A Self-Instructional Module on Mathematics of

Investment
2020 Edition

Dr. Floriza N. Laplap Dr. Glen M. Pesole

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A Self-Instructional Module on Mathematics of

Investment

Course Code
#XXXX-XXXX

Course Name/Title

Mathematics of Investment

Name/s of Faculty

Dr. Floriza N. Laplap

Professor VI

Dr. Glen M. Pesole

Professor IV
 2

A Self-Instructional Module on Mathematics of

Investment

Course Code
#XXXX-XXXX

Course Name/Title

Mathematics of Investment

Name/s of Faculty

Dr. Floriza N. Laplap

Professor VI

Dr. Glen M. Pesole

Professor IV
 3

I. Module Overview/ Introduction

Why is it important to know about interests, discounts, annuities, amortization,
depreciation ? Why is there a need to invest money, whether just to save for a phone or laptop?

This module aims to provide you with a basic understanding of the applications of
mathematical knowledge and skills on mathematics of investment to help in understanding the
basic concepts of the value of money using simple and compound interest as well as discounting,
variation of annuities, amortization, sinking fund and depreciation. It aids the learners in real world
applications in terms of saving, credit and debit as well as fund accumulations.

Mathematics of Investment is a 3-unit course which primarily deals with money

transactions associated with interest and time. This course equips you with mathematical tools in
the practical applications of mathematical concepts in finance and guides you in looking into
situations that will enable wise financial planning and utilization.

II. Desired Learning Outcomes

At the end of the semester, you must be able to:

1. Identify various financial concepts as well as types of investment problems and relate
these to other curricular areas if possible

2. Solve financial problems involving simple and compound interest, discounting,

annuities, amortization and others using appropriate teaching and assessment
methods/techniques

3. Develop critical thinking in making sound plans and decisions pertaining to financial
matters

4. Apply the financial knowledge and skills in real-world situations

5. Develop the sense of responsibility and commitment in the world of finances

III. Learning Contents, Tasks and Assessments

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Chapter 4: AMORTIZATION AND SINKING FUND

Amortization

Amortization commonly refers to the process of paying off debt in equal periodic installments of
interest and principal in order to repay the loan in full up to date of maturity. These equal
payments form an annuity of the simple case. In a loan amortization payment, a higher percentage
of the flat monthly payment goes toward interest early in the loan. While in a later succeeding
payment, a greater percentage of the payment goes toward the loan's principal.

In an amortized loan, the present value can be thought of as the amount borrowed or the original
principal obligation, n is the number of periods the loan lasts for, i is the interest rate per period,
and payment is the loan payment that is made.

Steps in the Construction of an Amortization Schedule

Firstly: Find R, the periodic payment that form an annuity of the simple case, by:

Ordinary Annuity
A
R= 1−(1+i)−n
[ i
]

Annuity Due
Adue
R= 1−(1+i)−n
[ i
](1+i)

Deferred Annuity
𝐴𝑑
R= 1−(1+𝑖) −(𝑛+𝑑) 1−(1+𝑖)−𝑑
{[ ]−[ ]}
𝑖 𝑖

Secondly, Construct the Amortization Schedule

Amortization schedule is a table that shows how much is applied to reduce the principal and how
much is paid to interest to indicate the remaining obligations on the outstanding principal after
each payment period.
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❑ We can observe the following from the table:

❑ Size of each payment remains the same.

❑ However, Interest payment declines each year as the amount owed declines and
more of the principal is repaid.

Example 21. Mr. Nasar purchase a lot worth PhP2,500,000 to be repaid on a semi-annual
installment for 15 years. Find the semi-annual payment if money is worth 12% compounded semi-
annually.

Solution:

Given:

A= PhP2,500,000

m = number of conversions per year = 2

t= 15 years

n= number of conversion periods =tm= 30

j = interest rate per year = 12% =0.12

i= compounding interest rate per conversion period = j/m = 0.12 / 2 = 0.06

Find R= periodic payment?

A
R=
1 − (1 + i)−n
[ ]
i

2.500,000
R=
1 − (1 + 0.06)−30
[ ]
0.06

2,500,000
R=
13.76483115

R = PhP181,622.28
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Example 22. Suppose you plan to get a PhP9,000 loan from a cooperative bank at 18% annual
interest with annual payments that you will pay off in over (5) five years.

a. What will your annual payments be on this loan?

b. Construct an amortization schedule.

