11

GENERAL
MATHEMATICS
Guided Learning Activity Kit
Simple and General Annuity
Quarter 2 - Week 3 - 4

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 SIMPLE AND GENERAL
ANNUITY

Introduction

Mathematics is very important in our daily lives. It is used in

commercial institutions, in business transactions and investments. Bank
deposits and withdrawals, loans, retirement pensions, educational plans and
mortgages are just a few of the very important applications of Business
Mathematics or Mathematics of Investment.
In this Guided Learning Activity Kit (GLAK), you will be studying these
important applications of Mathematics; Simple and General Annuity which
deals with investments, loans, retirement planning and saving money.
Solving for present and future values of simple and general annuities will also
help you to decide for your future plans, to know how much your money is
worth and how to spend them wisely.
For a better understanding of the concepts presented in this Guided
Learning Activity Kit, you are advised to review your lessons on Simple and
Compounded Interest. A quick look at your knowledge of sequence and series
will also help you understand better the concepts presented.

Learning Competencies

1. Illustrates simple and general annuities (M11GM-IIc-1);

2. Distinguishes between simple and general annuities (M11GM-IIc-
2);
3. Finds the future value and present value of both simple annuities
and general annuities (M11GM-IIc-d-1);
4. Calculates the fair market value of a cash flow stream that includes
an annuity (M11GM-IId-2); and
5. Calculates the present value and period of deferral of a deferred
annuity (M11GM-IId-3).

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 Objectives

At the end of this learning activity kit, the students are expected to:
1. describe simple annuity and general annuity;
2. differentiate simple annuity and general annuity;
3. solve for the future value and present value of simple annuities and
general annuities;
4. find the fair market value of a cash flow stream that includes
annuity; and
5. calculate the present value and period of deferral of a deferred
annuity.

Review

For your opening activity, let’s try this!

Answer each of the following problems.

1. You deposited Php 20,000 in a bank that earns an interest
rate of 4% compounded monthly for 5 years. Give the value
of each variable in the formula.
im
F = P ( 1 + j )n where j = and n = mt
m
a. P =
b. im =
c. j =
d. n =

2. If Php 50,000 is invested for 10 years that earns an interest

of 7.5% compounded semi-annually, how much would it be
at the end of that time? (Use the formula provided in problem
number 1.)
Given: P=
im =
j=
n=
Find: F=
im
Formula: F= P(1 +j) n where j = and n = mt
m

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 3. What is the present value of Php 80,000 after 2 years if money
earns 6% compounded monthly?
Given: F=
im =
j=
n=
Find: P=
im
Formula: P = F ( 1 + j )-n where j = and n = mt
m

Discussion

What is Annuity?

Annuity is a common practice of payments in business. It literally

means payment made annually. It refers to the equal payments made at
different segmental payment period. Examples of which are rent payments
or house rentals, pension plans, monthly payments of car loans or mortgage,
life insurance premiums, and bond dividends.

Annuity also refers to the sequence of payments made at equal or fixed

intervals or periods of time. It can be classified according to payment interval
and interest period; that is, it can be Simple Annuity or General Annuity.

Simple Annuity is an annuity where the payment interval is the same

as the interest period. Example of which is monthly payments of loans whose
interest is compounded monthly. General Annuity, on the other hand, refers
to an annuity where the payment interval is not the same as the interest
period. A car loan whose interest is compounded monthly, but payments are
done quarterly is classified as general annuity.

Annuity can also be classified according to time of payment and

duration. For annuity according to payment, we have Ordinary Annuity (or
Annuity Immediate) or Annuity Due. Ordinary Annuity is a type of annuity
in which the payments are made at the end of each payment interval, while
Annuity Due is a type of annuity in which the payments are made at the
beginning of each payment interval. For annuity according to duration, these
are Annuity Certain and Contingent Annuity. Annuity Certain is an annuity
in which payments begin and end at definite times while Contingent Annuity
is an annuity in which the payments extend over an indefinite length of time.

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 This learning activity kit will focus on Simple Annuities and General
Annuities, Ordinary Annuities and Annuity Certain.

For the purpose of our discussions, we will define terms that we are
going to use throughout our lesson.

Definition of terms:

1. Term of an annuity, t - time between the first payment interval and

last payment interval.
2. Regular or Periodic payment, R – amount of each payment.
3. Amount (Future Value) of an annuity, F – sum of future values of
all the payments to be made during the entire term of the annuity.
4. Present value of an annuity, P – sum of present values of all the
payments to be made during the entire term of the annuity.
5. Number of conversions m – number of times interest is compounded
annually.
6. Interest rate per annum im – annual interest rate.
7. interest per period j – interest rate per payment period and can be
solved using the formula
𝐢𝐦
j=
𝐦
8. Number of payments n – total number of payments in the annuity.

Simple Ordinary Annuity

A simple ordinary annuity refers to an annuity where the length of the

payment interval is the same as the length of the interest compounding period
and is made at the end of the payment interval.

Example 1. Try to consider this problem. Suppose your family plan to

save Php 10,000 a month for your house renovation in a fund that gives 9%
annual interest compounded monthly. How much is the amount or future
value of your savings after 6 months?

