Course Outline: CEE 109 – Engineering Economics
Course Coordinators:
Jerald M. Mutoc, REE
Jobenilita R. Cuñado
Email:
jeraldmutoc0517@gmail.com,
jobenilita_cunado@umindanao.edu.ph
Student Consultation:
By appointment
Mobile:
(+63)9306691873
(+63) 9360860118
Phone:
None
Effectivity Date:
June 2020
Mode of Delivery:
Blended (Online with face-to-face or virtual sessions)
Time Frame:
54 Hours
Student Workload:
Expected Self-Directed Learning
Requisites:
None
Credit:
3 Units
Attendance Requirements: A minimum of 95% attendance is required for all scheduled faceto-face virtual sessions
Course Policy
Areas of Concern
Overview
Contact Hours
Assessment Tasks
Assessment Schedule
Submissions
Details
This course instruction manual is designed for blended
learning – an approach where opportunities for virtual
interaction is combined with classroom-based instruction
delivery. The learning units are organized to be completed in
order as this module was designed to capitalize on prerequisite knowledge gained from preceding units.
This 3-unit course module is expected to be completed in 54
hours. A certain number of hours is scheduled for each
learning unit; however, a student has control over his/her
pace provided assessment activity requirements are
submitted as scheduled. Virtual and place-based learning
sessions, including assessments, comprise course contact
hours.
A combination of theory and practice-based assessments is
to be expected.
Periodical assessments are scheduled based on the school
calendar. Other assessment schedules will be posted as
deemed necessary.
All assessment tasks posted online via Learning Management
System are REQUIRED to be submitted as scheduled. A
Dishonesty Penalties
Return of Assessment
Results
Grading System
classwork will be created for submission purposes.
Classroom-based assessments are to be submitted within set
time limits.
Students are expected to work on individual assessments by
themselves.
Cheating incidences in any form are
DISCOURAGED, however, are beyond the control of the
course coordinator, especially that resources can be easily
sourced and shared online. As such, shared credits shall be
given to students who will submit identical works.
The use of other people’s work without acknowledging the
author shall be dealt with accordingly based on the
appropriate provisions in the student handbook.
The results of any assessment, if not readily available after
completion such as those of multiple-choice types, will be
returned within two weeks, at most, from the date of
submission.
Grades shall be computed based on the following:
Course Exercises (Quizzes, Assignments, Participation)
25%
Research
15%
First Major Assessment
10%
Second Major Assessment
10%
Third Major Assessment
10%
Final Assessment
30%
*Base 15 grading policy will be followed.
Mode of Communication
All communication shall be done on the virtual classroom via
streams, comment boxes or emails. Queries will only be
entertained during our class schedule or during special
meetings when one is scheduled.
Contact Details of the Dean Dr. Gina Fe G. Israel
Email:
Phone:
Contact Details of the Engr. Reena Ross E. Pusta
Program Head
Email:
Phone:
Students with Special Needs Students with special needs shall communicate to the course
coordinator the nature of his or her special needs. Depending
on the nature of the need, the course coordinator with the
approval of the program coordinator, may provide
alternative assessment tasks or extension of the deadline of
submission of assessment tasks. However, the alternative
assessment tasks should still be in the service of achieving the
desired course learning outcomes.
Course Information – View/Download the course syllabus in/from the LMS
CC’s Voice:
Hello! Welcome to CEE 109 – Engineering Economics. This course will expose you
to the fundamentals of engineering economics which you will find useful in
managing engineering projects.
CO
Engineering economics is a field that addresses the dynamic environment of
economics involving the evaluation of costs and benefits of proposed projects.
Engineers play an important role in investments, especially in making decisions
based on economic analysis and design considerations. One of the expected
outcomes of this course is for you to apply the various principles of engineering
economy to various engineering problems, engineering economy equations and
techniques to solve problems related to the economic aspect of engineering
projects. You will also be required to propose engineering projects based on
economic factors which necessitates the application of engineering economics
fundamentals.
Let us begin!
BIG PICTURE
Week 1 – 3: Unit Learning Outcomes (ULO)
At the end of the unit, you are expected to:
a. discuss the basic principles of engineering economy;
b. classify cost items based on the basic cost terminologies and concepts; and
c. apply the concept of interest in solving engineering problems.