Solution:

Given:

A= PhP9,000.00

m = number of conversions per year = 1

t= 5 years

n= number of conversion periods =tm= 5

j = interest rate per year = 18% =0.18

i= compounding interest rate per conversion period = j/m = 0.18 / 1 = 0.18

a. Find the annual payment of this loan, can be computed to R= periodic payment?
A
R=
1 − (1 + i)−n
[ ]
i

9,000
R=
1 − (1 + 0.18)−5
[ ]
0.18

9,000
R=
3.127171021

R = PhP2,878.00
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b. The loan amortization schedule constructed as follows:

LOAN AMORTIZATION SCHEDULE (in pesos)

Year Amount Annuity Interest Portion Repayment of Outstanding

Owed on Payment of the Annuity the Principal Loan Balance
Principal at (2) (3) = (1) × 18% Portion of the at Year end,
the Annuity After the
Beginning of (4) = (2) –(3) Annuity
the Year Payment
(1) (5)=(1) – (4)

1 9,000 2878 1,620.00 1,258.00 7,742.00

2 7,742 2878 1,393.56 1,484.44 6,257.56

3 6257.56 2878 1,126.36 1,751.64 4,505.92

4 4505.92 2878 811.07 2,066.93 2,438.98

5 2,438.98 2878 439.02 2,438.98 0.00

Example 23. Mr. Nasar approved salary loan worth of PhP100,000 is to be amortized in 60
payments every 6 months. Find the semi-annual payment and the outstanding balance after the
25th payment. Money is worth 20% compounded semi-annually.

Solution:

Given: A= PhP100,000.00, n = 60 m =2

J = 20% i = j/m = 0.20/2 = 0.10

Find: a. The semi-annual payment, R=?

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A
R=
1 − (1 + i)−n
[ ]
i

100,000.00
R=
1 − (1 + 0.10)−60
[ ]
0.10

100,000.00
R=
1 − (1.10)−60
[ ]
0.10

100,000.00
R=
[9.967157297]

R = 10,032.95

b. The remaining liability after the 25th payment is the present value of the remaining periodic
payment / outstanding balance, OP=?

where; R = 10,032.95 n= 60 k=25 i = 0.10

1−(1+𝑖)−(𝑛−𝑘)
𝑂𝑃 = 𝑅[ ]
𝑖

1−(1+0.10)−(60−25)
𝑂𝑃 = 10,032.95[ ]
0.10

𝑂𝑃 = 10,032.95[9.644158973]
𝑂𝑃 = 96,759.36
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Activity 14. Amortization

Solve each of the following.

1. Joshua’s debt in an XYZ savings bank worth of Php100,000.00 with interest at 12% computed
quarterly must be paid at the end of every 3 months for 2 years. a.) How much is the amount for
each quarter payment? b.) Find the remaining liability just after making the 6th payment. c.)
Construct the amortization schedule.

2. A PhP300,000.00 loan is to be repaid on installment semi-annually for 5 years with the

agreement that the money 24% compounded semi-annually

Find:

a.) The semi-annually payment;

b.) The amount of outstanding principal after the 5th payment;

c.) Construct the amortization schedule.

Sinking Fund

Sinking Fund denotes to a fund created by investing equal periodic deposits to anticipate the
need of paying a large amount of money at some specific dates. Sinking fund is used to pay off
other debts, replace worn-out equipment, or provide money for the purchase of new equipment.

The amount in the fund at any time is the total value of an ordinary annuity accumulated by equal
periodic payments at equal intervals of time. Together with, the amount of interest earned.

In solving for ordinary annuity problems, we can use the following formulas:
[(𝟏+𝒊)𝒏 −𝟏]
𝑺=𝑹
𝒊
𝑺(𝒊)
𝑹=
(𝟏+𝒊)𝒏 −𝟏

Where:

S= final amount or future value of an ordinary annuity

R= 𝑒𝑞𝑢𝑎𝑙 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑎𝑦𝑚𝑒𝑛𝑡𝑠

n= number of payments/conversion periods

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i= interest rate per conversion period

Example 24. A savings fund is created by making equal quarterly deposits of Php 10,000 at 12%
compounded quarterly.

(a) Determine the sum at the end of 2 years.

(b) What is the amount in the fund after the 5th deposit?

(c) Construct the sinking fund schedule for 2 years.

Solution:

Given:

R=PhP10,000, j= 12%, m=4 , I = j/m= 3%

Find S:

a.) the sum at the end of 2 years

[(1+i)n −1] [(1+.03)8 −1] [(1.03)8 −1]
S=R = 10,000 = 10,000
i .03 .03

S = 10,000(8.89336046) → S = Php88,923.36

b.) the amount in the fund after the 5th deposit

[(1+𝑖)𝑛 −1] [(1+.03)5 −1] [(1.03)5 −1]
𝑆 =𝑅 𝑖
= 10,000 .03
= 10,000 .03

𝑆 = 10,000(5.30913581) → 𝑆 =Php53,091.36

c.) the sinking fund schedule for 2 years

Sinking Fund Schedule

No. of In Fund at Interest at 12% Payments to In Fund at the

Quarterly Beginning of Compounded Fund at the End of Interval
Payments Interval Quarterly End of Interval
received

1 Php10,000.00 Php10,000.00

2 Php10,000.00 Php300.00 Php10,000.00 Php20,300.00

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3 Php20,300.00 Php609.00 Php10,000.00 Php30,909.00

4 Php30.909.00 Php927.27 Php10,000.00 Php41,836.27

5 Php41,836.27 Php1,225.09 Php10,000.00 Php53,091.36

6 Php53,091.36 Php1,592.74 Php10,000.00 Php64,684.10

7 Php64,684.10 Php1,940.52 Php10,000.00 Php76,624.62

8 Php76,624.62 Php2,298.74 Php10,000.00 Php88,923.36

Example 25. Five years from now, Alex needs Php400,000 to replace an office equipment. Money
is at 16% compounded quarterly. How much must he deposit at the end of every 3 months to pay
the debt in full?

