Chapter 3
Mathematics of
Finance
Section 4
Present Value of an
Annuity; Amortization
Learning Objectives for Section 3.4
Present Value of an Annuity;
Amortization
 The student will be able to calculate the present value of an
annuity.
 The student will be able to construct amortization schedules.
 The student will be able to calculate the payment for a loan.
 The student will have developed a strategy for solving
mathematics of finance problems.
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Present Value of an Annuity
 In this section, we will address the problem of determining
the amount that should be deposited into an account now at
a given interest rate in order to be able to withdraw equal
amounts from the account in the future until no money
remains in the account.
 Here is an example: How much money must you deposit
now at 6% interest compounded quarterly in order to be
able to withdraw $3,000 at the end of each quarter year for
two years?
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Derivation of Formula
 We begin by solving for P in the compound interest
formula:
A  P 1  i  
n
P  A(1  i )
n
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Present Value of the
Eight Payments
 Interest rate each period is 0.06/4=0.015
 0.06 
P1  3000 1 

4 

P2  3000 1.015 
1
2
3
P3  3000(1.015)
4
P4  3000(1.015)
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Derivation of Short Cut Formula
 We proceed to calculate the eight payments. We could
simply find the total of the 8 payments. There are 8
payments since there will be 8 total withdrawals:
(2 years)  (four withdrawals per year) = 8 withdrawals.
This method is tedious and time consuming so we seek a
short cut method.
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Present Value of an Ordinary Annuity
 1  (1  i )  n 
PV  PMT 

i




i
PMT  PV 
n 
 1  (1  i ) 
PV = present value of all payments
PMT = periodic payment
i = rate per period
n = number of periods
Note: Payments are made at the end of each period.
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Understanding the Concept of
Present Value
Interest Rates, Compounding, and
Present Value
 In economics, an interest rate is known as the yield to
maturity.
 Compounding is the process that gives us the value of a
sum invested over time at a positive rate of interest.
 Present value is the process that tells us how much an
expected future payment is worth today.
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Compounding
 Assume you have $1 which you place in an account paying
10% annually.
 How much will you have in one year, two years, etc?
An amount of $1 at 10% interest
Year
1
2
3
n
o
$1.10
$1.21
$1.33
$1(1 + i)n
 Formula: FV = PV(1 + i)
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Compounding over Time
 Extending the formula over 2 years
FV = PV(1 + i) (1 + i) or FV = PV(1 + i)2
 3 years
FV = PV(1 + i) (1 + i) (1 + i) = PV(1 + i)3
 n years
FV = PV(1 + i)n
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Present Value
 Present value tells us how much an expected future
payment is worth today.
 Alternatively, it tells us how much we should be willing to
pay today to receive some amount in the future.
• For example, if the present value of $1.10 at an interest
rate of 10% is $1, we should be willing to spend $1
today to get $1.10 next year.
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Present Value Formula
 The formula for present value can be found by rearranging
the compounding formula.
FV = PV(1 + i)
solve for PV
o FV/(1 + i) = PV
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Present Value over Time
 Extending the formula over 2 years
FV = PV(1 + i)2
PV = FV/(1 + i)2
 3 years
FV = PV(1 + i)3
PV = FV/(1 + i)3
 n years
FV = PV(1 + i)n
PV = FV/(1 + i)n
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Things to Notice
 An increase in the interest rate causes present
value to fall.
Higher rates of interest mean smaller amounts can grow
to equal some fixed amount during a specified period of
time.
 A decrease in the interest rate causes present
value to rise.
Lower rates of interest mean larger amounts are needed
to reach some fixed amount during a specified period of
time.
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Example:
How much must I invest today to get $10,000 in five years
if interest rates are 10%?
PV = FV/(1 + i)n
PV = $10,000/(1 + .10)5 = $10,000/1.6105 = $6,209.2
How much must I invest today to get $10,000 in five years if
interest rates are 5%?
PV = FV/(1 + i)n
PV = $10,000/(1 + .05)5 = $10,000/1.2763 $7,835.15
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More Things to Notice
 Present value is always less than future value.
(1 + i)n is positive so FV/(1 + i)n < FV
 In addition, PV4 < PV3 < PV2 < PV1
(1 + i)1 < (1 + i)2
o The longer an amount has to grow to some fixed
future amount, the smaller the initial amount
needs to be.
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Time Value of Money
 The longer the time to maturity, the less we need to set
aside today. This is the principal lesson of present value.
It is often referred to as the “time value of money.”
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Example:
If I want to receive $10,000 in 5 years, how much do I have
to invest now if interest rates are 10%?
$10,000 = PV(1 + .10)5
$10,000/1.5105 = $6209.25
If I want to receive $10,000 in 20 years, how much do I have
to invest now if interest rates are 10%?
$10,000 = PV(1 + .10)20
$10,000/6.7275= $1486.44
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Back to Our Original Problem
 How much money must you deposit now at 6% interest
compounded quarterly in order to be able to withdraw
$3,000 at the end of each quarter year for two years?
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Back to Our Original Problem
 How much money must you deposit now at 6% interest
compounded quarterly in order to be able to withdraw
$3,000 at the end of each quarter year for two years?
 Solution: R = 3000, i = 0.06/4 = 0.015, n = 8
 1  (1  i )  n 
P  R

