Bicol University College of Engineering

Geodetic Engineering Department

BU East Campus, Legazpi City, 4500

Engineering Economics

Aprilla M. Gualvez
BSGE-4A
aprillamagallones.gualvez@bicol-u.edu.ph

INTRODUCTION

Engineering economics, merging economics with precise engineering

analysis, guides strategic decision-making for engineers. This term paper
covers fundamental topics starting with Simple and Compound Interest,
explaining interest calculation, pivotal for investment and borrowing
decisions in projects. Annuities like Uniform Series and Payments are
indispensable for analyzing cash flows in credits, investments, and project
revenues. Bonds, a means of long-term capital raising, demand clear
comprehension of types, yields, and pricing mechanisms. Additionally,
Depreciation, concerning asset cost depreciation over their life span, impacts
budgeting and project evaluation. Amortization elucidates gradual debt
repayment, aiding engineers in assessing various loan structures and project
financial feasibility. This knowledge equips engineers with robust tools to
make informed, strategic, and economically sound decisions, ensuring
ongoing project success.
I. Interest accrued interest over time. These
differences hold substantial
Understanding the dynamics of
importance for borrowers and
interest forms the cornerstone of
lenders, shaping the way interest
financial transactions. Interest
accumulates and influences
serves as the cost of borrowing
money or the earnings on
deposited funds. When we lend DEFINITION 1.1
money, we receive interest, while
borrowing entails paying it. Simple interest is the
Interest isn't confined to calculation used to determine the
traditional banking systems; it interest charged on a loan's
permeates various financial principal amount. It involves
instruments and transactions. From multiplying the principal by the
personal loans and mortgages to daily interest rate and the
savings accounts and investments, number of days between
interest manifests in diverse forms. payments. The formula for simple
It essentially reflects the time value interest, expressed as
of money, acknowledging that a
sum of money today holds different PrT
SI =
100
value than the same amount in the
future due to factors like inflation, Where P for the principal amount,
opportunity costs, and risk. r for the interest rate, and T for
Additionally, interest comes in the time or duration of the loan.
various forms, each determined by
distinct calculation methods.
Simple interest is computed solely https://www.cuemath.com/commercial-math/
simple-interest/
based on the initial principal
financial results across time.
amount borrowed or invested,
without factoring in any previously A. Simple Interest
accumulated interest. In contrast,
compound interest encompasses
Principal Amount (P): This is the
not just the principal balance but
initial sum of money borrowed or
also includes any interest
invested. For instance, if you take
previously earned or accrued.
out a loan of $1,000, then $1,000 is
Typically, loans tend to employ
your principal amount.
simple interest calculations,
whereas savings accounts
Rate of Interest (r): This is the
commonly utilize compound
percentage of the principal amount
interest computations to determine
charged as interest over a certain
period. For example, if the interest Simple Interest Problems
rate is 5%, then r=0.05 when
expressed as a decimal. Example 1.1

Time Period (T): This is the Jovelyn’s uncle borrowed

duration for which the money is $800 from a lending institution at
borrowed or invested, typically in an interest rate of 6%. Calculate
years. If the time is given in the simple interest for durations of
months, it should be converted to 1 year, 2 years, 3 years, and 10
years (months ÷ 12) to maintain years, as well as the total amount
consistency. to be repaid in each case.

In order to calculate the total Solution:

amount, the following formula is Principal Amount = $800
used: Rate of Interest = 6%= 0.06

Duratio Simple Interest (SI)

Amount (A) = Principal (P) +
n
Simple Interest (SI)
1 year SI =( 800 ×0.06 × 1 )=P 48
2 years SI =( 800 ×0.06 × 2 )=P 96
Where,
3 years SI =( 800 ×0.06 × 3 )=P 144
Amount (A) is the total 10 years SI =( 800 ×0.06 × 10 )=P 480
money paid back at the end of the
time period for which it was
Total Amount to be repaid:
borrowed.
The total amount formula in Simple
case of simple interest can also be Duratio Interest Total
written as: n (SI) Amount
(A)
A = P (1 + RT) 1 year SI =$ 48 A= 800+48
= P848
Here, 2 years SI =$ 96 A= 800+96
A = Total amount after the given = P896
time 3 years SI =$ 144 A=
period 800+144 =
P = Principal amount or the initial P944
loan 10 years SI =$ 480 A=
amount 800+480 =
R = Rate of interest (per annum) P1280
T = Time (in years)
Interest Rate
I I I
P= R= T=
The interest rate refers to the RT PT PR
Time
percentage charged or earned on a Principal Rate
principal amount over a specified
period of time. The formula for Example 1.3
finding the interest rate on simple for Principal (P) and Time (T)
interest is expressed as:
P5,000 loan for 48 days at 6%
Simple Interest ×100 interest.
Interest Rate=
Principal ×Time

Example 1.2 (
SI = 5000 × 0.06 ×
48
365 )
=¿P39.45

I 39.45
P= = ≈ P5000
Romelo borrowed P600 from
the bank at some rate per annum
RT
0.06 ( )
48
365
and that amount becomes double
in 3 years. Calculate the rate at I 39.45
T¿ = ≈ 0.13
which James borrowed the money. PR 5000(0.06)
Solution: 365 days × 0.13 ≈ 48 days