Solution:

Given: S=PhP400,000.00, j= 16%, m=4 , i= 4%, t=5years , n = 20

Find: R
𝑺(𝒊) 𝟒𝟎𝟎,𝟎𝟎𝟎(.𝟎𝟒) 𝟒𝟎𝟎,𝟎𝟎𝟎(.𝟎𝟒) 𝟏𝟔,𝟎𝟎𝟎
(𝒂) 𝑹 =
(𝟏+𝒊)𝒏 −𝟏
= (𝟏+.𝟎𝟒)𝟐𝟎 −𝟏
= (𝟏..𝟎𝟒)𝟐𝟎 −𝟏
= 𝟏.𝟏𝟗𝟏𝟏𝟐𝟑𝟏𝟒𝟑 = 𝑷𝒉𝒑𝟏𝟑, 𝟒𝟑𝟐. 𝟕𝟎

Activity 15. Sinking Fund

Solve each of the following.

1. How much must Mr. Shabe place in the fund at the beginning of each 6 months in order to
have PhP150,000.00 at the end of 7 years, if money is worth 4.5% compounded semi-annually?

2. A fund is being created by annual deposits of Php5,000 invested at 18% compounded semi-
annually.

(a) Find the amount of the fund after the 8th deposit.

(b) Find the amount of the fund after the 9th deposit.

(c) How much increase occurs in the fund at the end of 8th deposit?
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Quiz 5. Amortization and Sinking Fund

Directions. Use your scientific calculator to solve each of the problems below and show your
solutions.

1. A loan of PhP20,000 is to be amortized with 10 equal quarterly payments. If the interest rate is
6%, compounded quarterly, what is the periodic payment?

2. A newly married couple who borrow PhP90,000 for 30 years at 7.2% compounded monthly,
must make monthly payments of PhP610.91.

a.) Find their unpaid balance after 2 years.

b) During the first 2 years, how much interest do they pay?

3. A debt of PhP150,000.00 with interest rate at 10% payable monthly will be amortized by equal
payments at the beginning of each month for 10 years. Construct the amortization table for the
first five periodic payments.

4. The vehicle owner provide PhP10,000.00 for the repair and maintenance of his Mitsubishi car,
He decides to place equal deposits in a fund at the end of each six months. If the money is invested
at 8% compounded semi-annually.

a.) Find the periodic deposits.

b.) Construct a sinking fund schedule to show the accumulated savings fund into PhP10,000.00 at
the end of 2 years.
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Chapter 5: BONDS VALUATION AND STOCKS

INTRODUCTION
What is a bond?
Bond, in finance, is an instrument or certificate of indebtedness of the bond issuer to the bond holder.
It is a fixed income instrument that represents a loan made by an investor to a borrower which is typically
corporate or government. Bonds are transferable by endorsement or delivery, which can be used as collateral
for loans.
How does a stock differ from a bond?
A stock is an instrument or certificate to describe the ownership in corporation. It is a portion of the
ownership held by an individual or a group of individuals.

ACQUISITION OF CAPITAL

How will corporations and other business entities put up their capital to do their business?
A restaurant owner may need to open branches in other parts of the city or outside of the city, a
company may need a new factory to manufacture more products like tables, chairs, etc., and other business
ventures.
Have you heard that even governments need to borrow money to finance some of its operations and
projects?
When a business enterprise or government wishes to borrow money from the public and other
investors, it usually does it by issuing or selling debt securities that is called bonds.
Capital could be in two forms: debt capital and equity capital. To raise debt capital, the companies
sell bonds to the public while to raise equity capital, the corporation sells the stock of the company. Both stock
and bonds are financial instruments and they have certain intrinsic values.
How would the selling go?
Instead of selling directly to the investors, a corporation usually sells its stock and bonds through an
intermediary like an investment bank. An investment bank acts as an agent between the corporation and the
public. They, as underwriters, raise the capital for the business organization and charge a fee for their
services. The investors could also buy and sell the stock in the secondary markets, such as the Philippine
Stock Exchange (PSE).
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BONDS

