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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Table of Content




Present Value of an Annuity
Amortization
Amortization Schedules
General Problem-Solving Strategies
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Terms
 annuity
 amortization
 equity
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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
First Four Payments
 Interest rate each period is 0.06/4=0.015
 0.06 
P1  3000 1 

4 

P2  3000 1.015 
2
P3  3000(1.015)
3
P4  3000(1.015)
4
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1
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Derivation of Short Cut Formula
 We could proceed to calculate the next four payments and
then 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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Derivation of Short Cut Formula
Suppose we need to find out how much should you deposit in
an account paying 6% compounded semiannually in order to
be able to withdraw $1,000 every 6 months for the next 3
years?
 We are interested in finding the present value of each
$1,000 that is paid out during the last 3 years.
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Derivation of Short Cut Formula
Let’s look at it in terms of a time line using P = A(1 + i)-n.
Years
Number of periods
Payments
Now
0
1 yr
1
$1000
2
$1000
2 yr
3
$1000
4
$1000
3 yr
5
6
$1000
$1000
Present Value
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Derivation of Short Cut Formula
1
Sum of the present values:
P = 1,000(1.03)-1 + 1,000 (1.03)-2 + ….. +1,000 (1.03)-6
2
Multiply each side by 1.03:
1.03P = 1,000 + 1,000(1.03)-1 + ….. +1,000 (1.03)-5
Now subtract equation 1 from equation 2:
1.03P – P = 1,000 - 1,000(1.03)-6
0.03P = 1,000[1 – (1 + 0.03)-6]
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Derivation of Short Cut Formula
 In general, if R is the periodic payment, i the rate period, and
n the number of periods, then the present value of all
payments is given by
P = R(1 + i)-1 + R(1 + i)-2 + ….. + R(1 + i)-n
 And proceeding as the above example, we obtain the general
formula of the Present Value of the an Ordinary Annuity:
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Derivation of Short Cut Formula
 Returning to the example, we get:
 It is common to use PV (present value) for P and PMT
(payment) for R. Then we have …
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Present Value of an Ordinary Annuity
1 (1 i)
PV  PMT
i
n
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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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 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 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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Present Value of an Ordinary Annuity
Example 1 – Present Value of an Annuity
 Problem: What is the present value of an annuity that pays
$200 per month for 5 years if money is worth 6%
compounded monthly?
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Present Value of an Ordinary Annuity
Example 1 – Present Value of an Annuity
 Solution: We have PMT = $200, i = 0.06/12 = 0.005, and
n = 12(5) = 60.
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Present Value of an Ordinary Annuity
Example 2 – Retirement Planning
 Problem: Lincoln Benefit Life offered an ordinay annuity
that earned 6.5% compounded annually. A person plans to
make annual deposits into this account for 25 years and
then make 20 equal annual withdrawals of $25,000,
reducing the balance in the account to zero. How much
must be deposit annually to accumulate sufficient funds to
provide for these payments? How much total interest is
earned during this 45-year process?
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Present Value of an Ordinary Annuity
Example 2 – Retirement Planning
 Solution: This problem involves both future and present
value. The flow of money is illustrated:
Years
Payments
1
$PMT
2
$PMT
…
25
1
$PMT
$PMT
2
$PMT
…
20
$PMT
FV = PV
 Given the required withdrawals, we find first the present
value necessary to provide for this withdrawals. We have
then:
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Present Value of an Ordinary Annuity
Example 2 – Retirement Planning (cont.)
 PMT = $25,000; i = 0.065, and n = 20:
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Present Value of an Ordinary Annuity
Example 2 – Retirement Planning (cont.)
 Now it is necessary to find the deposits that will produce a
future value of $275,462.68 in 25 years. We have: FV =
$275,462.68; i = 0.065, and n = 25:
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Present Value of an Ordinary Annuity
Example 2 – Retirement Planning (cont.)
 Thus, depositing $4,677.76 annually for 25 years will
provide for 20 annual withdrawals of $25,000. The interest
earned during the entire 45 years is:
interest = (total withdrawals) – (total deposits)
= 20(25,000) – 25($4,677.76)
= $383,056
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Amortization
 It is an important use of the Present Value Formula for an
Ordinary Annuity [Mort means “death”, literally you “kill”
a loan].
 In general, amortizing a debt means that the debt is
retired in a given length of time by equal period payments
that include compound interest.
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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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Amortization Problem
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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Amortization Problem
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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Amortization Problem
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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Amortization Formula
 In general, we are interested in computing the equal
periodic payments to retire a debt.
 Solving the Present Value Formula for PMT in terms of the
other variables, we obtain:
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Amortization Example
Example 3 – Monthly Payment and Total Interest on an
Amortized Debt

