Workbook in MOI

UNIVERSITY OF SOUTHERN MINDANAO
COLLEGE OF BUSINESS, DEVELOPMENT ECONOMICS AND
MANAGEMENT
KABACAN, COTABATO

WORKBOOK

IN

MATH OF INVESTMENTS

ACCOUNTANCY DEPARTMENT

Compiled by:
REINZON JAMES G. JUANITEZ
RESTY MYRRH B. PURCA
ELLA MAE P. VILLANUEVA

2019 Edition

No part of this book may be reproduced or utilized in any form or by any means, electronic or mechanical,
including photocopying, recording, or any information storage and retrieval system,
without permission in writing from the University of Southern Mindanao.

About the University of Southern Mindanao

USM Mission

Help accelerate the socio‐economic development, promote harmony among the diverse
cultures in Southern Philippines, and improve the quality of life through instruction, research,
extension and production.

USM Vision

Quality and relevant education for its clientele to be globally competitive, culture sensitive
and morally responsive human resources for development.

USM Core Values

Goodness
Responsiveness
Excellence
Assertion of Right and
Truth

Workbook in Math of Investments ii

USM Quality Policy Statement

The University of Southern Mindanao, as a premier university, is committed to provide
quality instruction, research development and extension services and resource generation
that exceed stakeholders’ expectations through the management of continual
improvement efforts on the following initiatives.
1. Establish key result areas and performance indicators across all mandated
functions;
2. Implement quality educational programs;
3. Guarantee competent educational service providers;
4. Spearhead need‐based research outputs for commercialization, publication,
patenting, and develop technologies for food security, climate change mitigation
and improvement in the quality of life;
5. Facilitate transfer of technologies generated from research to the community for
sustainable development;
6. Strengthen relationship with stakeholders;
7. Sustain good governance and culture, sensitivity; and Comply with customer,
regulatory and statutory requirements.

USM Institutional Outcomes

The graduates of USM shall:
1. provide leadership in various development programs both public and private,
2. be equipped with technical, conceptual and human resource skills,
3. engage in entrepreneurial activities,
4. be able to adapt to diverse culture, and
5. pursue advanced studies in emerging related fields.

Workbook in Math of Investments iii

About the College of Business, Development
Economics and Management

CBDEM Goal

To produce competent and responsible professionals who will provide leadership in
business, development economics and management.

CBDEM Objectives

The College of Business, Development Economics and Management particularly the
Department of Accountancy aims to:
1. Develop professionally competent and socially responsible graduates equipped with
theoretical knowledge, sound principles & techniques to serve the needs of the
increasingly complex, rapidly changing global business environment and aware of the
profound magnitude of their role in nation building and in the stewardship of business
resources;
2. Provide adequate training to prepare graduates for the various types of assessments
(including professional licensures and certifications) and career as professional
accountants in public practice, commerce and industry, government and academe
and for entrepreneurship;
3. Produce competent professional accountants with appropriate values, ethics and
attitudes capable of making a positive contribution over their lifetimes to the
profession and society; and
4. Develop the qualities that enhance the students’ professional and research
competence to undertake theoretical and empirical researches in the various
disciplines of accounting that will advance the frontiers of knowledge in these fields
that will promote the development of the nation.

Workbook in Math of Investments iv

PREFACE

This workbook in Mathematics of Investment is designed for USM Accountancy and
Management Accounting students that take up the course subject ACC 112 – Math of
Investments. This workbook covers the topics on simple interest and bank discount,
compounding or future value of amounts, discounting or present value amounts, simple
ordinary annuity, simple annuity due, simple deferred annuity, simple perpetuity, sinking
fund, amortization, depreciation, stocks and bonds, and complex or general annuity‐ all of
which are tools in preparation for the higher finance/accounting/business subjects.

Each chapter includes brief discussion of the topics, illustration problems, and
exercise sheets for students to practice what they’ve learned on the chapter. If supplemented
with discussion materials and similar discussions that are available, the students will develop
strong conceptual and computational background in mathematics of investment which will
help them develop critical thinking for the business organization. Proficiency to the topics
presented in this workbook will be attained if the students be guided and facilitated well,
especially in this no face‐to‐face learning.

Computations are made with the use of tables and formulas. And step‐by‐step
explanation of the calculations are also presented for illustration. Also, the students are
encouraged to use both the basic (ordinary) and scientific calculators while taking this course
subject in preparation for the actual board examination.

May you enjoy the journey towards working with the growth and decline of money,
preparation of different table for financial accounting and reporting purposes, and the
knowledge on how stocks and bonds work through the financial market.

REINZON JAMES JUANITEZ
RESTY MYRRH PURCA
ELLA MAE VILLANUEVA

Workbook in Math of Investments v

TABLE OF CONTENTS
CHAPTER 1. SIMPLE INTEREST AND BANK DISCOUNT
Simple Interest
Definition of Terms 1
Formulas 1
Manipulating the Simple Interest Formula 3
Bank Discount
Formulas 4
Promissory Notes
Formulas 8
Discounting of Notes 9
Exercises 11
CHAPTER 2. COMPOUND AMOUNT OR FUTURE VALUE
Time Value of Money
Underlying ideas of Time Value of Money 13
General‐Problem Solving Techniques 13
Compound Interest 13
The Compound Amount Formula 14
Fractional Part of Compounding Periods 15
Manipulating the Compound Amount Formula 16
Effective and Nominal Interest Rates 17
Exercises 18
CHAPTER 3. PRESENT VALUE
Present Value 20
Fractional Part of Compounding Periods 20
Equivalent Values 21
Exercises 22
Workbook in Math of Investments vi

CHAPTER 4. SIMPLE ORDINARY ANNUITY
Annuities 23
Classification of Annuities 23
Ordinary Annuity 24
Future Value of an Ordinary Annuity 24
Future Value using the Table 24
Present Value of an Ordinary Annuity 24
Present Value using the Table 24
Finding the Size of Each Periodic Payment 25
Present Value is known 25
Future Value is known 25
Finding the Interest Rate per Period and Nominal Interest Rate 26
Present Value is known 26
Finding the Term 27
Future Value is known 27
Exercises 29

CHAPTER 5. SIMPLE ANNUITY DUE
Future Value using the Formula 31
Future Value using the Formula 31
Additional Problems
Finding the Annuity Payment when Present Value is known 32
Finding the Annuity Payment when Future Value is known 32
Finding the Interest Rate per Period 33
Finding the Term 34
Exercises 35

CHAPTER 6. SIMPLE DEFERRED ANNUITY AND SIMPLE PERPETUITY
Deferred Annuity 37
Present Value using the Table 37
Additional Problems
Finding the Annuity Payment when Present Value is known 38
Workbook in Math of Investments vii

Finding the Term when Present Value is known 38
Perpetuity 39
Simple Ordinary Perpetuity 39
Simple Perpetuity Due 40
Exercises 41

CHAPTER 7. SINKING FUND AND AMORTIZATION
Sinking Fund 43
Sinking Fund Payment using the Table 43
Sinking Fund Payment using the Formula 43
Sinking Fund Schedule 44
Amortization 45
Amortization Payment using the Table 45
Amortization Payment using the Formula 46
All Periodic Payments are Equal 46
All Periodic Payments Except the Final Payment are Equal 47

Exercises 48
CHAPTER 8. DEPRECIATION
Methods Used for Financial Statement Reporting
Straight‐line Method 50
Sum‐of‐the‐Years’ Digit Method 51
Declining‐Balance Method 52
Units‐of‐Production Method 53
Compound Interest Method
Annuity Method 54
Sinking Fund Method 55

Exercises 56

CHAPTER 9. STOCKS, BONDS AND MUTUAL FUNDS
STOCKS
Stock Quotation 69
Dividends on Preferred and Common Stock 60
Workbook in Math of Investments viii

Current Yield for a Stock 62
Price‐Earnings Ratio of a Stock 62
Cost, Proceeds and Gain (Loss) on a Stock Transaction 62
Return on Investment 63
BONDS
Cost, Proceeds and Gain (Loss) on a Stock Transaction 63
Current Yield for a Bond 64
MUTUAL FUNDS
Definition and the Net Asset Value 64
Return on Investment 65

Exercises 66

APPENDIX – Present Value/Future Value Tables
Present Value Interest Factor 68
Present Value Interest Factor for an Annuity 70
Future Value Interest Factor 71
Future Value Interest Factor for an Annuity 72

Workbook in Math of Investments ix

Chapter 1
SIMPLE INTEREST AND BANK DISCOUNT

SIMPLE INTEREST

Money

Lender Borrower
Money
+

interest
Lender/creditor – the one who invests the money
Borrower/debtor ‐ the one who owes the money lent
Interest – the income earned by the lender; the cost for the use of money by the borrower

At the end of the time period, the borrower repays the amount originally owed plus the interest.

Interest can be computed by two methods:
1. Simple interest is an interest computed on the amount the borrower received at the time
the Ioan is obtained and is added to that amount when the loan becomes due.
‐Thus, simple interest is computed only once for the entire time period of the loan.
‐Usually applied to loans whose time period is less than a year.
2. Compound interest are the interest computed more than once during the time period of the
loan.
‐These are generally for time periods of a year or longer.
FORMULAS
Interest
𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒙 𝑹𝒂𝒕𝒆 𝒙 𝑻𝒊𝒎𝒆

Illustration: Find the interest to be paid by A for a loan of 200,000 for one year at a simple interest
rate of 4%.
Principal is the amount or deposit made by a depositor or the face amount lent to the borrower on
loan date. Ex. 200,000
Simple interest rate (expressed in percentage) is converted to a decimal far computation purposes.
Unless otherwise stated, the simple interest rate is an annual rate. Ex. 4% or 0.04
Time is the length of time for which the money is borrowed or lent. The time expressed in years or
fractional part of a year is the period between the loan date and maturity date.
date when the loan
date when the loan was obtained becomes due ex. 1 yr.

Workbook in Math of Investments 1

The simple interest may now be computed using the formula I = PRT. Substituting the given in the
sample:
Interest = Principal x Rate x Time
Interest = 200,000 x 0.04 x 1
Interest = P8,000

Maturity value
Maturity value is the sum of the principal and interest. MV = P + I or
𝑴𝒂𝒕𝒖𝒓𝒊𝒕𝒚 𝑽𝒂𝒍𝒖𝒆 𝑷 𝒙 𝟏 𝑹𝑻
The maturity value may now be computed using the formula given above. Substituting the given in
the sample:
Maturity Value = Principal + interest
Maturity Value = 200,000 + 8,000
Maturity Value = 208,000
or
Maturity Value = P x (1 + RT)
Maturity Value = 200,000 x [1 + 0.04(1)]
Maturity Value = 200,000 (1.04)
Maturity Value = 208,000

Time
‐ the period between the loan date and the maturity date. This is converted to decimal to
facilitate the computation, if practicable. (e. g. 1 ½ years = 1.5; 2 years and 3 months = 2.25;
3 years and 9 months = 3.75 years; 1 year and 7 months = 1 ).
Loan date TIME
Maturity date

NOTE: Since the simple interest rate is given as an annual rate, the time should be prorated
accordingly in 12 months or 4 quarter or 2 semi‐annuals or 6 bimonthly payments.
Instances of time:
1. Maturity date is unknown
If the time T is given in months and only the loan date is stated, the maturity date shall
coincide with the loan date. Thus, a loan obtained on June 13, 2018 payable in 4 months will
mature on October 13, 2018.
2. Year is unknown
If either the loan date and maturity date does not mention the year, it shall be assumed that
these dates fall on the same year. For example, a loan that was granted on February 14,
2018 and to mature on September 20 would mature on September 20, 2018.
3. T is stated as a certain number of days
1. Exact interest method – uses 365 as denominator (366 days in a leap year)
2. Ordinary interest method – uses 360 days

Note: Ordinary interest method yielded a higher interest than exact interest method.
4. Only loan date and maturity date are given
a. Actual time – counting every day excluding the loan date until the maturity date
b. Approximate time – assuming that each month has 30 days
Four possible combinations (also known as Banker’s Rule) to compute T where only the
loan date and maturity date are given:

Illustration: Count the time and approximate time from April 8, 2018 to September 20, 2018.

So, which of the Banker’s Rule has lower or higher yield of interest?

MANIPULATING THE SIMPLE INTEREST FORMULA
1. Principal is unknown (In pesos)
𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕
𝐏𝐫𝐢𝐧𝐜𝐢𝐩𝐚𝐥
𝑹𝒂𝒕𝒆 𝒙 𝑻𝒊𝒎𝒆
Illustration: A bank loaned Anna money at 8% simple interest for 90 days. If the interest is P4, 000,
find the principal amount borrowed.
Principal = 𝐼

𝑅𝑥𝑇
Principal = 4,000

90
8%𝑥
360
Principal = 4,000

0.02
Principal = P200,000
2. Rate is unknown (In percentage)
𝐑𝐚𝐭𝐞
𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒙 𝑻𝒊𝒎𝒆

Illustration: If Anna applies for a P175, 000 loan in a bank the interest of which is P5, 810 for 125
days. What interest rate is being charged?
Rate = 𝐼

𝑃𝑥𝑇
Rate = 5,810

125
175,000 𝑥
360
Rate = 5,810

60,763.8888
Rate = 0.095616
Rate = 9.56%
3. Time is unknown (In years, months or days)
𝐓𝐢𝐦𝐞
𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒙 𝑹𝒂𝒕𝒆
Illustration: What would be the time period of Anna’s loan for P266, 000, at 11% ordinary interest, if
the amount of interest is P10, 150?
Time = 𝐼

𝑃𝑥𝑅
Time = 10,150

266,000 𝑥 0.11
Time = 10,150

29, 260
Time = .3468899 year x 360
Rate = 124.8 or 125 days


BANK DISCOUNT

Proceeds

Lender Borrower

Maturity

value

Bank discounts are interest deducted in advance.
A bank discount is an interest computed on the maturity value of the loan and is deducted from
that amount at loan date to determine the net amount to be received by the borrower.
‐The amount of loan applied for at loan date is the maturity value of the loan.