Given: periodic payment R = Php 10000

term t = 6 months
interest rate per annum i(12) = 0.09
number of conversions per year m = 12
0.09
interest rate per period j = = 0.0075
12
Find: amount (future value) at the end of the term, F = ?

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 Let us illustrate the cash flow in a table.

Amount of Savings in
Time (month) Amount in Pesos
Exponential Form
1 10,000( 1 + 0.0075)5 Php 10,380.67
2 10,000( 1 + 0.0075)4 Php 10,303.39
3 10,000( 1 + 0.0075)3 Php 10,226.69
4 10,000( 1 + 0.0075)2 Php 10,150.56
5 10,000( 1 + 0.0075) Php 10,075.00
6 10,000 Php 10,000.00
Total amount after 6 months Php 61,136.31

So, at the end of 6 months, you will have a total of Php 61, 136.31.

Amount (Future Value) of a Simple Ordinary Annuity

The Amount (Future Value) of a Simple Ordinary Annuity refers to the

total accumulation of the equal payments and interest earned. This value
measures how much you would have in the future given a specified interest
rate.

The Amount (Future Value) of an ordinary annuity can be found using

the formula:

( 𝟏 + 𝐣 )𝐧 − 𝟏
F = R
𝐣

where F = amount or future value

R = the regular payment
j = interest rate per period
n = number of payments

Applying the formula in the previous example, you will have

Given: R = Php 10,000

j = 0.0075
n=6
Find: F=?

( 1 + j )n − 1
Formula: F = R
j

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 ( 1 + 0.0075 )6 − 1
F = Php 10,000
0.0075

F = Php 61,136.31

Example 2. A father who wants to prepare for his son’s college

education, saves Php 2,000 a month through a bank that offers 3% annual
interest compounded monthly. How much will his money be at the end of 12
years?
Given: R = Php 2 000
i12 = 3% = 0.03
m = 12 (compounded monthly or 12 times a year)
i12 0.03
j= = = 0.0025
m 12
t = 12 years
n = mt = 12(12) = 144

Find: F=?
( 1 + j )n − 1
Formula: F = R
j
( 1 + 0.0025 )144 − 1
F = Php 2000
0.0025

F = Php 346,148.51

So, at the end of 12 years, he will earn Php 346,148.51.

Example 3. Bobet and Lyn are best friends. After graduating and
being able to get a decent job, they plan for their retirement. Bobet, starting
at age 24, deposits Php 12,000 at the end of each year for 36 years, while
Lyn, who started later at age 42, deposits Php 24,000 at the end of each year
for 18 years. Who will have the greater savings if both annuities earn 9% per
year compounded annually?

Solution:
Given:
For Bobet’s plan For Lyn’s plan
R = Php 12,000 R = Php 24,000
i = 9% =0.09 i = 9% = 0.09
m = 1 (annually) m = 1 (annually)
i 0.09 i 0.09
j= = = 0.09 j= = = 0.09
m 1 m 1
t = 36 years t = 18 years
n = 36 n = 18

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 Find: F = ? Find: F = ?

Formula:
( 1 + j )n − 1 ( 1 + j )n − 1
F = R F = R
j j
(1 + 0.09 )36 −1 ( 1 + 0.09 )18 − 1
F = Php 12,000 F = Php 24,000
0.09 0.09

F = Php 2,833,496.67 F = Php 991,232.11

Hence, at the end of the period, Bobet has Php 2,833,496,67 on his
retirement while Lyn has Php 999,232.11 so Bobet has the greater savings.

Notice how the example shows the difference in their savings, so that
the value of time and the advantage of saving early affects the amount of
money.

Present Value of a Simple Ordinary Annuity

If today’s value of a future payment of a simple ordinary annuity is so

desired, the quantity needed is the present value of the annuity. The present
value of a simple ordinary annuity refers to the principal amount that must
be invested or paid today to provide the regular payments of an annuity. So,
think of this as a discounted value for your payments.

Example 4. Consider the situation in example 1 where your family

plans to save Php 10,000 a month for your house renovation in a fund that
gives 9% annual interest compounded monthly. How much is the present
value of your savings at the end of 6 months?

Given: periodic payment R = Php 10000

term t = 6 months
interest rate per annum i(12) = 0.09
number of conversions per year m = 12
0.09
interest rate per period j = = 0.0075
12

Find: present value of the annuity P = ?

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 Let us illustrate the discounted payments in a table.

Discounted Payments in Discounted Payments

Time (month)
Exponential Form in Pesos
1 10,000( 1 + 0.0075)-1 Php 9,925.56
2 10,000( 1 + 0.0075)-2 Php 9,851.67
3 10,000( 1 + 0.0075)-3 Php 9,778.33
4 10,000( 1 + 0.0075)-4 Php 9,705.54
5 10,000( 1 + 0.0075)-5 Php 9,633.29
6 10,000( 1 + 0.0075)-6 Php 9, 561.58
Present Value of the Annuity Php 58,455.97

Thus, the cost of the savings at the beginning of the term is

Php58,455.97 which is the present value of the annuity.