Big Picture in Focus: ULOa. Discuss the basic principles of engineering economy.
Metalanguage
1. Deviation Analysis. It is the measure of the difference between an observed variable’s
value in a set and some other value which is often the mean of the set.
2. Risk. In financial terms, it is defined as the probability that an investment’s actual gains
are different from the expected returns.
3. Uncertainty. It refers to the unpredictability of an economy’s future outlook.
4. Capital. It refers to the money required in the production of goods or services.
5. Warranty Cost. It is the cost of repairing or replacing previously sold products during their
warranty periods.
6. First Cost. It is the sum of the initial expenditures involved in capitalizing a property or
building a project; includes items such as transportation, installation, preparation for
service, as well as other related costs.
7. Salvage Value. It is the estimated resale value of an asset at the end of its useful life.
8. Annual Operating Cost. It is a yearly expense associated with the maintenance and
administration of a business on a day-to-day basis.
9. Subcontract Cost. It is the cost associated with outsourcing part of the obligations and
tasks under a contract to another party.
10. Income. It is money (or some equivalent value) that an individual or business receives,
usually in exchange for providing a good or service or through investing capital.
11. Cash Flow. It is the net amount of cash and cash-equivalents being transferred into and
out of a business.
12. Time Value of Money. It is the concept that explains the change in the amount of money
over time for both owned and borrowed funds.
13. Tax. It is an involuntary fee imposed on individuals or corporations and enforced by a
government entity—whether local, regional or national—in order to finance government
activities.
14. Inflation. It is an economic term that refers to an environment of generally rising prices
of goods and services within a particular economy. As general prices rise, the purchasing
power of consumers decreases.
15. Rate of Return. It is the net gain or loss of an investment over a specified time period,
expressed as a percentage of the investment's initial cost.
Essential Language
INTRODUCTION
Engineering economy involves the systematic evaluation of the economic merits of
proposed solution to engineering problems. For a solution to be economically acceptable, it must
exhibit a positive balance between long-term benefits and long-term costs.
The need for engineering economy is primarily motivated by the work that engineers do
in performing analyses and synthesizing as the work on project of all sizes comes to a conclusion.
Engineers play a major role in investment by making decisions based on economic analysis and
design considerations. These decisions often reflect the engineer’s choice of how to best invest
funds by choosing the proper alternative out of a set of alternatives.
Principles of Engineering Economy
The development, study, and application of any discipline begin with a basic foundation.
In engineering economy, this foundation is the set of principles that provide a comprehensive
doctrine for developing methodologies.
The foundation of the discipline of Engineering Economy can be seen in terms of seven
principles. These seven principles are:
1. Develop the alternatives.
Alternates need to be developed from which the decision has to come from.
Brainstorming sessions have been found to be productive in coming up with these
alternatives.
2. Focus on the differences.
In this principle, deviation analysis is used. However, only the differences
among the expected future outcomes among the alternatives are considered relevant
in decision making.
3. Use a consistent viewpoint.
The viewpoint for the particular decision need to be first defined. The
perspective or viewpoint of the decision maker, which is often that of the owners of
the firm, would normally be used. It is then used consistently in the description,
analysis, and comparison of the alternatives.
4. Use a common unit of measure.
In measuring the economic consequences, a monetary unit is the common
measure. Other outcomes should be translated into monetary units in order to not
complicate the overall project analysis.
5. Consider all relevant criteria.
The decision maker will normally select the alternative that will best serve the
long-term interests of the owners of the organization. In engineering economic
analysis, the primary criterion relates to the long-term financial interests of the
owners. This is based on the assumption that available capital will be allocated to
provide maximum monetary return to the owners. Often, though, there are other
organizational objectives you would like to achieve with your decision, and these
should be considered and given weight in the selection of an alternative.
6. Make risk and uncertainty explicit.
Risk and uncertainty are inherent in estimating the future outcomes of the
alternatives and should be recognized. The analysis of the alternatives involves
projecting or estimating the future consequences associated with each of them. The
magnitude and the impact of future outcomes of any course of action are uncertain.