The bank discount is deducted from that amount to arrive at the proceeds‐the amount the borrower
is to receive. This is in contrast to simple interest, which is altogether paid with the principal at
maturity date.
In computing for the bank discount, three factors are being considered: maturity value, bank
discount rate and time. In this section, bank discount rate is referred to as discount rate or rate and
bank discount as discount.

Formulas
𝑩𝒐𝒏𝒅 𝒅𝒊𝒔𝒄𝒐𝒖𝒏𝒕 𝑴𝒂𝒕𝒖𝒓𝒊𝒕𝒚 𝒗𝒂𝒍𝒖𝒆 𝒙 𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕 𝑹𝒂𝒕𝒆 𝒙 𝑻𝒊𝒎𝒆
Illustration: Allen availed of a P245, 000 loan at 14% discount rate for 9 months. Find the bank
discount and proceeds of the loan.
Maturity value – the amount applied for by the borrower on loan date. Ex. 245,000
Bank discount rate – expressed as a percentage, is converted to decimal and is an annual rate. Ex.
0.14
Time – expressed in years or fractional part of a year, is a period between the loan and maturity
date. Ex. 9 months or 9/12 or 0.75.

The bank discount may now be computed using the formula BD = MV x R x T. Substituting the given
in the sample:
Bank discount = Maturity value x Discount Rate x Time
Bank discount = 245,000 x .14 x .75
Bank discount = P25,725

Proceeds
Proceeds is the amount the borrower is to receive. P = MV ‐ BD or
𝑷𝒓𝒐𝒄𝒆𝒆𝒅𝒔 𝑴𝑽 𝒙 𝟏 𝑹𝑻

The proceeds may now be computed using the formula given above. Substituting the given in the
sample:
Proceeds = Maturity value – bank discount
Proceeds = 245,000 – 25,725
Proceeds = 219,275
or
Proceeds = MV x (1 ‐ RT)
Proceeds = 245,000 x [1 ‐ 0.14(.75)]
Proceeds = 245,000 (1.105)
Proceeds = 219, 275

PROMISSORY NOTES
‐ A note evidencing indebtedness and commitment to pay.
‐ an unconditional promise in writing made by one person to another, signed by the maker
engaging to pay on demand or at a fixed or determinable future time a sum certain in money
to order or to bearer (Negotiable Instruments Law).
‐ A document stating the details of a loan is a negotiable instrument which, when properly
endorsed, can be transferred or sold to another person or a bank which is nor a party to the
original loan.

Two parties involved:
Maker – who makes the promise and who signs the instrument
Payee – to whom the promise is made and to whom the instrument is payable

Parts of the promissory note:
1. Bank discount Note

Face
Value (FV)
225, 000 Agdao, Davao City June 13, 2018 Issue

Date
Term
75 days after date, I promised to pay to the order of Upside Payee

Down, two hundred twenty five thousand and xx/100 pesos for

the value received with bank discount at 8% per annum. Bank

discount
1. rate
2.
Maturity 3. Due Aug 27, 2018 Thessa Sales
date Maker
4.


2. Simple interest note
Face
Value (FV) 175, 000 Kabacan, Cotabato May 13, 2018 Issue
Date

Term 90 days after date, I promised to pay to the order of Amplayo Payee

Grocery store, one hundred seventy five thousand and xx/100
pesos for the value received with an interest at 12% per annum. Simple
interest

rate
Maturity Due Aug 11, 2018 Thalia Domingo
date Maker
Types of Notes: (interest‐bearing or non‐interest‐bearing)
1. Simple interest note (FV = Principal) and (MV > FV) Answer: Maturity Value = P180, 250
2. The bank discount note (FV = MV) and (FV > Proceeds) Answer: Proceeds = P221, 250

EFFECTIVE RATE OF A BANK DISCOUNT NOTE
In a simple interest note, the borrower receives the full face value, whereas with a bank discount
note the borrower receives only the proceeds. Because proceeds is less than the face value, the
stated discount rate is not the true or effective rate of the note.
Formula
𝑩𝒂𝒏𝒌 𝒅𝒊𝒔𝒄𝒐𝒖𝒏𝒕 𝑩𝑫
𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒓𝒂𝒕𝒆
𝑷𝒓𝒐𝒄𝒆𝒆𝒅𝒔 𝑷 𝒙 𝑻𝒊𝒎𝒆 𝑻
Where:
BD = FV x DR x T
P = FV ‐ BD
Illustration: What is the effective interest rate of a bank discount note for P350, 000, at a bank
discount rate of 14% for a period of 6 months? To find the effective interest rate, the bank discount
and proceeds should be computed first. The following procedures are observed:
1. Solve for the bank discount
Bank discount = Maturity value x Discount Rate x Time
Bank discount = 350,000 x .14 x .5
Bank discount = P24,500
2. Solve for the proceeds
Proceeds = 350,000 – 24,500
Proceeds = 325, 500


3. Solve for the effective interest rate
Effective interest rate = 𝐵𝑎𝑛𝑘 𝑑𝑖𝑠𝑐𝑜𝑢𝑛𝑡

𝑃𝑟𝑜𝑐𝑒𝑒𝑑𝑠 𝑥 𝑇𝑖𝑚𝑒
Effective interest rate = 24,500

325,500 𝑥 .5
Effective interest rate = 24,500

162,750
Effective interest rate = .15053 or 15.05 %


DISCOUNTING
Discounting Notes before Maturity
For a company or individual to cash in a note at any time before maturity, the payee (lender)
may take the note to a bank and sell it. This process is known as discounting a note.
In discounting a note, the original payee receives the proceeds of the discounted note, and
the bank (the new payee) receives the maturity value of the note when it matures. The time used to
compute the proceeds is from the date the note is discounted to the maturity date known as the
discount period.

Discounting a Simple Interest Note
Illustration: XYZ Distributors received a P150, 000 simple interest note for 5 months at 12% simple
interest from one of its customers. After 3 months, XYZ distributors needed cash so it discounted the
note at the ABC bank at a discount rate of 14%. What are the proceeds XYZ will receive from the
discounted note?
The following procedures are observed for simple interest note:
1. Solve for the maturity value of the original note.
Maturity Value = Principal (1 + RT)
Maturity Value = 150,000 [1 + (.12 x 5/12)]
Maturity Value = 150,000 (1 + .05)
Maturity Value = P 157, 500
2. Count the number of months or days of the discount period: In this illustration, the discount
period is 2 months – 5 months less the 3 months that had lapsed.
3. Solve for the bank discount.
Bank discount = Maturity value x Discount rate x Time
Bank discount = 157, 500 x .14 x 2/12
Bank discount = P 3, 675

4. Solve for the proceeds.
Proceeds = 157,500 – 3,675
Proceeds = P 153, 825

Discounting a Bond Discount Note
Illustration: XYZ Distributors received a P350, 000 bank discount note for 6 months from one of its
customers. After 3 months, XYZ distributors discounted the note at the ABC bank at a discount rate
of 14%. What are the proceeds XYZ will receive from the discounted note?
The following procedures are observed for bank discount note: (same with simple interest note
except for step 1: No need to compute for maturity value because this is the face value)

1. Count the number of months or days of the discount period. In this illustration, the discount
period is 3 months – 6 months less the 3 months that had lapsed
2. Solve for the bank discount.
Bank discount = Maturity value x Discount rate x Time
Bank discount = 350,000 x .14 x 3/12
Bank discount = P 12, 250

3. Solve for the proceeds.
Proceeds = 350, 000 – 12,250
Proceeds = P 337, 750


CHAPTER 1 EXERCISES
I. SIMPLE INTEREST
1. Find the simple interest on P8, 000 at an annual interest rate of 12% for two years.
2. Find the simple interest on a P30, 000 loan due in 5 years when the annual interest rate
on the loan is 16%. What is the maturity value of this loan?
3. Find the simple interest on a P50, 000 loan at 14 ½ % for 7 months.

II. CONCEPT OF TIME
a. Find the time, in days, of each of the following notes using approximate and actual time:
1. January 10, 2000 to February 18, 2000

2. March 6, 2000 to November 15, 2000

3. May 30, 2001 to August 16, 2013

b. Find the due dates of each of the following notes:
Date of the loan Time Due Date
1. April 10, 2011 3 months

2. August 18, 2014 6 months

3. July 31, 2019 90 days

c. Find the Exact Interest and Ordinary Interest
Principal Rate Time Exact Ordinary
Interest Interest
1. 450, 000 13% 100 days
2. 9, 000 10 ¼ % 60 days
3. 504, 700 9 1/8 % 58 days

III. MATURITY VALUE
Principal Rate Time
1. 540,000 11.9% 2 years
2. 1,250,000 12 ½ % 5 months
3. 750,000 14% 6 months
4. 186,200 10 ½ % 30 months
5. 7,500,000 13.35% 11 months

IV. MANIPULATING THE FORMULAS
Principal Rate Time Interest Interest Maturity Value
Method
1. 9.5% 100 days Exact 3,400
2. 150,000 14% Ordinary 9,600

3. 36,000 160 days Exact 2,250

V. BANK DISCOUNT
Face Value Discount Date Of Term Maturity Bank Proceeds
Rate Note Date Discount
1. 50,000 14.7% Apr 16 July 9
2. 8,750,000 9 ½ % Oct 25 87 days
3. 8,000 12.1% Sep 3 109 days

VI. PROMISSORY NOTE (1‐3 SIMPLE INTEREST NOTE AND 4‐6 BANK DISCOUNT
NOTE)
Face Int. Note Term Maturity Maturity Date of Disc. Disc. Proceeds
value rate date date value disc. period Rate
1. 40,000 10.4% Dec 12 50 days Jan 19 15%
2. 55,000 12% Mar 4 70 days Apr 15 13%
3. 15, 000 15% May 13 80 days July 10 17%
4. 80,000 11.4% Jan 13 60 days Feb 20 16%
5. 50,000 13% Apr 5 80 days May 16 14%
6. 30,000 16% Feb 27 68 days Apr 7 18 %

VII. WORD PROBLEMS
1. Find the (a) exact and (b) ordinary simple interest on a 120‐day loan of P1, 450,000 that
has an annual interest rate of 19 ¼ %. (c) Which gives the lender a greater return on the
investment and by how much?
2. Find the exact time from Feb 4 to Apr 21 of the year 2016. Then, using the Banker’s rule,
find the interest on P18, 000 at 17 1/6 %.

3. Alex borrowed money at 9% interest for 125 days. If the interest charge was P56, 000,
use the ordinary interest method to compute the amount of principal of the loan.
4. Alison goes to the bank and borrows P150, 000 at 9 ½ % for 250 days. If the bank uses
the ordinary interest method, how much will Alison have to pay?

5. Angelica signed a P240, 000 bank discount note at the CocoBank. The discount rate is
14%, and the note was made on February 19, for 50 days. (a) What proceeds will
Angelica receive on the note and its maturity value?

Chapter 2
COMPOUND AMOUNT OR FUTURE VALUE
Time value of money – a peso received today is worth more than a peso received tomorrow.
The difference between a present value and a future value is the interest that is included in the
future amount. Interest accrues over time.

PV – FV = 2 factors:
1. Interest rate. The greater the rate, the larger the interest, so is Future Value.
2. Length of time. The longer the time, the larger the interest, so is Future Value.
Underlying ideas of time value of money:
 A present value is always less than a future amount
 A future amount is always greater than a present value
 A peso available today is always worth more than a peso that does not become available
until a future date
 A peso available at a future date is always worth less than a peso that is available today
GENERAL PROBLEM‐SOLVING TECHNIQUES
The following steps will help categorize which investment math problem is at hand.
1. Determine if 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.
2. If a single payment is involved, determine if simple or compound interest is used. Simple interest
is usually used for durations of a year or less and compound interest for longer periods.
3. If a sequence of periodic payments is involved, determine if the payments are being made into an
account that is increasing in value‐a future value problem‐or if payments are being made out of an
account that is decreasing in value‐a present value problem.