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Amortization Example
Example 3 – Monthly Payment and Total Interest on an
Amortized Debt
 Solution (A): Using the amortization formula, we have PV
= $800, i = 0.015 = 0.005, and n = 18.
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Amortization Example
Example 3 – Monthly Payment and Total Interest on an
Amortized Debt
 Solution (B):
Total interest paid = (amount of all payments) – (initial loan)
= 18($51.04) - $800
= $118.72
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Amortization Schedule
 Allows us to examine the effect each payment has on the
debt.
 It helps to understand what happens in situations when you
are amortizing a debt with equal periodic payments and
later decide to pay off the remainder of the debt in one
lump-sum payment.
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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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Constructing an
Amortization Schedule
Example 4 – Constructing an Amortization Schedule
 Problem: 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 the monthly payment is used
for interest and how much is used to reduce balance?
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Amortization Schedule Example
Example 4 – Constructing an Amortization Schedule
 Solution: First, we compute the required monthly payment.
We have PV = $500, i = 0.01, and n = 6.
 At the end of the first month, the interest due is:
$500(0.01) = $5.00
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Amortization Schedule Example
Example 4 – Constructing an Amortization Schedule
 Solution (cont.): The amortization payment is divided into
two part:
Monthly payment
Interest due
Unpaid balance reduction
$86.72 = $5.00
+ $81.27
The unpaid balance for the next month is;
Previous unpaid balance
$500.00
Unpaid balance reduction
-
$81.27
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New unpaid balance
=
$418.73
39
Amortization Schedule Example
Example 4 – Constructing an Amortization Schedule
 Solution (cont.): At the end of the 2nd month, we have:
 Interest due on the unpaid balance of $418.73:
$418.73(0.01) = $4.19
 The monthly payment of $86.27 covers interest and unpaid
balance reduction of:
$86.27 = $4.19 + $82.08
 And the unpaid balance for the 3rd month is:
$418.73 - $82.08 = $336.66
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Amortization Schedule Example
Example 4 – Constructing an Amortization Schedule
 Solution (cont.): Continuing with this process until all
payments have been made and the unpaid balance is reduced
to zero, we get the following amortization schedule:
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Amortization Schedule Example
Example 4 – Constructing an Amortization Schedule
Payment
Number
Payment
Interest
Unpaid Balance
Reduction
0
Unpaid
Balance
$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
253.75
4
86.27
2.54
83.73
170.02
5
86.27
1.70
84.57
85.45
6
86.27
0.85
85.45
0.00
Totals
$517.65
$17.65
$500.00
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Amortization Schedule Example
Example 5 – Equity in a Home
 Problem: A family purchase a home 10 years ago for
$80,000. The home was financing by paying 20% down and
signing a 30-year mortgage at 9% on the unpaid balance.
The net market value of the house is now $120,000, and the
family wishes to sale the house. How much equity does the
family has in the house now after making 120 monthly
payments?
[Equity = (current net market value) – (unpaid loan balance)]
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Amortization Schedule Example
Example 5 – Equity in a Home
 Solution: We need to find out the unpaid balance of the
loan. This is equal to the present value of the remaining
payments. Three step are necessary to get the equity:
 Step 1 - Find the monthly payment.
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Amortization Schedule Example
Example 5 – Equity in a Home
 Solution (cont):
 Step 2 - Find the present value of a $514.96 per month, 20year annuity.
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Amortization Schedule Example
Example 5 – Equity in a Home
 Solution (cont):
 Step 3 - Find the equity.
equity = (current net market value) – (unpaid loan balance)
= $120,000 - $57,235
= $62,765
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Amortization Schedule Example
Example 6 – Automobile Financing
 Problem: You have negotiated a price of $25,200 for a new
Ford pickup truck. Now you must choose between 0%
financing for 48 months or a $3,000 rebate. If you choose the
rebate, you can obtain a credit union loan for the balance at
4.5% compounded monthly for 48 months. Which option
should you choose?
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Amortization Schedule Example
Example 6 – Automobile Financing (cont.)
 Solution: Choosing 0% financing, the monthly payment
will be:
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Amortization Schedule Example
Example 6 – Automobile Financing (cont.)
 Solution: Choosing the $3,000 rebate, and borrow $22,200
at 4.5% compounded monthly for 48 months, the monthly
payment will be, with PV = $22,200; i = .045/12 =
0.00375, and n = 48 :
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Amortization Schedule Example
Example 6 – Automobile Financing (cont.)
 Solution: You should choose the $3,000.
• You will save: 525 – 506.24 = $18.76 monthly or
• 48(18.76) = $900.48 over the life of the loan.
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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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Strategy - Selecting the correct
formula for a problem
Single
payment
Sequence
of
payment
Simple
interest
A = P(1 + rt)
Compound interest
(or continuous
compund interest
A = P(1 + i)n
(or A = Pert)
Future
value of an
ordinary annuity
(1 + 𝑖)𝑛 −1
𝐹𝑉 = 𝑃𝑀𝑇
𝑖
Present
value of an
ordinary annuity
1 − (1 + 𝑖)−𝑛
𝑃𝑉 = 𝑃𝑀𝑇
𝑖
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Chapter 3
Mathematics of
Finance
Section 4
Present Value of an
Annuity; Amortization
END
Last Update: Abril 9/2013