BES 02 - Engineering

Economics

This is a property of
PRESIDENT RAMON MAGSAYSAY STATE UNIVERSITY
NOT FOR SALE

BES 02 – Engineering Economics

First Edition, 2021

Copyright. Republic Act 8293 Section 176 provides that “No copyright shall subsist in any work of
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wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or
office may, among other things, impose as a condition the payment of royalties.

Borrowed materials included in this module are owned by their respective copyright holders. Every
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copyright owners. The University and authors do not claim ownership over them.

Learning Module Development Team

Assigned
Title Author
Chapter
Chapter 1: The Economic Environment
Chapter 2: Interest and Money-Time Relationship Dionisio M. Martin Jr.
Chapter 3: Depreciation
Chapter 4: Capital Financing
Chapter 5: Selections in Present Economy

Evaluators:

(First Name, Middle Initial, Last Name), Position

(First Name, Middle Initial, Last Name), Position
(First Name, Middle Initial, Last Name), Position
Course Overview
Introduction

Engineering Economics is a three-unit basic engineering science course, that is common to all
engineering disciplines. This course discusses the economics environment that affects the
money value in the daily operations in engineering fields.

This course explores the effects of inflation and deflation of money on the market where
engineering plays a part of this important factors in economics. It also involves the systematic
evaluation of the economic benefits of proposed solutions to engineering problems. The
engineering economics involves technical analyzing with emphasis on the economic aspects
and has the objective of assisting decisions.

Course General Objectives

At the end of the semester, 85% of the students have attained 90% level of understanding for
being aware in the engineering economics, locally and globally.

1. Understand the economics environment in engineering field perspective.

2. Solve problems involving interest and the time value of money.
3. Evaluate project alternatives by applying engineering economic principles and
methods and select the most economically efficient one.
4. Deal with risk and uncertainty in project outcomes by applying the basic economic
decision-making concept.
5. Perform mathematically solution to perform decision-making in a certain economic
condition.

Course Details:

 Course Code: BES 02

 Course Title: Engineering Economics
 No. of Units: 3-unit lecture
 Classification: Lecture-based
 Pre-requisite / Co-Requisite: 2nd Year Standing
 Semester and Academic Year: 1st Semester, AY 2021-2022
 Schedule: BSCpE 2A – Monday, Wednesday and Friday, 9:30AM-10:30 AM
 Name of Faculty: Dionisio M. Martin Jr.
 Contact Details
Email: dmmartinjr@yahoo.com
Mobile Number: 0939-906-0585
FB Account: Dionisio Martin Jr.
 Consultation
Day: TTH
Time: 8:00-9:30AM
Learning Management System

The University LMS will be used for asynchronous learning and assessment. The link and
class code for LMS will be provided at the start of class through the class’ official Facebook
Group.

 Edmodo
 Google Classroom
 University LMS

Assessment with Rubrics

Students will be assessed in a regular basis thru quizzes, assignments, individual/group

outputs using synchronous and/or asynchronous modalities or submission of SLM exercises.
Rubrics are also provided for evaluation of individual/group outputs.

Major examinations will be given as scheduled. The scope and coverage of the examination
will be based on the lessons/topics as plotted in the course syllabus.
 0323

Module Overview
Introduction

This module aims to introduce economics to engineering students particularly the computer
engineering students as to their chosen field of specialization in engineering. It will let them
to meaning and usage of interest in engineering field. The money time value relationship for
certain applied condition and the depreciation effect to the organization where he/she
belongs. The capital financing suited to start new business or project related in some cases
and situation where the best selection is needed.

On the later part of this module, the application of different methods for different economic
study for engineering will be discussed. The comparing of solved alternatives and the
replacement studies in dealing engineering project study. The break-even analysis as well as
the benefit/cost analysis in establishing new concept and ideas in engineering application
related in computer field.

The students will learn how to make decision using a mathematical approach from solving
different case studies at the end of each lesson/chapter.

Table of Contents

Chapter 1: The Economic Environment

Chapter 2: Interest and Money-Time Relationship
Chapter 3: Depreciation
Chapter 4: Capital Financing
Chapter 5: Preparing for an Engineering Career
Engineering Economics

Chapter 1

The Economic
Environment
Chapter 1

The Economic Environment

Introduction

The economy simply states the stability of a country.

Specific Objectives

At the end of the lesson, the students should be able to:

- define the economics and its importance in engineering

- differentiate necessities and luxuries, supply and demand, and consumer and
producers in engineering
- discuss and explain the flow of economy
- discuss and explain the supply and demand curved
- identify the different factors influencing/affecting the supply and demand
- enumerate the types of engineering economic decision
- discuss the engineering economic decision process

Duration

Chapter The Economic Environment = 3 hours

1: (2.5-hours discussion;
0.5-hour assessment)

_____________________________________________

INTRODCUTION OF ENGINEERING ECONOMICS

Economics – is the science that deals with the production and consumption of goods and
services and the distribution and rendering of these for human welfare. The following are
the economic goals:
a. A high level of employment
b. Price stability
c. Efficiency
d. An equitable distribution of income
e. Growth

Necessities – are those products or services that are required to support human life and
activities that will be purchased in somewhat the same quantity even the price varies
considerably.
Luxuries – are those products or services that are desired by humans and will be purchased if
money is available after the required necessities have been obtained.

Flow of Economy

Consumer goods and services – are those products or services that are directly used by
people to satisfy their wants.
Producer goods and services – are used to produce consumer goods and services or other
producer goods.

Law of Supply and Demand

“Under conditions of perfect competition the price at which a given product will be
supplied and purchased is the price that will result in the supply and the demand being
equal.”