COMPOUND INTEREST

Compound interest – in contrast to simple interest. Computes interest more than once during the
term of the loan or investment. It yields higher interest than simple interest because the interest are
earning interest. The relationships between the present value and the future amounts assume that
the interest earned on the investment is reinvested, rather than withdrawn. Reinvesting the interest
causes the investment to increase, thus causes more interest to be earned in each successive period.
𝑪𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝑭𝒖𝒕𝒖𝒓𝒆 𝑽𝒂𝒍𝒖𝒆 𝒍𝒆𝒔𝒔 𝑷𝒓𝒆𝒔𝒆𝒏𝒕 𝑽𝒂𝒍𝒖𝒆

Compound amount – the final sum of principal and accumulated interest at the end of the
borrowing period
Compounding period – the period for computing the interest which has regular intervals such as
annually, semi‐annually, quarterly, or monthly
Compounding – the process of finding a compound amount when the present value is known

The Compound Amount Formula
𝑨 𝑷 𝟏 𝒊 𝒏
Where:
A = compound amount of future value
P = present value or principal
i = interest rate per period, expressed as a decimal
n = total compounding periods

In practice, there are two methods for solving compound interest problems. The first uses the
compound amount table (see annex Table 1) and the second uses the compound amount formula.
Using the Table
𝑪𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝑽𝒂𝒍𝒖𝒆 𝑭𝑽 𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒙 𝑻𝒂𝒃𝒍𝒆 𝑭𝒂𝒄𝒕𝒐𝒓

In order to use the compound amount table, the number of compounding periods and the interest
rate per period must be known.
The chart below serves as a guide in finding the number of compounding periods per year.
Interest Compounding made Compounding

compounded periods per year
Annually Every year 1
Semi‐annually Every 6 months 2
Quarterly Every 3 months 4
Monthly Every months 12
Daily Every day 365

The chart below is a useful tool in determining the interest rate per period.
Nominal Interest Compounding Interest rate

interest rate compounded periods per per period
year (Nominal rate
/ periods per
year)
9% Annually 1 9%
12% Semi‐annually 2 6%
16% Quarterly 4 4%
18% Monthly 12 1.5%
21% Quarterly 4 5.25%

Illustration: Bea invested P12, 000, at 8% interest compounded quarterly, for 5 years. Use the
compound amount table or Table 1 to find the compound amount of Bea’s investment.
Interest rate per period = 8% ÷ 4 = 2%

Compound periods = 5 x 4 = 20
Compound amount = Principal x Table factor
Compound amount = P12, 000 x 1.485947 (Table 1)
Compound amount = P17, 831.40
Compound interest = Compound amount – Principal
Compound interest = 17, 831.40 – 12, 000
Compound interest = P5, 831.40

Note: For the periods beyond the table, a new table factor can be provided by multiplying the
factors for any two periods that add up to the number of periods required. (e. g. Table factor for 72
periods would be the product of the factors for 40 and 30 periods, 35 and 37, or any other
combination that adds up to 72.) Same rule will apply for the table factors for present value to be
discussed in next chapter.

Steps for Fractional Part of Compounding Periods
Illustration: Find the compound amount and the compound interest when P10, 000 is invested for 3
years and 2 months at 6% compounded semi‐annually, the procedures will be used:
1. Find the compound amount for the whole compounding periods. (A = P X Table factor)
Compound amount = Principal x Table factor
Compound amount = P10, 000 x 1.194052
2. Solve for the simple interest for the remaining fractional period based on the compound
amount on step 1. (SI = A X R X T)
Simple interest = Principal x Rate x Time
Simple interest = 11, 940.52 x .06 x 2/12
Compound amount = P119.40
3. Compute for the final compound amount by adding the A and SI.
Final compound amount = Compound amount + simple

interest
Final Compound amount = 11, 940.52 + 119.40
Final Compound amount = P12,059.92
The compound interest is found by subtracting the principal from the final compound amount.
Compound interest = Final compound amount – Principal
Compound interest = 12, 059.92 – 10,000


Manipulating the Compound Amount Formula
1. Nominal interest rate or time is unknown
𝒏
𝐴
𝟏 𝒊
𝑃
In finding the value of i, sometimes the factor cannot be located. It’s between two rates. To get the
exact interest rate, interpolation method can be used.

Illustration: At what nominal interest rate compounded semi‐annually for 10 years will P30, 000
accumulate to P89, 000?

𝟏 𝒊 𝒏
= 𝐴

𝑃
1 𝑖 = 89,000

30,000
1 𝑖 = 2.966667
Exponent n is equal to 20 (2 compounding periods per year x 10 years). The factor cannot be located
in the table. It is in between of 5 ½% and 6%.

Interest period per Table factor
period
6% 3.207135 (1)
i 2.966667 (2)
5 1/2% 2.917757 (3)
(2)‐(3) x – 5 ½ % 0.048910 (4)

=
(1)‐(3) ½ % 0.289378 (5)

Solve for i from the proportion formed by the differences on lines (4) and (5)
= 1 0.048910
i – 5 ½ % %
2 0.289378
i – 5 ½ % = 0.000845
i = 0.000875 + .055
i = 0. 055845
The desired value of i is .055845 or 5.58 % per semi‐annual period. So, the nominal rate per annum is
11.17%.
Interpolation method can also be used for the value of T if the table factor cannot be found.

Effective and Nominal Interest Rate

Nominal rate – the advertised or stated interest rate
Effective rate –the real rate of return of an investment

𝐓𝐨𝐭𝐚𝐥 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝 𝐢𝐧𝐭𝐞𝐫𝐞𝐬𝐭 𝐞𝐚𝐫𝐧𝐞𝐝 𝐢𝐧 𝐨𝐧𝐞 𝐲𝐞𝐚𝐫
𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝑹𝒂𝒕𝒆
𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍
or
𝒊𝑵𝒐𝒎 𝒎
𝒊𝑬𝒇𝒇 𝟏 𝟏
𝒎
Where:
𝑖 = nominal rate
𝑚 = number of compounding periods for one year
i = interest rate per period (𝑖 𝑚
𝑖 = effective rate


CHAPTER 2 EXERCISES

I. COMPOUNDING AMOUNT
a. Convert the Nominal Interest Rate to the Interest Rate per Period and Determine the Total
Number of Interest Periods:
Nominal Term Of Loan Interest Compounded Periodic Rate Number Of

Rate Periods
1. 14% 4 years Annually
2. 12% 8 years Quarterly
3. 13% 3 years Monthly
4. 12% 9 months Quarterly
5. 14% 4 years Semi‐annually

b. Find The Compound Amount And Compound Interest
Principal Term Of Nominal Interest Compounded Compound Compound

Investment Rate Amount Interest
1. P 50,000 1 year 12% Quarterly
2. 80,000 3 years 8% Semi‐annually
3. 400,000 1 ½ years 8% Semi‐ annually
4. 142,500 10 years 16% Quarterly
5. 60,000 2 years 24% Bi‐monthly

c. Creating a New Table Factor and Finding the Compound Amount
Principal Term Of Nominal Interest Compounded New Table Compound

Investment Rate Factor Interest
1. 190,000 60 years 9% Annually
3. 347, 000 20 years 16% Quarterly

d. Find the Compound Amount with Fractional Part of Compounding Periods
Principal Interest Interest Compounded Time Compound

Rate Amount
1. 4,000 18% Monthly 10 ½ months
2. 17,000 8% Quarterly 4 years, 2 months
3. 36,000 10% Annually 20 years, 5 months

e. Find the missing figure of interest rate and number of compounding periods
Principal Future Interest Compounded Number Of Interest Rate

Value Compounding Per Period
Periods
1. 50,000 114,860 Monthly 3 ½ years
2. 25,000 29,600 Quarterly 4 years
3. 400,000 916,800 Quarterly 4 ¼ years
4. 30,000 46,890 Monthly 18%
5. 20,000 32,380 Quarterly 14%
6. 50,000 87,050 Quarterly 8%

II. EFFECTIVE INTEREST
Principal Nominal Interest Compounded Compound Interest Effective
Rate In One Year Interest Rate
1. 50,000 10% Semi‐annually
2. 85,000 12% Monthly
3. 10,000 8% Quarterly
4. 30,000 46,890 Monthly

III. WORD PROBLEM
1. (a) Find the simple interest on P60, 000 for one year at 10%. (b) Find the compound
interest if the same investment is compounded semi‐annually? (c) How much is the
compound interest greater than the simple interest?
2. Ace deposited P500, 000 in an account earning 12% compounded monthly. This account
is intended to pay for the construction of a new warehouse. How much will be available
for the project in 2 ½ years?

3. Accumulate P300, 000 for eight years at 7% compounded quarterly. How much is the
interest?
4. Find the difference between two investments: (a) 150,000 is invested for the years at 8%
compounded monthly; and (b) 150,000 is invested for five years at 8% simple interest.

5. On Jan 1, 1998, Abram borrowed P12, 000 and agreed to repay it with P4, 650.70
interest. If the interest is at 6% compounded quarterly, what amount must repay and on
what date?

6. Agatha invested P30, 000 at the Prime Bank, at 6% interest compounded quarterly. (a)
What is the effective interest rate of this investment? (b) What will Agatha’s investment
be worth after 6 years?


Chapter 3
PRESENT VALUE
PRESENT VALUE – the amount that must be invested today in order to accumulate with compound
interest to the future value given. This value is crucial because it provides financial people with a
basis for comparing the profitability of different projects or investments over a period of time. This is
the cash value of future returns or income once a discount rate has been applied to it.

𝒏
𝑷𝑽 𝑨 𝟏 𝒊
Where:
PV = present value
A = compound amount
i = interest rate per period, expressed in decimal
n = total compounding periods

The PV factors are less than one, because the original investment is less than the compound amount.
(unlike the table factors of compound interest which are greater than one)

Fractional Part of Compounding Periods
There are cases when the time are with a fractional part. The steps to compute for the final present
value are as follows:
Illustration: Find the present value of a non‐interest bearing note of P10, 000 for three years and
two months at 6% compounded semiannually.
1. Find the present value for the whole compounding periods plus 1 period. Refer to Table 2.
PV = Compound amount x Table Factor
PV = 10,000 x 0.813092 = P8, 130.92

2. Compute simple interest for the difference between the period in step 1 and the subject period.
SI = PRT = 8,130.92 x .06 x 4/12 = P162.62

3. Add the simple interest to the resulting value in step 1 to arrive at the present value.
Final PV = PV + SI
Final PV = 8,130.92 + 162.62 = P8, 293.54

4. Additionally, compound amount less final present value equals compound interest.
CI = Compound amount – Final PV
CI = 10,000 – 8,293.54 = P1, 704.46


Equivalent Values
As long as interest is involved, a sum of money can have different values at a different times. A single
or set of obligations may be replaced by another single or set of obligations set on different due
date. Let’s look at the equation of value that makes the original obligations and the new obligations
be equal value on a comparison date.
Illustration: Jude owes P30, 000 due in three years and P40, 000 due in eight years. He and his
creditor have agreed to settle the debts by two equal payments in five and six years, respectively.
Find the size of each payment if money is worth 6% compounded semiannually. Let x be each
payment and the comparison date be the six years from now. The values on the comparison date are
computed as follows:
1. The value of the old debt of 30,000 becomes P35, 821.56 on the comparison date. Consider
PV = 30,000, i = 3% (6% / 2), n = 6 (3 years – from due date to comparison date x 2
compounding periods per year). The compound amount table or Table 1 is used.
Compound amount = Present value x Table factor
Compound amount = 30,000 x 1.194052 = P35,821.56

2. The value of the old debt of 40,000 becomes P35, 539.48 on the comparison date. Consider
A = P40, 000, i = 3%, n = 4 (2 years from the comparison date to the due date x 2
compounding periods per year). The present value table or Table 2 is used.
Present value = Compound amount x Table factor
Present value = 40,000 x 0.888487= P35,539.48

3. The value of the new debt, which is the first payment due in 5 years, becomes 1.0609x on
the comparison date. Consider PV = x, i = 3%, n = 2 (1 year from the 5th year – the date on
which the first equal installment is made to the 6th year – the comparison date x 2
compounding periods per year). Refer to Table 1.
Compound amount = Present value x Table factor
Compound amount = X (1.0609) = 1.0609x

4. The value of the second payment due in six years is x. it does not change since the
comparison date is also six years. The equation of value based on the comparison date is
given below:
New debts = Old debts
x + 1.0609x = 35,821.56 + 35,539.48
x + 1.0609x = 71,361.04
2.0609x = 71,361.04
X = 34,626.15

Hence, Jude should pay 34,626.15 at the end of the 5th and 6th year, respectively.


CHAPTER 3 EXERCISES
I. PRESENT VALUE
a. Find the Present Value and The Compound Interest Using Table and Using Formula
Compound Amount Term Of Nominal Interest Present Compound

Investment Rate Compounded Value Interest
2. 500,000 25 years 10.50% Annually
3. 2,500 1 year 18% Monthly
5. 100,000 2 years 12% Bi‐monthly

b. Creating a New Table Factor and Finding the Present Value
Principal Term Of Nominal Interest New Factor Present

Investment Rate Compounded Value
1. 120,000 30 years 16% Quarterly
2. 1,100,000 55 years 8% Semi‐annually

II. WORD PROBLEMS
1. Angelo wants to renovate his house in 3 years. He estimates the cost to be P300, 000.
How much must Angelo invest now at 8% interest compounded quarterly, in order to
have P300, 000, 3 years from now?
2. Anne is planning a vacation in Europe in 4 years, after graduation. She estimates that she
will need P350, 000 for the trip. (a) If her bank is offering 4‐year certificates of deposit
with 8% interest compounded quarterly, how much must Anne invest now in order to
have the money for the trip? (b) How much compound interest will be earned on the
investment?

3. If P360, 000 is due in seven years from now and money is worth 5% compounded
annually, find the present value and the compound interest.
4. What principal will accumulate to P320, 000 in 4 years at 4% compounded quarterly?

5. Find the present value of P140, 000 due at the end of nine years if money is worth (a) 5%
compounded quarterly and (b) 7% compounded semiannually. How much is the
compound interest in each case?