The probability is high that today’s estimates of, for example, future cash receipts and
expenses will not be what eventually occurs. Thus, dealing with uncertainty is an
important aspect of engineering economic analysis.
7. Revisit your decisions.
A good decision-making process can result in a decision that has an
undesirable outcome. Other decisions, even though relatively successful, will have
results significantly different from the initial estimates of the consequences. Learning
from and adapting based on experience are essential and are indicators of a good
organization.
The Engineer’s Role in Decision Making
Decision is made routinely to choose one alternative over another in everyday life. Most
decisions involve money, called capital or capital fund, which is usually limited in amount. The
decision of where and how to invest this limited capital is motivated by a primary goal of adding
value as future, anticipated results of the selected alternative are realized.
People make decisions with the aid of computers, mathematics, concepts, and guidelines
in the decision-making process. Since most decisions affect what will be done, the time frame of
engineering economy is primarily the future.
Example 1:
An engineer is performing an analysis of warranty costs for drive train repairs within the
first year of ownership of luxury cars purchased in the United States. He found the average cost
(to the nearest peso) to be Php 28,490 per repair from data taken over a 5-year period.
Year
Average
Cost(Php/)Repair
2010
2011
2012
2013
2015
26,250
21,500
30,950
32,500
31,250
Table 1. Five-Year Average Cost per Repair Example
What range of repair costs should the engineer use to ensure that the analysis is sensitive
to changing warranty costs?
Solution:
At first glance the range should be approximately -30% to + 15% of the Php 28,490
average cost to include the low of Php 21,500 and high of Php 32,500. However, the last 3 years
of costs are higher and more consistent with an average of Php 31,566.67. The observed values
are approximately ±3% of this more recent average. If the analysis is to use the most recent data
and trends, a range of, say, ±5% of Php 31,566.67 is recommended. If, however, the analysis is to
be more inclusive of historical data and trends, a range of, say, ±25% or ±30% of Php 28,490 is
recommended.
WRITESHOP 1
Answer the following questions:
1. Why is the study of engineering economy important to the practicing engineer?
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2. You have discussed with a coworker in the engineering department the importance
of explicitly defining the viewpoint (perspective) from which future outcomes of a
course of action being analyzed are to be developed. Explain what is meant by a
viewpoint or perspective.
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Engineering Economy and the Design Process
An engineering economy study is accomplished using a structured procedure and
mathematical modeling techniques. The economic results are then used in a decision situation
that involves two or more alternatives and normally includes other engineering knowledge and
input.
Figure 1. The Steps in an Engineering Economy Study
The engineering economic analysis procedure is as follows:
1. Problem recognition, definition, and evaluation.
A problem must be well understood and stated in an explicit form before the
project team proceeds with the rest of the analysis. Recognition of the problem is
2.
3.
4.
5.
normally stimulated by internal or external organizational needs or requirements. An
operating problem within a company (internal need) or a customer expectation about
a product or service (external requirement) are examples.
As an illustration, assume the problem is that a coal-fueled power plant must
be shut down by 2015 due to the production of excessive sulfur dioxide. The
objectives may be to generate the forecasted electricity needed for 2015 and beyond,
plus to not exceed all the projected emission allowances in these future years.
Development of the feasible alternatives.
Alternatives are stand-alone descriptions of viable solutions to problems that
can meet the objectives. Words, pictures, graphs, equipment and service descriptions,
simulations, etc. define each alternative. The best estimates for parameters are also
part of the alternative. Some parameters include equipment first cost, expected life,
salvage value (estimated trade-in, resale, or market value), and annual operating cost
(AOC), which can also be termed maintenance and operating (M&O) cost, and
subcontract cost for specific services. If changes in income (revenue) may occur, this
parameter must be estimated.
The two primary actions in this step are:
a. Searching for potential alternatives; and
b. Screening them to select a smaller group of feasible alternatives for
detailed analysis.
Development of the outcomes and cash flows for each alternative.
All viable alternatives at this stage is crucial. For example, if two alternatives
are described and analyzed, one will likely be selected and the implementation
initiated. If a third, more attractive method that was available is later recognized, a
wrong decision was made.