The present value of a Simple Ordinary Annuity can be found using the
formula:
𝟏 − ( 𝟏 + 𝐣 )−𝐧
P = R
𝐣

where P = present value of the annuity

R = the regular payment
j = interest rate per period
n = number of payments

Applying the formula in the previous example, you will have

Given: R = Php 10,000

j = 0.0075
n=6
Find: P=?
1 − ( 1 + j )−n
Formula: P = R
j

1− ( 1 + 0.0075 )−6
P = Php 10,000
0.0075

P = Php 58,455.97

So, the present value of the annuity is Php 58,455.97.

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 Example 5. Jason paid Php 300,000 as down payment for his car. The
remaining balance is to be paid in installment at Php 17,500 at the end of
each month for 5 years. If the interest is 10.2% annually compounded
monthly, what is the cash price of his car?
Given: down payment = Php 300,000
R = Php 17,500
i12 = 10.2% = 0.102
m = 12 (compounded monthly or 12 times a year)
i12 0.102
j= = = 0.0085
m 12
t = 5 years
n = mt = 12(5) = 60
Find: P=?

1 − ( 1 + j )−n
Formula: P = R
j
1− ( 1 + 0.0085 )−60
P = Php 17,500
0.0085

P = Php 819,841.61

Cash value of car = Down payment + present value

= Php 300,000 + Php 819,841.61
Cash value of car = Php 1, 119,841.61

Therefore, the cash value of Jason’s car is Php 1,119,841.61.

Example 6. Ryan borrows money from a friend to buy a new

motorcycle in cash. He will repay his loan by making monthly payments of
Php 2,500 for the next 24 months at an interest rate of 6% compounded
monthly. How much did Ryan borrow? How much interest does he pay?

Given: R = Php 2,500

i12 = 6% = 0.06
m = 12 (compounded monthly or 12 times a year)
i12 0.06
j= = = 0.005
m 12

t = 2 years
n = mt = 12(2) = 24

Find: P=?

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 1 − ( 1 + j )−n
Formula: P = R
j
1− ( 1 + 0.005 )−24
P = Php 2,500
0.005

P = Php 56,407.17

Hence, Ryan borrowed Php 56,407.17 for his motorcycle. He paid an

interest of Php 3,592.83 for his loan.

Example 7. Jessica is planning to buy a living room set. It is on sale

at Php 45,000 in cash or on terms, Php 4,000 each month for 12 months, at
9% interest compounded monthly. Which is lower, the cash price or the
present value of the installment?
Given: Cash Price: Php 45,000
R = Php 4,000
i12 = 9% = 0.09
m = 12 (compounded monthly or 12 times a year)
i12 0.09
j= = = 0.0075
m 12
n = mt = 12 (months)
Find: P=?

1 − ( 1 + j )−n
Formula: P = R
j
1− ( 1 + 0.0075 )−12
P = Php 4,000
0.0075

P = Php 45,739.65

Therefore, buying the living room set in cash is lower than the present
value in its installment term.

Example 8. George borrowed Php 200,000. He agrees to pay the

principal plus interest for 3 years. What should be his monthly payments if
the interest is 6% compounded monthly?
Given: P = Php 200,000
i12 = 6% = 0.06
m = 12 (compounded monthly or 12 times a year)
i12 0.06
j= = = 0.005
m 12
t = 3 years
n = mt = 12(3) = 36

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 Find: R=?

1 − ( 1 + j )−n
Formula: P = R
j

1 − ( 1 + j )−n
Php 200,000 = R [ ]
j
1− ( 1 + 0.005 )−36
Php 200,000 = R [ ]
0.005

Php 200,000 = R ( 32.87101624 )

Php 200,000
= R
32.87101624

Php 6,084.38 = R

Thus, George must pay Php 6,084.38 every month for 3 years.

General Ordinary Annuity

However, not all annuities are simple ordinary annuities. There are
annuities where the length of the payment interval is not the same as the
length of the interest compounding period. Such annuities are called General
Annuities. General Ordinary Annuity is an annuity where the length of the
payment interval is not the same as the length of the interest compounding
period and is made at the end of the payment interval.

Examples of General Ordinary Annuity

1. Monthly installment of a car with an interest rate that is
compounded quarterly
2. Paying for a lot semi-annually when the interest rate is
compounded monthly
3. Making a monthly deposit while the interest is compounded
annually

Note that there is a difference in the payment period to the interest

compounding period.

For General Ordinary Annuity, you can use the formula for finding the
Future Value and the Present Value as follows:

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Amount (Future Value) of a General Ordinary Annuity:

( 𝟏 + 𝐣 )𝐧 − 𝟏
F = R
𝐣

where F = amount or future value

R = the regular payment
j = interest rate per period
n = number of payments

Present Value of a General Ordinary Annuity:

𝟏 − ( 𝟏 + 𝐣 )−𝐧
P = R
𝐣

where P = present value of the annuity

R = the regular payment
j = interest rate per period
n = number of payments

Note that the formulas are the same for a simple ordinary annuity. The
difference will only lie in finding the value of j: the given interest rate per
period must be converted to an equivalent rate per payment interval.