Chapter 4
SIMPLE ORDINARY ANNUITY
Annuities – series of equal periodic payments or receipts, rather than lump sums which is the
concern of compound interest. Originally, this only refers to annual equal payments, but is now
applies to payment intervals of any length of time.
The subject of annuities affects business firms as well as household, such as periodic savings, life
insurance premiums, interest payments on bonds and purchases of cars, houses, or home appliances
on installment payment plans.

Payment interval – the period of time between two successive payments dates

Term – the time between the beginning of the first payment interval and the end of the last
payment interval

Classifications of Annuities
1. By term
a) Annuity certain – term begins and ends on definite dates(e.g. 5‐year term from Jan 1,
2019 – Jan 1, 2024)
b) Perpetuity – term begins on a definite date but never ends (e.g. drawing interest)
c) Contingent annuity – term begins on a definite date, but the ending date is not fixed in
advance which depends on some conditions happening in the future. (e.g. life insurance
premium)
2. By dates of payment
a) Ordinary annuity – periodic payments are made at the end of each payment interval
(e.g. On Jan 1, 1 year interest payment is made quarterly. So, April 1 is the first payment)
b) Annuity due – periodic payments are made at the beginning of each payment interval
(e.g. 1 year interest payment quarterly starting Jan 1)
c) Deferred annuity – same with ordinary except the term of annuity does not begin until
after a designated period of time. (e.g. debtor with P20,000 loan on Jan 1, 2019 agrees
to three equal payments starting 2022)
3. By length of payment interval and interest compounding period
a) Simple annuity – payment interval coincides with the interest compounding period. (e.g.
payment interval is monthly, thus, the interest is compounded monthly)
b) Complex annuity or general annuity ‐ payment interval does not coincides with the
interest compounding period. (e.g. payment interval is monthly, the interest is
compounded quarterly)

ORDINARY ANNUITY
Future Value of Ordinary Annuity
Periodic payments are made at the end of each payment interval.
𝒏
𝒏 𝟏 𝒊 𝟏
𝑭𝑽𝑶𝑨
𝒊
Where:
𝐹𝑉 = Future value of an ordinary annuity
Pmt = Annuity payment
i = interest rate per period (nominal rate ÷ periods per year)
n = Number of periods (years x periods per year)

Using the Table
Illustration: Bea deposited P30, 000 at the end of each year for 8 years in her savings account. If her
bank paid 5% interest compounded annually, find the future value of Bea’s account. Use the Table 3
for the future value of annuity.
Future value = Annuity payment x table

factor
Future value = 30,000 x 9.549109
Future value = P286, 473.27

Present Value of Ordinary Annuity
𝒏
𝟏 𝟏 𝒊
𝑷𝑽𝑶𝑨
𝒊
Where:
𝑃𝑉 = Present value of an ordinary annuity
Pmt = Annuity payment
i = interest rate per period (nominal rate ÷ periods per year)
n = Number of periods (years x periods per year)

Using the Table
Illustration: How much must be deposited now, at 9% interest compounded annually, to yield an
annuity payment of P50, 000 at the end of each year, for 10 years? Use the Table 4 for the present
value of annuity.
Present value = Annuity payment x table factor

Present value = 50,000 x 6.417658
Present value = P320, 882.90


Finding the Size of Each Periodic Payment
Present value is known
To find the size of each payment, either of the formula may be used:
𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐕𝐚𝐥𝐮𝐞
𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝑻𝒂𝒃𝒍𝒆 𝟒
𝐓𝐚𝐛𝐥𝐞 𝐅𝐚𝐜𝐭𝐨𝐫
or
𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐕𝐚𝐥𝐮𝐞 𝐗 𝐓𝐚𝐛𝐥𝐞 𝐅𝐚𝐜𝐭𝐨𝐫 𝑻𝒂𝒃𝒍𝒆 𝟓

Illustration: The present value of an annuity for ten years is P10, 000. Find the size of the quarterly
payment if the interest rate is 8% compounded quarterly.
Substitute the value of the following: PV = P10, 000; i = 2%; n = 40
Annuity payment = Present Value

Table Factor
Annuity payment = 10,000

27.355479
Annuity payment = 365.56
Or
Annuity payment = Present Value x Table Factor
Annuity payment = 10,000 x 0.036556

Future value is known
To find the size of each payment, either of the formula may be used:
𝐅𝐮𝐭𝐮𝐫𝐞 𝐕𝐚𝐥𝐮𝐞
𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝑻𝒂𝒃𝒍𝒆 𝟑
or
𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝐅𝐮𝐭𝐮𝐫𝐞 𝐕𝐚𝐥𝐮𝐞 𝐗 𝐓𝐚𝐛𝐥𝐞 𝐅𝐚𝐜𝐭𝐨𝐫 𝐢 𝑻𝒂𝒃𝒍𝒆 𝟓

Illustration: The future value of an annuity for ten years is P10, 000. Find the size of the quarterly
payment if the interest rate is 8% compounded quarterly.
Substitute the value of the following: FV = P10, 000; i = 2%; n = 40

Annuity payment = Future Value

Table Factor

60.401983
or
Annuity payment = (Future Value x Table Factor 𝑖 )
Annuity payment = 10,000 x (0.036556 – 0.02)

Finding the Interest Rate per period and the Nominal Interest Rate
Present value is known
To find the interest rate when the present value is known, the formula may be used:
𝑻𝒂𝒃𝒍𝒆 𝟒 𝒇𝒂𝒄𝒕𝒐𝒓
𝐀𝐧𝐧𝐮𝐢𝐭𝐲 𝐩𝐚𝐲𝐦𝐞𝐧𝐭
Illustration: The present value of an annuity is 2,000 payable at the end of every six months for ten
years is P30, 000. Find the nominal rate compounded semi‐annually.
Substitute the value of the following: PV = P30, 000; n = 20, Pmt = P2, 000
Factor = Present Value

Annuity payment
Factor = 30,000

20
Factor = 15
Follow the line for n = 20 in table 4 to find the value/s of or closest to 15. This is between 2 % and
3% with factor of 15.227252 and 14.877475 respectively.
To get the more accurate value of the interest rate, the interpolation method is used:
Interest period per period Table factor
2 % 15.227252 (1)
x 15.000000 (2)

3% 14.877475 (3)
(2)‐(3) x – 3% 0.122525 (4)

(1)‐(3) = 0.349777 (5)
‐ %

Solve for x from the proportion formed by the differences on lines (4) and (5)
= 0.122525
x – 3% ‐ %
0.349777
x – 0.03 = ‐0.0025 x .350295
x – 0.03 = ‐0.000876
x = ‐0.000876 + 0.03
X = 0.29124
The desired value of i is 0.029124 or 2.91% per semi‐annual period. So, the nominal rate per annum
is 5.82%.
Finding the term
To find the term of an annuity when the future value is known, the formula may be used:
𝑻𝒂𝒃𝒍𝒆 𝟑 𝒇𝒂𝒄𝒕𝒐𝒓
𝐀𝐧𝐧𝐮𝐢𝐭𝐲 𝐏𝐚𝐲𝐦𝐞𝐧𝐭

Illustration: If P3, 000 is deposited at the end of each month, how many months will be required for
the deposits to amount to P122, 000, if the interest is 6% compounded monthly?
Substitute the value of the following: FV = P122, 000; i = 1/2%; Pmt = P3, 000
Factor = Future Value

Annuity Payment
Factor = 122,000

3,000
Factor = 40.666667

In the ½% column of Table 3, find the two values closest to 40.666667. This is between n = 37 and
38 with 40.532785 and 41.735449 factor respectively.

By trial‐and‐error method, we will compute the corresponding future value. The 37 term has future
value of P121, 598.36 which shows an insufficient payment. Therefore, 38 months is the answer with
the last deposit of less than P3, 000 to arrive at the desired future value.


CHAPTER 4 EXERCISES
I. ORDINARY ANNUITY
a. Finding the Size of Payment
Future Value Present Value Payment Term Compound

Interval Interest Rate
1. 1,000 1 year 5 years 10% annually
2. 7,000 1 month 7 years, 1 month 18% monthly
3. 6,000 1 quarter 3 years 14% quarterly

b. Finding the Future Value
Annuity Payment Time Nominal Interest Future Value Of The

Payment Frequency Period Rate Compounded Annuity
1. 10,000 Every 3 months 4 years 8% Quarterly
2. 15, 000 Every year 7 years 16% Quarterly
3. 2,000 Every 6 months 2 years 12% Monthly

c. Finding the Present Value
Annuity Payment Time Nominal Interest Present Value Of The

1. 3,000 Every 6 months 7 years 10% Semiannually
2. 85,000 Every year 3 years 16% Quarterly
3. 10,000 Every month 1 3/4 years 6% Annually

d. Finding the Interest Rate per Period and the Nominal Rate
Future Value Present Value Payment Term Interest – Compounding Period

1. 3,160 200 annually 10 years 1 year
2. 6,642 270 quarterly 7 years 1 quarter
3. 530 80 annually 8 years 1 year

e. Finding the Term
Future Value Present value Payment Compounding Interest Rate

1. 7,500 600 annually 10% annually
2. 4,092 120 quarterly 5% quarterly
3. 18,400 2,000 annually 8 ½ % annually


II. WORD PROBLEMS
1. If Bred borrowed P750, 000 and agreed to repay it by paying P80, 000 at the end of each
year. If the interest rate was 9% compounded annually, how many payments was he
required to make?
2. At what nominal interest rate compounded semiannually will an annuity of P2, 200
payable at the end of every 6 months amount to P25, 300 in 5 years?

3. A debt of P25, 000 was repaid in ten equal quarter payments. If the rate of the interest
was 7% compounded annually, what was the size of each payment?
4. Bon Savings and Loans is paying 6% interest compounded annually. How much must be
deposited now in order to withdraw P4, 000 at the end of each month for 2 years?

5. Bamboo is planning for his retirement. She deposits P30, 000 at the end of each year
into an account paying 5% interest compounded annually. (a) How much would the
account be worth after 10 years? (b) How would the account be worth after 20 years?
(c) When Bamboo retires in 30 years, what will be the total worth of the account? (d) If
Bamboo found a bank that paid 6% interest compounded annually, rather than 5%, how
much more would she have in the account after 30 years?


Chapter 5
SIMPLE ANNUITY DUE
Annuity Due ‐ periodic payments are made at the beginning of each payment interval
Future Value using the Formula

The formula is the same except that it is multiplied by (1 + i).
The formula to be used for the future value of annuity due states:
𝒏
𝟏 𝒊 𝟏
𝐅𝐕 𝑷𝒎𝒕 𝒙 𝒙 𝟏 𝒊
𝒊
Where:
FVAD = future value of an annuity due
Pmt = annuity payment
i = interest rate per period
n = number of periods (years x periods per year)

Using the Table
Illustration: Andrey deposited P6, 000 at the beginning of each month, for 2 years at her credit
union. If the interest rate was 12% compounded monthly, use the table 3 but with some
modifications to calculate the future value of annuity due.
To solve, add 1 period and deduct 1.000000 from the table factor to get the annuity due factor.
Then, substitute the following values: Pmt = P6, 000; Table factor = (28.243200 – 1)
Future value = Annuity payment x table factor

Future value = 6,000 x 27.243200
Future value = P163, 459.20

Present Value using the Formula
The formula is the same except that it is multiplied by (1 + i).
The formula to be used for the Present value of annuity due states:
𝒏
𝟏 𝟏 𝒊
𝐅𝐕 𝑷𝒎𝒕 𝒙 𝒙 𝟏 𝒊
𝒊
Where:
PVAD = Present value of an annuity due
Pmt = annuity payment
i = interest rate per period
n = number of periods (years x periods per year)


Using the Table
Illustration: How much must be deposited now, at 10% compounded semi‐annually, to get yield an
annuity payment of P20, 000 at the beginning of each period for 7 years? Use the table 4 but with
some modifications to calculate the Present value of annuity due.
To solve, deduct 1 period and add 1.000000 from the table factor to get the annuity due factor.
Then, substitute the following values: Pmt = P6, 000; Table factor = (9.395973 + 1)
Present value = Annuity payment x table factor

Present value = 20,000 x 10.395973
Present value = P207,871.46

ADDITIONAL PROBLEMS
Finding The Annuity Payment when Present value is known
To find the size of each payment, the formula may be used:
𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝑻𝒂𝒃𝒍𝒆 𝟒
Illustration: A washing machine that sells for P6, 000 can be bought under the terms of 20 equal
monthly payments starting now. If the money is worth 21% compounded monthly, what is the size
of each payment?
Substitute the value of the following: PV = P6, 000; i = 1 ¾ %; n = 19 (20 ‐1)
Annuity payment = Present Value

Table Factor

17.046057

Finding The Annuity Payment when Future value is known
To find the size of each payment, the formula may be used:
𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝑻𝒂𝒃𝒍𝒆 𝟑
Illustration: Angelo wishes to receive P20, 000 five years from now. How much must he invest at the
beginning of each year if the first payment starts now and the interest is 10% compounded annually?