Demand and Supply Curved

Demand – is the quantity of a certain commodity that is bought at certain price at a given
place and time.
 The shape of the demand curve is influenced by the following factors:
 Income of the people
 Prices of related goods
 Tastes of consumers

Supply – is the quantity of a good that the producer plans to sell in the market.
The shape of the supply curve is affected by the following factors:
 Cost of the inputs
 Technology
 Weather
 Prices of related goods

Competition – occurs in a situation where a commodity or service is supplied by a number of

vendors and there is nothing to prevent additional vendors entering the market.
Monopoly – is the opposite of competition.
– exists when a unique product or service is available from a single vendor and
that vendor can prevent the entry of all others into the market.
Oligopoly – exists when there are so few suppliers of a product or service that action by one
will almost inevitably result in similar action by the others.

Engineering economics – is the analysis and evaluation of the factors that will affect the
economic success of engineering projects to the end that a recommendation can be made
which will insure the best use of capital.
– involves formulating, estimating, and evaluating the expected economic
outcomes of alternatives designed to accomplish a defined purpose.
– deals with the methods that enable one to take economic decisions towards
minimizing costs and/or maximizing benefits to business organizations

PRINCIPLES OF ENGINEERING ECONOMICS

Fundamental Principles of Engineering Economics

The following are fundamental principles to follow in engineering economics:

 Principle 1: The time value of money. A money earned today is worth more than
a money earned in the future.
 Principle 2: Differential (incremental) cost and revenue. The only thing that
matters is the difference between alternatives.

 Principle 3: Marginal cost and revenue. Marginal revenue must exceed marginal
cost.

 Principle 4: The trade-off between risk and reward. Additional risk is not taken
without the expected additional return.
 COST CONCEPTS FOR DECISION MAKING

Engineering economic decision – refers to any investment decision related to an engineering

project.

Thus, the facet of an economic decision that is of most interest from an engineer’s point
of view is the evaluation of costs and benefits associated with making a capital
investment.

Types of Engineering Economic Decision

The five main types of engineering economic decisions are:
1. equipment or process selection
2. equipment replacement
3. new product or product expansion
4. cost reduction
5. improvement in service or quality

Engineering Economic Decision Process

The following are the rational decision making process:
1. Recognize problem
2. Define the goal or objective
3. Assemble relevant data
4. Identify feasible alternatives
5. Select the criterion to determine the best alternative
6. Construct a model
7. Predict each alternative’s outcomes or consequences
8. Choose the best alternative
9. Audit the result

Engineering decision making – refers to solving substantial engineering problems in which

economic aspects dominate and economic efficiency is the criterion for choosing from
among possible alternatives.

Cash-flow diagrams – is simply a graphical representation of cash flows drawn on a time

scale.
 a. Cash inflows – are the receipts, revenues, incomes, and savings generated by project
and business activity.

(positive cash flow or cash inflow)

Examples:
Income – money received, especially on a regular basis, for work or through
investments. Ex.: ₱50,000 per year from sales of solar-powered watches
Receipt – an amount of money received during a particular period by an
organization or business. Ex.: ₱750,000 received on large business loan
plus accrued interest
Savings – money that people and businesses receive in exchange for working,
producing a product or service, or investing capital. Ex.: ₱150,000 per year
saved by installing more efficient air conditioning
Revenue – the total amount of income generated by the sale of goods or services
related to the company's primary operations. Ex.: ₱50,000 to ₱75,000 per
month in sales for extended battery life iPhones

b. Cash outflows – are costs, disbursements, expenses, and taxes caused by projects and
business activity.

(negative cash flow or cash outflow)

Example:
Operating costs – the ongoing expenses incurred from the normal day-to-day of
running a business. Ex.: ₱230,000 per year annual operating costs for
software services
First cost – the initial expendetures involved in capitalizing a property; includes
items such as transportation, installation, preparation for service, as well as
other related costs. Ex.: ₱800,000 next year to purchase replacement
earthmoving equipment
Expense – the cost of operations that a company incurs to generate revenue. Ex.:
₱20,000 per year for loan interest payment to bank
Initial cost – the moneys required for the capital construction or renovation of a
facility or the construction of an addition: P1 to P1.2 million in capital
expenditures for a water recycling unit

Sample Usage Cash Flow Diagram:

A loan of ₱100 at simple interest of 10% will become ₱150 after 5 years.
 Cash flow diagram on the viewpoint Cash flow diagram on the viewpoint
of the lender of the borrower

_____________________________________________

References/Additional Resources/Readings

C. Park (2013). Fundamentals of Engineering Economics, 3rd ed., Pearson Education, Inc.

H. Sta. Maria (n.d.). Engineering Economy 3rd ed., National Bookstore, Inc.

https://www.economicsdiscussion.net/engineering-economics/engineering-economics-
meaning-and-characteristics/21680
Activity Sheet
ACTIVITY 1

Name: ______________________Course/Year/Section: ___________ Score: _________

Direction: Identify the following.

________________ 1. The analysis and evaluation of the factors that will affect the
economic success of engineering projects to the end that a
recommendation can be made which will insure the best of capital.
________________ 2. The quantity of a certain commodity that is bought at certain price at
a given place and time.
________________ 3. The science that deals with the production and consumption of goods
and services and the distribution and rendering of these for human
welfare.
________________ 4. Occurs in a situation where a commodity or service is supplied by a
number of vendors and there is nothing to prevent additional vendors
entering the market.
________________ 5. The quantity of a good that the producer plans to sell in the market.