i


 1  (1.015) 8 
P  3000 
  22, 457.78
 0.015 
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Interest Earned
 The present value of all payments is $22,457.78. The total
amount of money withdrawn over two years is
3000(4)(2)=24,000.
Thus, the accrued interest is the difference between the two
amounts:
$24,000 – $22,457.78 =$1,542.22.
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Amortization Problem
 Problem: A bank loans a customer $50,000 at 4.5%
interest per year to purchase a house. The customer agrees
to make monthly payments for the next 15 years for a total
of 180 payments. How much should the monthly payment
be if the debt is to be retired in 15 years?
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Amortization Problem
Solution
 Problem: A bank loans a customer $50,000 at 4.5%
interest per year to purchase a house. The customer agrees
to make monthly payments for the next 15 years for a total
of 180 payments. How much should the monthly payment
be if the debt is to be retired in 15 years?
 Solution: The bank has bought an annuity from the
customer. This annuity pays the bank a $PMT per month at
4.5% interest compounded monthly for 180 months.
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Solution
(continued)
 We use the previous formula for present value of an
annuity and solve for PMT:
 1  (1  i )  n 
PV  PMT 

i




i
PMT  PV 
n 
 1  (1  i ) 
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Solution
(continued)
Care must be taken to perform
the correct order of operations.
1. enter 0.045 divided by 12
2. 1 + step 1 result
3. Raise answer to -180 power.
4. 1 – step 3 result
5. Take reciprocal (1/x) of step 4
result. Multiply by 0.045 and
divide by 12.
5. Finally, multiply that result by
50,000 to obtain 382.50


i
PMT  PV 

n 
 1  (1  i) 


0.045


12
  382.50
PMT  50, 000 
180
  0.045  
 1  1 
 
12

 

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Solution
(continued)
 If the customer makes a monthly payment of $382.50 to
the bank for 180 payments, then the total amount paid to
the bank is the product of $382.50 and 180 = $68,850.
Thus, the interest earned by the bank is the difference
between $68,850 and $50,000 (original loan) = $18,850.
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Constructing an
Amortization Schedule
If you borrow $500 that you agree to repay in six
equal monthly payments at 1% interest per month
on the unpaid balance, how much of each monthly
payment is used for interest and how much is used
to reduce the unpaid balance?
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Amortization Schedule
Solution
If you borrow $500 that you agree to repay in six
equal monthly payments at 1% interest per month
on the unpaid balance, how much of each monthly
payment is used for interest and how much is used
to reduce the unpaid balance?
Solution: First, we compute the required monthly
payment using the formula
i
PMT  PV
1  (1  i )  n
0.01
 500
1  (1.01) 6
 $86.27
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Solution
(continued)
At the end of the first month, the interest due is
$500(0.01) = $5.00.
The amortization payment is divided into two parts, payment
of the interest due and reduction of the unpaid balance.
Monthly Payment
$86.27
Interest Due
=
$5.00
Unpaid Balance Reduction
+
$81.27
The unpaid balance for the next month is
Previous Unpaid Bal
$500.00
Unpaid Bal Reduction
–
$81.27
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New Unpaid Bal
= $418.73
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Solution
(continued)
This process continues until all payments have been made
and the unpaid balance is reduced to zero. The calculations
for each month are listed in the following table, which was
done on a spreadsheet.
Payment Payment Interest Inpaid Bal Unpaid
Number
Reduction Balance
0
$500.00
1
86.27
$5.00
$81.27 $418.73
2
86.27
$4.19
$82.08 $336.65
3
86.27
$3.37
$82.90 $253.74
4
86.27
$2.54
$83.73 $170.01
5
86.27
$1.70
$84.57
$85.44
6
86.27
$0.85
$85.42
$0.03
Barnett/Ziegler/Byleen Finite Mathematics 12e
In reality, the
last payment
would be
increased by
$0.03, so that the
balance is zero.
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Strategy for Solving Mathematics
of Finance Problems
 Step 1. Determine whether the problem involves a single
payment or a sequence of equal periodic payments.
• Simple and compound interest problems involve a
single present value and a single future value.
• Ordinary annuities may be concerned with a present
value or a future value but always involve a sequence
of equal periodic payments.
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Strategy
(continued)
 Step 2. If a single payment is involved, determine whether
simple or compound interest is used. Simple interest is
usually used for durations of a year or less and compound
interest for longer periods.
 Step 3. If a sequence of periodic payments is involved,
determine whether the payments are being made into an
account that is increasing in value -a future value problem
- or the payments are being made out of an account that is
decreasing in value - a present value problem. Remember
that amortization problems always involve the present
value of an ordinary annuity.
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Amount that can be
paid off in 20 years,
or present value of
$917.18 for 20
years at 7.75%.
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