A bond, as a financial instrument, represents a contractual agreement between the corporation and
the buyers. It is a written promise made by the corporation to pay the investors a final amount at a specified
future date and the formal contract between the corporation and the bondholder is called indenture.
How does it differ from a promissory note?
It is different from a promissory note inasmuch as bond is a long term obligation which may be
obtained from several investors.
Bondholders have the option to sell his bonds to another person, usually to the highest bidder:
• Bonds purchased at face value or at 100% is said to be bought at par.
• Bonds purchased at more than 100% of its face value, it is considered bought at a premium.
• If the bond is bought less than 100% of its face value, it is considered to be bought at a
discount.
Bonds are classified into registered and unregistered. Bonds that can be sold from one holder to
another with consent and endorsement of its underwriter are registered bonds whereas bonds that can be
sold to another person without the proper endorsement of its issuer are unregistered bonds.
The following are the important features of bonds:
Par Value refers to the face value of a bond which is the principal amount quoted in the bond.
Redemption Value is the value of the bond which is redeemable at the end of its maturity period.
Coupon is an agreement to pay a periodic payment on a designated date.
Redemption Date refers to the date when the bond is to be redeemed.
Redemption rate (j) is the interest rate used on the principal to compute the redemption value.
Bond rate ( r ) is the rate which the bond pays interest on its face value.
Yield rate ( y) is the interest rate realized by the seller of the bond on the invested principal.

Notations:
Fᵥ - face value or par value of the bond
Rᵥ - redemption value of the bond
Pp - purchase price or value of the bond
Bᵥ - book value
Pb - bond premium
Db - bond discount
rb - bond rate
c – coupon payment
j – redemption rate
y – yield rate
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b – bond rate per interest period rb/m

y
i – yield rate per interest period m

VALUATION OF BONDS

The bond valuation refers to the process of finding the price that an investor is willing to pay to get
the desired yield on the value of the bond. The amount that the investor is willing to pay is the purchase price
of the bond which is computed by adding the present value of the redemption value of the bond at periodic
yield rate and the sum of all coupon payments at periodic rate.

Example 26.

1. A ₱14,000 bond with interest at 16% converted quarterly at 114% in 15 years. Find the coupon
payment, redemption value, purchase price and bond premium that yields the buyer 12%
compounded quarterly.

Given: Fᵥ = ₱14,000 rb = 16% = 0.16 y = 12% = 0.12

j = 114% = 1.14 t = 15 years m=4
0.16 0.12
n = 4(15) = 60 b= = 0.04 i= = 0.03
4 4
Required: c, Rᵥ, Pp and Pb

Solution:
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a) Coupon payment: c = Fᵥ (b) = 14,000(0.04) = ₱560.00

b) Redemption value: Rᵥ = Fᵥ (j) = 14,000(1.14) = ₱15,960.00
1−(1+𝑖)⁻ⁿ
c) Purchase price Pp = Rᵥ(1 +i)⁻ⁿ + c( )
𝑖
1 −(0.13)⁻⁶⁰
= 15,960(1.03)⁻⁶⁰ + 560 ( )
0.03
= 2,705.94 + 15,498.32
= ₱18,207.26
d) Bond premium Pb = Pp - Rᵥ = 18,207.26 – 15,960 = ₱2,247.26

Thus, the bond was bought at a premium since the redeemable rate is more than 100% which
means that the purchase price is higher than the redemption value of the bond.

PREMIUM AND DISCOUNT BONDS

Premium bond is the one that its purchased or traded at a higher amount than its par value
whereas discount bonds are purchased at a price lower than the par value.

Bond premium (Pb)

1+(1+i)⁻ⁿ
Pb = Rᵥ ( b – i)( )
i
Bond discount (Db)
1+(1+i)⁻ⁿ
Db = Rᵥ ( i -b) ( )
i
Purchase price:
Bought at a premium: Pp = Rv + Pv
Bought at a discount: Pp = Rv – Db

Example 27:

1. Find the purchase price and discount of a ₱40,000-bond at 14% compounded semi-annually
maturing in 12 years to yield the buyer 15% compounded semi-annually and redeemable at par.

Given: Fᵥ = Rᵥ = ₱40,000 rb = 14% = 0.14 y = 15% = 0.15

j = 100% = 1.00 t = 12 years m=2
0.14 0.15
n = 2(12) = 24 b= = 0.07 i= = 0.075
2 4
Required: Purchase price (Pp) and bond discount (Db)

Solution:
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1+(1+𝑖)⁻ⁿ
Db = Rv (i – b)( )
𝑖
1 −(1.075)⁻₂₄
= 40,000 (0.075 – 0.07)( )
0.075
= 200 (10.98297)
= ₱2,196.59
Pp = Rᵥ - Db
= 40,000 – 2,196.59
= ₱37,803.41

Activity # 16

1. Compute the purchase price of the bond amounting ₱18,000 with interest at 10% payable semi-
annually and redeemable at 97% in 10 years. The bond yields 11% compounded semi-annually.

2.A bond amounting to P50,000 with interest at 15% payable quarterly is priced to yield 12% converted
quarterly. Find the bond premium and the value of the bond if it is redeemable at par at the end of 15 years.

Quiz # 6.