Substitute the value of the following: FV = P20, 000; i = 10%; n = 6 (5 + 1)
Annuity payment = Future Value

Table Factor

6.715610
Annuity payment = P2,978.14

Finding the Interest Rate per period
To find the interest rate and the term, the formula may be used:
(a) When Present value is known
Step 1 PV = Pmt x Table Factor (Table 4)
Step 2 PV = Pmt x Table Factor + Pmt (Table 4)
(b) When Future value is known
Step 1 FV = Pmt x Table Factor (Table 3)
Step 2 FV = Pmt x Table Factor + Pmt (Table 3)
Illustration: What is the nominal rate compounded quarterly if the present value of an annuity of
P3,000 payable at the beginning of each quarter for 6 years is P58,000?
Substitute the value of the following: PV = P58, 000; n = 23 (4 x 6 – 1), Pmt = P3, 000
PV = Pmt x Table factor + Pmt
PV – Pmt = Pmt x table factor
Factor = PV Pmt

Pmt
Factor = 58,000 3,000

3,000
Factor (Table 4) = 18.333333

Follow the line for n = 23 in table 4 to find the value/s of or closest to 18.33. This is between 1 %
and 2% with factor of 18.544215and 18.292204 respectively. So, the value of nominal rate is
between 7 ½ % and 8%.
To get the more accurate value of the interest rate, the interpolation method is used.


Finding the term
Illustration: If P1, 000 is deposited at the beginning of each month at an interest rate of 9%
compounded monthly, how many months will be required for the deposits to amount to at least
P76, 000?
Substitute the value of the following: FV = P76, 000; i = ¾ %; Pmt = P1, 000
FV = Pmt x Table factor ‐ Pmt
PV + Pmt = Pmt x table factor
Factor = 𝐹𝑉 𝑃𝑚𝑡

Pmt
Factor = 76,000 1,000

1,000
Factor = 77

In the ¾ % column of Table 3, find the two values closest to 77. This is between n = 61 and 62with
76.989818 and 78.567242 factor respectively.
By trial‐and‐error method, we will compute the corresponding future value of annuity due.
Therefore, 62, the larger number, is employed. Actual period is 5 years and 1 month (n = 62 – 1).


CHAPTER 5 EXERCISES

I. ANNUITY DUE
a. Finding the Future Value
Annuity Payment Time Nominal Interest Future value of the

payment frequency period Rate compounded Annuity
2. 44,000 Every 6 months 8 years 16% Monthly
3. 2,000 Every year 25 years 5% Annually

b. Finding the Present Value
Annuity Payment Time Nominal Interest Present Value Of The

1. 14,000 Every year 10 years 11% Quarterly
2. 40,000 Every month 7 years 18% Monthly

c. Finding the Unknown Values
Future Value Present Value Payment Term Compound Interest Rate
1. 5,800 P200 monthly 2 years ? monthly

2. ? 450 30 quarterly 5 years ? quarterly
3. 3,800 250 semiannually ? 6% semiannually
4. 6,000 ? semiannually 7 ½ years 4% semiannually
5. 3,200 120 monthly ? 7% monthly

II. WORD PROBLEMS
1. Cora is paying 6% interest compounded quarterly. Find the future value of P1, 000,
deposited at the beginning of every 3 months, for 5 years.
2. If P10, 000 must be withdrawn at the beginning of each 3‐month period for the next 3
years, how much must be deposited now, at 6% compounded quarterly, to yield the
annuity payment needed?

3. A house that sells for P900, 000 can be purchased under the terms requiring 100
monthly payments. Assume that the first payment begins now and the interest is 12%
compounded monthly. What is the size of each payment?

4. Chin made a New Year’s Resolution to put P1, 000 into the bank at the beginning of each
month, beginning January 2014. If the bank pays 6% interest compounded monthly on
the last day of each month, how much will Chin have one year later?

5. The monthly rent for the one‐bedroom apartment in JV Hotel is P15, 000, payable at the
beginning of the month. If the current interest rate is 9%, what would be a fair amount
to charge someone if they wish to pay their yearly rental in advance?


Chapter 6
SIMPLE DEFERRED ANNUITY AND SIMPLE PERPETUITY

DEFERRED ANNUITY
A deferred annuity is an annuity in which the first periodic payment is made several periods
after the beginning of the annuity.

Period of deferment ‐ The period between now and the beginning of the term of the annuity.

Future Value Using the Table
The future value of the deferred annuity is the same as the future value of the ordinary
annuity. Thus, the future value of an ordinary annuity table is used.

FV Pmt x Table factor

Illustration: What is the future value of an annuity of P3,000 payable at the end of each quarter for
six payments with the first payment due at the end of nine months. The interest rate is 5%
compounded quarterly.

Solution:
FV P3,000 x 6.190654
FV P18,571.96
Present Value Using the Table
In finding the present value of an annuity, a series of steps is followed. Let d = the number of
deferred payment intervals.
PV Pmt x Table factor

Illustration: What amount should you invest now if you want to receive payments of P3,000 at the
end of each quarter for six payments with the receipt of the first payment at the end of nine
months? The interest rate is 5% compounded quarterly.

Solution:
Given, Pmt P3,000; i 1 ¼%; n 6; d 2

Step 1. Find the present value assuming that payments were made even during the period of
deferment.
Using d + n as the total number of compounding periods, the period to be used in locating the table
factor is 8 (2+6).
PV P3,000 x 7.568124
PV P22,704.37

Step 2. Find a present value assuming only d or 2 – the period of deferment as the total number of
compounding periods.
PV P3,000 x 1.963115
PV P5,889.35
Step 3. Subtract the present value in step 2 from the present value in step 1. The difference is the
present value of the deferred annuity.
PV P22,704.37 – P5,889.35
PV P16,815.02
Alternatively, the final present value may be solved as:
PV Pmt x Difference in table factors
PV P3,000 7.568124-1.963115
PV P16,815.02

ADDITIONAL PROBLEMS

Finding the Annuity Payment when Present Value is Known

Illustration: Belle purchased on account a computer for P90,000 on June 1, 2015 and agreed to pay
in 14 annual payments plus interest at 9.5% compounded annually, with the first payment due on
June 1, 2018. What is the size of each payment?

Solution:
Given, PV P90,000; i 9.5%; n 14; d 2
𝑃𝑉
𝑃𝑚𝑡
Table factor
90,000
𝑃𝑚𝑡
8.062260 1.747253
𝑃𝑚𝑡 P14,251.77

Finding the Term when Present Value is Known

Illustration: If P10,000 is deposited at the end of each month and the interest rate is 6%
compounded monthly, how many months will be required for the deposits to equal a present value
of P500,000? The first deposit is made at the end of six months.

Solution:
Given, PV P500,000; Pmt P10,000; i .5%; d 5
Find: Factor AD; Factor d n
Factor AD PV/Pmt
Factor AD 500,000/10,000
Factor AD 50
Factor d n Factor AD Factor d

Factor d n 90 4.925866
Factor d n 94.925866
To find n; find the first entry greater than 94.925866 in the .5% column of Table 4. That is the factor
for 130 periods which is 95.421606. Therefore, n= 130‐5 = 125.

PERPETUITY
A perpetuity is an annuity where the periodic payments continue indefinitely.

Simple ordinary perpetuity ‐ a perpetuity in which the periodic payments are made at the end of
each interest period. Present value of which is computed as

PV∞ I/R
Where:
I = periodic payment made at the end of each period and
R = interest rate per period

Illustration: If money is worth 8% compounded quarterly, compare the present values of the
following:
1. an annuity of P2,000 payable quarterly for 50 years;
2. an annuity of P2,000 payable quarterly for 100 years;
3. a perpetuity of P2,000 payable quarterly.
Solution:
1. Pmt 2,000; i 2%; n 200
PV P2,000 x 49.047345
PV P98,094.69
2. Pmt 2,000; i 2%; n 400

PV P2,000 x 49.981849
PV P99,963.70
3. I 2,000; r 2%;
PV∞ I/R
PV∞
2,000/2%
PV∞
P100,000
Observe that the present value of an annuity, whose term is increasing, gets closer to the
present value of perpetuity.


Simple perpetuity due ‐ is a perpetuity in which the periodic payments are made at the beginning of
each interest period. Present value of which is computed as

PV∞ I/R I
Where:
I = periodic payment made at the beginning of each period and
R = interest rate per period

Illustration: at 2% interest per quarter, what will be the present value of the simple perpetuity if
payment of P2,000 is made at the beginning of each quarter?

Solution:
Given, I 2,000; r 2%;
PV∞ I/R
PV∞ 2,000/2% 2, 000
PV∞ P102,000


CHAPTER 6 EXERCISES
I. Deferred Annuity

Finding the unknown values.
Future Present Payment Number of Period of Compound
Value Value Payments Deferment Interest Rate
1. P1,600 125 quarterly ? 7 14% quarterly
2. P1,000 ? annually 10 3 10% annually
3. ? ? 600 monthly 14 12 18% monthly

Word Problems.

1. Find the future value and the present value of an annuity of P20,000 payable at the
end of every three months for 20 payments. Interest rate is 11% compounded
quarterly. The first payment is due at the end of six years.

2. A set of computer costs P70,000 if paid in cash. On the installment plan, a purchaser
should pay P20,000 down payment and the balance to be paid in 10 quarterly
installments, the first due at the end of the first year after purchase. If money is
worth 15% compounded quarterly, determine the size of each payment.

3. Baker Company owns a plot of land on which buried toxic wastes have been
discovered. Since it will require several years and a considerable sum of money
before the property is fully detoxified and capable of generating revenues, Baker
wishes to sell the land now. It has located two potential buyers: Buyer A, who is
willing to pay $320,000 for the land now, and Buyer B, who is willing to make 20
annual payments of $50,000 each, with the first payment to be made 5 years from
today. Assuming that the appropriate rate of interest is 9%, to whom should Baker
sell the land?

4. Bethany borrowed P65,000 at 7% interest compounded quarterly, and agreed to pay
the loan in quarterly payments of P5,000 each. The first payment is due in two years.
Find the number of payments.


II. Perpetuity
Word Problems.

1. It costs P40,000 at the end of each year to maintain a section of Diversion road in
Davao City. If money is worth 10% compounded annually, find the present value of
a simple perpetuity.

2. What will be the present value of the simple perpetuity if payment of P4,000 is
made at the beginning of each quarter at 5% interest per quarter.

3. Find the present value of a simple perpetuity of P15,000 payable semi‐annually if the
interest rate is 2% per six months and the first payment is due (a) six months from
now, and (b) now.


Chapter 7
SINKING FUND AND AMORTIZATION

SINKING FUND
Putting up a sinking fund means setting aside equal amounts of money at the end of each
period, at a compound interest, for the purpose of saving for a future project or obligation.

Sinking Fund Payment Using the Table
𝐅𝐮𝐭𝐮𝐫𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐢𝐧𝐤𝐢𝐧𝐠 𝐟𝐮𝐧𝐝
𝐒𝐢𝐧𝐤𝐢𝐧𝐠 𝐟𝐮𝐧𝐝 𝐩𝐚𝐲𝐦𝐞𝐧𝐭
𝐅𝐮𝐭𝐮𝐫𝐞 𝐯𝐚𝐥𝐮𝐞 𝐭𝐚𝐛𝐥𝐞 𝐟𝐚𝐜𝐭𝐨𝐫

Illustration: What sinking fund payment is required at the end of each 6‐month period at 6% interest
compounded semiannually, to amount to P120,000 in 4 years?

Solution:
Future value of the sinking fund
Sinking fund payment
Future value table factor
Sinking fund payment P120,000 / 8.892336
Sinking fund payment P13,494.77
Sinking Fund Payment Using the Formula
𝒊
𝐒𝐢𝐧𝐤𝐢𝐧𝐠 𝐟𝐮𝐧𝐝 𝐩𝐚𝐲𝐦𝐞𝐧𝐭 𝑭𝑽 𝐱 𝒏

𝟏 𝒊 𝟏
where
FV amount needed in the future
i interest rate per period
n number of periods
Illustration: HBS Corporation needs P1,200,000 in 6 years to pay off a bond issue. What sinking fund
payment is required at the end of each month, at 12% interest compounded monthly, to meet this
financial obligation?

Solution:
Sinking fund payment 𝐹𝑉 x

.01
Sinking fund payment 1,200,000 x
1 .01 1

.
Sinking fund payment 1,200,000 x
.


Sinking Fund Schedule

Illustration 1: A P40,000 debt is to be repaid at the end of 1.5 years. Interest charged is 15% payable
at the end of every 3 months. The debtor established a sinking fund that earns 12% interest
compounded quarterly. Construct a sinking fund schedule.

Solution:

Sinking fund payment 𝐹𝑉 x

.03
Sinking fund payment 40,000 x
1 .03 1

Sinking fund payment 40,000 x .1545975

(1) (2) (3) (4) (5) (6)
At end of period Interest Income Periodic Deposit Periodic Sinking Fund Book Value
on Sinking Fund in Fund Increase in Fund Accumulated P40,000 - (5)
3% x (5)* (2)+(3) (4)+(5)*
1 6,183.90 6,183.90 6,183.90 33,816.10

2 185.52 6,183.90 6,369.42 12,553.32 27,446.68
3 376.60 6,183.90 6,560.50 19,113.82 20,886.18
4 573.41 6,183.90 6,757.31 25,871.13 14,128.87
5 776.13 6,183.90 6,960.03 32,831.16 7,168.84
6 984.93 6,183.90 7,168.83 40,000.00 0.00
Total 2,896.60 37,103.40 40,000.00

Alternatively, the information in the columns of the sinking fund can be obtained without
constructing a schedule.