Study these examples below:

Example 1. Jericho started to deposit Php 2,000 monthly in a fund

that pays 4.5% compounded quarterly. How much money will be in his
account at the end of 15 years?
Given: R = Php 2,000
t = 15 years
n = 12(15) = 180 payments
i(4) = 0.045
m = 4 (compounded quarterly)

Note that the payment period is different from the interest rate period.

Steps:
1. Convert 4.5% compounded quarterly to its equivalent interest rate
for monthly payment interval.

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 im mt
Recall: F = P ( 1 + )
m
Let:
i12 12t i4 4t
F1 = P ( 1 + ) F2 = P ( 1 + )
12 4

(for monthly payments) (for compounded quarterly)

F1 = F2
i12 i4 4t
P(1+ )12t
= P(1+ )
12 4
i12 12 0.045 4
(1+ ) = (1+ )
12 4
i12 12
(1+ ) = ( 1.01125 )4
12
i12
1+ = [(1.01125)4]1/12
12
i12
= ( 1.01125 )1/3 – 1
12
i12
= 0.0037360247 = j
12
Thus, the interest rate per monthly payment interval is 0.0037360247
or 0.37360247%.

2. Apply the formula in finding the future value of an ordinary annuity

using the computed equivalent rate j.
( 1 + j )n − 1
Formula: F = R
j
( 1 + 0.0037360247 )180 − 1
F = Php 2000
0.0037360247

F = Php 512,119.30

Therefore, at the end of the period, Jericho will have Php 512,119.30 in
his account.

Note. When solving for the equivalent rate j, use seven or more decimal
places or the exact value.

Example 2. A teacher saves Php 15,000 from his bonus every 6

months in a bank that pays 3% annual interest compounded monthly. How
much would be his savings after 20 years?

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 Given: R = Php 15,000
t = 20 years
n = 2(20) = 40 payments
i(12) = 0.03
m = 12 (compounded monthly)
Steps:
1. Convert 3% compounded monthly to its equivalent interest rate for
semi-annually payment interval.
im mt
Recall: F = P ( 1 + )
m
Let:
i2 2t i12 12t
F1 = P ( 1 + ) F2 = P ( 1 + )
2 12

(for semi-annually payments) (for compounded monthly)

F1 = F2
i2 2t i12 12t
P(1+ ) = P(1+ )
2 12
i2 2 0.03 12
(1+ ) = (1+ )
2 12
i2 2
(1+ ) = ( 1.0025 )12
2
i2
1+ = [(1.0025)12]1/2
12
i2
= ( 1.0025 )6 – 1
2
i2
= 0.015094063 = j
2
Thus, the interest rate per semi-annual payment interval is
0.015094063 or 1.5094063%.

2. Apply the formula in finding the future value of an ordinary annuity

using the computed equivalent rate j.
( 1 + j )n − 1
Formula: F = R
j
( 1 + 0.015094063 )40 − 1
F = Php 15,000
0.015094063

F = Php 815,640.22

So, the total amount of his savings is Php 815,640.22.

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Present Value of a General Ordinary Annuity

Example 3. Jack borrowed an amount of money from Jill. He agrees

to pay the principal plus interest by paying Php 25,000 every 6 months for 4
years. How much money did Jack borrow if the interest is 10% compounded
quarterly?
Given: R = Php 25,000
t = 4 years
n = 2(4) = 8 payments
i(4) = 0.10
m = 4 (compounded quarterly)
Steps:
1. Convert 10% compounded quarterly to its equivalent interest rate
for semi-annually payment interval.
im mt
Recall: F = P ( 1 + )
m
Let:
i2 2t i4 4t
F1 = P ( 1 + ) F2 = P ( 1 + )
2 4

(for semi-annually payments) (for compounded quarterly)

F1 = F2
i2 i4 4t
P(1+ )2t = P ( 1 + )
2 4
i2 2 0.10 4
(1+ ) = (1+ )
2 4
i2 2
(1+ ) = ( 1.025 )4
2
i2
1+ = [(1.025)4]1/2
2
i2
= ( 1.025 )2 – 1
2
i2
= 0.050625 = j
2
Thus, the interest rate per semi-annual payment interval is 0.050625
or 5.0625%.

2. Apply the formula in finding the present value of an ordinary annuity

using the computed equivalent rate j.
1 − ( 1 + j )−n
Formula: P = R
j

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 1 − (1 + 0.050625 )−8
P = Php 25,000
0.050625

P = Php 161,172.87

Hence, Jack borrowed Php 161,172.87.

Example 4. A family wants to refurnish their house by buying new

appliances payable for 1 year starting at the end of each month. How much
is the cost of refurnishing their house if their monthly payment is Php20,0000
and interest is 9% compounded quarterly?
Given: R = Php 20,000
t = 1 years
n = 12(1) = 12 payments
i(4) = 0.09
m = 4 (compounded quarterly)
Steps:
1. Convert 9% compounded quarterly to its equivalent interest rate for
monthly payment interval.
im mt
Recall: F = P ( 1 + )
m
Let:
i12 12t i4 4t
F1 = P ( 1 + ) F2 = P ( 1 + )
12 4

(for monthly payments) (for compounded quarterly)

F1 = F2
i12 12t i4 4t
P(1+ ) = P(1+ )
12 4
i12 12 0.09 4
(1+ ) = (1+ )
12 4
i12 12
(1+ ) = ( 1.0225 )4
12
i12
1+ = [(1.0225)4]1/12
12
i12
= ( 1.0225 )1/3 – 1
12
i12
= 0.007444443 = j
12
Thus, the interest rate per monthly payment interval is 0.007444443
or 0.7444443%.