1.At P21,000 bond with interest at 14% is converted quarterly at 112% in 10 years. Compute for the value of
the coupon payment, redemption value, purchase price and bond premium that yields the buyer 11%
compounded quarterly.

2.A bond amounting P75,000 at 12% semi-annual coupons is priced to yield 11% converted semi-annually.
If it is redeemable at 105% at the end of 12 years, find the coupon payment, redemption value, purchase
price and bond premium.

3.Find the purchase price and discount of a P50,000 bond at 9% effective maturing in 10 years to yield the
buyer 10% converted annually and redeemable at par.
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STOCKS

After knowing what a bond is, shall we tackle the stock and its ownership.
A stock is a portion of the ownership of a company held by an individual or a group of individuals.
The corporations sell shares of stock to raise equity capital for the company. A share gives the owner of the
stock a stake in the company and its profits. If a corporation has issued 100 stocks in total, then each stock
represents a 1% ownership of the company. Stocks are bought and sold on what is called stock exchange.

STOCK OWNERSHIP

If you are a stockholder, you will receive a stock certificate which bears the name of your company
and your name as the investor. The certificate also indicates the certificate number, the number of shares
owned and the par value.
A dividend is the stockholder’s share of corporate earnings which is paid in the form of cash or as
additional share of stocks.
A stakeholder’s ownership or equity (Eo) in a company is equal to the ratio of the number of shares
owned (So) to the total number of shares (St) issued by the corporation. In formula:

Eo = So/St
Example 28:

1.Mr. De Leon bought 400 shares of stock of FGN Corporation, which has issued 70,000 shares.
What part of the equity of the firm did he acquire?

: So = 400 St = 70,000.00

uired: Part of equity (Eo)

Solution: Eo = So/Eo = 400/70000 = 1/175

Mr. De Leon owns 1/175 of the FGN Corporation.

TYPES OF STOCKS

There are two different types of stocks: common stock and preferred stock.
The common stock entitles the shareholders with voting rights and dividends. However, the dividends
are paid at company’s discretion and are not guaranteed. Common stock gives higher returns in the form of
appreciation in share prices. When investors buy common stock they can just hope that the share prices will
go up and earn them good profits. If the company goes bankrupt, a common stockholder will get money only
after creditors/bondholders/preferred stockholders have been paid.
 19

A preferred stock allows the shareholders different voting rights, although, many preferred stocks
are issued with no voting rights at all. The preferred stocks are usually sold to financial institutions or large
corporations. One of the advantages of a preferred stock over common stock is that preferred stockholders
are guaranteed a dividend even if the company is not making any profit. The dividend amount per share in a
preferred stock is fixed and is decided at the time of issue. If the company goes bankrupt, preferred
stockholders are paid before common stockholders.

DIVIDENDS

The portion of the profit distributed to shareholders is in the form of dividend payments.

Example 29.

1. In 2010, a company declared ₱30 million dividends on ts common stocks. If there were 700,000
such shares, how much was the dividend per share?

Dividend per share = 30,000,000/700,000 = P42.86 per share

Activity # 17

1. On December 2013, Bill Gates held 358 million shares of Microsoft Corporation. If his share represents
roughly 5% of the total shares outstanding, what is the total number of shares issued by the company?

2.The CCL Cooperative declared a 5.5% dividend on stock with a par value of P5,000. You own 500
shares, how much dividend will you receive?

Quiz # 7

1.If you have bought 3,000 shares of stocks of Monde Nissin Corporation, which has issued 600,000
shares, what part of the equity of the firm have you acquired?

2.How many shares of market value P250 can be purchased for P157,750?

3.The cooperative declared a dividend on stock with a par value of P500. You own 500 shares. How much
dividend will you receive?
 20

Chapter 6: DEPRECIATION

Depreciation

Depreciation or depreciation expense refers to the decrease or loss in value of an asset from the
original cost of a long-termed asset distributed accordingly to its useful life.

Some Basic Concepts

Total Cost or Original cost refers to the cost of an asset plus the freight cost; handling and set-
up charges when shouldered by the buyer.

Salvage value refers to the value of an asset at the time it is taken out of service. Salvage value is
also known as residual, scrap or trade-in value. An asset cannot be depreciated lower the salvage
value.

Useful Life refers to the length of time an asset is expected to generate revenue. It is expressed
in terms of time or unit of productions.

Book Value is the value of an asset at any given time. To compute the book value, find the
difference between the original cost and the accumulated depreciation expense at a given point
of time.

Depreciation schedule is a chart showing the depreciation activity of an asset of each year in
each useful life. It shows the amount of depreciation per year, the accumulated depreciation at a
given time and the book value of an asset.

Notations:

TD = total depreciation
TC = total cost of an asset
SV = Salvage Value, scrap value, residual value or trade-in value
AD = annual depreciation
Ad = accumulated depreciation at a point of time
OC = original cost
n = useful life of an asset in years, units of production
BV = book value
Dr = Depreciation rate = AD/TD
 21

Depreciation Methods

There are different methods in solving depreciation expense such as: straight-line method; sum-
of-the-year’ digits; declining balance; and units-of-production. In financial reporting, any of these
methods can be used depending on the choice of the company. However, once a certain method
is already utilized or implemented it cannot be changed anymore.

