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Transcript
Chapter 5 Mathematics of Finance
• Compound Interest
• Annuities
• Amortization and Sinking Funds
• Arithmetic and Geometric Progressions (optional)
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Section 5.1 Compound Interest
Simple Interest Formula
Simple Interest - interest that is compounded on the original principal only.
Interest:
I = Prt
Accumulated amount:
A = P(1 + rt)
A = Accumulated amount
P = Principal
r = Interest rate per year
t = Number of years
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. $800 is invested for 9 years in an account that pays 12% annual simple
interest. How much interest is earned? What is the accumulated amount in
the account?
Solution: note P = $800, r = 12%, and t = 9 years.
Interest: I = Prt
= (800)(0.12)(9)
= 864
or $864
Accumulated amount = principal + interest
= 800 + 864 = 1664
or $1664
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Compound Interest
Compound Interest – Interest is added to the original principal and then earns
interest at the same rate.
A  P (1  i ) n
where i 
r
and n  mt
m
A = Accumulated amount after n periods
P = Principal
r = Nominal interest rate per year
m = Number of conversion periods per year
t = Term (number of years)
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. Find the accumulated amount A, if $4000 is invested at 3% per year for 6
years, compounded monthly.
Solution: note P = $4000, r = 3%, t = 6, and m = 12.
So
i
r 0.03

 .0025 and n  mt  12(6)  72
m 12
A  P (1  i ) n
 4000(1  0.0025)72
 4787.79
or $4787.79
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Effective Rate of Interest
Effective Rate – the simple interest rate that would produce the same
accumulated amount in 1 year as the nominal rate compounded m times per
year.
m
reff
r

 1    1
 m
where
reff = Effective rate of interest
r = Nominal interest rate per year
m = Number of conversion periods per year
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. Find the effective rate that corresponds to a nominal rate of 6% per
year compounded quarterly.
Solution: note r = 6% and m = 4.
m
reff
r

 1    1
 m
4
 0.06 
 1 
 1
4 

 0.06136
So about 6.136% per year.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Present Value (Compound Interest)
Present Value (principal) – the amount required now to reach the desired
future value.
P  A(1  i ) n
where i 
r
and n  mt
m
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. Jackson invested a sum of money 10 years ago in an account that paid
interest at a rate of 8% per year compounded monthly. His investment has
grown to $5682.28. How much was his original investment?
A = $5682.28, r = 8%, t = 10, and m = 12
i
r 0.08

and n  mt  12(10)  120
m 12
 0.08 
P  5682.28 1 

12 

 2560.00
120
or $2560
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Section 5.2 Annuities
Annuity
Annuity – a sequence of payments made at regular time intervals.
Ordinary Annuity – payments made at the end of each payment
period.
Simple Annuity – payment period coincides with the interest
conversion period.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Future Value of an Annuity
The future value S of an annuity of n payments of R dollars each, paid at
the end of each investment period into an account that earns interest at the
rate of i per period is
 (1  i ) n  1 
S  R

i


Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. Find the amount of an ordinary annuity of 36 monthly payments of
$250 that earns interest at a rate of 9% per year compounded monthly.
R = 250, n = 36, and
i
0.09
12
 (1  i ) n  1 
S  R

i


  0.09 36
 1 
 1
12


S  250 
0.09


12

S  10288.18






or $10,288.18
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Present Value of an Annuity
The present value P of an annuity of n payments of R dollars each, paid at
the end of each investment period into an account that earns interest at the
rate of i per period is
1  (1  i )  n 
P  R

i


Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. Paige’s parents loaned her the money to buy a car. They required that
she pays $150 per month, for 60 months, with interest charged at 2% per
year compounded monthly on the unpaid balance. What was the original
amount that Paige borrowed?
1  (1  i )  n 
P  R

i


  0.02  60
1   1 

12


 150 
0.02


12

 8557.85






or $8557.85
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Section 5.3 Amortization and Sinking Funds
Amortization Formula
The periodic payment R on a loan of P dollars to be amortized over n
periods with interest charged at a rate of i per period is
R
Pi
1  (1  i)  n
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. The Kastners borrowed $83,000 from a credit union to finance the
purchase of a house. The credit union charges interest at a rate of 7.75%
per year on the unpaid balance, with interest computations made at the end
of each month. The Kastners have agreed to repay the loan in equal
monthly installments over 30 years. How much should each payment be if
the loan is to be amortized at the end of the term?
Solution: P = 83000, n = (30)(12) = 360, and i 
R
Pi
1  (1  i )  n
0.0775
12
 0.0775 
83000 

12 


360  594.62
0.0775


1  1 

12 

So the monthly installment is $594.62.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. A bank has determined that the Radlers can afford monthly house
payments of at most $750. The bank charges interest at a rate of 8% per
year on the unpaid balance, with interest computations made at the end of
each month. If the loan is to be amortized in equal monthly installments
over 15 years, what is the maximum amount that the Radlers can borrow
from the bank?
0.08
i

Solution: note R = 750, n = (15)(12) = 180, and
12
1  1  i  n 

P  R
i


  0.08  180 
1   1 


12


  78480.44
 750 
0.08




12


So they can borrow up to about $78480.44.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Sinking Fund Payment
The periodic payment R required to accumulate S dollars over n periods
with interest charged at a rate of i per period is
iS
R
(1  i ) n  1
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Ex. Max has decided to set up a sinking fund for the purpose of
purchasing a new car in 4 years. He estimates that he will need $25,000.
If the fund earns 8.5% interest per year compounded semi-annually,
determine the size of each (equal) semi-annual installment that Max
should pay into the fund.
Solution: note S = 25000, n = 4(2) = 8, and i 
R
iS
(1  i ) n  1
0.085
2
 0.085 

 25000
2 

 2689.12
8
 0.085 
1 
 1
2 

So the semi-annual payment is about $2689.12.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.