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 2. Apply the formula in finding the present value of an ordinary annuity
using the computed equivalent rate j.
1 − ( 1 + j )−n
Formula: P = R
j
1 − (1 + 0.007444443 )−12
P = Php 20,000
0.007444443

P = Php 228,779.12

So, the total amount in refurnishing their house is Php 228,779.12.

Example 5. To accumulate a fund of Php 600,000 for 4 years, how

much should an employee deposit in his account quarterly if it pays 7.5%
compounded annually?
Given: F = Php 600,000
t = 4 years
n = 4(4) = 16 payments
i(1) = 0.075
m = 1 (compounded annually)
Steps:
1. Convert 7.5% compounded annually to its equivalent interest rate
for quarterly payment interval.
im mt
Recall: F = P ( 1 + )
m
Let:
i4 4t i1 1t
F1 = P ( 1 + ) F2 = P ( 1 + )
4 1

(for quarterly payments) (for compounded annually)

F1 = F2
i4 4t i1 t
P(1+ ) = P(1+ )
4 1
i4 4
(1+ ) = ( 1.075 )
4
i4
1+ = (1.075)1/4
4
i4
= ( 1.075 )1/4 – 1
4
i4
= 0.0182446 = j
4
Thus, the interest rate per quarterly payment interval is 0.0182446 or
1.82446%.

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 2. Find the periodic deposit R of an ordinary annuity using the
equivalent rate j.
( 1 + j )n − 1
F = R
j
( 1 + 0.0182446 )16 − 1
Php 600,000 = R [ ]
0.0182446
Php 600,000 = R ( 18.387310085 )
Php 600,000
= R
18.387310085
Php 32,631.20 = R

So, the employee should deposit Php 32,631.20 quarterly.

Cash Flow Stream and Fair Market Value or Economic Value

Cash flow refers to cash inflows (payments received) or cash outflows

(deposits made, or payments made) whenever a transaction is made. It is
simply the transfer of cash from one person or company to the other. Cash
inflows can be represented positively or by positive numbers while cash
outflows are represented by negative numbers. Literally, cash inflow is what
comes in and what comes out is cash outflow.
The fair market value or economic value of a cash flow (payment stream)
on a particular date refers to a single amount that is equivalent to the value
of the payment stream made at that date. We call this particular date the
focal date.

Example 1. Mr. Tan is in the property development business. He

received two offers on a lot that he wants to sell. First offer is Php100,000
and a Php2,000,000 lump sum payment after 5 years. The second offer is
Php100,000 and Php90,000 every quarter for 5 years. Compare the fair
market values of the two offers if money can earn 6% compounded quarterly.
Which offer has a higher market value?

Solution: Find the fair market value for each offer. Choose a focal date or
the date the term starts then determine the present values of the two offers
at that focal date. Assume that focal date is the start of the contract.
Since the initial offer of Php100,000 is on the focal date, then its
present value is still the same as Php100,000.

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For the First offer:

For the present value of the Php 2,000,000 offered 5 years from now,
recall the formula:
P = F ( 1 + j )-n where F = future value
im
j =
m
n = mt
P = present value
Given: F = Php 2,000,000
t = 5 years
m=4
n = 20 payments
i4 0.06
j= = = 0.015
4 4
Find: P=?
Formula: P = F ( 1 + j )-n
P = Php 2,000,000 ( 1 + 0.015 )-20
P = Php 1,484,940.84

Fair Market Value = Down Payment + Present Value

= Php 100,000 + Php 1,484,940.84
= Php 1,584,940.84

For the Second offer:

Since the payment interval is the same as the interest period, we will
use simple annuity.
Solve for the present value of the simple annuity.
Given: R = Php 90,000
t = 5 years
n = 4(5) = 20 payments
i(4) = 0.06
m = 4 (compounded quarterly)
0.06
j= = 0.015
4
Find: P=?

19 | P a g e
 1 − ( 1 + j )−n
Formula: P = R
j
1− ( 1 + 0.015 )−20
P = Php 90,000
0.015

P = Php 1,545,177.49

Fair Market Value = Down Payment + Present Value

= Php 100,000 + Php 1,545,177.49
= Php 1,645,177.49

Therefore, the second offer has a greater fair market value at the given
focal date. The difference in the fair market values of the two offers at the
start of the term is:
Php 1,645,177.49 - Php 1,584,940.84 = Php 60,236.65

What if the focal date is at the end of the term? Let’s find out.