1. Straight-line Method

The simplest and most widely used in business nowadays. It provides an equal charge
distributed over the estimated useful life of an asset. For instance, the total depreciation is
Php45,000 with the estimated life of 5 years. The annual depreciation expense is Php9,000. When
the amount of annual depreciation expense already determined, we can set already a depreciation
schedule.

Steps in Solving Depreciation Using Straight-line Method

1. Identify the given in the problem. Determine the total cost if not given.

2. Find the total depreciation. TD= TC - SV

𝐓𝐃
3. Solve for annual depreciation 𝐀𝐃 = 𝐧

4. Solve for the book value. BV for first year = TC – AD

5. Set up depreciation schedule

The depreciation schedule consists of the following: the year , annual depreciation, accumulated
depreciation and the book value at the end of the year.

Year Annual Depreciation Accumulated Depreciation Book Value

(AD) (Ad) (BV)

 22

Example 30. MaxSpeed Corporation purchases a machine worth Php250,000, the freight cost is
Php5,000 and the set-up charges amounting to Php15,000. the machine is expected to last for 5
years and has a salvage value of Php25,000. The MaxSpeed Corporation chose to utilize the
straight-line method, find the total cost, total depreciation, annual depreciation, depreciation rate,
and the book value after 3 years. Prepare a depreciation schedule for its useful life.

Solution:

Given:

Cost of the machine = Php250,000

freight cost = Php5,000

Set up charges = Php15,000

Salvage value = Php25,000

Useful life of an asset = n = 5 years

Find:

a.) Total Cost of an asset =TC?

b.) Total Depreciation =TD?

c.) Annual Depreciation =AD?

d.) Accumulated Depreciation =Ad?

e.) Depreciation Rate = Dr?

f.) Book Value = BV?

g.) Depreciation Schedule

To solve for:

a) Total Cost of an asset = TC = Php250,000 + Php5,000+Php15,000

TC = Php270,000

b) Total Depreciation =TD =( TC-SV) = Php270,000 – Php25,000

TD = 245,000
𝑇𝐷 𝑃ℎ𝑝245,000
c) Annual Depreciation = AD = = = Php49,000
𝑛 5
𝐴𝐷 𝑃ℎ𝑝49,000
d) Depreciation rate = Dr = 𝑇𝐷 = 𝑃ℎ𝑝245,000 = 0.20 or 20%

e) Book value after 3 years

 23

A𝑑3 = AD x 3

A𝑑3 = Php49,000 x 3

A𝑑3 = Php147,000

B𝑉3 = TC - A𝑑3 = Php270,000 – Php147,000 = Php123,000

g.) Depreciation Schedule

Depreciation Schedule Using Straight Line Method

MaxSpeed Corporation Depreciation Schedule for 5 years

Year Annual Depreciation Accumulated Book value

Depreciation
(AD) BV
(Ad)

Php270,000 (Original Cost)

1 Php49,000 Php49,000 Php221,000

2 Php49,000 Php98,000 Php172,000

3 Php49,000 Php147,000 Php123,000

4 Php49,000 Php196,000 Php74,000

5 Php49000 Php245,000 Php25,000 (salvage value)

2. Units-of-Production Method

The useful life of an asset is not expressed in terms of year, rather the useful life is given
in terms of units of production, the total amount of work performed, or number of hours of
operation. Using the Units-of-Production Method to determine, if a machine can produce an
output product of 12,000 units, then the corresponding machine has useful life is 12,000 units
products output.
 24

Example 31. Top Shape company purchases a machine for developing a new product that will
costs Php10,000 and has an estimated salvage value of Php1,000 when the machine can produce
12,000 units. The distribution of the number of units produced are as follows:

3,000 units for the first year, 2,500 units for the second year, 4,000 units for its third year of
operation, 1,500 units and 1,000 units for fourth year and fifth year respectively. Find the annual
depreciation and construct a depreciation schedule.

Solution:

Given:

Total Cost= TC= Php10,000;

Salvage Value =SV= Php1,000

Useful life of an asset = n= 12,000 units

Find: Annual Depreciation= AD?

Steps in Solving Depreciation Using Units-of-Production Method

Step1: Find the depreciation per unit using the formula

TC −SV
Depreciation per unit =
total estimated units produced

Php10,000−Php1,000 Php9,000
= =
12,000 12,000

Depreciation per unit = Php0.75 per unit

Step2: Find the depreciation amount per year by multiplying the number of units produced

per year and the depreciation per unit.