Simple Interest= P1200 – P600 =

600
Time = 3 years

600× 100
Interest Rate=
600× 3
I
60 000
Interest Rate=
1800 T

Interest Rate=33.33 %

Solving for Principal, Rate and

Time

I I
P R T P R T P R
 compound interest amount by
B. Compound Interest deducting the principal amount
from the total accrued amount.
Compounding Period
Compound Interest Problems
The compound interest is
calculated at regular intervals like Example 1.4
annually (yearly), semi-annually
(half-yearly), quarterly (4 times in a Jihyo invest P5000 in an account
year), monthly (12 times in a year), with an annual interest rate of 6%,
weekly and daily. compounded quarterly. How much
money will she have after 3 years?
Period Formula
Annually A=P ( 1+r )
t

DEFINITION 1.2
Semi-
( )
2t
r
Annually A=P 1+
2
Compound interest is the
( )
Quarterly r
4t
A=P 1+
4 total interest earned on the
initial principal amount and the
( )
Monthly r
12 t
A= 1+
12 accumulated interest from

( )
Weekly r
52t
previous periods, often referred
A=P 1+
52 to as "interest on interest". This

( )
Daily r
365t
interest is calculated based on
A=P 1+
365 the interest rate and the initial
principal amount over a given
Note: period.

In these equations, A
( )
nt
r
represents the total compounded A=P 1+
n
amount, encompassing both the
principal and the compound Alternatively, we can write the
interest. To calculate only the formula as given below:
compound interest, we can find it
by subtracting the principal CI = A – P
amount 'P' from the entire formula.
For instance, in the case of And,
compound interest compounded
( ) −P
nt
r
monthly, the formula becomes: CI =P 1+
n

( )
12t
r where P is the initial amount, r is
CI =P 1+ −P
12
the interest rate (in decimal
form), n is the number of times
This formula allows us to
the interest is compounded
specifically determine the
annually, and t is the total time.
https://www.cuemath.com/commercial-
math/compound-interest/
 Annual interest rate (r) = 9% or
Solution: 0.09
Given: Compounding frequency (n) = 2
Principal amount (P) = P5000 (semi-annually)
Annual interest rate (r) = 6% or Time (t) = 3 years
0.06 Compounding frequency (n) =
4 (quarterly)
( )
2t
r
A=P 1+
Time (t) = 3 years 2

A=25000 (1+
2 )
2 (3 )
0.09
Formula for compound interest
compounded quarterly: A=25000 ( 1+ 0.045 )
6

( )
4t
r A=25000 ( 1.345375 )
A=P 1+
4 A ≈ 33634.38
A=5000 (1+
4 )
4(3)
0.06
To find the compound interest,
12
A=5000 (1.015 ) subtract the principal amount from
A ≈ 5000(1.19718) the total amount:
A ≈ 5985.90
After 3 years, the total amount will Compound Interest (CI) =
be approximately $5985.90. Total Amount (A) – Principal
Amount (P)

Example 1.5 CI = A-P

CI = 33634.38 - 25000
What is the compound CI = 8634.38
interest to be paid on a loan of Therefore, approximately P8634.38
P25,000 for 3 years at an annual is the compound interest of a
interest rate of 9%, compounded P25,000 loan compounded semi-
semi-annually? annually over a 3-year period.

Solution:
Given: II. Annuities
Principal amount (P) = P25000 Based on the published notes of
Engr. Menes (2016), Annuities
DEFINITION 2.1 come into play in a few situations.
First off, they're used when
Annuities represent a someone needs to pay off a debt
sequence of consistent payments by making a series of equal
happening at regular intervals payments over regular intervals.
over a defined period.
 Another way is when someone This means if you're making
wants to build up a certain sum by payments monthly, the interest is
putting aside the same amount of also calculated and compounded
money regularly. on a monthly basis. Essentially, the
Lastly, annuities step in when timing of your payments lines up
someone chooses to receive a set with how often the interest is
amount of money at consistent calculated or added to the account.
intervals instead of taking a lump This simplicity in
sum at their retirement. synchronization makes it
So, they're basically about straightforward to calculate and
Payment time interval understand the growth or payouts
Interval between each of the annuity.
consecutive
payment. i. Basic Concepts of Simple
entire duration Annuity
spanning from the
Term of beginning of the 2. General Annuity
the initial payment to D. Classification of Annuities
Annuity the end of the final (Based on Payment Schedule)
payment interval
Periodic Periodic annuity 1. Ordinary Annuity
Rent payment
Future It's the total
Value of an compounded value
Annuity of all payments by
the end of the
term.
Present The lump sum
DEFINITION 2.2
Value required at the
beginning
In a simple annuity, payment
managing payments or savings in a
period is the same as the interest
steady, organized way over time.
DEFINITION 2.3 period. These payments can occur
at any specified frequency.
C. Classification of Annuities
A general annuity involves
(Based
payments onmade
Payment Intervalat
or received
and Interest
varying Period)
intervals and amounts
over a specified period, unlike a
1. Simple Annuity
fixed-payment simple annuity.
i. Simple Ordinary Annuity r =4%=0.04
Formulas t =3 years
n = mt = 12(3) = 36
Future Value:

[ ]
n
(1+i) −1
F= A
i DEFINITION 2.4
An ordinary annuity refers to a
Present Value:
series of equal payments made or

[ ]
−n
1−(1+i) received at regular intervals, with
P= A
i payments occurring at the end of
each period. (e.g., salaries,
Periodic Payment: subscription services etc.)

