Section
1
1
Ordinary Annuity
Section Objectives:
1. Define the basic terms involved with annuities.
2. Find the amount of an annuity via Table Factors and Future Value Formula
3. Use the formula to find the present value of an ordinary annuity.
OBJECTIVE 1 Define the basic terms involved with annuities.
In Module 2, we discussed lump sums that were invested for periods of time. In
this module, we talk about an annuity, or a series of equal payments made at regular
intervals. Monthly mortgage payments, quarterly payments by a company into an
employee retirement account, and monthly checks paid by Social Security to a retired
couple are examples of annuities. There are two basic types of annuities. One type
grows as regular payments are made and it accumulates compound interest. Regular
payments into a retirement account or college savings account are examples of this type
of annuity. A second type of annuity involves regular payments made out of an
accumulated sum. Monthly payments from a retirement fund are an example.
Retirement funds can in fact represent both types, with money being added on a regular
basis while an employee is working, then with payments being made from the fund
after the employee retires will include a discussion of annuities from which payments
are made. The basic terms involved are the same for both types.
An ordinary annuity is one in which payments are made at the end of each
period, such as at the end of each month. The payment period is the length of time
between payments and the term of the annuity is the total time needed for all
payments. Interest calculations for annuities are done using compound interest. The
total amount in an annuity on a future date is the amount, compound amount, or
future value of the annuity. These terms are used interchangeably.
The amount an annuity grows to can be found using compound interest techniques
from the previous module. For example, suppose a firm makes deposits of P30,000 at
the end of each year for 6 years into an investment earning 8% compounded annually.
Use the compound interest tables in module 2 for 5 years and 8% to find the future
value of the first payment as follows:
P 30,000 * 1.46933 = P44,079.9
The future value of the annuity is the sum of the compound amounts of all six
payments. The annuity ends on the day of the last payment. Therefore, the last
payment, which is made at the end of year 6, earns no interest.
The future value of the annuity is P 220, 075.9.
Find the total amount deposited in the annuity and interest earned as follows:
Total deposits = 6 years * P 30,000 per year = P180,000
Interest earned = Future value of annuity - Total deposits
=P 220,075.9 – P 180,000
= P 40,075.9
OBJECTIVE 2 Find the amount of an annuity.
The amount of an annuity can also be found using the amount of an annuity table
on the next page. The number from the table is the amount or future value of an annuity
with a payment of P 1. The amount of an annuity with any payment is found as follows.
Finding Amount of an Annuity
Amount = Payment * Table Factor from amount of an annuity table
As a check, reconsider the annuity of P 30,000 at the end of each year for 6 years
at 8% compounded annually. Locate 8% at the top of the table and 6 periods in the far
left (or far right) column to find 7.33593.
Amount = P 30,000 * 7.33593 = P 220,077.9
This amount is identical to the amount calculated earlier, but sometimes the estimates
from the table differ slightly from those found using a calculator.
Table Factors for an Amount/Future Value of Annuity
-
Example 1
A father deposits P10,000 every quarter for 5 years in a firm that pays 12%
compounded quarterly. Assuming no withdrawals are made, how much would be in his
account at the end of five years?
Solution
Deposits per quarter = 10,000
Interest earned per quarter
for 5 years x 4 = 20 quarters. Look across
the top of the table for 3, and down the side for 20 periods to find 26.87037
Amount = P 10,000 x 26.87037= P 268,703.7
Total deposits = 20 quarters * P 10,00 per year = P 200,000
Interest earned = Future value of annuity - Total deposits
= P 268,703.7 – P 200,000
= P 68,703.7
Quick Check 1
At the end of every quarter, P 2000 is put into a educational plan that earns 6%
compounded quarterly. Find the future value in 5 years.