Direction: Match the items in column A to their descriptions in column B. write only the
letter of your choice on the space provided.
A B
_____ 1. Efficiency a. Achieved when all available resources are used to produce
goods and services.
_____ 2. Employment b. Achieved when income and wealth are fairly distributed
within a society.
_____ 3. Equity c. Achieved by avoiding or limiting fluctuations in production,
employment, and prices.
_____ 4. Growth d. Achieved by increasing the economy's ability to produce
goods and services.
_____ 5. Stability e. Achieved when society is able to get the greatest amount of
satisfaction from available resources.

Direction: Place a Check () mark on the corresponding column if the given situation is
either Supply factor or Demand factor.
Supply Demand
1. During a recession (economic) when there are fewer jobs
available and there is less money to spend, the price of
homes tends to drop.
2. Spare parts price of delivery trucks increases causing the
delivery goods to increase.
3. When college students learn that computer engineering jobs
pay more than English professor jobs, the graduate of
 students with majors in computer engineering will increase.
4. When consumers start paying more for cupcakes than for
donuts, bakeries will increase their output of cupcakes and
reduce their output of donuts in order to increase their
profits.
5. When your employer pays time and a half for overtime, the
number of hours you are willing to supply for work
increases.
6. Samsung recently releases their latest model of mobile
phone because of their high sales from the previous model.
7. Promotional grocery pricing frequently offers discounted
prices on the condition that a certain number of items are
purchased.
8. The Nintendo company might supply 1 million systems if
the price is P20,000 each, but if the price increases to
P30,000, they might supply 1.5 million systems.
9. Because of the super typhoon that hits the rice granary of
the Philippines, the government needs to import rice from
other nearby country in Asia.
10. The global shortage of pineapple causes price to rise tends
to consumer to find substitute products such as other fruits.
Assignment
ASSIGNMENT 1

Name: ______________________Course/Year/Section: ___________ Score: _________

Direction: Write your answer clear and concise.

1. How economics affects the decision making in engineering?

2. Choose one factor that influences the shape of demand curve and explain how it
influenced the demand.

3. Choose one factor that affects the shape of the supply curve and explain how it affects
the supply.
Assessment (Rubrics)
Each question will be graded based on this five (5) point rubric.

LEVEL DESCRIPTION

Well written and very organized.

Excellent grammar mechanics.
5 - Outstanding Clear and concise statements.
Excellent effort and presentation with detail.
Demonstrates a thorough understanding of the topic.

Writes fairly clear.

Good grammar mechanics.
4 - Good
Good presentation and organization.
Sufficient effort and detail.

Minimal effort.
Minimal grammar mechanics.
3 - Fair
Fair presentation.
Few supporting details

Somewhat unclear.
Shows little effort.
2 - Poor Poor grammar mechanics.
Confusing and choppy, incomplete sentences.
No organization of thoughts.

Very poor grammar mechanics.

Very unclear.
1 - Very Poor
Does not address topic.
Limited attempt.
 Engineering Economics

Chapter 2

Interest and
Money-Time Relationship
Chapter 2

Interest and Money-Time Relationship

Introduction

Time value of money is the idea that money that is available at the present time is worth more
than the same amount in the future, due to its potential earning capacity. This core principle
of finance holds that provided money can earn interest, any amount of money is worth more
the sooner it is received. One of the most fundamental concepts in finance is that money has a
time value attached to it. In simpler terms, it would be safe to say that a peso was worth more
yesterday than today and a peso today is worth more than a peso tomorrow.

Interest can be a charge or an income, depending on whether you are borrowing money or
lending/investing money. It is stated as a percentage over a specific period of time. There are
five (5) variables need to know:
1. Present value – This is the current starting amount. It is the money you have in your
hand at the present time, your initial investment for your future.
2. Future value – This is your ending amount at a point in time in the future. It should be
worth more than the present value, provided it is earning interest and growing over
time.
3. The number of periods – This is the timeline for your investment (or debts). It is
usually measured in years, but it could be any scale of time such as quarterly, monthly,
or even daily.
4. Interest rate – This is the growth rate of your money over the lifetime of the
investment. It is stated in a percentage value, such as 8% or .08.
5. Payment amount – These are a series of equal, evenly-spaced cash flows.

Specific Objectives

At the end of the lesson, the students should be able to:

- know the importance of time value of money

- understand the difference between simple interest and the compound interest
- understand the difference between the nominal interest rate and the effective interest
rate
- solve problems involving different type of interest as well as in annuity and
amortization.
- solve problems how loans are structured in terms interest and principal payments.
- understand the basics of investing in financial assets.

Duration

Chapter Interest and Money-Time Relationship = 6 hours

2: (5.5-hours discussion;
0.5-hour assessment)
_____________________________________________
INTEREST

Interest – is the amount of money paid for the use of borrowed capital or the income
produced by money which has been loaned.
– have many types and forms of interest. It is critical you know the terminology.
Here are the most commonly used terms:
a. Simple interest – is computed on the original amount as the return on that
principal for one-time period.
b. Compound interest – is computed on the original amount as the return on that
principal plus all unpaid interest accumulated to date.
c. Fixed interest rate – is a straight forward rate that remains constant during the
life of the loan or investment.
d. Variable interest rate – is changes during the life of the loan and is usually
tied to the prime rate. It can go up or down depending on the prime rate set
forth by the economy.
e. Mixed interest rate – is changes from fixed to variable or from variable to
fixed.