For the First offer:

Find the present value of Php 100,000 after 5 years at 6% compounded
quarterly.
Given: P = Php 100,000
0.06
j= = 0.015
4
n = 20
Find: F=?
Formula: F = P ( 1 + j )n
F = Php 100,000 ( 1 + 0.015)20
F = Php 134,685.50

Fair Market Value = Php 134,685.50 + Php 2,000,000.00

= Php 2,134,685.50

For the Second offer:

Find the future value of this simple Annuity
Given: R = Php 90,000
t = 5 years
n = 4(5) = 20 payments
i(4) = 0.06
m = 4 (compounded quarterly)
0.06
j= = 0.015
4
20 | P a g e
 Find: F=?
( 1 + j )n − 1
Formula: F = R
j
( 1 + 0.015 )20 − 1
F = Php 90,000
0.015
F = Php 2,081,130.04

Fair Market Value = Php 134,685.50 + Php 2,081,130.04

= Php 2,215,815.54

So, even though we have different focal dates, the second offer still has
a higher fair market value. The difference between the market values of the
two offers at the end of the term is:
Php 2,215,815.54 - Php 2,134,685.50 = Php 81,130.04

Example 2. Which company offers a better investment? Company A

offers Php 400,000 at the end of 3 years plus Php 600,000 at the end of 5
years. Company B offers Php 50,000 at the end of each quarter for the next
5 years. Assume that the money earns 8% compounded semi-annually.

Solution:
Assume that the selected focal date is the start of the term. So, since
the focal date is at the start of the term, compute for the present value for
each investment offer.
Company A offer:

The present value of Php 400,000 three years from now is:

P1 = F ( 1 + j )-n
P1 = Php 400,000 ( 1 + 0.04)-6
P1 = Php 316,125.81

The present value of Php 600,000 five years from now is

P2 = F ( 1 + j )-n
P2 = Php 600,000 ( 1 + 0.04 )-10
P2 = Php 405,338.50
Fair market Value = P1 + P2
= Php 316,125.81 + Php 405,338.50
= Php 721,464.31

21 | P a g e
Company B offer:
Compute the present value of a general annuity with quarterly
payments but with semi-annual compounding at 8%.

Solve the equivalent rate of quarterly payments compounded semi-

annually.
i4 4t i2 2t
F1 = P ( 1 + ) F2 = P ( 1 + )
4 2

(for quarterly payments) (for compounded semi-annually)

F1 = F2
i4 4t i2 2t
P(1+ ) = P(1+ )
4 2
i4 4 0.08 2
(1+ ) = (1+ )
4 2
i4 4
(1+ ) = ( 1.04 )2
4
i4
1+ = [(1.04)2]1/4
4
i4
= ( 1.04 )1/2 – 1
4
i4
= 0.0198039 = j
4

Solve for the present value of the annuity.

1 − ( 1 + j )−n
Formula: P = R
j
1 − (1 + 0.0198039 )−20
P = Php 50,000
0.0198039

P = Php 819,120.97

Therefore, Company B offer is preferable since its market value is

larger, that is Php 819,120.97, while that of Company A’s fair market value
is Php 721,464.31.

Deferred Annuity
Deferred annuity refers to an annuity that does not begin until a given
interval has passed. For a deferred annuity, payments could be set on a later
or future date as agreed upon by the contract.

22 | P a g e
 Period of deferral refers to the time between the purchase of an annuity
and the start of the payments for the deferred annuity.

To determine the present value of a deferred annuity, find the present

value of all k + n payments, then subtract the present value of all artificial
payments. The artificial payments each equal to R but not actually paid are
those made during the period of deferral.

Present Value of a Deferred Annuity.

The present value of a deferred annuity is given by

𝟏 − ( 𝟏+𝐣 )−(𝐤+𝐧) 𝟏 − ( 𝟏+𝐣 )−𝐤

P=R - R
𝐣 𝐣
where
R = regular payment
j = interest rate per period
n = number of payments
k = number of conversion periods in the deferral

Example 1. A man decided to buy a pension plan for his retirement.

The plan allows him to claim Php 20,000 quarterly for 5 years starting 3
months after his 60th birthday. What one-time payment should he make on
his 45th birthday to pay off his pension plan if the interest rate is 10%
compounded quarterly?
Given: R = Php 20,000
m=4
i4 = 0.10
k = 4(15) = 60 (number of artificial payments)
n = 4(5) = 20 (number of actual payments)
0.10
j= = 0.025
4
Find: P=?

1 − ( 1+j )−(k+n) 1 − ( 1+j )−k

Formula: P=R - R
j j
1 − ( 1+0.025 )−80 1 − ( 1+0.025 )−60
P = Php20,000 - Php20,000
0.025 0.025

P = Php 689,036.34 - Php 618,173.13

P = Php 70,863.21

Therefore, the present value of his pension is Php 70,863.21.

23 | P a g e
 Example 2. In purchasing appliances, a credit card company offers
deferred payment options. A card holder plans to buy a smart television set
with monthly payments of Php 7,000 for 1 year. The payments will start at
the end of 3 months. How much is the cash price of the television set if the
rate is 8.1% compounded monthly?
Given: R = Php 7,000
m = 12
i12 = 0.081
k=2 (number of artificial payments)
n = 12 (number of actual payments)
0.081
j= = 0.00675
12
Find: P=?

