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Chapter 5Mathematics of Finance
Section 5.1 Simple Interest
Section 5.2 Compound Interest
Section 5.3 Annuities and Sinking Funds
Section 5.4 Present Value of an Annuity and
Amortization
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Simple Interest
Simple interest is most often used for loans of shorter
duration.
The money borrowed in a loan is called theprincipal.
The number of dollars received by the borrower is the
present value.
In a simple interest loan, the principal and present value
are the same. The interest rate is the fee for a simple interest loan and
usually is expressed as a percent of the principal.
Simple interest is paid on the principal borrowed and
not paid on interest already earned.
Section 5.1
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Simple Interest
I=Prtwhere P= principal (amount borrowed)
r= interest rate per year (in decimal form)
t= time in years.
The simple interest of a loan can be calculated using the
following formula.
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Example
An individual borrows $300 for 6 months at 1% simple
interest per month. How much interest is paid?
SOLUTIONRemember to check that rand tare consistent in time
units. Here, it will be months.
I=Prt= 300 0.01 6 = $18
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Another Example
Jane borrowed $950 for 15 months. The interest was
$83.13. Find the interest rate.
SOLUTION
First, determine the information we have been given.
P= 950
t= 15/12 = 1.25 years
I= 83.13
UsingI=Prt, we need to find r
83.13 950 (1.25)1187.5
83.130.07
1187.5
r
r
r
The interest rate was 7%.
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Future Value
A loan made at simple interest requires that the borrower pay
back the sum borrowed (principal) plus the interest. This total
is called thefuture value, oramountand is equal toP+I.
A =P+I
= P + Prt
= P(1 + rt)
where P= principal or present value
r= annual interest rate (in decimal form)
t= time in years
A = amount or future value
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Total Number of Compounding Periods
Annual Interest rate expressed as a percent
Present Value: What it is worth to you right now
Amount of the payment being made if any.
Future Value: What it will be worth to you
in the future
The number of payments per year
The number of compounds per year.
Is Interest/Payment due at the beginning or
end of the compound period?
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Example
Find the amount (future value) of a $2400 loan for 9 months
at 11% interest.
SOLUTIONWe need to findA which equalsP+I=P+Prt.
Now, determine the information we have been given.9
122400, 0.11, and 0.75 yearsP r t
FindI: I= 2400(0.11)(0.75) = 198Now, findA: A = 2400 + 198 = 2598
A could have also been found
using the formulaA =P(1 + rt)
2400(1 0.11(0.75))2400(1 0.0825)2400(1.0825)2598
A
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Simple Discount
The simple discount loan differs from the simple interest
loan in that the interest is deducted from the principal and
the borrower receives less than the principal.
This type of loan is referred to as asimple discount note.
The interest deducted is the discount.
The amount received by the borrower is theproceeds.
The discount rate is the percentage used.
The amount repaid is the maturity value.
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Simple Discount Note
D =Mdt
PR = MD
= M
Mdt= M(1dt)
where M= maturity value (principal)
d= annual discount rate (in decimal form)t= time in years
D = discount
PR = proceeds or amount the borrower receives
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Example
Find the discount and the amount a borrower receives (proceeds)
on a $1500 simple discount loan at 8% discount rate for 1.5
years.
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Example
A bank paid $987,410 for a 90-day $1 million treasury
bill. What was the simple discount rate?
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Example
A bank wants to earn 7.5% simple discount interest on a
90-day $1 million treasury bill. How much should it bid?
SOLUTIONFirst, determine what information is given.
90
3601,000,000, 0.075, and 0.75 yearsM d t
We need to find the proceeds which equalsPR.
90360
1,000,000 1 (0.075)
1,000,000 1 0.01875
1,000,000 0.98125
981,250
PR
The bank should bid $981,250.
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Example
How much should a bank bid on a 30-day $2 million
treasury bill if the bank wants to earn 5.125% on its money.
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HW 5.1
Pg 348-350 1-32, 34-64 even
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Compound Interest
Section 5.2
Suppose you deposit money into a savings account, the bank
will typically pay you interest for the use of your money at a
specific period of time, say every three months. The interest
is usually credited to your savings account at each timeperiod. At the next time period, the bank will pay interest on
the new total, this is called compound interest.
Amount of Annual Compound Interest
WhenPdollars are invested at an annual interest rate randthe interest is compounded annually, the amountA at the end
oftyears is
A =P(1 + r)t
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Example
Suppose $800 is invested at 6%, and it is compounded
annually. What is the amount in the account at the end
of 4 years?
SOLUTION `
First, determine the information that is given.
P= 800, r= 0.06, and t= 4
4800(1.06) 800(1.26248) 1009.98A
There will be $1009.98 in the account after 4 years.