Illustration 2: Given the same information in Illustration 1, find (a) the amount in the sinking fund at
the end of the fourth period, (b) the sinking fund interest income for the fifth payment period, and
(c) the book value of the debt at the end of the fourth period.

(a) The amount in the sinking fund at the end of the fourth period is the future value an annuity of
P6,183.90 payable quarterly at 12% compounded quarterly for four periods.

FV P6,183.90 x 4.183627
FV P25,871.13
(b) The principal at the beginning of the fifth payment period is the amount in the sinking fund at
the
end of the fourth period.
Interest income P25,871.13 x .03

Interest income P776.13
(c) The book value is the net obligation, which equals the original debt less the accumulated amount
in
the fund at that time.

Book value 40,000 – 25,871.13
Book value P14,128.87
Illustration 3: Using illustration 1, except that interest rates for different periods vary. During the
first and second periods, interest rate was 3%, 3 ½ % during the third and fourth periods , and 2 ½%
during the fifth and sixth periods.
(1) (2) (3) (4) (5) (6) (7)

At end of Interest Actual Interest Adjusted Periodic Sinking Fund
period Income on Interest Discrepancy Deposit Increase in Accumulated
Sinking Income Schedule Fund (3) +
Fund 3% x P6,183.90 + (5)
(5)* (4)
1 - 6,183.90 6,183.90
2 185.52 185.52 - 6,183.90 6,369.42 12,553.32
3 376.60 439.37 (62.77) 6,121.13 6,560.50 19,113.82
4 573.41 668.98 (95.57) 6,088.33 6,757.31 25,871.13
5 776.13 646.78 129.36 6,313.26 6,960.03 32,831.16
6 984.93 820.78 164.16 6,348.06 7,168.83 40,000.00
Total 2,896.60 2,761.42 135.18 37,238.58 40,000.00
AMORTIZATION
Amortization is a method of repaying debt, the principal and interest included, usually by a
series of equal payments at equal interval of time.

Amortization Payment Using the Table
𝐎𝐫𝐢𝐠𝐢𝐧𝐚𝐥 𝐚𝐦𝐨𝐮𝐧𝐭 𝐨𝐟 𝐨𝐛𝐥𝐢𝐠𝐚𝐭𝐢𝐨𝐧
𝐀𝐦𝐨𝐫𝐭𝐢𝐳𝐚𝐭𝐢𝐨𝐧 𝐩𝐚𝐲𝐦𝐞𝐧𝐭
𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐯𝐚𝐥𝐮𝐞 𝐭𝐚𝐛𝐥𝐞 𝐟𝐚𝐜𝐭𝐨𝐫
Illustration: What amortization payments are required each month, at 15% interest, to pay off a
P15,000 loan in 2 years?
Original amount of obligation

Amortization payment
Present value table factor
15,000
Amortization payment
20.624235
Amortization payment P 727.30


Amortization Payment Using the Formula

Illustration: Tyndale purchased on account a machine worth P50,000. What amortization payment is
required each month, at 18% interest to pay off his debt in 3 years?

𝑖
Amortization payment Present value x
1 1 𝑖
.
Amortization payment 50,000 x
.
.
.
Amortization payment P1,807.62

All Periodic Payments are Equal

Illustration: A P40,000 debt is to be amortized by equal payments at the end of every quarter for 1.5
years. Interest charged is 12% compounded quarterly. The debtor established a sinking fund that
earns 12% interest compounded quarterly. Construct an amortization schedule.
𝑖
Amortization payment Present value x
1 1 𝑖
.
.
.
.
Amortization payment P7,383.90

(1) (2) (3) (4) (5)
Period Outstanding Interest Due at Equal Payment Portion of
Principal at End of Period at End of Each Principal
Beginning of (2) x 3% Period Reduced by
Each Period Each Payment
(2)-(5)** (4)-(3)
1 40,000.00 1,200.00 7,383.90 6,183.90
2 33,816.10 1,014.48 7,383.90 6,369.42
3 27,446.68 823.40 7,383.90 6,560.50
4 20,886.18 626.59 7,383.90 6,757.31
5 14,128.87 423.87 7,383.90 6,960.03
6 7,168.84 215.07 7,383.90 7,168.83
Total 4,303.40 44,303.40 40,000.00

Alternatively, the outstanding principal can be obtained using the present value of an
annuity. For example, the outstanding principal after the fourth payment is made is the present
value of an annuity formed by the two remaining unpaid payments.

Outstanding principal Periodic payment x Present value table factor
Outstanding principal 7,383.90 x 1.913470
Outstanding principal P14,128.87

All Periodic Payments Except the Final Payment are Equal

Illustration: A P60,000 debt is to be discharged by payments of P10,000 at the end of every month.
Interest charged is 12% compounded monthly. Construct an amortization schedule.

(1) (2) (3) (4) (5)
Period Outstanding Interest Due at Equal Payment Portion of
Principal at End of Period at End of Each Principal
Beginning of (2) x 1% Period Reduced by
Each Period Each Payment
(2)-(5)** (4)-(3)
1 60,000.00 600.00 10,000.00 9,400.00

2 50,600.00 506.00 10,000.00 9,494.00
3 41,106.00 411.06 10,000.00 9,588.94
4 31,517.06 315.17 10,000.00 9,684.83
5 21,832.23 218.32 10,000.00 9,781.68
6 12,050.55 120.51 10,000.00 9,879.49
7 2,171.06 21.71 2,192.77 2,171.06
Total 2,192.77 62,192.77 60,000.00


CHAPTER 7 EXERCISES
I. Finding the Unknown.

a. Finding the sinking fund payment.
Sinking Fund Payment Time Nominal Interest Future
Payment Frequency Period Rate Compounded Value
1.__________ every 6 months 8 years 10% semi‐annually 500,000
2.__________ every year 14 years 9% annually 2,500,000
3.__________ every 3 months 5 years 12% quarterly 15,000

In eaI
b. Finding the unknown.
In each of the following problems, find (a) the interest payment for each interest period, (b)
the size of deposits to the sinking fund, (c) the amount in the sinking fund at the end of the
nth period, (d) the book value of the debt at the end of the nth period, and (e) the sinking
fund schedule interest income for the (n + 1)th payment period. Do not construct a sinking
fund schedule in finding your answers.
Debt Interest Rate No. of Deposits Interest Rate on nth Period

on the Debt in Sinking Fund Sinking Fund
1. P6,000 24%, monthly 20, monthly 18%, monthly 6th
2. P10,000 6%, semi‐annuall 8, semi‐ 5% semi‐ 5th

annually annually
3. P450 7%, annually 5, annually 10%, annually 3rd

c. Finding the amortization payment.

Loan Payment Term of Loan Nominal Present Value
Payment Period Rate (Amount of
Loan)

1.__________ every month 1 ¾ years 18% 100,000

2.__________ every year 12 years 9% 300,000
3.__________ every month 1 ½ years 12% 8,500


d. Finding the unknown.
In each of the following problems, find (a) the number of payments, (b) the outstanding
principal at the indicated time, (c) the interest and the principal included in the next
payment after the indicated time in (b), and (d) the size of the final payment and the total
cash payments. Do not construct an amortization schedule in finding your answers.
Debt Payments Compound Interest Required Outstanding

Rate Principal
1. P4,000 P200 every 3 months 8%, quarterly After 20th payment
2. P5,000 P1,000 every month 24%, monthly After 3rd payment
3. P1,400 P50 every month 12%, monthly After 6th payment

II. Word Problems.
1. Ana wants to accumulate P80,000 in 5 years for a trip. If her bank is paying 12%
interest compounded quarterly, how much must Ana deposit at the end of each 3‐
month period to reach her desired goal?
2. A condominium association wants to establish a sinking fund to accumulate
2,500,000 in 3 years to repair the concrete roof. The fund earns 9% interest
compounded monthly. If there are 200 units in the condominium, how much should
each unit owner be assessed each month as a fair contribution to the fund? Assume
that all units are of equal size and with equal assessment.

3. A P50,000 debt is to be repaid at the end of one year. The debtor establishes a
sinking fund that earns 8% interest compounded quarterly. Construct a sinking fund
schedule.

4. Referring to no. 1 above, assume that the interest rate on the sinking fund was 8%
compounded quarterly during the first ½ years and 12% compounded quarterly
during the second ½ year. Construct a sinking fund schedule.

5. Find the monthly payment of an auto loan of P200,000 to be amortized over a 5‐
year period at a rate of 9%.

6. Khan purchased a new motorcycle for P130,000. He made a P20,000 down payment
and financed the balance at his bank for 7 years. What amortization payments are
required every 3 months, at 16% interest, to pay off the boat loan?
7. A debt of P80,000 is to amortized with P25,000 being paid at the end of every six
months. The interest rate is 6% compounded semi-annually. Construct an
amortization schedule.
8. A debt of P40,000 is to be amortized with P8,000 being paid at the end of each
quarter. The interest rate is 16% compounded quarterly. Construct an amortization
schedule.


Chapter 8
DEPRECIATION

Definition of terms

Original cost or total cost – this amount includes items such as purchase price, freight, handling and
set‐up charges and other costs that are directly attributable in bringing the asset to its present
location and condition.

Salvage value/ Scrap value/ Residual value/ Trade‐in value– this amount is the estimated value of
the asset at the time it is taken out if service.

Estimated useful life – this is the length of time that an asset is expected to generate revenue.

Depreciation ‐ is the systematic allocation of the cost of the assets over their useful life. This is done
to properly match expense with revenue earned.

METHODS USED FOR FINANCIAL STATEMENT REPORTING

Straight‐Line Method
It is simple and is more widely used than any other method. It provides a uniform annual
charge and is calculated as:

𝐓𝐨𝐭𝐚𝐥 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐛𝐥𝐞 𝐜𝐨𝐬𝐭
𝐃𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧
𝐄𝐬𝐭𝐢𝐦𝐚𝐭𝐞𝐝 𝐮𝐬𝐞𝐟𝐮𝐥 𝐥𝐢𝐟𝐞

Illustration 1: A contractor purchased a gantry crane for P250,000. Freight and insurance charges
amounted to P18,000; customs’ broker’s fees, P8,500; taxes, permits and other expenses, P25,000.
The contractor estimates the life of the gantry crane to be 6 years with a salvage value of P20,000.
(a) Determine the annual depreciation charge and (b) prepare a depreciation schedule.

Total depreciable cost
Depreciation
Estimated useful life

, , , , ,
Depreciation

,
Depreciation

Depreciation P56,300


Depreciation Schedule
Annual Accumulated
End of Year Depreciation Depreciation Book Value
301,500
1 56,300 56,300 245,200
2 56,300 112,600 188,900
3 56,300 168,900 132,600
4 56,300 225,200 76,300
5 56,300 281,500 20,000

Sum‐of‐the‐Years’ Digits Method
The basic assumption for this method is that the value of the property decreasing at a
decreasing rate. It provides very rapid depreciation during the early years of its useful life. To
determine the depreciation, the following equations will be used:

𝒏 𝒏 𝟏
𝑺𝒀𝑫
𝟐

𝐘𝐞𝐚𝐫𝐬 𝐨𝐟 𝐮𝐬𝐞𝐟𝐮𝐥 𝐥𝐢𝐟𝐞 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠
𝑺𝒀𝑫 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐫𝐚𝐭𝐞 𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧
𝐒𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐝𝐢𝐠𝐢𝐭𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐮𝐬𝐞𝐟𝐮𝐥 𝐥𝐢𝐟𝐞

Illustration 2: Refer to Illustration 1. Determine the depreciation charges per year using SYD. Prepare
a depreciation schedule.

5 5 1
𝑆𝑌𝐷 15
2

Years of useful life remaining
𝑆𝑌𝐷 depreciation rate fraction
Sum of the digits of the useful life

Year SYD Depreciation rate fraction
1 5/15
2 4/15
3 3/15
4 2/15
5 1/15

End of Total Depreciation Annual Accumulated Book Value
Year Depreciation Rate Fraction Depreciation Depreciation
301,500
1 281,500 5/15 93,833 93,833 207,667
2 281,500 4/15 75,067 168,900 132,600
3 281,500 3/15 56,300 225,200 76,300
4 281,500 2/15 37,533 262,733 38,767
5 281,500 1/15 18,767 281,500 20,000
Declining‐Balance Method
The depreciation cost in any year is a constant ratio of the book value at the beginning of the
year. Salvage value, if any, is ignored. The most frequently used multiples are 1.25, 1.5 and 2, known
as 125% declining balance, 150% declining balance and 200% double‐declining balance.
To calculate for the declining balance rate, the following equation is used:
𝟏
𝐃𝐞𝐜𝐥𝐢𝐧𝐢𝐧𝐠 𝐛𝐚𝐥𝐚𝐧𝐜𝐞 𝐫𝐚𝐭𝐞 𝐱 𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐞
𝐔𝐬𝐞𝐟𝐮𝐥 𝐥𝐢𝐟𝐞

Illustration 3: JCF Shipping bought an equipment worth P200,000. It is estimated to have a useful life
of 5 years and a salvage value of P20,000. Determine the depreciation charges and prepare a
depreciation schedule using the double‐declining balance method.