1. Simple Interest – is calculated using the principal only, ignoring any interest that had
been accrued in preceding period.
– in practice, it is paid on short-term loans in which the time of the loan is measured
in days.
Formula:
I =Pni
F=P+ I
F=P+ Pni
F=P ( 1+ ¿ )
Where: I – interest
P – principal or present worth
n – number of interest periods
i – rate of interest per interest period
F – accumulated amount or future worth

Classification of Simple Interest

 a. Ordinary simple interest is computed on the basis of 12 months of 30 days each or
360 days a year. 1 interest period = 360 days
b. Exact simple interest is based on the exact number of days in a year, 365 days for an
ordinary year and 366 days for a leap year. 1 interest period = 365 or 366 days

Sample Problem:

Problem 1.) Determine the ordinary simple interest on ₱700 for 8 months and 15 days if the
rate of interest is 15%.
Find: ordinary simple interest ( I )
Solution:
Number of days = 8 months and 15 days
= (8)(30) + 15
= 255 days
I =Pni
255
¿ 700 x x 0.15
360
I =₱ 74.38

Problem 2.) Determine the exact simple interest on ₱500 for the period from January 10 to
October 28, 1996 at 16% interest.
Find: exact simple interest ( I )
Solution:
January 10–31 = 21 (excluding January 10)
February = 29
March = 31
April = 30
May = 31
June = 30
July = 31
August = 31
September = 30
October = 28 (including October 28)
Total = 292 days
I =Pni
292
¿ 500 x x 0.16
366
I =₱ 63.83

Problem 3.) What will be the future worth of money after 14 months, if a sum of ₱10,000 is
invested today at a simple interest rate of 12% per year?
Find: future worth ( F )
Solution (a):
 Number of days = 14 months
= (14)(30)
= 420 days
I =Pni
420
¿ 10,000 x x 0.12
360
I =₱ 1,400

F=P+ I
¿ 10,000+1,400
F=₱ 11,400

Solution (b):
F=P(1+¿)

¿ 10,000 1+
( ( )( ))
420
360
0.12 ∨10,000 1+
14
12 ( ( ) )
( 0.12 )

F=₱ 11,400

2. Compound Interest – is calculated on the principal plus total amount of interest

accumulated in previous periods.
– means “interest on top of interest.”
Borrower’s Viewpoint:
Principal at Interest Earned Amount at End of
Interest Period
Beginning of Period During Period Period
1 P Pi P+ Pi
¿ P(1+i)
2 P(1+i) P(1+i)i P ( 1+ i )+ P ( 1+i ) i
2
¿ P(1+i)
3 P(1+i )
2
P(1+i ) i
2 2 2
P ( 1+ i ) + P ( 1+i ) i
3
¿ P(1+i)
˸ ˸ ˸ ˸
˸ ˸ ˸ ˸
N P(1+i )
n−1
P(1+i )
n−1
i P(1+i )
n

Formula:
n
F=P(1+i )

F=P ( FP .i % . n)
−n
P=F (1+i)

P=F ( PF .i % . n)
Rate of Interest
 a. Nominal rate of interest specifies the rate of interest and a number of interest periods
in one year.
Formula:
r
i=
m
Where: i – rate of interest per interest period
r – nominal interest rate
m – number of compounding periods per year
b. Effective rate of interest is the actual or exact rate of interest on the principal during
one year.
Formula:
Effective Rate ¿ F−1
m
¿ ( 1+i ) −1

Sample Problem:

Problem 1.) If the nominal rate of interest is 10% compounded quarterly, what is the rate of
interest per interest period?
Find: rate of interest (i )
Solution:
r
i=
m
10 %
¿
4
¿ 2.5 %

Problem 2.) If ₱1 is invested at a nominal rate of 15% compounded quarterly, what will be
the total earnings after one year?
Find: total earnings after one year ( F )
Solution:
n
F=P ( 1+ i )

( )
4
0.15
F=1 1+
4
¿ ₱ 1.1586

Problem 3.) Suppose you deposit ₱1,000 in a bank savings account that pays interest at a
rate of 10% compounded annually. Assume that you don’t withdraw the interest earned at
the end of each period (one year), but let it accumulate. How much would you have at the
end of year 3?
Find: total earnings after three year ( F )
Solution:
For compound interest: i=10 % per year
 n
F=P ( 1+ i )
3
F=1,000 ( 1+0.10 )
¿ ₱ 1,331.00

Problem 4.) Find the amount at the end of two years and seven months of ₱1,000 is invested
at 8% compounded quarterly and using simple interest for anytime less than a year
interest period.
Find: amount at the end of two years and seven months ( F )
Solution:
For compound interest:
8%
i= =2% and n=( 2 ) ( 4 )=8
4
For simple interest:
7
i=8 % and n=
12

Amount at the end of two years:

n 8
F 1=P (1+i ) =1,000 ( 1+0.02 ) =₱ 1,171.66

Amount at the end of two years and seven months:

(7
)
F 2=F1 ( 1+¿ )=1,171.66 1+ (0.08) =₱ 1,226.34
12

Problem 5.) A ₱2,000 loan was originally made at 8% simple interest for 4 years. At the end
of this period the loan was extended for 3 years, without the interest being paid, but the
new interest rate was made 10% compounded semi-annually. How much should the
borrower pay at the end of 7 years?
Find: amount to be pay at the end of seven years ( F )
Solution:
 F 4=P ( 1+¿ ) =2,000 ( 1+(4)(0.08) )=₱ 2,640

n 6
F 7=F 4 ( 1+ i ) =2,640 ( 1+ 0.05 ) =₱ 3,537.86

DISCOUNT
Discount – is the difference between the present worth and the worth at some time in the
future.
– is the interest paid in advance.
Discount=Future Worth−Present Worth

Rate of Discount – is the discount on one unit of principal for one unit of time.
Formula:
−1
d=1−( 1+i )
d
i=
1−d

Where: d – rate of discount for the period involved

i – rate of interest for the same period

Sample Problem:

Problem 1.) A man borrowed ₱5,000 from a bank and agreed to pay the loan at the end of 9
months. The bank discounted the loan and gave him ₱4,000 in cash.
(a) What is the rate of discount?
(b) What was the rate of interest?
(c) What was the rate on interest for one year?
Solution (a):
discount
d=
principal
1,000
¿
5,000
d=0.20 or 20 %