Future value of an ordinary annuity can also be determined using its formula and it is
given by:
Amount of an Annuity or Future Value of an Ordinary Annuity (FVOA)
𝑭𝑽𝑶𝑨
𝑷𝒎𝒕 (
𝒊 𝒏
𝒊
𝟏
𝟏
)
Where 𝑭𝑽𝑶𝑨 𝑭𝒖𝒕𝒖𝒓𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒓 𝒂𝒎𝒐𝒖𝒏𝒕
𝑷𝒎𝒕 𝒑𝒆𝒓𝒊𝒐𝒅𝒊𝒄 𝒅𝒆𝒑𝒐𝒔𝒊𝒕 𝒐𝒓 𝒑𝒂𝒚𝒎𝒆𝒏𝒕
𝒏 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒅𝒆𝒑𝒐𝒔𝒊𝒕𝒔 𝒐𝒓 𝒑𝒂𝒚𝒎𝒆𝒏𝒕𝒔 𝒎𝒂𝒅𝒆
𝒊 𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒓𝒂𝒕𝒆 𝒑𝒆𝒓 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅𝒊𝒏𝒈 𝒑𝒆𝒓𝒊𝒐𝒅
Example 2
Mark Ezekiel wants to put up his Art Studio 5 years from now. If he deposits P
5,000 from his monthly
salary for the next 5 years in an account that yields 12% compounded monthly. How
much does he have by that time?
Solution
Amount deposited at the end of each month is P 5000 for 5 years x 12 = 60
months
Using Pmt= P5,000, n=60 and
(
)
(
)
Quick Check 2
Verify Quick Check 1 using the formula.
Objective 3. Use the formula to find the present value of an ordinary annuity.
The present value of an ordinary annuity is the total of the present values of all the
payments of the annuity. To get the present value, assume an annuity of n number of
payments at rate i per period. Calculate the present value of each payment to the start of
the annuity and take their sum. The total is the present value of the annuity.
There is also a corresponding table factor for Present Value of an Ordinary
Annuity given below.
Exa
mpl
e3
An
alu
mn
us
in a
cert
ain
uni
vers
ity wants to provide a P 250,000 research fellowship fund at the end of each year for
the next five years. If the University can invest the money at 10% compounded
annually, how much should a man give now to setup the fund for the scholarship?
Solution
Annual fellowship fund = P 250,000
Interest earned per year
for 5 years x 1 = 5.
Look across the top of the table for 10, and down the side for 5 periods to find
3.79079
Present Value = P 250,000 x 3.79079= P 947,697.5
The amount P 947, 697.5 is the lump sum need to be deposited in an investment
earning 10% compounded annually to be able to provide P 250,000 pesos every end of
the year for 5 years. At the end of the 5th year of the scholarship, the fund is fully
exhausted. The fund was able to provide P 250,000 x 5= P 1,250,000 by investing P
947, 697.50
The present value of an ordinary annuity formula can also be used and it is given
by:
Present Value of an Ordinary Annuity (FVOA)
𝟏 𝒊 −𝒏
𝑷𝑽𝑶𝑨
)
𝒊
Where 𝑷𝑽𝑶𝑨 𝑷𝒓𝒆𝒔𝒆𝒏𝒕 𝑽𝒂𝒍𝒖𝒆 𝒐𝒇 𝒂𝒏 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒚 𝑨𝒏𝒏𝒖𝒊𝒕𝒚
𝑷𝒎𝒕 𝒑𝒆𝒓𝒊𝒐𝒅𝒊𝒄 𝒅𝒆𝒑𝒐𝒔𝒊𝒕 𝒐𝒓 𝒑𝒂𝒚𝒎𝒆𝒏𝒕
𝒏 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒅𝒆𝒑𝒐𝒔𝒊𝒕𝒔 𝒐𝒓 𝒑𝒂𝒚𝒎𝒆𝒏𝒕𝒔 𝒎𝒂𝒅𝒆
𝒊 𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒓𝒂𝒕𝒆 𝒑𝒆𝒓 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅𝒊𝒏𝒈 𝒑𝒆𝒓𝒊𝒐𝒅
𝟏
𝑷𝒎𝒕 (
Example 4
A company wants to purchase a machine the will require a payment of
P100,000 at the end each 6 months for the next four years. How much should the
company invest at present to cover for the semiannual payment if the money can be
invested 12% compounded semiannually?