F (i) P(i) r 0.04

A= ; A= i= = =0.00333
n
(1+i) −1 1−(1+i)−n m 12
F =?
where,
Now, substitute to the formula:

[ ]
n
F = Future Value of annuity (1+i) −1
F= A
P = Present Value of annuity i
A = Periodic Payment
[ ]
36
(1+0.0033) −1
F=500
period ¿)
i = rate of interest per conversion 0.0033

[ ]
36
(1+0.0033) −1
n = number of payments (n=mt) F=500
0.0033
r = rate of interest F=500 [ 38.16 ]
t = term of the annuity F ≈ P 19079.46
m = payment interval
Thus, Dylan will have P19,079.46
at the end of 3 years.
Example 2.1

Dylan deposits P500 at the Example 2.2

end of each month for 3 years at
an annual interest rate of 4% Sofia wants to accumulate
compounded monthly. How much P50,000 by making equal deposits
will she have at the end of 3 years? at the end of every 6 months into
an investment account that earns
Solution: 6% interest compounded semi-
Given: A =P500 annually. If he makes the first
deposit on January 1, 2023, and the A
FV =
i
[ (1+i) −1 ]− A
n +1

last deposit on January 1, 2030,

how
much should each deposit where,
be? PV = represents the present
value
Solution: or the initial sum of
Given: F = P50,000 money.
r = 6% = 0.06 FV = stands for the future
t=7 value
n = mt + 1 = 2(6) + 1 = 15 or the total sum of
*note: +1 refers to the initial deposit money at some later
r 0.06 time.
i= = =0.03
m 2 A = Periodic Payment
A=? i = rate of interest per
conversion
Now, to find A: period ¿)
F (i) n = number of payments
A=
(1+i)n−1 (n=mt)
50000 (0.03) r = rate of interest
A=
(1+0.03)15 −1
1500
A=
0.5579674166 Example 2.3
A ≈ P 2688.33
Suppose Ms. Erlano receives
Thus, Sofia needs to make a
P800 at the beginning of each year
regular deposit of P 2688.33 semi-
for 15 years as a retirement
annually.
pension. We need to determine the
present value of her pension at an
interest rate of 5% compounded
2. Annuity Due annually.

i. Annuity Due Formulas DEFINITION 2.5

Present Value:
Annuity Due is a type of annuity

[ ]
n−1
A (1+i) −1 where equal payments are made
PV = A+
i (1+ i)n−1 at the beginning of each period.

Future Value:
Solution:
Given: Example 2.4
A =800
r 0.05 Under a college savings plan,
i= = =0.05
m 1 Mr. Smith deposits $100 at the
beginning of each month for 3
DEFINITION 2.6 years. The plan ensures an
accumulation at 8% compounded
monthly. How much will be
A deferred annuity is an available in the savings at the end
annuity where the initial regular of 4 years?
payment starts after a specific
waiting period, known as the Solution:
deferral period. Given
A= $100
This deferral period indicates r 0.08
i= = =0.0066
the duration between the present m 12
time and the beginning of the t = 12
annuity payments. n = mt = 4(12) = 48
PV= ?

[ ]
-https://www.scribd.com/document/430564875/Deferred- 48−1
Annuity 100 (1+ 0.0066 ) −1
PV =100+
0.0066 ( 1+0.0066 )48−1
t = 15
100
n = mt = 1(15) = 15 PV =100+ [ 0.265951974 ]
0.0066
PV = ?
PV ≈ $ 4,129.58
Using the formula:
Hence, Mr.Smith savings will

[ ]
n−1
A ( 1+i ) −1 amount to $4, 129.58 at the end of 4
P V = A+
i ( 1+i )n−1 years.

[ ]
15−1
800 (1+0.05) −1
P V =800+
0.05 (1+0.05)15−1
3. Deferred Annuity
800
P V =800+ [ 0.494932047 ]
0.05
P V ≈ P 8,718.91 Deferred Annuities

Therefore, Ms. Erlano’s present Ordinary Deferred

Deferred Annuity Due
pension value amounts to
Annuities
P8,718.91 compounded annually.
Deferral Deferral
Period ends Period ends at
one payment the first
interval periodic
before the payment.
first periodic
payment.
 where,

Pdue = Annuity payment due

r = Effective rate of interest
n = No. of periods
t = Deferred periods

The key difference between

the two lies in when the payments Example 2.5
begin after the deferral period:
ordinary deferred annuities initiate Suppose Sarah agrees to
payments at the end of each lend P80,000 today, and in return,
period, while deferred annuities she will receive twenty annual
due start payments at the payments of P8,000 each. The
beginning of each period. Both annuity will commence five years
types allow for a waiting period from now, and the effective
before the regular annuity interest rate is 5%. Determine
payments commence, offering whether this deal is viable for
flexibility in timing according to Sarah if the payment remains an
individual financial needs or goals. ordinary annuity and annuity due,
just like in the initial scenario.
i. Deferred Annuity Formulas
Solution:
Ordinary Annuity Given:
Pordinary ∧Pdue =8000
−n
[1−( 1+ r) ] r = 5% = 0.05
Pordinary ×
[(1+ r )t × r ] n = 20 years
t=5
where,
Using the formula for Ordinary
Pordinary = Ordinary annuity payment Deferred Annuity:
r = Effective rate of interest
n = No. of periods [1−( 1+ r) ]
−n
Pordinary ×
t = Deferred periods [(1+ r )t × r ]
−20
[1−(1+ 0.05) ]
8000 ×
[(1+ 0.05)5 × 0.05]
Annuity Due
−n
8000 ×[9.764467895]
[1−(1+r ) ] P 78,115.74
Pdue ×
[(1+ r)t −1 × r ]
 Under these circumstances, Example 2.6
John should refrain from lending
the money as the value of the Let's say you receive P50,000
deferred annuity falls below annually from a perpetuity, and the
P80,000. interest rate for this period remains
at 5%.
While, The present value of a
growing perpetuity can be
Using the formula for Deferred
Annuity Due:
−n
DEFINITION 2.7
[1−(1+r ) ]
Pdue ×
[(1+ r)t −1 × r ]
[1−(1+0.05) ]
−20 Perpetuity is an annuity is a
8000 × financial arrangement
[(1+ 0.05)5−1 ×0.05 ]
characterized by periodic
8000 ×[10.25269129]
payments that persist indefinitely
P 82 ,021.53
or extend infinitely over time.
In this case, John should
https://www.scribd.com/document/
proceed with lending the money 535213415/ANNUITY
since the value of the deferred
calculated as follows:
annuity exceeds P80,000.