Solution (b):
 d
i=
1−d
0.20
¿
1−0.20
i=0.25 or 25 %
Or:
I
i=
P
1,000
¿
4,000
i=0.25 or 25 %

Solution (c):
I
i=
Pn
1,000
¿
4,000 ( )
9
12
i=0.3333 or 33.33 %

INFLATIONS
Inflation – is the increase in the prices for goods and services from one year to another, thus
decreasing the purchasing power of money.
Formula:
n
F=P ( 1+ f )
Where: P – present cost of commodity
F – future cost of the same commodity
f – annual inflation rate
n – number of years

In an inflationary economy, the buying power of money decreases as costs increased

Formula:
P
F=
( 1+ f )n
Where: P – present worth
F – future worth
f – annual inflation rate
n – number of years

Sample Problem:
Problem 1.) An item presently costs ₱1,000. If inflation is at the rate of 8% per year, what
will be the cost of the item in two years?
Find: cost of the item in two years ( FC )
Solution:
n
F=P ( 1+ f )
2
¿ 1,000 ( 1+0.08 )
F=₱ 1,166.40

Problem 2.) An economy is experiencing inflation at an annual rate of 8%. If this continue,
what will ₱1,000 be worth two years from now?
Find: worth two years from now ( F )
Solution:
P
F=
( 1+ f )n
1,000
¿
( 1+ 0.08 )2
F=₱ 857.34

ANNUITIES
Annuity – is a series of equal payments occurring at equal periods of time.
Meaning: P – value or sum of money at present
F – value or sum of money at some future time
A – a series of periodic, equal amounts of money
n – number of interest period
i – interest rate per interest period

Ordinary Annuity – is one where the payments are made at the end of each period.
Formula:
Finding for A :

[ ] [ ( 1+i )n−1
]
n
1− ( 1+i )
P= A or ¿ A n
i i ( 1+ i )
P
P= A ( , i% , n)
A
Finding for F :

F= A [
( 1+i )n−1
i ]
F
F= A( ,i % , n)
A
Finding A when P is given:
 A=P
[ i
1− ( 1+i )
n
]
A
A=P( , i% , n)
P

Finding A when F is given:

A=F
[ i
( 1+i )n−1 ]
A
A=F( ,i % , n)
F
Thus:
A A
, i % , n+1= ,i % , n
F P

Sample Problem:

Problem 1.) What are the present worth and the accumulated amount of a 10-year annuity
paying ₱10,000 at the each of each year, with interest at 15% compounded annually?
Find: present worth (P) and the accumulated amount (F)
Solution:
A=₱ 10,000
n=10
i=15 %

(Illustration here)

P= A ( PA , i % , n)=10,000( PA ,15 % ,10)

[ ]
10
1− (1+ 0.15 )
P=10,000
0.15
P=₱ 50,188

F= A ( FA ,i % ,n)=10,000 ( FA , 15 % , 10)
F=10,000 [ ( 1+ 0.15 )10−1
0.15 ]
F=₱ 203,037

Problem 2.) What is the present worth of ₱500 deposited at the end of every three months
for 6 years if the interest rate is 12% compounded semiannually?
Find: present worth ( P)
Solution:
 Interest rate per quarter:

( )
2
( 1+i )n−1= 1+ 0.12 −1
2
0.3
1+i=( 1.06 )
i=0.0296 or 2.96 % per quarter

P= A ( PA , i % , n)=500( PA ,2.96 % , 24)

[ ]
−24
1− (1+ 0.0296 )
P=500
0.0296
P=₱ 8,504

Problem 3.) A businessman needs ₱50,000 for his operations. One financial institution is
willing to lend him the money for one year at 12.5% interest per annum (discounted).
Another lender is charging 14% with the principal and interest parable at the end of one
year. A third financier is willing to lend him ₱50,000 payables in 12 equal monthly
installments at ₱4,600. Which offer is best for him?
Find: what is the best offer (find the lowest effective rate)
Solution:
First Offer:
Rate of discount, d=12.5 %
d 0.125
Rate of interest, i= = =14.92 %
1−d 1.0125
Effective rate ¿ 14.29 %
Another solution:
Amount received ¿ 50,000(0.875)=₱ 43,750
50,000−43,750
Rate of interest ¿ =14.29 %
43,750
Effective rate ¿ 14.29 %

Second Offer:
Effective rate ¿ 14 %

Third Offer:

P= A
P
A( ,i %,n )
[ ]
12
1−( 1+i )
50,000=4,600
i
12
1−( 1+i ) 50,000
=
i 4,600
12
1−( 1+i )
=10.8696
i
 Try i=1 %:
12
1−( 1+0.01 )
=11.2551
0.01
Try i=2 % :
12
1−( 1+0.02 )
=10.5753
0.02

1%
[[ i% 10.8696
2% 10.5753
]
x 1 % 11.2551 0.3855
0.6798
]
x 0.3855
=
1% 0.6798
x=0.57
i=1 %+ 0.57 %=1.57 % per month
Effective rate ¿ ( 1+0.0157 )12−1=20.26 %

Therefore, the second offer is the best.

AMORTIZATION
Amortization – is any method of repaying a debt, the principal and interest included, usually
by a series of equal payments at equal interval of time.