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Amount (Future Value)
The general formula for finding the amount after a specified number
of compound periods is
A =P(1 + i)n
where r= annual interest rate
m = number of times compounded per year
i = r/m = interest rate per periodn = mt, the number of periods, where tis the number of years
A = amount (future value) at the end ofn compound periods
P= principal (present value)
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Example
Suppose $800 is invested at 12% for 2 years. Find the amount
at the end of 2 years if the interest is compounded (a)
annually, (b) semiannually, and (c) quarterly.
SOLUTION
First, determine what information is given then useA =P(1 + i)n.
P= 800, r= 0.12, and t= 2 years.
2 2800(1 0.12)
a) 1, so
8
0.12/1 0.
00(1.12) 800(1.2544) 1003.5
12, 2;
2A
m i n
4 4800(1 0.06) 800(1.06) 800(1.26248)b) 2, so 0.12/ 2 0.06, 4;
1009.98m
A
i n
8 8800(1 0.03) 800(1.03) 800(1.26677)
c) 4, so 0.12/4 0.03, 8;
1013.42
m
A
i n
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Total Number of Compounding Periods
Annual Interest rate expressed as a percent
Present Value: What it is worth to you right now
Amount of the payment being made if any.
Future Value: What it will be worth to you
in the future
The number of payments per year
The number of compounds per year.
Is Interest/Payment due at the beginning or
end of the compound period?
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(compounds per year)(# of years)
Annual Interest rate expressed as a percent
Present Value: What it is worth to you right now
0
Future Value: What it will be worth
Compounds per year
Always on End
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Example
Suppose $800 is invested at 12% for 2 years. Find the amount at
the end of 2 years if the interest is compounded (a) annually
1(2) 1 compound every year for two years
12 Yearly interest rate
-800 The amount being invested todayPresent Value
0 No regular payments are being made
What we want to solve forAlpha Enter
1There is 1 compound per year
1
1003.52
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Example
Suppose $800 is invested at 12% for 2 years. Find the amount at
the end of 2 years if the interest is compounded (b) semiannually
2(2) 2 compound every year for two years
12 Yearly interest rate
-800 The amount being invested todayPresent Value
0 No regular payments are being made
What we want to solve forAlpha Enter
2There are 2 compound per year
2
1009.98
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Example
Suppose $800 is invested at 12% for 2 years. Find the amount at
the end of 2 years if the interest is compounded (c) quarterly.
4(2) 4 compound every year for two years
12 Yearly interest rate
-800 The amount being invested todayPresent Value
0 No regular payments are being made
What we want to solve forAlpha Enter
4There is 4 compound per year
4
1013.42
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1(5) 1 compound every year for five years
What we want to solve forAlpha Enter
-100 We do not know the present value so use $100
0 No regular payments are being made
If we started with $100 and we want to double it?
1There is 1 compound per year
1
14.8698355
200
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Vocabulary
Nominal Rate
Compound rate thatreturns the same total
value as a simpleinterest investment.
Effective Rate
simple interest rate
that produces thesame total value ofinvestment per yearas the compoundinterest.
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Effective Rate
The effective rate of an annual interest rate rcompounded m
times per year is the simple interest rate that produces the same
total value of investment per year as the compound interest.
If money is invested at an annual rate rand compounded m
times per year, the effective rate,x, in decimal form is
x = (1 + i)m1 where i = r/m
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Example
The Mattson Brothers Investment Firm advertises Certificates of
Deposit paying a 7.2% effective rate. Find the annual interest
rate, compounded quarterly, that gives the effective rate.
SOLUTIONIf we let i = quarterly rate, then
4
4
4
0.072 (1 ) 11.072 (1 )
1.072 11.017533 10.017533
i
i
ii
i
The annual rate = 4(0.017533) = 0.070133 = 7.013% (rounded).
The annual rate just found is also called the nominal rate.
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HW 5.2
Pg 359-360 1-17 odd, 18-63 every 3rd
Section 5 3
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Ordinary Annuity
An annuity refers to equal payments paid at equal time
intervals.
The time between successive payments is called the
payment period.
The amount of each payment is theperiodic payment.
The interest on an annuity is compound interest.
An ordinary annuity is an annuity with periodic payments
made at the end of each payment period.
Section 5.3
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Future Value (Amount)
Payments are made at the end of each period.
where i = interest rate per period
n = number of periods
R = amount of each periodic paymentA = future value of amount
(1 ) 1n
iA R
i
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Example
How much money will you have when you retire if you save $20
each month from graduation until retirement? Lets assume you
start saving at age 22 until age 65, 43 years, and the interest rate
averages 6.6% annual rate compounded monthly.SOLUTION
The periodic rate = 0.066/12 = 0.0055 since payments are
made monthly. The number of periods is n = 12(43) = 516.
The periodic payments areR = 20. Substitute these valuesinto the formula for future value.516(1 ) 1 (1.0055) 1
20 57,997.300.0055
ni
A Ri
You will accumulate $57,997.30 in 43 years.
How much money will you have when you retire if you save $20 each
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How much money will you have when you retire if you save $20 each
month from graduation until retirement? Lets assume you start saving at
age 22 until age 65, 43 years, and the interest rate averages 6.6% annual
rate compounded monthly.