Solution
Amount to be paid at the end of each month is P 100,000 for 4 years x 2 = 8
Using Pmt= P100,000, n=8 and
(
`
−
)
−
(
)
The value can verified using the table factor for present value. Given i=6% and
n=8, the
corresponding table factor is 6.20979
Present Value = P 100,000 x 6.20979= P 620, 979
The difference in decimal places is due to the limit decimal places used by the
table factor.
Section Exercises
Find the amount of the following ordinary annuities rounded to the nearest cent. Find the total
interest earned.
Amount of
Each Deposit
Deposited
Rate
Time
(Years)
Amount of
Annuity
Interest
1. P 900
annually
5%
18
__________
__________
2. P 2900
annually
8%
5
__________
__________
Earned
3. P 7500
semiannually 6%
10
__________
__________
4. P 9200
semiannually 8%
5
__________
__________
5. P 3500
quarterly
7
__________
__________
10%
Find the present value of the following annuities. Round to the nearest cent
Amount per
Payment
Payment at
End of Each
Time
(Years)
Rate of
Investment
P 1800
year
18
10%
P 4100
year
7
Compounded
annually
6% annually
______________
______________
Present
Value
P 2000
6 months
12
8%
semiannually
______________
P 1700
6 months
14
5%
semiannually
______________
P 894
quarter
6 4% quarterly
______________
Solve the following application problems.
1. Roman Rodriguez would like to know if he can retire in 35 years at age 60,
when he plan to fish a lot. Assume the total deposit into his retirement account
at the community college is P 3800 at the end of each year and that the fund
earns 6% per year. Find (a) the amount of the annuity and (b) the interest
earned.
2. Monique Chaney places P 250 of her quarterly child support check into an
annuity for the education of her child. She does this at the end of each quarter
for 8 years into an account paying 8% per year, compounded quarterly. Find the
amount of the annuity and (b) the interest earned.
3. In 4 years, Jennifer Videtto will need to purchase a delivery van for her
plumbing company. She estimates it will require a down payment of P 10,000
with payments of P 950 per month for 48 months. (a) Find the total amount
needed in 4 years assuming 12% compounded monthly. (b) Will she have
enough if she invests P 2200 at the end of every quarter for 4 years and earns
6% compounded quarterly?
4. Jessica Thames expects to receive P 18,400 per year based on her deceased
husband’s contributions to Social Security. Assume that she receives payments
for 14 years and a rate of 4% per year, and find the present value of this annuity.
Section
2
1
Manipulating the Ordinary Annuity Formula
Objective 4. Solve the Size of the
Periodic Payment for Ordinary Annuity
The size of each periodic payment (Pmt) must be determined when the future
value or present value of an annuity is known. In this type of problem, the interest rate
and the term of the annuity are usually given.
Present value is known
(Pmt)=
(
)
Future value is known
(
)
Example 5: (Present Value is Given)
To help finance the purchase of house and lot, couple borrows P 350,000. The
loan is to be repaid in equal monthly installment over a period of 8 years. If the interest
rate is 15 compounded monthly, how much is the monthly payment?
Solution:
Amortization of loans are examples of application of present value, for this case,
P350,000 is a Present value. The number of payments, n, is 8 years times 12
compounding periods per year is 96 monthly payments. The interest compounding
period is 15 divided by 12 compounding periods is 1.25%. Then the size of each
payment is,
−
(
)
(
−
)
=P 6,280.89
Example 6: (Cash Value and Down payment given)
A man wants to buy a car worth P 450,000. He pays P150,000 down payment and
agrees to [ay the rest by paying in monthly installments for 5 years. If the money is
worth 12% compounded monthly, how much is the monthly payment?