Sample Problem:

Problem 1.) A debt of ₱5,000 with interest at 12% compounded semiannually is to be

amortized by equal semiannual payments over the next 3 years, the first due in 6 months.
Find the semiannual payment and construct an amortization schedule.
Find: semiannual payment ( A) and construct an amortization schedule
Solution:
P
A=
P
,6 % , 6
A
5,000
A=
4.9173
A=₱ 1,016.82

Amortization Schedule

Outstanding
Interest due at Principal
principal at
Period the end of Payment repaid at end
beginning of
period of period
period
 1 5,000.00 300.00 1,016.82 712.82
2 4,283.18 256.99 1,016.82 759.83
3 3,523.35 211.40 1,016.82 805.42
4 2,717.93 163.08 1,016.82 853.74
5 1,864.19 111.85 1,016.82 904.97
6 959.22 57.55 1,016.82 959.27
TOTALS 1,100.87 6,100.92 5,000.05

Problem 2.) A debt of 10,000 with interest at the rate of 20% compounded semiannually is
to be amortized by 5 equal payments at the end of each 6 months, the first payment is to
be made after 3 years. Find the semiannual payment and construct an amortization
schedule.
Find: semiannual payment ( A) and construct an amortization schedule
Solution:

A=P ( AP , 10 % , 5)( FP ,10 % ,5)

A=10,000 ( 0.2638 ) (1.1605 )
A=₱ 4,248.50

Amortization Schedule

Outstanding
Interest due at Principal
principal at
Period the end of Payment repaid at end
beginning of
period of period
period
1 10,000.00 1,000.00
2 11,000.00 1,100.00
3 12,100.00 1,210.00
4 13,310.00 1,331.00
5 14,641.00 1,464.10
6 16,105.10 1,610.51 4,248.50 2,637.99
7 13,467.11 1,346.71 4,248.50 2,901.79
8 10,525.32 1,056.53 4,248.50 3,191.97
9 7,373.35 737.34 4,248.50 3,511.16
10 6,862.19 386.22 4,248.50 3,862.28
TOTALS 11,242.41 21,242.50 16,105.19

Uniform Arithmetic Gradient

An arithmetic gradient cash flow is one wherein the cash flow changes (increases or
decreases) by the same amount in each cash flow period. The general equation to find the
present worth of an arithmetic gradient cash flow series is:
Formula:
P=P A + PG
 [ ]
n
1−( 1+i )
P A =A
i
P
P A =A ( ,i % , n)
A

[ ][ ]
n
G ( 1+i ) −1 1
PG = −n n
i i (1+i )
P
PG =G( , i% ,n)
G

P= A ( PA , i % , n)+ G( GP ,i % ,n)
Where: A – amount of money in period 1
G – change in amount between periods 1 and 2
i – interest rate per period
n – number of periods
If the gradient cash flow decreases from one period to the next instead of increases, the
only change in the general equation is that the plus sign becomes a minus sign.

Sample Problem:

Problem 1.) The MIS Department expects the cost of maintenance for a particular piece of
computer equipment to be ₱5,000 in year 1, ₱5,500 in year 2, and amounts increasing by
₱500 through year 10. At an interest rate of 10% per year, find the present worth of the
maintenance cost.
Find: present worth ( A)
Solution:

P= A ( PA , i % , n)+ G( GP ,i % ,n)
P=5,000 ( 6.1446 ) +500(22.8913)
P=₱ 42,168.55

Problem 2.) A loan was to be amortized by a group of four end-of-year payments forming an
ascending arithmetic progression. The initial payment was to be ₱5,000 and the
difference between successive payments was to be ₱400. But the loan was renegotiated to
provide for the payment of equal rather than uniformly varying sums. If the interest rate
of the loan was 15%, what was the annual payment?
Find: annual payment ( A)
Solution:
A=5,000 G=400 n=4 i=15 %
−4
P 1− (1+ 0.15 )
, 15 % , 4= =2.8550
A 0.15
 [ ][ ]
4
P 1 ( 1+0.15 ) −1 1
, 15 % , 4= −4 4
=3.7865
G 0.15 0.15 ( 1+0.15 )

P= A ( PA , 15 % , 4)+G( GP , 15 % , 4)
P=5,000 ( 2.8550 )+ 400(3.7865)
P=₱ 15,789.60

A ( PA ,i % ,n)=P
P
A=
( PA , i% , n)
15,789.60
A=
2.8550
A=₱ 5,530.51

Problem 3.) Find the equivalent annual payment of the following obligations at 25% interest.
End of year Payment
1 8,000
2 7,000
3 6,000
4 5,000
.

Find: annual payment ( A)

Solution:
A=8,000 G=1,000 n=4 i=20 %
−4
P 1− (1+ 0.20 )
, 15 % , 4= =2.5887
A 0.20

[ ][ ]
4
P 1 ( 1+0.20 ) −1 1
, 15 % , 4= −4 4
=3.2986
G 0.20 0.20 ( 1+0.20 )

P= A ( PA , 20 % , 4)−G( GP , 20 % , 4)
P=8,000 (2.5887 )−1,000(3.2986)
P=₱ 17,411

A ( PA ,i % ,n)=P
P
A=
( PA , 20 % , 4)
 17,411
A=
2.5887
A=₱ 6,725.77

_____________________________________________

References/Additional Resources/Readings

C. Park (2013). Fundamentals of Engineering Economics, 3rd ed., Pearson Education, Inc.

H. Sta. Maria (n.d.). Engineering Economy 3rd ed., National Bookstore, Inc.

http://engineering.utep.edu/enge/EE/02/05/1.htm
 Activity Sheet
ACTIVITY 2

Name: ______________________Course/Year/Section: ___________ Score: _________

Direction: Identify the following.