Also, all cash flows are estimated for each alternative. Since these are future
expenditures and revenues, the results of this step usually prove to be inaccurate
when the alternative is actually in place and operating. When cash flow estimates for
specific parameters are expected to vary significantly from a point estimate made
now, risk and sensitivity analyses (step 5) are needed to improve the chances of
selecting the best alternative. Sizable variation is usually expected in estimates of
revenues, AOC, salvage values, and subcontractor costs.
Selection of a criteria.
The criterion used to select an alternative in engineering economy for a
specific set of estimates is called a measure of worth. The measures of worth account
for the fact that money makes money over time. This is the concept of time value of
money.
Analysis and comparison of the alternatives.
In this step, techniques and computations involving cash flow estimates, time
value of money and a selected measure of worth are used. The result of the analysis
will be one or more numerical values – this can be in one of several terms, such as
money, an interest rate, number of years, or a probability.
Before an economic analysis technique is applied to the cash flows, some
decisions about what to include in the analysis must be made. Two important
possibilities are taxes and inflation as these will impact the costs of every alternative.
An after-tax analysis includes some additional estimates and methods compared to a
before-tax analysis. If taxes and inflation are expected to impact all alternatives
equally, they may be disregarded in the analysis. However, if the size of these
projected costs is important, taxes and inflation should be considered. Also, if the
impact of inflation over time is important to the decision, an additional set of
computations must be added to the analysis.
6. Selection of the preferred alternative.
The measure of worth is a primary basis for selecting the best economic
alternative. For example, if alternative A has a rate of return (ROR) of 15.2% per year
and alternative B will result in an ROR of 16.9% per year, B is better economically.
However, there can always be non-economic or intangible factors that must be
considered and they may alter the decision. Some of these non-economic factors
include:
 Market pressures, such as need for an increased international presence
 Availability of certain resources, e.g., skilled labor force, water, power, tax
incentives
 Government laws that dictate safety, environmental, legal, or other
aspects
 Corporate management’s or the board of director’s interest in a particular
alternative
 Goodwill offered by an alternative toward a group: employees, union, etc.
7. Performance monitoring and post-evaluation results.
Performance monitoring helps in determining if a project is on track and when
changes may be needed. Monitoring and evaluation forms the basis for modification
of interventions and assessing the quality of activities being conducted.
A sound engineering economic analysis procedure incorporates the seven basic
principles. The seven-step procedure is also used to assist decision making within the engineering
design process.
Engineering Economic Analysis Procedure
Step
1. Problem recognition, definition, and
evaluation.
Engineering Design Process
Activity
1. Problem/Need definition.
2. Problem/Need formulation and
evaluation.
2. Development of the feasible
alternatives.
3. Development of the cash flows for
each alternative.
4. Selection of a criterion (or criteria).
5. Analysis and comparison of the
alternatives.
6. Selection of the preferred alternative.
7. Performance monitoring and postevaluation results.
3. Synthesis of possible solutions
(alternatives).
4. Analysis, optimization, and
evaluation.
5. Specification of preferred alternative.
6. Communication.
Source: Middendorf, W.H., Design of Devices and Systems (New York: Marcel Dekker, Inc., 1986)
Table 2. The Engineering Economic Analysis Procedure vis-à-vis the Engineering Design Process
Example 1:
Bad news – you have just wrecked your car! An automobile wholesaler offers you Php 100,000
for the car “as is”. Also, your insurance company’s claims adjuster estimates that there is Php
100,000 in damages to your car. Because you have collision insurance with a Php 50,000
deductibility provision, the insurance company mails you a check for Php 50,000. The odometer
reading on your wrecked car is 58,000 miles.
You need another car immediately. What should you do?
Solution:
(a) Develop your alternatives (Principle 1):
1. Sell the wrecked car to the wholesaler and spend this money, the Php 50,000
insurance check, and all of your Php 350,000 life savings on a newer car. Total amount
paid out of savings is Php 350,000, and the car will have 28,000 miles.
2. Spend the Php 50,000 insurance check and Php 50,000 of savings to fix the car. Total
amount paid out of savings is Php 50,000, and the car will have 58,000 miles.