12(43) 12 compounds every year for 43 years
6.6 Yearly interest rate
0 The amount being invested todayPresent Value-20 You are putting 20 per month in the account
What we want to solve forAlpha Enter
12
There is 12 compound per year12
57997.30
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1(20) 1 compounds every year for 20 years
8.5 Yearly interest rate
0 The amount being invested todayPresent Value-2000 You are putting 2000 per year in the account
What we want to solve forAlpha Enter
1
There is 1 compound per year1
96754.03
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Sinking Funds
A sinking fund refers to a fund that is created when an amount of
money will be needed at some future date. For example, a family
may need a new car in 3 years, or a company may expect to
replace a piece of equipment in the future.
Formula for Periodic Payments of a Sinking Fund
where A = value of the annuity after n payments
n = number of payments
i = periodic interest rate
R = amount of each periodic payment
(1 ) 1
n
AiR
i
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ExampleDarden Publishing Company plans to replace a piece of
equipment at an expected cost of $65,000 in 10 years. The
company establishes a sinking fund with annual payments.
The fund draws 7% interest, compounded annually.
What are the periodic payments?
SOLUTION
First, determine the information that is given.
10
65,000, 0.07, and 10
65000(0.07) 45504704.54
(1.07) 1 0.9671513
A i n
R
The annual payments are $4704.54.
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HW 5.3
Pg 372-374 1-47 odd
Section 5.4
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Present Value
Section 5.4
The present value of an annuity is the lump sum payment
that yields the same total amount as that obtained through
equal periodic payments made over the same period of time.
The present value of an annuity is
where i = periodic rate
n = number of periods
R = periodic payments
P= present value of the annuity
(1 ) 1 1 (1 )
(1 )
n n
n
i iP R R
i i i
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Example
Find the present value of an annuity with periodic payments
of $2000, semiannually, for a period of 10 years at an interest
rate of 6% compounded semiannually.
SOLUTIONUse the values given,R = 2000, i = 0.06/2 = 0.03 and n = 20
in the formula for present value.20 20
20
(1 ) 1 (1.03) 1 1 (1 0.03)2000 2000
(1 ) 0.03(1.03) 0.03
2000(14.8774749) 29754.95
n
n
iP R
i i
The present value of the annuity is $29,754.95. This lump sum
will accumulate the same amount in 10 years as investing $2000
semiannually for 10 years.
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Equal Periodic Payments
The amount needed to provide equal periodic
payments can be found using the formula
or equivalently,
where P= amount needed in the fund
R = amount of periodic payments
i = periodic interest rate
n = number of payments
(1 ) 1
(1 )
n
n
iP R
i i
1 (1 ) niP R
i
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Example
Find the present value of an annuity (lump sum investment)
that will pay $1000 per quarter for 4 years. The annual
interest rate is 10%, compounded quarterly.
SOLUTION
16
0.01Given 1000, 0.025, and 16 quarters,
4
1 (1.025)
1000 0.025
1000(13.0550027) 13,055
R i n
P
A lump sum investment of $13,055 will provide $1000 per
quarter for 4 years.
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Amortization
The amortization of a debt (repayment of a debt) requires no
new formula because the amount borrowed is just the
present value of an annuity.
Amortization of a Loan (Debt Payments)The amount borrowed,P, is related to the periodic payments,R,
by the formula
where i = periodic interest rate
n = number of payments
Note: This is the present value formula for an ordinary annuity.
(1 ) 1 1 (1 )
(1 )
n n
n
i iP R R
i i i
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Example
A student obtained a 24-month loan on a car. The monthly
payments are $395.42 and are based on a 12% interest rate.
What was the amount borrowed?
SOLUTION
Since the amount borrowed is the present value of the
annuity, we have
395.420.12
0.0112
24
R
i
n
(Monthly rate)
(Number of months)
241 (1.01)
395.42
0.01
395.42(21.2433873)
8400.06
P
The amount borrowed was $8400.
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Balance of an Amortization
The balance aftern periods is the amount of compound
interest minus the amount of an annuity. Mathematically we
can find the balance using the formula
(1 ) 1Balance (1 )
nn i
P i Ri
where P= the amount borrowed
i = periodic interest raten = number of time periods elapsed
R = monthly payments
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ExampleA family borrowed $60,000 to buy a house. The loan was for
30 years at 12% interest rate. The monthly payments were
$617.17. What is the balance of their loan after 2 years?
SOLUTION 1212
60,000, 1% per month, 617.17, 24 monthsP i R n 24
24 (1.01) 1Balance 60,000(1.01) 617.17
0.01
60,000(1.26973) 617.17(26.9734649)76,184.08 16,647.21
59,536.87
The part of the loan repaid is the equity which after
2 years is is 60,00059,536.87 = $463.13.
This is the balance of the
loan after two years.
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HW 5.4
Pg 389-391 1-47 Odd