Solution
Down payment= P150,000
Cash Value = P450,000
Interest compounding period, i = 12/12= 1%
n=(5 years)(12)=60 months
Present Value= Cash value – Down payment= P450,000-P150,000=P300,000
(
)
(
)
= P6,673.33
Example 7: Future Value is given
A young accountant wishes to accumulate a fund amounting to 1 million pesos for
future business ventures ten years from now. How much should be invested at the end
of each quarter in an investment firm earning 8% compounded quarterly?
Solution:
FV= P 1,000,000
i= 8%/4= 2%
n= (4) (10 years)=40
(
)
(
)
=P16,555.75
Finding the interest rate in an ordinary annuity
The interest rate per compounding period, I can be solved by means of
interpolation using table factors since there is no direct formula can be derived.
Example:
At what nominal rate compounded monthly is P 200,000 the present value of
P 10,000 payable every end of each month for 4 years?
Solution:
PV= P200,000
Pmt= P10,000
n= (12)(4 years)=48
From
−
(
−
)
(
)
From the present value table factor, for n= 48 with unknown i,
the (
−
)=20 which is between i=5%(21.19513) and i=4%(18.07716)
Linear interpolation states that, the
if the two known points are given by the coordinates (x0 ,
y0) and (x1 , y1) , the linear interpolant is the straight line
between these points. For a value x in the interval (x0 , x1) , the value y along the straight
line is given from the equation of slopes
(
)
(
)
which can be derived geometrically from the figure on the right. Solving this equation for y,
which is the unknown value at x, gives
which is the formula for linear interpolation in the interval (x0 , x1) .
Transforming the given data of the problem into coordinates, as how in the table below,
Table factor
18.07716
20
21.19513
interest
5%
i
4%

(x0 , y0)
(x,y)
(x1 , y1)
(18.07716, 5% )
(20, i)
(21.19513, 4% )
Therefore, y is the unknown value of i.
Solving for the value of y
(
(
)
)
Which implies that the nominal rate is,
compounded monthly.
Example 8:
Payments of P10,000 each are made every 6 months. At what rate compounded
semi-annually will these payments amount to P143,000 in 5 years?
Solution:
Pmt= P10,000
FV= P 143,000
n=mt=(2)(5)=10
a
From
(
−
(
−
),
)
−
On the Future Value table for n=10, the value of
interest which is between 6% and 8%.
=14.3 with unknown
Transforming the given data of the problem into coordinates, as how in the table below,
Table Factor
13.18079
14.3
14.48656
Interest
6%
I
8%
(x0 , y0)
(x,y)
(x1 , y1)

(
(
Which implies that the nominal rate is,
compounded semi-annually.
Finding the term of an ordinary annuity
(13.18079, 6% )
(14.3, i)
(14.48656, 8% )
)
)
The number of payments n of an ordinary annuity can be determined if the future
value or present value, size of each periodic payment and interest rate is given. When
the integral number of payments is not exactly the equivalent to the original amount or
present value, a smaller concluding or final payment can be made on period after the
last full payment, but a certain amount of money is to be accumulated, this smaller final
payment may not be necessary because the interest after the last payment will equal or
exceed the balance needed.
Formula of n if Future value is given:
Formula of n if Present value is given:
Example 9:
If P 1750 is deposited in a fund at the end of 3 months, when will the fund
amount at least P75,000 if the interest rate is 12% compounded quarterly?
Solution:
Pmt=P 1750
Future Value=P 75,000
m=4
Using the formula for n where FV is given,
This means that the fund will be less than P75,000 just after the 27 th payment
and greater than P 75,000 just after the 28th payment. Thus the fund will amount
at least P P75000 at the end of 28 quarters or equivalent to 7 years (28/4=7).
It should be remembered that n should be an integral number of payments.
Example 10
A man borrows P 80,000 at an interest rate of 24% compounded monthly. He
agrees to pay P 5000 at the end of each month until the final payment date. How long
must he pay?
Solution:
Pmt=P 5000
PV= P80,000
m=12
To find t, we use formula for n given a present value.