________________ 1. A series of equal payments occurring at equal periods of time.
________________ 2. Interest computed on the basis of 12 months of 30 days each or 360
days a year.
________________ 3. Any method of repaying a debt, the principal and interest included,
usually by a series of equal payments at equal interval of time.
________________ 4. The amount of money paid for the use of borrowed capital or the
income produced by money which has been loaned.
________________ 5. The actual or exact rate of interest on the principal during one year.
________________ 6. Calculated on the principal plus total amount of interest accumulated
in previous periods.
________________ 7. The increase in the prices for goods and services from one year to
another, thus decreasing the purchasing power of money.
________________ 8. The interest paid in advance.
________________ 9. Interest based on the exact number of days in a year, 365 days for an
ordinary year and 366 days for a leap year.
________________ 10. Calculated using the principal only, ignoring any interest that had
been accrued in preceding period.

Direction: Solve the following.

1. What is the annual rate of interest if ₱265 is earned in four months on an investment
of ₱15,000? (Ans. 5.3%)

2. A loan of ₱2,000 is made for a period of 13 months, from January 1 to January 31 the
following year, at a simple interest rate of 20%. What future amount is due at the end of the
loan period? (Ans. ₱2,433.33)

3. Determine the exact simple interest on ₱5,000 for the period from January 15 to
November 28 if the rate of interest is 22%.

4. If you borrow money from your friend with simple interest of 12%, find the present
worth of ₱20,000 which is due at the end of nine months. (Ans. ₱18,348.62)

5. A person deposits a sum of ₱20,000 at the interest rate of 18% compounded annually
for 10 years. Find the future worth after 10 years.
 6. A man wishes his son to receive ₱200,000 ten years from now. What amount should
be investing if it will earn interest of 10% compounded annually during the first 5 years and
12% compounded quarterly during the next 5 years? (Ans. ₱68,758.67)

7. A person wishes to have a future sum of ₱1,000,000 for his son’s education after 10
years from now. What is the single payment that he should deposit now so that he gets the
desired amount of 10 years? The bank gives 15% interest rate compound annually.

8. Jones Corporation borrowed ₱9,000 from Brown Corporation on Jan, 1, 1978 and
12,000 on Jan 1, 1980. Jones Corporation made a partial payment of ₱7,000 on Jan. 1, 1981.
It was agreed that the balance of the loan would be amortized by two payments, one on Jan.1,
1982 and the other on Jan. 1, 1983, the second being 50% larger than the first. If the interest
rate is 12%, what is the amount of each payment? (Ans. ₱9,136.91, ₱13,705.36)

9. Determined the present worth and the accumulated amount of an annuity consisting of
6 payments of ₱120,000 each, the payment is made at the beginning of each year. Money is
worth 15% compounded annually.

10.Calculate the capitalized cost of a project that has an initial cost of ₱3,000,000 and an
additional investment cost of ₱1,000,000 at the end of every ten years. The annual operating
cost will be ₱300,000 at the end of every year for the first four years and ₱160,000 thereafter.
In addition, there is expected to be a recurring major rework cost of ₱300,000 every 13 years.
Assume i=13 % . (Ans. ₱4, 281,960)
Engineering Economics

Chapter 3

Depreciation
Chapter 3

Depreciation
Introduction

On a project level, engineers must be able to assess how the practice of depreciating fixed
assets influences the investment value of a given project. To do this, the engineers need to
estimate the allocation of capital costs over the life of the project, which requires an
understanding of the conventions and techniques that accountants use to depreciate assets.

Engineering products and

applications are essential to everyday
activities in our life. These
include housing, transportations,
communications, and other industrial
processes. In order to
safeguard life, health and property,
and to promote public welfare, the
practice of engineering
is treated as a learned profession, and
its practitioners shall be held
accountable by high
professional standards in keeping
with the ethics and practices of other
learned professions.
These
include housing, transportations,
communications, and other industrial
processes.
Specific Objectives

At the end of the lesson, the students should be able to:

- know the meaning and types of depreciation

- know the effect of depreciation on income calculations
- determine the difference between ordinary gains and capital gains
- solve and determine the appropriate rate to use in project analysis
- solve to find the relationship of net income and net cash flows

Duration

Chapter Depreciation = 2 hours

3: (1.5-hours discussion;
0.5-hour assessment)

_____________________________________________

DEPRECIATION

Depreciation – is the decrease in the value of physical property with the passage of time.

Definition of Value
a. Value – is the present worth of all future profits that are to be received through
ownership of a particular property.
b. Market value – is the amount which a willing buyer will pay to a willing seller for the
property where each has equal advantage and is under no compulsion to buy or sell.
c. Utility or use value – is what the property is worth to the owner as an operating unit.
d. Fair value – is the value which is usually determined by a disinterested third party in
order to establish a price that is fair to both seller and buyer.
e. Book value – sometimes called depreciated book value, is the worth of a property as
shown on the accounting records of an enterprise.
f. Salvage value – or resale, is the price that can be obtained from the sale of the
property after it has been used.
g. Scrap value – is the amount the property would sell for if disposed of as junk.
Purpose of Depreciation
1. To provide for the recovery of capital which has been invested in physical property.
2. To enable the cost of depreciation to be charged to the cost of producing products or
services that results from the use of the property.

Types of Depreciation
1. Normal depreciation
a. Physical – is due to the lessening of the physical ability of a property to
produce results. Its common causes are wear and deterioration.
b. Functional – is due to the lessening in the demand for the function which the
property was designed to render. Its common causes are inadequacy, changes
in styles, population centers shift, saturation of markets or more efficient
machines are produced.
2. Depreciation due to changes in price levels
3. Depletion – refers to the decrease in the value of a property due to the gradual
extraction of its contents.

Physical life – is the length of time during which it is capable of performing the function for
which it was designed and manufactured.
Economic life – is the length of time during which the property may be operated at a profit.

Requirement of a Depreciation Method

1. It should be simple
2. It should recover capital
3. The book value will be reasonably close to the market value at any time.
4. The method should be accepted by the Bureau of Internal Revenue (BIR)..