Solution:
Given:
4. Perpetuity
A = P50,000
r = 5% = 0.05
From the formula of present value
PV =?
of an ordinary annuity:

To find P:
[ ]
−n
1−(1+i) A
P= A PV =
i i
50000
PV =
The expression (1+i)−n approaches 0.05
to zero when n approaches to PV =1,000,000
infinity. Thus, the present value of
a perpetuity becomes: In this case, the present
A value of your perpetuity would be
PV =
i P1,000,000.
i. Present Value of Growing 80000
PV =
Perpetuity 0.04−0.08

A growing perpetuity adjusts PV ≈ P 2,000,000

payments for inflation, maintaining
constant buying power. Its present Therefore, the value of the growing
value exceeds that of a fixed perpetuity amounts to
perpetuity. The higher growth rate P2,000,000.
leads to a greater present value.

A i. Other Types of Annuities

PV =
r−g (Fixed Annuity and Variable
Annuity)
Where,
PV = Present Value According to Mark Lewis (2023),
A = Cash Flow a fixed annuity ensures a set
r = discount rate interest rate on the contributed
g = growth rate principal, offering a steady income
stream over time.
An example of a fixed annuity
Example 2.7 could be an insurance contract
guaranteeing a 4% annual interest
Let's consider a perpetuity rate on a principal amount of
with an initial cash flow of P80,000, $100,000, providing a fixed annual
set to grow at a rate of 4%, with an income of $4,000.
interest rate of 8%. The present
value of this growing perpetuity On the other hand, a variable
would be? annuity's returns vary based on an
investment portfolio's
Solution: performance, leading to fluctuating
Given: income payments throughout the
A = 80,000 annuity period.
r = 4% = 0.04 For instance, a variable annuity
g = 8% = 0.08 might involve investing $100,000
PV= ? in a portfolio of stocks and bonds.
The returns and subsequent
To solve for PV, payments to the beneficiary
A fluctuate based on the
PV =
r−g performance of these investments,
potentially resulting in varying Coupon
annual income payments. periodic interest payment
that the bondholder receives
II. Bonds during the time between
purchase date and maturity date,
i. Bond Characteristics usually received semi-annually.
EC Math TV (m.me/EcMathTv)
ii. Value of a bond Using formula for present value of
annuity:
It is defined to be the present
value of all the amounts the Bond Characteristics
bondholder will receive through his Face The initial value of
possession of the bond. Value the bond, also
Derivation of the formula for the known as par
value of a bond: value
(from annuity and compound interest) The interest rates
Coupon Rate the issuer pays
Let Vn = value of the bond n
annually to the
periods before redemption
bondholder,
V n = P1 + P2
usually a
percentage of the
face value.
DEFINITION 3.1 Denoted as r
Maturity The date when the
Date issuer repays the
- A bond is an interest- bond's face value
bearing security that obligates to the bondholder
the issuer to pay the bondholder Yield The effective rate
a specified sum of money. of return for the
usually at specific intervals investor,
(known as a coupon/dividend), influenced by the
and to repay the principal bond's price and
amount of the loan at maturity. coupon payments
n
A [(1+i) −1]
- Interest-bearing security P 1=
i(1+i)n
which promises to pay;
 Stated amount of money n
Fr [(1+i) −1]
on the maturity date, and P 1= n
i(1+i)
 Regular interest payments
called coupons.
Using formula for present value of F = Php 250,000
compound interest: n = 7 years
F=P ( 1+ i )
n
i = 7.8% = 0.078
C=P2 (1+i )
n C(if not given) = Php200,000
C Vn = ?
P 2=
( 1+i )n
Therefore, Using the formula for bond
n value:
F r [(1+i) −1] C
V n= n
+ n
i(1+i) ( 1+i ) n
F r [(1+i) −1] C
V n= n
+ n
i(1+i) ( 1+i )
Where,
Vn = value of the bond n periods
7
before [(1+0.078) −1] 200,000
V n=200,000(0.08) +
redemption 0.078 (1+ 0.078) ( 1+ 0.078 )7
7

F = par value of the bond V n=200,000(0.08) [ 5.242173074 ] + [ 118222.1 ]

C = amount paid to the bondholder
at V n ≈ Php202 , 096.87
maturity of the bond which is
usually equal to F Therefore, the price of the
n = number of periods prior to bond is Php 202,096.87.
redemption
r = rate of interest on the bond per Find the yield:
period
i = actual rate of interest on the Example 3.2
amount
invested in the bond usually A Php 100,000, 10% bond,
called pays dividend every quarter for 8
yield. years. The bond is priced at par
Example 3.1 and is redeemable at 110% of the
par value. Find the yield to
What is the price of this bond maturity.
with a face value of Php250,000,
maturing in 7 years, and an 8% Solution:
coupon rate if the yield to maturity Given:
is 7.8%? F = Php 100,000
10 %
r= = 2.5% (quarterly) =
Solution: 4
Given: 0.025
r = 8% =0.08 n = 8(4) = 32 quarters
 Vn = F = Php 100,000 OP−DP
Discount %= × 100
C = 110% × F = Php 110,000 OP
i=?