1
Declining balance rate x 2
5

Declining balance rate 40%

End of Beginning Depreciation Annual Accumulated Ending Book
Year Book Value Rate Depreciation Depreciation Value
200,000
1 200,000 40% 80,000 80,000 120,000
2 120,000 40% 48,000 128,000 72,000
3 72,000 40% 28,800 156,800 43,200
4 43,200 40% 17,280 174,080 25,920
5 25,920 40% *5,920 180,000 20,000

*In year 5, although the estimated depreciation is P10,368 (25,920 x 40%), the allowable
depreciation is
limited to P5,920, because the book value has already reached the P20,000 salvage value. At this
point,
depreciation is complete.


Units‐of‐Production Method

This method accounts for depreciation on the basis of actual service rendered or actual units
produced. To calculate depreciation using this method, the depreciation per unit is first determined.

Cost salvage value
Depreciation per unit
units of useful life

Illustration 4: A textile company purchased an equipment for P100,000 with a salvage value of
P20,000. For depreciation purposes, the equipment is expected to have a useful life of 5,000 hours.
From the following estimate of hours of use, prepare a depreciation schedule for the equipment
using the units‐of‐production method.

Year Hours of Use
1 1,500
2 1,200
3 2,000
4 500

Cost salvage value
units of useful life

100,000 20,000
5,000

Depreciation per unit P16

End of Depreciation Hours Used Annual Accumulated Book Value
Year per hour Depreciation Depreciation
100,000
1 16 1,500 24,000 24,000 76,000
2 16 1,200 19,200 43,200 56,800
3 16 2,000 32,000 75,200 24,800
4 16 500 *4,800 80,000 20,000

*Maximum allowable to reach the salvage value.


COMPOUND INTEREST METHODS

Annuity Method
This method accounts for the periodic depreciation charges as equal and includes the
interest on the book value for each operating period. The periodic book value is assumed to be
earning the same interest as the amount would earn if it were invested elsewhere. The following
equations will be used to determine annual depreciation charges.

𝟏. 𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐬𝐚𝐥𝐯𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞 𝐒𝐚𝐥𝐯𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞 𝐱 𝐭𝐚𝐛𝐥𝐞 𝐟𝐚𝐜𝐭𝐨𝐫

𝟐. 𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐛𝐥𝐞 𝐜𝐨𝐬𝐭
𝐂𝐨𝐬𝐭 𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐬𝐚𝐥𝐯𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞

𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐛𝐥𝐞 𝐜𝐨𝐬𝐭
𝟑. 𝐀𝐧𝐧𝐮𝐚𝐥 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧 =
𝐓𝐚𝐛𝐥𝐞 𝐟𝐚𝐜𝐭𝐨𝐫

Illustration 5: JCF Company bought a machine worth P11,000. It is estimated to have a useful life of
5 years and a salvage value of P1,200. Determine the depreciation charges and prepare a
depreciation schedule using the double‐declining balance method. Assume that the effective
interest rate is 6%.

𝟏. Present value of salvage value Salvage value x table factor

Present value of salvage value 20,000 x .747258 P14,945.16

2. Present value of depreciable cost Cost present value of salvage value

Present value of depreciable cost 100,000 14,945.16 P85,054.84
3. Annual depreciation =

, .
Annual depreciation = = P20,191.71
.

(1) (2) (3) (4) (5) (6)
End of Year Annual Interest Income Net Accumulated Book Value
Depreciation (6) x 6% Depreciation Depreciation
Charges
100,000
1 20,191.71 6,000.00 14,191.71 14,191.71 85,808.29
2 20,191.71 5,148.50 15,043.21 29,234.92 70,765.08
3 20,191.71 4,245.90 15,945.81 45,180.73 54,819.27
4 20,191.71 3,289.16 16,902.55 62,083.28 37,916.72
5 20,191.72 2,275.00 17,916.72 80,000.00 20,000.00


Sinking Fund Method
This method assumes that a sinking fund is established for the purpose of replacing an asset
at the end of its useful life. It accounts for the periodic depreciation charges as exactly the same as
the periodic increases (including the periodic deposit and interest) in the sinking fund. The following
equations will be used to determine annual depreciation charges and periodic deposit in fund.

𝟏. 𝐓𝐨𝐭𝐚𝐥 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐚𝐭 𝐭𝐡𝐞 𝐞𝐧𝐝 𝐨𝐟 𝐮𝐬𝐞𝐟𝐮𝐥 𝐥𝐢𝐟𝐞
𝐂𝐨𝐬𝐭 𝐒𝐚𝐥𝐯𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞

𝐓𝐨𝐭𝐚𝐥 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧
𝟐. 𝐀𝐧𝐧𝐮𝐚𝐥 𝐝𝐞𝐩𝐨𝐬𝐢𝐭 =
𝐓𝐚𝐛𝐥𝐞 𝐟𝐚𝐜𝐭𝐨𝐫

Illustration 6: Assume that the effective interest rate is 6%. Use the data in Illustration 5 to find the
annual depreciation charges by the sinking fund method. Prepare a depreciation schedule.

1. Total depreciation at the end of useful life Cost Salvage value
Total depreciation at the end of useful life 100,000 20,000 P80,000

2. Annual deposit =

,
Annual deposit P14,191.71
.

(1) (2) (3) (4) (5) (6)
End of Year Peridic Deposit Interest Income Peridic Increase Accumulated Book Value
in Fund from Sinking in Fund = Sinking Fund =
Fund (5) x 6% Annual Accumulated
Depreciation Depreciation
Charges (2) + (3)
100,000
1 14,191.71 14,191.71 14,191.71 85,808.29
2 14,191.71 851.50 15,043.21 29,234.92 70,765.08
3 14,191.71 1,754.10 15,945.81 45,180.73 54,819.27
4 14,191.71 2,710.84 16,902.55 62,083.28 37,916.72
5 14,191.72 3,725.00 17,916.72 80,000.00 20,000.00


CHAPTER 8 EXERCISES
I. Finding the unknown.
a. Straight‐line method
Cost Freight Set‐up Total cost Salvage Useful life Total Annual
cost value Dep’n Dep’n
1. 45,000 150 500 ________ 3,500 10 years ________
2. 158,200 0 1800 ________ 20,000 ________ ________ 14,000
3. 88,600 625 2,500 ________ 9,000 7 years ________
4. 5,600 210 54 ________ 600 ________ ________ 658

b. Sum‐of‐the‐Years’ Digits Method
Depreciation rate fraction
Useful life SYD Year 1 Year 3 Year 5
1. 5 Years ________ ________ ________ ________
2. 7 years ________ ________ ________ ________
3. 10 years ________ ________ ________ ________

c. Declining‐balance method
Years Straight line rate Multiple Declining Balance
Rate
1. 4 ________ 125% ________
2. 5 ________ 200% ________
3. 10 ________ 150% ________
4. 9 ________ 125% ________
5. 6 ________ 200% ________

d. Units‐of‐Production Method
Cost Salvage Value Units of Useful life Dep’n per Unit
1. 45,000 5,000 250,000 units ________

2. 274,000 34,000 60,000 kms ________
3. 155,000 2,000 15,000 hours ________
4. 8,900 250 500,000 gallons ________
5. 3,900 0 160,000 copies ________


II. Preparing a Depreciation Schedule

1. MB purchased a new machine worth P570,000. Freight charges were P4,700 and
installation amounted to P5,000. Estimated useful life is 5 years with a salvage value
of P20,000. Prepare a depreciation schedule using straight‐line method.

2. A new unit of production‐line machinery was purchased for P445,000. Estimated
useful life of the machine is 6 years and a trade‐in value of P25,000. Prepare a
depreciation schedule using sum‐of‐the‐years’ digit method.

3. Demetrius Air Service bought a fleet of helicopter for P38,600,000. It is expected to
have a useful life of 4 years and a trade‐in value of P7,000,000. Prepare a
depreciation schedule using the 150% declining‐balance method.

4. A meat truck was purchased by Cleveland Marketing worth P545,000. It is expected
to have a useful life of 75,000 kms. Scrap value was set at P75,000. If the truck was
driven for the following number of kilometers per year, prepare a depreciation
schedule using the units‐of‐production method.

Year 1 2 3 4 5
Kms Driven 12,500 18,300 15,900 19,100 12,400

5. Anabel purchased an equipment for P24,000. Estimated useful life is 4 years and
salvage value is P3,000. Assume that the effective interest rate is 5%.
(a) Construct a depreciation schedule using the annuity method.
(b) Construct a depreciation schedule using the sinking fund method.


Chapter 9
STOCKS, BONDS AND MUTUAL FUNDS

Corporations need resources to finance their various business operations. These resources
can be in the form of debts, stocks and retained earnings. Commonly used though more expensive is
the issuance of additional shares of stock.

Definition of terms:

Stocks – represent ownership and rights of the holders in a corporation.

Certificate of stock – is a formal written evidence of the holder’s ownership of one or more shares
and is a convenient instrument for the transfer of title.

Scripless Trading – is the shift from physical transfer of stock certificates to electronic book‐entry
securities transactions.

Dividends – are earnings distributed to shareholders of the corporation on a pro‐rata basis. It may
be in the form of cash or shares.

For discussion purposes, a stock quotation table is from the Philippine Stock Exchange is shown in
the next page and each column entry in the stock quotation is explained as:

a & b. 52 Weeks High and Low. The highest and lowest traded prices of a stock for the past 52
weeks.
c. Name. The name of the listed company.
d. EPS. Earnings per share, calculated by dividing after‐tax income by number of shares outstanding.
e. Prev Close. The closing price of the previous trading day.
f. Open. The opening price of the stock for the day.
g. High. The highest traded price of a stock during a specific trading period.
h. Low. The lowest traded price of a stock during a specific trading period.
i. Close. The closing price of the trading day.
j. Volume. The total number of shares traded during a given period.
k. Value. The amount of transactions in pesos traded for a period.
l. % change. Calculated as (i‐e)/e. A negative value indicates that the closing price for the day is
lower than that for the previous day.
m. PE ratio. Price earnings ratio. Calculated as i/d. It indicates how much an investor pays for a
company’s earning power.


Dividends on Preferred and Common Stock

Cash dividends must first be paid to preferred shareholders before any common shareholders are
paid.

When preferred shares are non‐participating, the stockholders receive only the fixed dividend and
no more.

When preferred shares are participating, the stockholders may receive additional dividends if the
company does well.

When preferred shares are cumulative, any dividends in arrears must be paid to preferred
shareholders before allocating any dividends to common shareholders.

When preferred shares are non‐cumulative, only the current year’s dividend must be paid to
preferred shareholders before paying any dividends to common shareholders.

Convertible preferred means the stock may be exchanged for a specified number of common shares
in the future.

The steps to distribute dividends on preferred and common stock follow:

1. If the preferred stock is cumulative, any dividends that are in arrears are paid first; then the
preferred dividend is stated in pesos (no‐par) go to Step 2. When the dividend per share is stated as
a percent (par), multiply the par value by the dividend rate.

Dividend per share (preferred) = Par value x dividend rate

2. Calculate the total amount of the preferred stock dividend by multiplying the number of preferred
shares by the dividend per share.

Total preferred dividend = number of shares x dividend per share

3. Calculate the total common stock dividend by subtracting the total preferred stock dividend from
the total dividend declared.

Total common dividend = total dividend – Total preferred dividend

4. Calculate the dividends per share for the common stock by dividing the total common stock
dividend by the number of shares of common stock.

Common dividend per share =


Illustration1: FTF Corporation has 2,500,000 shares of common stock outstanding. If a dividend of
P4,000,000 was declared by the company directors last year, what is the dividend per share of
common stock?

, ,
Dividend per share (common) = = = P1.60 per share
, ,

Illustration2: The same with illustration 1, except that the corporation has 1,000,000 shares of
preferred stock that pay dividend of P0.50 per share. Calculate the amount due to preferred
shareholders and the dividend per stock of common stock.

1. Total preferred dividends = Number of shares x Dividend per share

Total preferred dividends = 1,000,000 x P0.50 = P500,000

2. Dividend per share (common) =
.

, , ,
Dividend per share (common) = = P1.40 per share
, ,

Illustration 3: AFV Company has 100,000 shares of P100 par value, 6% cumulative preferred stock
and 2,500,000 shares of common stock. Although no dividend was declared last year, a P5,000,000
dividend had been declared this year. Calculate the amount of dividends due the preferred
shareholders and the dividend per share of common stock.

Because the preferred stock is cumulative, and the company did not pay dividends last year, the
preferred shareholders are entitled to the dividends in arrears and the dividends for the current
period.

1. Dividend per share (preferred) = Par value x Dividend rate

Dividend per share (preferred) = 100 x 6% = P6.0 per share

2. Total preferred dividend (per year) = Number of shares x Dividend per share

Total preferred dividend = 100,000 x P6.0 x 2 = P1,200,000

3. Dividend per share (common) =
.

, , , ,
Dividend per share (common) = = P1.52 per share
, ,


Current Yield for a Stock
Current yield is a way of measuring the rate of return earned from dividends. When a stock
does not pay dividends, there is no current yield. The current yield for a stock is determined by the
equation:

Current Yield =

Illustration 4: Determine the current yield for PTP Corporation stock, which pays a dividend of P1.70
per year and is currently selling at P34.50 per share.

.
Current Yield = = .0493 = 4.93%
.

Price‐Earnings Ratio of a Stock
Price earnings ratio measures the relationship between market price per share and earnings
per share. This ratio reflects “buyer confidence” in a particular stock compared with the market as a
whole. The price‐earnings ratio of a stock is expressed in the equation:

Price‐earnings ratio =

Illustration 5: FCA stock is currently selling at P104.75. If the company had earnings per share of P4.6
last year, calculate the price‐earnings ratio of the stock.

