(book uses the
€ second formula)
r nt
)
n
Present Value Formulas
€ A be the balance wanted in the future and r be the annual interest rate
Let
(written as€a decimal) compounded n times a year for t years. Then the
present value P of the account is:
(
P=A 1+
r
n
– nt
)
=
A
(1 +
r nt
)
n
Solve the following:
Ex. 6 € Find the amount that Juanita should invest in an account paying
€
8% annual
interest so that after 20 years, she will have $50,000
if the interest is compounded quarterly.
Solution:
If interest is compounded quarterly, then n = 4, A = $50,000,
r = 0.08, and t = 20. Thus,
(
P=A 1+
r
n
– nt
)
(
= 50000 1 +
0.08
4
– 4•20
)
= 50000(1.02)– 80
≈ 50000(0.2051097282) ≈ $10,255.49.
Objective #3:
€
Calculating Effective Rates of Return.
€
When comparing several different accounts, it is useful to determine what
the equivalent interest rate would be if the money was invested an account
paying simple interest for one year. This equivalent interest rate is known
as the effective rate of interest.
Effective Rate of Interest
The effective rate of interest Y of an investment earning an interest rate r
(
compounded n times a year is Y = 1 +
decimal.
€
r
n
n
)
– 1 where r is converted to a
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Solve the following:
Ex. 7
Suppose Bank A pays 4.52% annual interest compounded
annually while Bank B pays 4.5% annual interest compounded
quarterly and Bank C pays 4.47% annual interest compounded
daily. Which one is the best deal?
Solution:
Bank A
Bank B
Bank C
r = 0.0452; n = 1
r = 0.045; n = 4
r = 0.0447; n = 360
(
(
= 1+
€
n
)
r
–1
n
1
0.0452
–
1
Y= 1+
)
(
(
1
= (1.0452) – 1
€
= 0.0452 = 4.52%
Bank B is the best deal.
= 1+
n
)
(
r
–1
n
4
0.045
–1
4
Y= 1+
4
)
(
= 1+
= (1.01125) – 1
€
= 0.04576… ≈ 4.58%
€
n
)
)
r
–1
n
360
0.0447
–1
360
Y= 1+
= 0.04571…
€
≈ 4.57%
€
When investing money, you will want to look for the account with the
highest effective rate of interest. When borrowing money, you will want to
look for the loan that has the lowest effective rate of return. The effective
rate of return in borrowing money is also called the annual percentage
rate or APR.