Using the same formula:

DEFINITION 3.2
n
F r [(1+i) −1] C
V n= n
+ n - The dictionary meaning of
i(1+i) ( 1+i )
32 discount is a deduction from
[(1+i) −1] 110,000
100,000=100,000(0.025) + the usual cost of something,
i(1+ i)32 ( 1+i )32
typically given for prompt or
i=0.027=2.7 % ( every quarter )
advance payment or to a special
So, category of buyers.
i=2.7 % ×4=10.8 % every year
Defined by the formula,
Then, the yield to maturity is
10.8% every year. D=F–P

- A discount is the difference

between the future worth of a
certain commodity and its
present worth.

https://www.cuemath.com/commercial-math/
discounts/

Where,

Original Price (OP): The amount

displayed on the product's tag as
its initial price.
iii. Additional Notes
Discounted Price (DP): The price
Discount at which an item is actually
purchased or sold.

Given the idea of a discount, Discount %: The percentage

the associated rate used to representing the reduced
calculate the current value of amount from the list price.
future cash flows is known as the
discount rate.
 Discount rate considers that
future money is worth less due to
factors like inflation and potential
DEFINITION 4.1
investment returns.

Depreciation is the decrease

Example 3.3 in the value of physical property
with the passage of time.
While shopping, Dani finds a
pair of shoes listed at $80, but due https://www.civilengineermag.com/engineering-economy/

to a sale, the store offers a

discounted price of $64. Calculate
the discount rate applied to the
shoes.

Solution:
Given:
OP = $80
DP = $64
D% = ?

80−64
Discount %= ×100
80
80−64
Discount %= ×100
80
Discount %=20 %

Therefore, the discount rate

applied to the shoes is 20%.

III. Depreciation

Other terminologies include:

 The estimated It is stated in the document
duration an asset uploaded by Utto A. (2023) cited at
Useful will be used for https://www.studocu.com/ph/ that
Life business purposes depreciation serves two key
to generate income, purposes:
not its physical
lifespan but how Capital Recovery - It helps in
long it's expected to gradually recuperating the initial
be productive. investment made in physical assets
like machinery or buildings over
their useful lifespan.
The total expense to
Initial obtain an asset,
Investmen covering its Terms
t (Cost) purchase cost,
taxes, In a business context,
transportation, and Value value represents the
any necessary current worth of all
preparatory forthcoming profits
expenses for its use. expected from owning
a specific property.

or resale value
Salvage signifies the amount The recorded worth of
Value achievable from Book an asset in a
selling a property Value company's accounting
once it has been records. It's the original
utilized. cost of the asset minus
any allowable
depreciation
refers to the sum deductions.
Scrap obtained from
Value selling the property
if it's discarded or The price agreed upon
sold as junk. Market by a willing buyer and
Value
seller, assuming both
have equal advantage
i. Purpose of Depreciation and are not compelled
to make the
transaction.
 2. Cost Allocation for
Production - By spreading the L = represents the property's
cost of these assets across their useful life
usable life, depreciation allocates in years.
expenses to match the revenue Co = stands for the original cost.
generated from their use, aligning CL = denotes the value at the end
costs with the related income. of its
life, including any gain or loss
ii. Types of Depreciation due to
Understanding the various removal.
types of depreciation aids in d = signifies the annual
assessing the declining value of depreciation
assets and their impact on financial cost.
statements.
Types of Depreciation
Depreciation due to Changes in Normal Depreciation
Price Levels Physical Functional
Depreciation Depreciation
It occurs due to fluctuations in
the general price level within an results from the results from
economy. However, it's typically not deterioration or decreased
factored into economic studies due demand for an
wear and tear of
to its unpredictability. asset's intended
an asset, purpose. It
reducing its occurs due to
Depletion capacity to obsolescence
function from
refers to the reduction in the effectively. technological
value of an asset caused by the advancements,
gradual extraction or consumption shifts in
of its resources or contents. This consumer
often applies to natural resources, preferences, or
such as mineral reserves, where changes in
extraction diminishes the asset's market demand.
value.

Source:
https://www.studocu.com/ph/document/notre-dame-university/civil- Cn = indicates the book value at
engineering/12-depreciation-engineering-economy/21937710
the end
of n years.
iii. Methods of Depreciation Dn = represents the total
depreciation up
Let, to n years.
 salvage value of P10,000 after its

a. Straight Line Method

DEFINITION 4.2
Example 4.1
A method based on the
Rica who owns a printing
assumption that the decrease in
business, purchased a specialized
value is exactly proportionate to
printer for P150,000 with an estimated
salvage value of P20,000 at the end of the property’s age.
its 5-year useful life. What would be
the book value of the printer after The simplest depreciation
three years, utilizing the straight-line method assumes a consistent
method for depreciation? annual depreciation amount over
the asset's useful life.
Solution:
Given: It is defined by the formula:
Co = P150,000
C n=C o−Dn
CL = P20,000
L=5
https://www.civilengineermag.com/engineering-
n=3 economy/
Co −C L 150,000−20,000
d = = 5-year useful life. Calculate the
L 5
accumulated depreciation after 3
=26,000
years.
Dn = n(d) = 3(26,000) = 78,000
Cn = ? Solution:
Substitute the values to the formula: Given:
C n=C o−Dn Co = P80,000
C 3=150,000−78,000 CL = P10,000
C 3=P 72,000 L=5
n=3
Hence, after three years, following Dn = ?
the straight-line depreciation
method, the book value of Sarah's For Annual Depreciation (d),
printer would be P72,000. C −C L 80,000−10,000
d= o = =14,000
L 5