𝐃𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐚𝐦𝐨𝐮𝐧𝐭 = 𝐮𝐧𝐢𝐭𝐬 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝 𝐱 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐩𝐞𝐫 𝐮𝐧𝐢𝐭

 25

Depreciation amount for first year = 3.000 units X Php0.75 = Php2,250

Depreciation amount for 2nd year = 2,500 units X Php0.75 = Php1,875

Depreciation amount for 3rd year = 4.000 units X Php0.75 = Php3,000

Depreciation amount for 4th year = 1,500 units X Php0.75 = Php1,125

Depreciation amount for 5th year = 1,000 units X Php0.75 = Php750

. Depreciation Schedule Using Units-of-Production Method

Top Shape Company Machine’s Depreciation Schedule for 5 years

Year Annual Depreciation Accumulated Book Value

Depreciation
(AD) (BV)
(Ad)

- - Php10,000(Original Cost

1 Php2,250 Php2,250 Php7,750

2 Php1,875 Php4,125 Php5,875

3 Php3,000 Php7,125 Php2,875

4 Php1,125 Php8,250 Php1,750

5 Php750 Php9,000 Php1,000(Salvage value)

Accelerated Depreciation

There are two commonly used methods of accelerated depreciation, the Sum-of-the-Years Digit
and Declining Balance Method. In these methods, it is an assumption that the larger amount of
depreciation will be taken place at the early part of the useful life of an asset and will decrease in
the later years.
 26

a.) Sum-of-the-Years Digit Method

Using the straight-line method, the total amount of depreciation will be distributed
equally on the useful life of an asset. In the sum -of-the-year’s digit method, the larger the
amount of depreciation will be allocated at the beginning of the useful life of the asset and
decreases each year at a proportionate rate.

To solve depreciation using sum-of-the-years’ digit method, follow the given steps.

Step1. Find the total depreciation. TD = TC –SV

Step2. Find the sum-of- the-digits of useful life of an asset; that is:

If the expected useful life of an asset is 4 years, then the sum of the digits is 4+3+2+1 =10.

Step3. Make a fraction using the year numbers in the useful life as the numerator in the reverse
order, the denominator will be the sum of the digits of the useful life.
4
Year 1: 10
3
Year 2: 10
2
Year 3: 10
1
Year 4:
10

Step4. Find the amount of depreciation per year by multiplying the total depreciation and each
fraction formed in step3.

Example 32. MaxSpeed Corporation purchases a machine worth Php250,000, the freight cost is
Php5,000 and the set-up charges amounting to Php15,000. the machine is expected to last for 5
years and has a salvage value of Php25,000. The MaxSpeed Corporation chose to utilize the Sum-
of-the-Years Digit Method. Prepare a depreciation schedule for its useful life.

Given:

Cost of the machine = Php250,000

freight cost = Php5,000

Set up charges = Php15,000

Salvage value = Php25,000

Useful life of an asset = n = 5 years

 27

Find: Depreciation schedule for its useful life.

Solution:

Depreciation Schedule Using Sum-of –the-Years Digit Method

MaxSpeed Corporation Depreciation Schedule for 5 years

Year Total Depreciatio Annual Accumulated Book value

Depreciation n Rate
Depreciation Depreciation (BV)
(TD) (DR)
(AD) ( Ad)

Php270,000

(Original Cost)

1Php245,000 5 Php81,666.67 Php81,666.67 Php183,333.33

15

2Php245,000 4 Php65,333.33 Php147,000.00 Php123,000.00

15

3Php245,000 3 Php49,000.00 Php196,000.00 Php74,000.00

15

4Php245,000 2 Php32,666.67 Php228.666.67 Php41,333.33

15

5Php245,000 1 Php16,333.33 Php245,000.00 Php25,000.00

15
(Bank Value)
 28

b.) Declining Balance Method

Declining balance method of accelerated depreciation in business is similar to sum-of-the-digits

method wherein the large depreciation amount will be allocated at the beginning of the useful
life and decreases for each succeeding year.

Steps in solving the depreciation expense using declining balance method:

1.) find first the annual depreciation rate, also known as straight line rate (SLR). Straight line
1
rate can be determined by n,

where n is the useful life of an asset. When straight line rate is being utilized to get the depreciation
expense it is called declining balance method.

2.) find the declining balance rate or get the straight line rate (SLR) then multiply by the multiple.
The most commonly used multiple of straight line rate 1.25 or 125%, 1.5 or 150%. When 2 or 200%
is being used as a multiple it is called double declining balance method.

1
SLR = straight line rate; SLR = 𝑛
1
DBR =declining balance rate; DBR = 𝑛 x multiple

N= useful life of an asset

Multiple = 1.25, 1.5, or 2

Example 33. ABC Company purchases 10 computer units at a total Php350,000. The expected
life time of the computer units is 4 years. Using the straight-line declining balance method, find
the annual depreciation and construct the depreciation schedule.
 29

Solution

Given:

TC or OC = Php350,000,
1 1
n = 4 years; SLR = 𝑛 = 4 or 25%

Find: Annual Depreciation = AD

a) Use straight-line declining balance method to find the annual depreciation.