Depreciation Methods
The following symbols for the different depreciation methods:
L = useful life of the property in years
C O = the original cost
C L = the value at the end of the life, the scrap value (including gain or loss due to
removal)
d = the annual cost of depreciation
C n = the book value at the end of n years
Dn = depreciation up to age n years

1. Straight Line Method – assumes that the loss in value is directly proportional to the age
of the property.
Formula:
C O−C L
d=
L
 n ( C O−C L )
D n=
L
C n=C O−Dn

Sample Problem:

Problem 1.) An electronic balance costs 90,000 and has an estimated salvage value of 8,000
at the end of its 10-year life time. What would be the book value after three years, using
the straight line method in solving for depreciation?
Find: book value after three years (C n)
Solution:
C O=90 , 000 C L =8 , 0 00 L=10 n=3
C O−C L
d=
L
90,000−8,000
¿
10
¿ ₱ 8,200

Dn=nd
D3=3 ( 8,200 )
¿ ₱ 24,600

C n=C O−Dn
C 3=90,000−24,600
¿ ₱ 65,400

2. Sinking Fund Method – assumes that a sinking fund is established in which funds will
accumulate for replacement. The total depreciation that has taken place up to any given
time is assumed to be equal to the accumulated amount in the sinking fund at that time.
Formula:
CO −C L
d=
F
,i %, L
A

Dn=d ( FA , i% , n)
C n=C O−Dn

Sample Problem:

Problem 1.) A broadcasting corporation purchased an equipment for 53,000 and paid 1,500
for freight and delivery charges to the job site. The equipment has a normal life of 10
 years with a trade-in value of 5,000 against the purchase of a new equipment at the end of
the life.
(a) Determine the annual depreciation cost by the straight line method.
(b) Determine the annual depreciation cost by the sinking fund method. Assume
interest at 6.5% compounded annually.
Find: annual depreciation cost (d )
Solution:
C O=53 , 000+1,500
¿ 5 4 ,5 00
C L =5 , 00 0

C O−C L
(a) d=
L
54 , 5 00−5 , 000
¿
10
¿ ₱ 4 , 95 0

CO −C L
d=
(b) F
,i %, L
A
54,500−5,000
¿
F
,6.5 % ,10
A
49 , 500
¿
18.3846
¿ ₱ 3 , 668

Problem 2.) A firm bought an equipment for 56,000. Other expenses including installation
amounted to 4,000. The equipment is expected to have a life of 16 years with a salvage
value of 10% of the original cost. Determine the book value at the end of 12 years by (a)
the straight line method and (b) sinking fund method at 12% interest.
Find: book value (C n)
Solution:
C O=5 6 , 000+ 4 , 0 00
¿ 60 , 0 00

C L =60 , 00 0 ( 0.10 )
¿ 6 , 00 0

L=1 6 n=12 i=12 %

C O−C L
(a) d=
L
 60 , 000−6 , 000
¿
16
¿ ₱ 3 , 375

Dn=nd
D12=12 ( 3 ,375 )
¿ ₱ 40 ,5 00

C n=C O−Dn
C 12=6 0,000−40 ,5 00
¿ ₱ 19 , 5 00

CO −C L
d=
(b) F
,i %, L
A
60 , 000−6 , 000
¿
F
,12 % ,1 6
A
54 , 0 00
¿
42 .7533
¿ ₱ 1 , 263

( FA , i% , n)
D n=d

D =1,263 ( ,12 % ,12 )

F
12
A
¿ 1,263 ( 24.1331 )
¿ 30,480

C n=C O−Dn
C 12=CO −D12
¿ 60,000−30,480
¿ 29,520

_____________________________________________

References/Additional Resources/Readings

C. Park (2013). Fundamentals of Engineering Economics, 3rd ed., Pearson Education, Inc.

H. Sta. Maria (n.d.). Engineering Economy 3rd ed., National Bookstore, Inc.
http://engineering.utep.edu/enge/EE/02/05/1.htm
Activity Sheet
ACTIVITY 3

Name: ______________________Course/Year/Section: ___________ Score: _________

Direction: Identify the following.

________________ 1. A paid occupation, especially one that involves prolonged training
and a formal qualification.
________________ 2. Sector concerns the delivery of services.
________________ 3. The profession in which a knowledge of the mathematical and natural
sciences is applied with judgment to develop ways to utilize,
economically, the materials and forces of nature for the benefit of
mankind.
________________ 4. These service industries perform routine support activities for the day-
to-day operations of other organizations.
________________ 5. These industries produce equipment and machinery needed for
transporting people and goods.

Direction: Match the items in column A to their descriptions in column B. write only the
letter of your choice on the space provided.
A B
_____ 1. Information a. Structural Engineer
_____ 2. Utilities b. Quality Assurance Engineer
_____ 3. Construction c. Network Engineer
_____ 4. Administrative and support d. Safety Engineer
_____ 5. Electronic product manufacturing e. Electrical Engineer

Direction: Place a Check () mark on the corresponding column if the given sector is either
A-Manufacturing Sector or B-Service Sector.
A B
1. Transportation equipment manufacturing
2. Technical services
3. Electronic manufacturing
4. Information
5. Mining

Direction: Give 3 each of the following:

1. Skills needed for engineers
2. Sectors of services

3. Sectors of manufacturing

4. Engineering job functions

5. Engineering positions for manufacturing sector.

Learner’s Feedback Form

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Program : ___________________________________________________
Year Level : ___________ Section : ____________
Faculty : ___________________________________________________
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___________________________________________________________________________
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NOTE: This is an essential part of course module. This must be submitted to the subject
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