.
Price‐earnings ratio = = 29.0972 or 29:1
.

The ratio shows that investors are currently willing to pay 29 times the earnings for 1 share
of FCA stock.

Cost, Proceeds and Gain (Loss) on a Stock Transaction
The cost of purchasing stock includes not only the purchase price but also brokerage
commission. Brokerage commission rates are competitive, and vary from broker to broker. Rates
range from 0.25% to 1.5%.

Proceeds from selling the stock = selling price – brokerage commission

Gain or Loss = Proceeds from sale – the cost of purchasing stock

Illustration 6: Claviel decides to buy a stock whose market price is P10.00 and with a par value of
P1.00. The minimum number of shares he can buy at a regular transaction is 1,000 shares. Compute
for the required cash flows.

Market Price Per Share P10,000
Multiply by Minimum No. of Shares 1,000
P10,000
Add: Broker’s Commission (1.5%) 150
Total Cash Outlay P10,150

After a year, Claviel opts to sell all the shares he previously bought. Current market price is P13.00
per share. Compute for the cash inflows.

Market Price Per Share P13.00
Multiply by Minimum No. of Shares 1,000
P13,000
Less: Broker’s Commission (1.5%) 195
Net Cash Receivable P12,805

Return on Investment
Return on investment measures the total monetary gain on a stock for an investor. It is
expressed as:

𝐍𝐞𝐭 𝐠𝐚𝐢𝐧 𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐯𝐢𝐝𝐞𝐧𝐝𝐬
ROI =
𝐓𝐨𝐭𝐚𝐥 𝐜𝐨𝐬𝐭 𝐨𝐟 𝐬𝐭𝐨𝐜𝐤 𝐩𝐮𝐫𝐜𝐡𝐚𝐬𝐞

Illustration 7: Using the data in the previous illustration and assuming that a total of P1,300 in
dividends was received for the year, Claviel’s ROI is calculated as follows:

, , ,
ROI =
,

ROI = 0.3897 or 38.97%

BONDS
A formal unconditional promise made under seal to pay a specified sum of money at a
determinable future date, and to make periodic interest payments at a stated rate until the principal
sum is paid.

Cost, Proceeds and Gain (Loss) on a Bond Transaction

Cost of bonds includes the purchase price, broker’s commission, taxes and other charges incurred in
their acquisition.

Gain (or loss) from a bond transaction is the difference between the proceeds from the sale and the
cost of purchase.

Current Yield for a Bond
Current yield of a bond is a simple measure of the return on investment based on the
current market price. When bonds are purchased at face, the current yield is equal to the coupon
rate. Current yield for a bond is expressed as:

𝐀𝐧𝐧𝐮𝐚𝐥 𝐢𝐧𝐭𝐞𝐫𝐞𝐬𝐭
Current Yield =
𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐦𝐚𝐫𝐤𝐞𝐭 𝐩𝐫𝐢𝐜𝐞

Illustration 8: Calculate the current yield for JJV Centrics bond with face of P1,000 with a coupon
rate of 6.75%, and currently selling at a premium of 107.25.

Annual interest = Face value x Coupon rate = 1,000 x .0675 = P67.50
Current price = Face value x Price percent = 1,000 x 1.0725 = P1,072.50

Current Yield =

.
Current Yield = = .0629 = 6.29%
, .

MUTUAL FUNDS
A mutual fund is a fund managed by an investment company. It means that someone works
on your behalf. There are funds having high potential growth but high risk while others have lower
potential growth but less or tolerable risk.

Load mutual fund – charges a fee when you buy or sell shares.

No‐load mutual fund – do not charge a fee when you buy or sell shares.

Common mutual fund charges:
Load – 0.25% to 3%;
Early redemption fee – 1%;
Management, advisory and distribution fees – 0.75 to 2% per annum based on the net asset value of
the fund

𝑵𝒆𝒕 𝑨𝒔𝒔𝒆𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒇𝒖𝒏𝒅
𝐍𝐞𝐭 Asset Value
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑺𝒉𝒂𝒓𝒆𝒔 𝑶𝒖𝒕𝒔𝒕𝒂𝒏𝒅𝒊𝒏𝒈
For illustration purposes, a sample mutual fund quotation is shown below:

Fund NAV NET CHG YTD % RET
Fixed Income Fund 179.47 0.08 11.05
Equity Fund 255.82 0.03 30.31
Balanced Fund 135.67 0.01 21.34


Looking at the equity fund, the net asset value for today is at P255.82 per share. The fund closed
P0.03 higher that yesterday’s closing price, that means yesterday’s closing price is P255.79. The fund
has a 30.31% return since January 1 of the year.

Return on Investment
To check whether an investor made the right decision to invest in a certain fund, computing
the return on investment from such fund can be a tool. The return on investment for a mutual fund
depends on the increase in net asset value and on the dividends paid from the fund expressed in the
equation:

𝐄𝐧𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐲𝐞𝐚𝐫 𝐍𝐀𝐕 𝐃𝐢𝐯𝐢𝐝𝐞𝐧𝐝 𝐃𝐢𝐬𝐭𝐫𝐢𝐛𝐮𝐭𝐢𝐨𝐧 – 𝐁𝐞𝐠.𝐘𝐞𝐚𝐫 𝐍𝐀𝐕
ROI =
𝐁𝐞𝐠. 𝐘𝐞𝐚𝐫 𝐍𝐀𝐕

Illustration 9: A certain fund net asset value on January 1 was 104. Dividend distributions during the
year amounted to P25 per share. Net asset value at the end of the year was set at P100. Calculate
ROI.
ROI 20.19%


CHAPTER 9 EXERCISES
I. Finding the unknown.
a. Find the Preferred and Common Dividend per Share
Common Preferred Shares Dividend Arrears
Shares Shares Div. or Par Cum Declared
1. 5,000,000 None P3,000,000 None
2. 10,000,000 3,000,000 P5.50 No 25,000,000 None
3. 8,000,000 2,000,000 P100, 6% No 10,000,000 None
4. 4,000,000 1,000,000 P100, 4% Yes 14,000,000 1 year
5. 20,000,000 4,000,000 P6.25 Yes none 1 year

b. EPS, CY Ratio, PE Ratio, Dividends and Market Price
Earnings per Annual Dividend Current Price Current Yield Price‐Earnings
Share per Share Ratio
1. 32.95 8.00 230.63 _________ _________
2. 3.85 1.20 88.13 _________ _________
3. _________ 2.25 122.50 _________ 21
4. 24.60 8.00 _________ 2.5% _________
5. _________ _________ 211.88 0.7% 30

c. Total Cost, Proceeds, and Gain (Loss) and ROI
No. of Purchase Selling Dividend Buying Total Cost Proceeds Gain (Loss) ROI
Shares Price Price per and
Share Selling
Rate

1. 100 142.50 169.50 5.09 1% ________ ________ ________ ________
2. 900 85.87 108.75 5.44 ¾% ________ ________ ________ ________
3. 500 77.63 77.63 3.11 1% ________ ________ ________ ________

d. Annual Interest and Current Yield
Assume that the bonds issued by the following companies have a uniform face value
of P1,000.
Coupon Rate Annual Interest Market Price Current Yield
1. 6.625 __________ 91.13 __________
2. 9.25 __________ 108 __________
3. 7.5 __________ 125.25 __________
4. 11.875 __________ 73.5 __________
5. 5.375 __________ 84.38 __________

e. Mutual Fund Quotation
FUND NAV NET CHG YTD % RET
Balance Fund 14.64 ‐0.45 11.35
1. Give interpretation to the mutual fund quotation above.

II. Word Problems

1. Agile Pharmaceuticals has 300,000 shares of P100 par value, 7.5%, cumulative
preferred stock and 5,200,000 shares of common stock. Although no dividend was
declared last year, a P7,000,000 dividend has been declared for this year. Calculate
the amount of dividends due the preferred shareholders and the dividend per share
pf common stock.

2. You purchased 650 shares of Amore, Inc. common stock at P132.75 per share. A
few months later you sold the shares at P153.38. Your stockbroker charges *¾ %
commission.
(a) What is to total cost of purchase?
(b) What are the proceeds on the sale?
(c) What is the gain or loss on the transaction?
*Selling and Buying Rate

3. Using the answers obtained in Problem # 2 compute for the ROI of Amore, Inc.
Dividends received during the year =P 7.8 dividend per share

4. On January 1, Peso Bond Fund’s total assets were P350 and total liabilities were
P190. During the year the fund distributed P20 per share to investors. At the end of
the year, total assets were P382 and total liabilities were P220. Calculate the return
on investment.

5. The Elm Industries is currently selling at P186.88 per share which pays a dividend of
P6.80 per share. Earnings per share for the past 12 months is P33.50.
(a) What is the current yield on the stock?
(b) What is the price‐earnings ratio for Elm?


Appendix 1 – Present Value Interest Factor
Workbook in Math of Investments

68

Appendix 2– Present Value Interest Factor for an Ordinary Annuity

69

Appendix 3 – Future Value Interest Factor


Appendix 4 – Future Value Interest Factor on an Ordinary Annuity

71

Workbook in MOI

Uploaded by

Copyright:

Available Formats

Workbook in MOI

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Workbook in MOI

Uploaded by

Copyright:

Available Formats

𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒙 𝑹𝒂𝒕𝒆 𝒙 𝑻𝒊𝒎𝒆

Effective interest rate = 𝐵𝑎𝑛𝑘 𝑑𝑖𝑠𝑐𝑜𝑢𝑛𝑡

𝑪𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝑭𝒖𝒕𝒖𝒓𝒆 𝑽𝒂𝒍𝒖𝒆 𝒍𝒆𝒔𝒔 𝑷𝒓𝒆𝒔𝒆𝒏𝒕 𝑽𝒂𝒍𝒖𝒆

𝑪𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝑽𝒂𝒍𝒖𝒆 𝑭𝑽 𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒙 𝑻𝒂𝒃𝒍𝒆 𝑭𝒂𝒄𝒕𝒐𝒓

Interest Compounding made Compounding

Nominal Interest Compounding Interest rate

Final compound amount = Compound amount + simple

i 2.966667 (2)

5 1/2% 2.917757 (3)

(2)‐(3) x – 5 ½ % 0.048910 (4)

Nominal Term Of Loan Interest Compounded Periodic Rate Number Of

Principal Term Of Nominal Interest Compounded Compound Compound

Principal Term Of Nominal Interest Compounded New Table Compound

Principal Interest Interest Compounded Time Compound

Principal Future Interest Compounded Number Of Interest Rate

Compound Amount Term Of Nominal Interest Present Compound

Principal Term Of Nominal Interest New Factor Present

Future value = Annuity payment x table

Present value = Annuity payment x table factor

𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐕𝐚𝐥𝐮𝐞 𝐗 𝐓𝐚𝐛𝐥𝐞 𝐅𝐚𝐜𝐭𝐨𝐫 𝑻𝒂𝒃𝒍𝒆 𝟓

Annuity payment = Present Value

𝑨𝒏𝒏𝒖𝒊𝒕𝒚 𝒑𝒂𝒚𝒎𝒆𝒏𝒕 𝐅𝐮𝐭𝐮𝐫𝐞 𝐕𝐚𝐥𝐮𝐞 𝐗 𝐓𝐚𝐛𝐥𝐞 𝐅𝐚𝐜𝐭𝐨𝐫 𝐢 𝑻𝒂𝒃𝒍𝒆 𝟓

Factor = Present Value

(2)‐(3) x – 3% 0.122525 (4)

Factor = Future Value

Future Value Present Value Payment Term Compound

Annuity Payment Time Nominal Interest Future Value Of The

Annuity Payment Time Nominal Interest Present Value Of The

Future Value Present Value Payment Term Interest – Compounding Period

Future Value Present value Payment Compounding Interest Rate

Future value = Annuity payment x table factor

Present value = Annuity payment x table factor

Annuity payment = Present Value

Annuity payment = Future Value

Factor = 58,000 3,000

Factor = 76,000 1,000

Annuity Payment Time Nominal Interest Future value of the

Annuity Payment Time Nominal Interest Present Value Of The

Future Value Present Value Payment Term Compound Interest Rate

1. 5,800 P200 monthly 2 years ? monthly

2. Pmt 2,000; i 2%; n 400

1. P1,600 125 quarterly ? 7 14% quarterly

2. P1,000 ? annually 10 3 10% annually

3. ? ? 600 monthly 14 12 18% monthly

Sinking fund payment P6,183.90

1 6,183.90 6,183.90 6,183.90 33,816.10

(1) (2) (3) (4) (5) (6) (7)

Original amount of obligation

Amortization payment P 727.30

Amortization payment P1,807.62

Amortization payment P7,383.90

1 60,000.00 600.00 10,000.00 9,400.00

1.__________ every 6 months 8 years 10% semi‐annually 500,000

2.__________ every year 14 years 9% annually 2,500,000

3.__________ every 3 months 5 years 12% quarterly 15,000

Debt Interest Rate No. of Deposits Interest Rate on nth Period

2. P10,000 6%, semi‐annuall 8, semi‐ 5% semi‐ 5th

3.__________ every month 1 ½ years 12% 8,500

Debt Payments Compound Interest Required Outstanding