Example 4.2 For Accumulated Depreciation (Dn),

Dn = n(d)
John purchased a computer Dn = 3(14,000)
system for his office at a cost of Dn = P42,000
P80,000, expecting it to have a
Therefore, using the straight-line end of their 8-year expected
lifespan is projected to be P10,000.
Determine the original cost of the
DEFINITION 4.3
tools.
In the sinking fund method,
money is saved up over time to Solution:
replace an asset when it's no Given:
longer useful. The depreciation Dn = P30,000
happening until a certain point
CL = P10,000
matches the amount saved up in
the fund at that time. This L=8
method ensures there's enough n=4
money set aside to replace the Co = ?
asset as its value declines.
To find Co:
Formula in Sinking Fund Method:
(C ¿ ¿ o−C L )i
d= ¿ Co = D n + CL
(1+i) L−1 Co = 30,000 + 10,000
Also, Co = P40,000
(C ¿ ¿ o−C n )i
d= n
¿
(1+i) −1 Hence, using the straight-line
Depreciation at the end of nth method, the original cost of Mr.
years; Minatozaki’s specialized tools was
n
(1+ i) −1 P40,000.
D n=d [ ]
i
Book Value at the end of nth
years;
C n=C o−Dn

https://www.civilengineermag.com/engineering-economy/

method, the accumulated

depreciation after 3 years for John's
computer system is P42,000.
Example 4.3
b. Sinking Fund Method
Mr. Minatozaki, who owns a
construction firm, bought a set of Example 4.4
specialized tools. After 4 years of
use, the accumulated depreciation A dump truck was bought P30,000
for these tools amounted to six years ago. It will have a salvage
P30,000. The salvage value at the value of P3,000 four years from
now. It is sold now for P8,000. What
is its sunk cost if the depreciation Example 4.5
used is the sinking fund method at
6%? An industrial generator was
purchased for P60,000, and it's
Solution: expected to have a lifespan of 8
Given: years with an estimated salvage
Co = P30,000 value of P5,000 at the end of its
CL = P3,000 useful life. The sinking fund
L = 10 years method, with an interest rate of
C6 = P8,000 7.5%, is being used to calculate its
i = 6% = 0.06 depreciation.
n=6
Cn =? Solution:
Given:
To solve, substitute the given Co = P60,000
values to: CL = P5,000
(C ¿ ¿ o−C L ) i L = 8 years
d= ¿
L
(1+i) −1 i = 7.5% = 0.075
(¿ 30000−3000)(0.06) d=?
d= ¿
(1+ 0.06)10−1
d=P 2 048.43 Using the formula:
Then, (C ¿ ¿ o−C L ) i
d= L
¿
n
(1+ i) −1 (1+i) −1
D n=d [ ] (60000¿¿ −5000)(0.075)
i d= ¿
(1+ 0.06) −1
6 (1+0.075)8−1
D6=(2048.43) d ≈ P5,264.99
0.06
D6=P 14 ,288. 49
Therefore, the depreciation
Lastly, for this industrial generator is
C n=C o−Dn approximately P5,264.9.
C n=30,000−14288.49 – 8000
C n ≈ P 7711.5
c. Declining Balance Method
Therefore, the truck sunk cost at
the end of 6 years, considering the
sinking fund depreciation, is
approximately P7,711.5
 By the “mth year”
Example 4.6
m−1
d n=Co (1−k ) k
A piece of machinery at ALJL

[ ]
m
Construction Services costs CL
m L
C n=C o (1−k) =c o
P300,000 and has a salvage value CO
of P30,000 at the end of its life of
L
10 years. Find the 7th year C L =C o (1−k)
depreciation using a constant
percentage of the declining Where,
balance value. d n = depreciation/year
C o=¿ First Cost
Solution: C L = Salvage Value
Given: C n = Book Value
C o=¿ P300,000 k = rate of depreciation
C L = 30,000
L = 10 years https://www.civilengineermag.com/engineering-economy/

m = 7th year d7 = ?

DEFINITION 4.4 Using the formula:

This approach, also known as

the constant percentage method
k =1 –
√
L CL
Co

or Matheson Formula, operates

under the assumption that yearly
k =1−
10

√
30,000
300,000

( )
1
depreciation cost remains a 30,000 7
k =1−
consistent percentage of the 300,000
asset's salvage value at the start k =0.280
of each year.
Then, to find the 7th year
depreciation:
The relationship between the 7−1
depreciation for a given year and d 7=300,000(1−0.280) (0.280)
the asset's book value at the d 7=P 11,702.38
beginning of that year remains
constant throughout its lifespan, Therefore, in this new scenario, the
denoted by 'k', representing the 7th-year depreciation using a
depreciation rate. constant percentage of the
declining balance value is

√ √
Cn P11,702.38.
n L CL
K=1− =1−
Co Co
 ( )
depreciation at the “mth year”: CL 5
1
0.376=1−
530,000
m−1