3. Spend the Php 50,000 insurance check and Php 50,000 of savings to fix the car, then
sell the car for Php 225,000. Spend the Php 225,000 plus Php 275,000 of additional
savings to buy the newer car. Total amount paid out of savings is Php 325,000, and
the car will have 28,000 miles.
4. Give the car to a lesser mechanic, who will repair it for Php 55,000 (Php 50,000
insurance and Php 5,000 of your savings) but will take an additional month of repair
time, and rent a car for that time at Php 20,000/month (paid out of savings). Total
amount paid out of savings is Php 25,000, and the car will have 58,000 miles.
5. Same as 4, but then sell the car for Php 225,000 and use this money plus Php 275,000
of additional savings to buy the newer car. Total amount paid out of savings is Php
300,000, and the car will have 28,000 miles.
ASSUMPTIONS
1. The less reliable repair shop in Alternatives 4 and 5 will not take longer than the
additional month to repair the car.
2. Each car will perform at the normal operating condition as it was originally intended
and to the same total mileage before being sold or salvaged.
3. Interest earned on money remaining in savings is negligible.
(b) Focus on the differences (Principle 2):
1. Alternative 1 varies from all others because the car is not to be repaired at all but
merely sold. This eliminates the benefit of the Php 25,000 increase in value of the car
when it is repaired then sold. Also, this alternative leaves no money in the bank.
2. Alternative 2 varies from Alternative 1 because it allows the old car to be repaired.
Alternative 2 differs from Alternatives 4 and 5 since it utilizes a more expensive (Php
25,000 more) and less risky repair facility. It also varies from Alternatives 3 and 5
because the car will be kept.
3. Alternative 3 gains an additional Php 25,000 by repairing the car and selling it to buy
the same car as in Alternative 1.
4. Alternative 4 uses the same idea as Alternative 2 but involves a less expensive repair
shop. This repair shop is more risky in the quality of its end product but will only cost
Php 55,000 in repairs and Php 20,000 in an additional month’s rental of a car.
5. Alternative 5 is the same as Alternative 4 but gains an additional Php 25,000 by selling
the repaired car and purchasing a newer car as in Alternatives 1 and 3.
(c) Use a consistent viewpoint (Principle 3):
The viewpoint taken here is that of the owner of the wrecked car.
(d) Use a common unit of measure (Principle 4):
The peso represents the value of the car to the owner. Hence, peso is used as the
consistent value against which everything is to be measured. This reduces all decisions to
a quantitative level, which can then be reviewed later with qualitative factors that may
carry their own peso value (e.g. how much is a reliable repair shop worth?).
(e) Consider all relevant criteria (Principle 5):
1. Alternative 1 is eliminated because Alternative 3 gains the same end result and would
also provide the car owner with Php 25,000 more cash. This is experienced with no
change in the risk to the owner. Car value = Php 500,000, savings = 0.
2. Alternative 2 is a good alternative to consider because it spends the least amount of
cash, leaving Php 300,000 in the bank. However, Alternative 2 provides the same end
result as Alternative 4 but costs Php 25,000 more to repair. Therefore, Alternative 2
is eliminated. Car value = Php 200,000, savings = Php 300,000.
3. Alternative 3 is eliminated because Alternative 5 also repairs the acr but at a lower
out-of-savings cost (Php 25,000 difference), and both Alternatives 3 and 5 have the
same end result of buying the newer car. Car value = Php 500,000, savings = Php
25,000.
4. Alternative 4 would be a good alternative because it saves Php 25,000 by using a
cheaper repair facility, provided that the risk of a poor repair job is judged to be small.
Car value = Php 200,000, savings = Php 325,000.
5. Alternative 5 is the alternative accepted because it repairs the car at a lower cost (Php
25,000 cheaper) but eliminates the risk of breakdown by selling it to someone else at
an additional Php 25,000 gain. Car value = Php 500,000, savings = Php 50,000.
(f) Make uncertainty explicit (Principle 6):
Among the uncertainties that can be found in this problem, the following are the most
relevant to the decision. If the original car is repaired and kept, there is a possibility that
it would have a higher frequency of breakdowns. If a cheaper repair facility is used, the
chance of a later breakdown is even greater. Also, the newer car purchased may be too
expensive, based on additional price paid (which is at least Php 300,000/30,000 miles =
Php 10 per mile). Finally, the newer car may also have been in an accident and could have
a worse repair history than the presently owned car.