𝐴𝐷1= TC x SLR

= Php350,000 x 25% = Php350,000 x 0.25 = Php87,500

𝐵𝑉1 = TC - 𝐴𝐷1 = Php350,000 – Php87,500

𝐵𝑉1 = Php262,500

𝐴𝐷2 = 𝐵𝑉1 x SLR = Php262,500 x 25% = Php65,625

𝐵𝑉2 = 𝐵𝑉1 - 𝐴𝐷2 = Php262,500 – Php65,625 = Php196,875

𝐴𝐷3 = 𝐵𝑉2 x SLR = Php196,875 x 25% = Php49,218.75

𝐵𝑉3 = 𝐵𝑉2 - 𝐴𝐷3 = Php196,875 - Php49,218.75 = Php147,656.25

𝐴𝐷4 = 𝐵𝑉3 x SLR = Php147,656.25x 25% = Php36,914.06

𝐵𝑉4 = 𝐵𝑉3 - 𝐴𝐷4 = Php196,875 - Php36,914.06 = Php110742.19

Depreciation Schedule Using Declining Balance Method

ABC Company Depreciation Schedule for 4 years

Year Straight Annual Accumulated Book value

Line Rate
Depreciation Depreciation (BV)
(SLR)
(AD) ( Ad)

Php350,000
 30

(Original Cost)

1 25% Php87,500.00 Php87,500.00 Php262,500

2 25% Php65,625.00 Php153,125.00 Php196,875

3 25% Php49,218.75 Php202,345.75 Php147,656.25

4 25% Php36,914.06 Php239259.81 Php110742.19

Activity # 18. Depreciation

Solve each of the following. Show your solution.

1. Mr. Sy purchases sewing machine costs P6,000 and will have a salvage value of P500 after its
useful life of 4 years. Prepare a depreciation schedule for the sewing machine a.) by the straight-
line method; b) by the sum of the year’s digit method.

2. The Marc Angel Engineering Company purchased a machine that cost P50,000 and will last 5
years. A salvage value was not assigned to the asset. Determine the annual depreciation expense
using the declining balance method and prepare the depreciation schedule to record the expense.

Quiz # 8. Depreciation

Directions. Use your scientific calculator to solve each of the problems below and show your
solutions.

1. The Marc Angel Engineering Company purchased a machine that cost P50,000 and will be able
to produce 500,000 units of product before wearing out. Expected production by year will be: year
1 – 80,000 units; year 2 – 100,000; year 3 – 100,000; year 4 – 110,000 and year 5 – 110,000. A
salvage value was not assigned to the asset. Determine the annual depreciation expense using
the units-of-production method and prepare the depreciation schedule to record the expense.
 31

2. GVC Industries’ purchased equipment for P70,000. This equipment has a 5-year life and an
P8,000 residual value. Calculate depreciation for each of the five years using the declining balance
method at twice the straight-line rate.

3. Digital Universe Company purchased a machine that cost P50,000 and will last 5 years. A salvage
value was not assigned to the asset.

a.) Determine the annual depreciation expense using the straight-line method and prepare the
depreciation schedule to record the expense.

b.) Determine the annual depreciation expense using the sum of the year’s digit method and
prepare the depreciation schedule to record the expense.

IV. LEARNING ENHANCEMENT/EXTENSION

Think of situations where you have applied or plan to apply the concepts discussed in any of the lessons in
any form of your choice. You may use your creativity or ingenuity in addressing this activity.
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V REFERENCES
Arao, CoPo, Laddaran, Gabuyo, Villanueva (2015) Mathematics of Investment 2015 Edition
REX Bookstore, Manila Philippines
Mejia, L. DP, Y.A. Gabuyo, J.C. Ignacio, and J. Sajise (2008) Business Mathematics: A
Complete Textbook and Workbook. Books Atbp.Publishing Corp.
Arce, Ma. Teresa B. ,et al. (2010) Mathematics of Investment. REX Bookstore, Manila
Philippines
Caras, Madeleine S., et al. (2008) Mathematics of Investment. Bookstore Publishing 2008
Edition
Slater, J. (2008) Practical Business Math Procedures. Mc Graw Hill Irwin, Ninth Edition
Naval, Victoria C., et. al., (2007) Mathematics of Investment C& E Publishing Inc.,2007 Edition
Hart, William L. Mathematics of Investment. 5th edition. DC Health and Company, 1980
Sta. Maria, Antonina C., et. al., Mathematics of Investment National Bookstore 1988 Edition.
https://francisjosephcampena.weebly.com/uploads/1/7/8/6/17869691/chapter_1_mathe
matics_of_investment.pdf

https://www.youtube.com/watch?v=xWrAg5A7n4k&t=38s

https://www.youtube.com/watch?v=ZrWdptBwEPE&t=782s

VI MODULE EVALUATION
Kindly make use of this portion to express freely how you feel about this module as a basis for its
improvement and future revision.
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