( )
1
d n=Co (1−k ) k CL 5
=0.624
530,000
m−1
2 2 C L =0.946 (530,000)
d n=Co (1− )
n n C L =P 50,138
https://www.youtube.com/watch?
v=B5zGIrz0mdY&t=632s
Hence, the salvage value of the
equipment is P50,138.
Example 4.7
d. Double Declining Balance
A printing equipment cost Method
P500,000 and the cost of handling
and installation is P30,000. The life Example 4.8
of the equipment is 5 years. Book
Value at the end of 3 years is Let's consider an equipment
P128,556. Compute the purchased at P200,000 with a
salvage value of the equipment salvage value of P30,000 at the
using Matheson’s Method or DBM. end of its 8-year life. We need to

Solution: DEFINITION 4.5

Given:
C o=530,000 The double declining
L = 5 years balance (DDB) depreciation
C n = 128.556 method also known as the
CL = ? reducing balance method is an
m=3 accelerated approach to
3
accounting for long-lived assets.
C n=C o (1−k) It expenses the asset more rapidly
3
128,556=530,000(1−k ) compared to straight-line
128,556 3 depreciation, depreciating the asset
=( 1−k )
530,000 twice as fast as the standard
1
declining balance method.
(0.243) 3 =(1−k )
k =1−0.624
k =0.376 2
K=
n
Then,

√ CL C n at the end of “mth year”:

L
k =1 – m
C n=C o (1−k ) ∨¿
Co

[ ]
m
2
C n=c o 1−
n
find the book value at the end of 4 n−r +1
dr= × Dn
years using the double declining SYD
balance (DDB) method.
DEFINITION 4.6
Solution:
Given:
C o=¿ 200,000 The Sum-of-the-Years'-
L = 8 years Digits (SYD) method is an
C L = 30,000 approach used in accounting to
m=4 spread the cost of an asset over
C4 = ? its useful life. It operates by
summing the digits of the
Using the formula for DDB Method:
expected lifespan of the asset—
2 say, for a 5-year asset, the digits
K= sum up to 15.
n
2 2
K= = =0.25
n 8 Each year, the depreciation
4
C 4=200,000 ( 1−0.25 ) expense is calculated by taking
C 4=P 63,281.25 the remaining useful life of the
asset at the beginning of the
Therefore, at the end of the year and dividing it by this sum.
4th year, using the double (Kenton,2022)
declining balance (DDB) method,
the book value of the equipment is To get SYD:
P63,281.25.
n(n+1)
SYD =
2
d. The Sum of the Year’s Digit
Method

Consequently, this method

Where,
results in higher depreciation
d = depreciation cost/change
expenses in the earlier years,
r = remaining useful life
reflecting the greater wear and
n = economic life
tear typically experienced by new
Dn= Total depreciation cost
assets, and lower expenses as the
asset ages. SYD = Sum of the years’ digit

Formula for Depreciation cost (d r ):

Example 4.9
 or
Engr. Mirabueno purchased a Cn2020 = $278.58
high-end graphics card, the GTX
1080, in 2017 for $600. It was
estimated to maintain relevance Example 4.10
until 2023, and even then, its Same given problem but in Table
estimated value would be $150. Form:
Use the Sum-of-the-Years'-Digits
Method (SYDM) to find out the From here, we verified that in 2020
value of this graphics card in 2020. (Cn2020), the value of the graphics
card will be $278.58.
Solution: https://www.youtube.com/watch?v=FI0U2TLh37E&t=199s
Given:
n = 6 years (2017 to 2023) 5. Amortization
r = 3rd year (2020)
Dn = 600-150 = $450 Important Question:
6(6+ 1)
SYD = =21
2
Cn2020 = Cn3 =?

Cn3 = Co - D3
D3 = d 1 + d 2 + d 3
n−r +1
dr= × Dn
SYD How does Amortization differ
6−1+1 from Depreciation?
d 1= × 450=128.57
21
6−2+1
d 2= × 450=107.14
21
6−3+1
d 3= × 450=85.71
21

So, substituting the values of d to:

D3 = d1 + d2 + d3
D3 = (128.57) + (107.14) + (85.71) DEFINITION 5.1
D3 = $321.42
Amortization is the process
Then, of gradually allocating the cost of
Cn3 = Co - D3 an intangible asset, such as
Cn3 = 600 – 321.42 patents, copyrights, or franchise
agreements, over its useful
Cn3 = 278.58
lifespan.
 Type of Asset
It follows a straight-line
method, deducting a consistent Amortization is Depreciation is
for Intangible for physical
amount from the asset's value
assets like assets like
regularly. Typically, when an patents and
asset gets amortized, it does not plants,
trademarks,
have any resale or salvage value copyrights, machinery, land,
at the end of its useful life. franchise buildings,
agreements, furniture, etc.
So, the most important
etc.
amortization formula is the Method
calculation of the payment Amortization Depreciation
amount per period. applies the involves utilizing
Straight-Line either the
The formula Method to
Straight-Line
for spread the cost
of an asset Method or the
over time. Written Down
calculating the payment amount
Value Method to
is shown below.
reduce the value
of an asset over
its lifespan.
Salvage Value
where, There is no A physical asset
salvage or that gets
A = payment Amount per period
scrap value for depreciated can
P = initial Principal (loan amount) an intangible
r = interest rate per period have a salvage
asset that gets
n = total number of payments or amortized. or scrap value.
periods