1 − ( 1+j )−(k+n) 1 − ( 1+j )−k

Formula: P=R - R
j j
1 − ( 1+0.00675 )−14 1 − ( 1+0.00675 )−2
P = Php7,000 - Php7,000
0.00675 0.00675

P = Php 93,212.35 - Php 13,859.52

P = Php 79,352.83

Therefore, the present value of the TV set is Php 79,352.83.

Activities

Guided Practice 1. Simple Ordinary Annuity

Solve each problem completely following the guided pattern below.

A. Find the future value F of the following simple ordinary annuities.

1. Monthly payments of Php 5,000 for 4 years with interest rate of 3%
compounded monthly.
Given: R=
i12 =
m= (compounded monthly or 12 times a year)
i 12
j= =
m
t=
n=

24 | P a g e
 Find: F=?
( 1 + j )n − 1
Formula: F = R
j

2. Quarterly payments of Php 2,500 for 10 years with interest rate of

5% compounded quarterly
Given: R=
i4 =
m=
i4
j= =
m
t=
n=
Find: F=?
( 1 + j )n − 1
Formula: F = R
j

B. Find the present value P of the following simple ordinary annuities.

1. Monthly payments of Php 3,000 for 5 years with interest rate of 12%
compounded monthly
Given: R=
i12 =
m=
i12
j= =
m
t=
n=
Find: P=?
1 − ( 1 + j )−n
Formula: P = R
j

2. Semi-annual payment of Php 20,000 with interest rate of 7.5%

compounded semi-annually for 6 years
Given: R=
i2 =
m=
i2
j= =
m
t=
n=
Find: P=?
1 − ( 1 + j )−n
Formula: P = R
j

25 | P a g e
C. Find the periodic payments R of a simple ordinary annuity that pays
monthly payment of the future value of Php 150,000 for 2 years with an
interest rate of 7.2% compounded monthly.
Given: F=
i12 =
m=
i12
j= =
m
t=
n=

Find: R=?
( 1 + j )n − 1
Formula: F = R
j

Guided Practice 2. General Ordinary Annuity

Solve each problem completely following the guided pattern below.

A. Find the future value F of the following general ordinary annuities.

1. Monthly payments of Php 5,000 for 10 years with interest rate of 6%
compounded quarterly
Given: R=
t=
n = (number of payments)
i(4) =
m = (compounded quarterly)
Steps:
a. Convert 6% compounded quarterly to its equivalent interest rate for
monthly payment interval to solve for j.
Use:
i12 i4
F1 = P ( 1 + 12 )12t F2 = P ( 1 + 4 )4t
(for monthly payments) (for compounded quarterly)
F1 = F2
i12 12t i4 4t
P(1+ ) = P(1+ )
12 4

b. Apply the formula in finding the future value of an ordinary annuity

using the computed equivalent rate j.
( 1 + j )n − 1
Formula: F = R
j

26 | P a g e
 2. Semi-annual payments of Php 30,000 with interest rate of 8%
compounded annually for 15 years
Given: R=
t=
n=
i1 =
m=
Steps:
a. Convert 8% compounded annually to its equivalent interest rate for
semi-annual payment interval j.
Use:
i2 i1
F1 = P ( 1 + 2 )2t F2 = P ( 1 + 1 )1t
(semi-annual payments) (for compounded annually)

F1 = F2
i2 2t i1 1t
P(1+ ) = P(1+ )
2 1

b. Apply the formula in finding the future value of an ordinary annuity

using the computed equivalent rate j.
( 1 + j )n − 1
Formula: F = R
j

B. Find the present value P of the following general ordinary annuities.

1. Quarterly payment of Php 20,000 for 10 years with interest rate of
7.5% compounded annually.
Given: R=
t=
n = ( number of payments )
i(1) =
m = ( compounded annually )
Steps:
a. Convert 7.5% compounded annually to its equivalent interest rate
for quarterly payment interval j.
Use:
i4 4t i1 1t
F1 = P ( 1 + ) F2 = P ( 1 + )
4 1

(for quarterly payments) (for compounded annually)

F1 = F2
i4 i1 1t
P(1+ )4t = P ( 1 + )
4 1

27 | P a g e
 b. Apply the formula in finding the present value of an ordinary annuity
using the computed equivalent rate j.
1 − ( 1 + j )−n
Formula: P = R
j

2. Annual payments of Php100,000 with an interest rate of 5%

compounded quarterly for 20 years
Given: R=
t=
n=
i(4) =
m=
Steps:
a. Convert 5% compounded quarterly to its equivalent interest rate for
annual payment interval j.
Use:
i1 1t i4 4t
F1 = P ( 1 + ) F2 = P ( 1 + )
1 4

(for annual payments) (for compounded quarterly)

F1 = F2
i1 1t i4 4t
P(1+ ) = P(1+ )
1 4

b. Apply the formula in finding the present value of an ordinary annuity

using the computed equivalent rate j.
1 − ( 1 + j )−n
Formula: P = R
j

C. Fred received two offers for investment. The first offer is Php 180,000
every year for 5 years at 9% compounded annually. The other offer is
Php15,000 per month for 5 years with the same interest rate. Which fair
market value between these offers is preferable? (Compute the future value
using focal date as the end of the term.)