This means that the present value of 19 payments is less than P80,000 but if 20
payments are made, the present value will be over P80,000. Thus the debtor must pay
19 whole payments of P5000 plus amount less than P5000 at the end of the 20th month.
The next example will provide a computation for the size of the last payment
which
is
less
than
the
periodic
payment
size.`````````````````````````````````````````````````````````````````
Example 11:
A man wants to accumulate P50,000 by making deposits of P10,000 at the
end of each year. If he gets 3% of his money, how many regular payments will he
make and what is the size of the last payment if there’s any?
Solution:
F=P 50,000
Pmt=P 10,000 at the end of each year
m=1
Using the formula for n where FV is given,
The value 4.73 means that 4 regular payments of P10,000 and the 5th payment is
less than 10,000. To find the value of the last payment, we use the concept of
equation of values.
P41,836.27
P 50,0000
0
1
2
3
4
5
CD
P 10,000
P 10,000
P 10,000
P 10,000
x
Step 1: Find the future value of the 4 regular payments of P10,000
(
)
(
)
Step 2: Use the value of FV for n=4 and setup the equation of value where the
comparison date is at the 5th year as indicated in the timeline given above. Then
the equation of values
P41,836.27(1+.03)1 + x=P 50,000
x=P50,000- P41,836.27(1+.03)1
x= P6908.64
Thus, the man needs to make 4 regular deposits of P10,000 each end of the year
and P6908.64 at the end of the 5th year in order to accumulate the exact amount of
P50,000 in his account within 5 years.
Example 12:
A father left his son an inheritance of P500,000 which will be released in
monthly pension of P10,000 which accumulates at 12% compounded monthly.
How long will the inheritance last? Determine the size of last amount to be
received.
Solution:
PV=P 500,000
Pmt=P 20,000 at the end of each month
m=12
To find t, we use formula for n given a present value.
The pension will last 70 months, in which the son will receive P10,000 69 times and
the 70th will be smaller than P10,000. To solve for the last month’s pension,
P500,000
0
1
2
3
67
P986,894
69
68
70
CD
P 10,000
P 10,000
P 10,000
P 10,000
P 10,000 P 10,000
x
Step 1: Get the future value of 69 regular payment of P10,000
(
)
(
)
Step 2: Use the value of FV for n=69 and setup the equation of values where the
comparison date is at the 70th month as indicated in the timeline given above. Then
the equation of values
P 986, 894.42(1+.01)1 + x = P 500,000(1+.01)70
x= P 500,000(1+.01)70 - P 986, 894.42(1+.01)1
x=6,618.32
Therefore, the son will receive 69 times of P10,000 and P 6,618.32 on the 70 th
month as the last pension.
Section 2 Exercises
1. A P 300,000 loan is due in one year. To repay the loan, the debtor deposits
an amount every month in a fund earning 8% m= 12. How much should he
deposit monthly?
2. A couple would like to accumulate P 500,000 in 5 years by making deposits
at the end of each quarter in an account that pays 16% compounded
quarterly. What is the size of each deposit?
3. A man borrows P 50,000. He agree to settle by paying every 6 months in 2
years. If the interest is 8% compounded semi annually, how much should
he pay every 6 months?
4. A man left his wife P1 million insurance policy. What monthly income
would this provide for 10 years if the insurance company pays 9%
compounded monthly?
5. The cash price of a 32 inch LED TV is P25,000. A buyer who prefers
installment by paying P5000 as down payment and P1250 monthly
installment for 18 months. What is the nominal rate compounded monthly?
6. Find the rate of interest per period and the nominal rate compounded
semi-annually at which payments of P15000 at the end of each 6 months
will amount to P200,000 in 5 years?
7. A man borrows P60,000 at 24% compounded monthly. He will discharge
the debt by paying P 4500 monthly. Find the number of regular payments
and the size of the final payment.
8. A newlywed couple would like to accumulate P 500,000 by making P15,000
deposits at the end of each quarter in an account that pays 16%
compounded quarterly. How long will it take for them to save the said
amount and determine the size of the last deposit?