i. Amortization Problems
Amortization Depreciation
Definition
Example 5.1
Amortization Depreciation
involves involves
distributing the distributing the Suppose you've taken out a
expense of an expense of a car loan for $18,000 with a 6%
intangible tangible asset annual interest rate to purchase a
asset across its across its car originally priced at $20,000.
expected expected useful
You've made a down payment of
useful lifespan, lifespan,
typically typically $2,000. The loan term is 4 years.
covering a covering a What would be the monthly
specific period specific period payment for this car loan?
corresponding corresponding to
to its value. its value
 p = number of payment periods
Solution: per year
Given:
P = $18,000 Emily is considering taking
r = 6% per year / 12 months out a loan to finance her education.
= She's exploring different loan
0.005 options and comes across one with
n = 4 years * 12 months = a nominal annual interest rate of
48 6.5%. The lender mentions that the
total periods interest will compound quarterly.
A=? Additionally, the lender requires
loan repayments to be made semi-
Using the formula: annually. Emily wants to
r ( 1+r )
n
understand the effective interest
A=P n
( 1+r ) −1 rate on this loan to make an
0.005 ( 1+0.005 )48 informed decision about her
A=18,000 48 borrowing.
(1+ 0.005 ) −1
A=$ 422.73
Solution:
Given:
Hence, the monthly payment for
i = 6.5% = 0.065
the car loan would amount to
n=4
$422.73.
p=2
r=?

Example 5.2
To compute, use the formula:

( )
n
i
Calculating the Rate Period: r = 1+ p −1
n

DEFINITION 5.2

Where, Straight-line amortization is

a technique used to evenly
r = rate per payment period distribute the cost of intangible
i = nominal annual interest rate assets or bond interest expenses
across each accounting period
n = number of compounding
until the asset's life ends or the
periods per bond matures. (Wallstreetmojo
year Team,2022).
The straight-line amortization Solution:
formula is calculated below: Given:
IC n=¿ $30,000
(Intangible Asset Book value –
IC L =$ 6,000
Expected salvage value) ÷ number
of periods. L=6 years
A = annual amortization

IC n−IC L
A= Given the formula:
L IC n−IC L
Where,
L
IC n=Intangible Asset Book Value
(30,000)−(6,000)
A=
6
IC L =Intangible Asset SalvageValue
A=$ 4,000 per year
L=Useful Life /number of periods

https://www.stockmaster.com/straight-line-
DEFINITION 5.3
amortization

( )
4
0.065 2
r = 1+ −1 The unit of production
4
method is a commonly employed
r =0.0328 %
approach in amortization. It
or 3.28%
comes into play when an asset's
lifespan is determined by the
Therefore, the effective interest
quantity of output it can
rate for the loan Emily is
generate. With this method, the
considering is 3.28%.
asset's cost is allocated across
the entire anticipated number of
a. Straight Line Amortization
units it is projected to produce
throughout its useful life.
Example 5.2

The unit of production

Suppose a company in US
method calculates the
acquires a software license for
depreciation or amortization
$30,000, and it plans to retain
expense based on the number of
these rights for six years before
units produced. The formula for
considering any potential resale at
calculating the amortization
an estimated value of $6,000.
expense is:
Utilizing the straight-line
amortization formula, we can
Amortization Expense =
calculate the annual amortization.

Cost of Asset −Salvage Value

Total Units of Production
Hence, the annual amortization for To solve, use the formula for
this software license would be amortization expense:
$4,000 per year.
Ae =
Cost of Asset −Salvage Value
Again, straight line Total Units of Production
amortization means the same thing
as straight-line depreciation. 120,000−15,000
Ae =
However, the latter is used when 15,000
dealing with tangible assets, while Ae = $ 7
the former is used for intangible
assets. Therefore, utilizing the unit of
production method, the
amortization expense for each unit
b. Units of Production Method of the building component
Example 5.4 produced would be $7 per unit.

Imagine a construction firm

buys a specialized machine for Different assets and
$120,000. They estimate it will industries have their own styles for
produce 15,000 units of a amortization. If something like a
particular building component machine's usage directly links to
during its useful life. The expected how much it produces, the unit of
salvage value at the end of its life production method works best. But
is $15,000. Utilizing the unit of for assets without this direct
production method, find the connection, straight-line
amortization expense for each unit amortization might be the way to
of the building component go. Basically, amortization aims to
produced. divvy up the cost of something, like
a machine or equipment, over its
Solution: life in a way that shows how much
Given: it's worth to the company.

Cost of Asset =$120,000 Aside from the straight-line

Salvage Value = $15,000 amortization and unit of production
Total Units of Production = approach, there exist alternative
15,000 methods for amortization. These
Amortization expense (Ae) encompass accelerated
=? amortization, and sum – of – the –
years – digits amortization.
Accelerated amortization prioritizes TAGALOG!!! (2021). YouTube.
higher expenses in the initial years Retrieved December 13, 2023,
of the asset's life. Sum-of-the- from
https://youtu.be/RKKksoHRS48
years-digits amortization, similar to
?si=3K-BS8V_2yNMHzdl.
accelerated amortization, spreads
the expenses over a broader Compound interest meaning -
timeframe, giving less weight to definition, formulas and solved
the early years. examples. Cuemath. (n.d.-a).
https://www.cuemath.com/co
mmercial-math/compound-
interest/

Deferred annuity formula | how to

calculate PV of ... -
wallstreetmojo. (n.d.).
https://www.wallstreetmojo.co
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Discounts - definition, formula,

rate: Discount calculation.
Cuemath. (n.d.-b).
https://www.cuemath.com/co
mmercial-math/discounts/

Eaomar. (n.d.). Value of a bond.

MATHalino.
https://mathalino.com/forum/e
ngineering-economy/value-
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Depreciation. Retrieved from Depreciation Methods
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