(g) Revisit your decisions (Principle 7):
The newer car turned out after being “test driven” for 20,000 miles to be a real beauty.
Mileage was great, and no repairs were needed. The systematic process of identifying and
analyzing alternative solutions to this problem really paid off.
WRITESHOP 2
The Almost-Graduating Senior Problem. Consider this situation faced by a first-semester
senior in electrical engineering who is exhausted from extensive job interviewing and penniless
from excessive partying. Jose’s impulse is to accept immediately a highly attractive job offer to
work in his brother’s successful manufacturing company. He would then be able to relax for a
year or two, save some money, and then return to college to complete his senior year and
graduate. Jose is cautious about impulsive desire because it may lead to no college degree at all!
a. Develop at least two formulations for Jose’s problem.
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b. Identify feasible solutions for each problem formulation in part (a). Be creative!
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Big Picture in Focus: ULOb. Classify cost items based on the basic cost terminologies and
concepts.
Metalanguage
1. Cost. It is the monetary value of goods and services that producers and consumers
purchase.
2. Revenue. It is the income that a firm receives from the sale of goods or services to its
customers.
3. Depreciation. It is the measure of the decrease in monetary value of an asset over time
due to use, wear and tear or obsolescence.
Essential Knowledge
Cost Concepts for Decision Making
When conducting engineering economy studies and communicating results, it is
important to use consistent definitions for cost terms.
Fixed, Variable, and Incremental Costs
Fixed costs are those unaffected by changes in activity level over a feasible range of
operations for the capacity or capability available. Typical fixed costs include insurance and taxes
on facilities, general management and administrative salaries, license fees, and interest costs on
borrowed capital.
Variable costs are those associated with an operation that vary in total with the quantity
of output or other measures of activity level. If you were making an engineering economic
analysis of a proposed change to an existing operation, the variable costs would be the primary
part of the prospective differences between the present and changed operations as long as the
range of activities is not significantly changed. For example, the costs of material and labor used
in a product or service are variable costs – because they vary in total with the number of output
units – even though the costs per unit stays the same.
An incremental cost, or incremental revenue, is the additional cost, or revenue, that
results from increasing the output of a system by one (or more) units. Incremental cost is often
associated with “go/no go” decisions that involve a limited change in output or activity level.
Example 3:
A company considers increasing the number of goods that they produce. Their current
level of production and the added costs of producing additional units are listed below:

5,000 units cost Php 7.5M or Php 1,500 per unit.

7,000 units cost Php 8.75M or Php 1,250 per unit.
What is the incremental cost involved?
Solution:
The total incremental cost to produce the 2,000 additional units is given by:
Incremental Cost = Php 8.75 M – Php 7.5M = Php 1.25M
The incremental cost per unit is Php 1.25M/2,000 = Php 625.
The lower incremental cost per unit is due to certain costs, such as fixed costs remaining
constant. Although a portion of fixed costs can increase as production increases, usually, the cost
per unit declines since the company isn't buying additional equipment or fixed costs to produce
the added volume.
Recurring and Nonrecurring Costs
These two general terms are often used to describe various types of expenditures.
Recurring costs are those that are repetitive and occur when an organization produces
similar goods or services on a continuing basis. Variable costs are also recurring costs because
they repeat with each unit of output. But recurring costs are not limited to variable costs. For
example, office space rental – which is a fixed cost – is also a recurring cost.
Nonrecurring costs are those that are not repetitive, even though the total expenditure
may be cumulative over a relatively short period of time. Typically, nonrecurring costs involve
developing or establishing a capability or capacity to operate. For example, the purchase cost for
real estate upon which a plant will be built is a nonrecurring cost, as well as the cost of
constructing the plant itself.
Direct, Indirect, and Overhead Costs
These cost terms involve most of the previous cost elements.
Direct costs are those that can be reasonably measured and allocated to a specific output
or work activity. The labor and material costs directly associated with a product, service, or
construction activity are direct costs. For example, the materials needed to make a pair of scissors
would be a direct cost.