Solution:
First offer (Ordinary Annuity)
Given: R=
i1 =
m=
t=
n=

28 | P a g e
 Find: F=
( 1 + j )n − 1
Formula: F = R
j

Second offer (General Annuity)

Given: R=
i1 =
m=
t =
n=
Find: F=

Steps:
a. Convert 9% compounded annually to its equivalent interest rate for
monthly payment interval j.
i12 12t i1 1t
F1 = P ( 1 + ) F2 = P ( 1 + )
12 1
(for monthly payments) (for compounded quarterly)

b. Apply the formula in finding the present value of an ordinary annuity

using the computed equivalent rate j.
( 1 + j )n − 1
F = R
j

Independent Practice
Solve each problem completely and logically.

A. Find the future value F of the following simple and general ordinary
annuities.
1. Semi-annual payments of Php 8,000 with interest rate of 7%
compounded semi-annually for 15 years

2. Quarterly payments of Php 20,000 with interest rate of 9%

compounded quarterly for 8 years

3. Monthly payments of Php 7,500 for 3 years with an interest rate of

10.5% compounded quarterly

B. Find the present value P of the following simple and general ordinary
annuities.
1. Monthly payments of Php 2,000 for 1 year with an interest rate of
15% compounded monthly

29 | P a g e
 2. Annual payments of Php 50,000 with an interest rate of 7%
compounded annually for 20 years

3. Semi-annual payments of Php 30,000 for 2 years with an interest

rate of 6% compounded monthly

C. Solve for the quantity that is being asked for.

1. Find the periodic payments R of a simple ordinary annuity that
pays monthly payment of the future value of Php 250,000 for 4
years with an interest rate of 5.4% compounded monthly.

2. Find the periodic payments R of a general ordinary annuity that

pays quarterly payment of the future value of Php 1,000,000 for 10
years with an interest rate of 6% compounded monthly.

3. Maria received two offers for investment. The first offer is

Php200,000 every year for 5 years at 12% compounded annually.
The other offer is Php16,000 per month for 5 years with the same
interest rate. Which fair market value between these offers is
preferable? (Compute the future value using focal date as the end
of the term.)

4. Mae availed a loan from a bank that gave her an option to pay
Php30,000 monthly for 2 years. The first payment is due after 4
months. How much is the present value of the loan if the interest
rate is 10.5% compounded monthly?

5. Bonn decided to sell their farm and to deposit the fund in a bank.
After computing the interest, they learned that they may withdraw
Php500,000 yearly for 10 years starting at the end of 8 years when
it is time for him to retire. How much is the fund deposited if the
interest rate is 6.5% converted annually?

30 | P a g e
5|P age
Independent Practice Review
A. 1, P = Php20,000 i12 = 0.04
1. F = Php 412,981.42 j = 0.0033333 n = 60
2. F = Php 922,758.25 2. P = Php50,000 i2 = 0.075
3. F = Php 315,321.78 j = 0.0375 n = 20
B. F = Php104,407.60
1. P = Php 22,158.62 3. F = Php80,000 i12 = 0.06
2. P = Php 529,700.71 j = 0.005 n = 24
3. P = Php 111,412.36 P = Php70,974.85
C.
1. R = Php 4,677.43 Guided Practice 1
2. R = Php 18,397.83 A1. R=Php5000 i12 =0.03 m=12 j=0.0025
3. first offer t=4 n= 48 F=Php254,656.04
F = Php 1,270,569.47 2. R=Php2,500 i12 =0.05 m=4 j=0.0125
second offer t=10 n= 40 F=Php128,723.89
F = Php 1,285,460.34 B1. R=Php3000 i12 =0.12 m=12 j=0.01
second offer is better t=5 n= 60 P=Php134,865.12
4. P = Php 630,197.88 2. R=Php20000 i2 =0.075 m=2 j=0.0375
5. P = Php 2,315,028.46 t=6 n= 12 P=Php190,453.88
C. F=Php150000 i12 =0.072 m=12
Assessment j=0.006 t=2 n= 24 R=Php5,829.50
1. c
2. a Guided Practice 2
3. b A1. R=Php5000 t=10 n=120
4. d i4=0.06 m=4 j=0.00497521
5. d F=Php818,075.23
6. a 2. R=Php30000 t=15 n=30
7. c i=0.08 m=1 j=0.03923048
8. d F=Php1,661,082.43
9. b B1. R=Php20000 t=10 n=40
10. a i=0.075 m=1 j=0.0182446
P=Php564,338.00
2. R=Php100000 t=20 n=20
i4=0.05 m=4 j=0.05094534
P=Php1,236,292.16
C. a. first offer
R=Php180,000 i=0.09 m=1
t=5 n=5 F=1,077,247.91
b. second offer
R=Php15,000 i=0.09 m=1
t=5 n=60 j=0.00720732
F=1,120,992.93
Second offer is better
Key to Corrections