Indirect costs are those that are difficult to attribute or allocate to a specific output or
work activity. The term normally refers to types of costs that would involve too much effort to
allocate directly to a specific output. For example, the costs of common tools, general supplies,
and equipment maintenance in a plant are treated as indirect costs.
Overhead costs consist of plant operating costs that are not direct labor or direct material
costs. Examples of overhead include electricity, general repairs, property taxes, and supervision.
Administrative and selling expenses are usually added to direct costs and overhead costs to arrive
at a unit selling price for a product or service.
Standard Costs
Standard costs are representative costs per unit of output that are established in advance
of actual production or service delivery. They are developed from the direct labor hours,
materials, and support functions planned for the production or delivery process. For example, a
standard cost for manufacturing one unit of an automotive part such as a starter would be
developed as follows:
Standard Cost Element
Direct Labor
+
Direct Material
+
Factory Overhead Costs
Sources of Data for Standard Costs
Process routing sheets, standard times,
standard labor rates
Material quantities per unit, standard unit
material costs
Total factory overhead costs allocated based
on prime costs (direct labor plus direct
material costs)
= Standard Cost (per unit)
Table 3. Standard Cost and Sources
Standard costs play an important role in cost control and other management functions.
Some representative uses are the following:
1. Estimating future manufacturing or service delivery costs.
2. Measuring operating performance by comparing actual cost per unit with the
standard unit cost.
3. Preparing bids on products or services requested by customers.
4. Establishing the value of work-in-process and finished inventories.
Cash Cost versus Book Cost
A cost that involves payment of cash is called a cash cost (and results in a cash flow) to distinguish
it from one that does not involve a cash transaction and is reflected in the accounting system as
a noncash cost. This noncash cost is often referred to as a book cost. The most common example
of book cost is the depreciation charged for the use of assets such as plant and equipment.
Sunk Cost
A sunk cost is one that has occurred in the past and has no relevance to estimates of
future costs and revenues related to an alternative course of action.
Sunk costs result from past decisions and therefore are irrelevant in the analysis and
comparison of alternatives that affect the future which can be disregarded in an engineering
economic analysis.
The concept of sunk cost is illustrated in the following simple example. Suppose that John
Carlo finds a motorcycle he likes and pays Php 2,000 as a down payment, which will be applied
to the Php 65,000 purchase price but which must be forfeited if he decides not to take the cycle.
Over the weekend, John Carlo finds another motorcycle he considers equally desirable for a
purchase price of Php 61,500. For the purpose of deciding which motorcycle to purchase, the
Php 2,000 is a sunk cost and thus would not enter into the decision except that it lowers the
remaining cost of the first cycle. The decision then is between paying Php 63,000 (Php 65,000 –
Php 2,000) for the first motorcycle versus Php 61,500 for the second motorcycle.
Opportunity Cost
An opportunity cost is incurred because of the use of limited resources, such that the
opportunity to use those resources to monetary advantage in an alternative use is foregone.
Thus, it is the cost of the best rejected opportunity and is often hidden or implied.
Examples of opportunity costs:
 A business owns its building. If the company moves, the building could be rented to
someone else. The opportunity cost of staying there is the amount of rent the company
would get.
 David decides to quit working and got to school to get further training. The opportunity
cost of this decision is the lost wages for a year.
 Jill decides to take the bus to work instead of driving. It takes her 60 minutes to get there
on the bus and driving would have been 40, so her opportunity cost is 20 minutes.
Life-Cycle Cost
Life-cycle cost refers to a summation of all costs, both recurring and nonrecurring, related
to a product, structure, system, or service during its life span. Life cycle begins with the
identification of the economic need or want (requirement) and ends with retirement and disposal
activities.
Figure 2. Phases of the Life Cycle and their Relative Costs
The cumulative committed life-cycle cost curve increases rapidly during the acquisition
phase. In general, approximately 80% of life-cycle costs are “locked in” at the end of this phase
by the decisions made during requirements analysis and preliminary detailed design. In contrast,
as reflected by the cumulative life-cycle cost curve, only about 20% of actual costs occur during
the acquisition phase, with about 805 being incurred during the operation phase.
Thus, one purpose of the life-cycle concept is to make explicit the interrelated effects of
costs over the total life span for a product. An objective of the design process is to minimize the
life-cycle cost – while meeting other performance requirements – by making the right tradeoffs
between prospective costs during the acquisition phase and those during the operation phase.
The cost elements of the life cycle that need to be considered will vary with the situation.
Several basic life-cycle cost categories include:
1. Investment Cost. It is the capital required for most of the activities in the acquisition
phase. Acquiring a specific equipment is an investment cost. This cost is also called a
capital investment.
2. Working Capital. It refers to the funds required for current assets that are needed for
the start-up and subsequent support of operation activities. For example, products
cannot be made or services delivered without having materials available in inventory.
Functions such as maintenance cannot be supported without spare parts, tools,
trained personnel, and other resources.
3. Operation and Maintenance Cost. It includes many of the recurring annual expense
items associated with the operation phase of the life cycle. The direct and indirect
costs of operation associated with the five primary resource areas – people, machines,
materials, energy, and information – are a major part of the costs in this category.
4. Disposal Cost. It includes nonrecurring costs of shutting down the operation and the
retirement and disposal of assets at the end of the life cycle. A classic example of a
disposal cost is that associated with cleaning up a site where a chemical processing
plant had been located.
WRITESHOP 3
1. A company in the process industry produces a chemical compound that is sold to
manufacturers for use in the production of certain plastic products. The plant that
produces the compound employs approximately 300 people. Develop a list of six different
cost elements that would be fixed and a similar list of six cost elements that would be
variable.
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2. Refer to Problem (1) and your answer to it. Develop a table that shows the cost elements
you defined and classified as fixed and variable. Indicate which of these costs are also
recurring, nonrecurring, direct or indirect.
3. Classify each of the following cost items as mostly fixed or variable:
a. Raw materials
b. Direct labor
c. Depreciation
d. Supplies
e. Utilities
f. Property taxes
g. Administrative salaries
h. Payroll taxes
i. Insurance (building and equipment)
j. Clerical salaries
k. Sales commissions
l. Rent
m. Interest on borrowed money
Present Economy Studies
When the influence of time on money is not a significant consideration, cost analyses are
usually called present economy studies. Typical situations involving present economy studies are
as follows:
1. There is no initial investment of capital; only immediate operating costs and other
factors are involved. For example, assume that you are employed by Company A and
are making plans for a business trip. You can travel by commercial aircraft, which will
require 3 hours of travel time and the rental of the car at your destination. The
alternative is to travel by automobile, which will take 7 hours. Here, the basic
considerations are the immediate costs, the value of your time, and nonmonetary
factors (e.g. fatigue).
2. There is an initial investment of capital, but after this investment cost, the remaining
life-cycle cost is estimated to be the same, or directly proportional to the initial
investment. Thus, the alternative with the lowest investment cost will be the most
economical. Consider the construction of an interstate highway bridge overpass.
Whether a longitudinally reinforced concrete slab or a precast concrete design is used,
the maintenance and other life-cycle costs of the two designs are usually assumed to
be proportionally the same. However, if one or more steel design alternatives were
also being considered, a present economy study probably would not be appropriate.
Maintenance and other related costs would be expected to vary among the
alternatives, and the cost analysis should be based on the life cycle of the structure.
3. The differences in revenues and costs among the alternatives all occur within a limited
time period (1 year or less is a general guideline), or any future differences are
estimated to remain proportional to those in the first time period. This is often the
case when the decision is between alternative materials in manufacturing. For
example, if estimations show that using low-alloy/high-yield strength steel in
particular application gives better revenue and cost results than using low-carbon
steel, this relative advantage would be expected to remain in future time periods.
In engineering practice, situations that give rise to present economy studies are quite
common. Recognizing these situations often saves considerable analysis effort.
Big Picture in Focus: ULOc. Apply the concepts of interest in solving engineering problems.
Metalanguage
1. Interest. It is the price paid for the use of credit or money.
2. Profit. It is the financial benefit realized when revenue generated from a business activity
exceeds the expenses, costs, and taxes involved in sustaining the activity in question.
3. Interest Period. It is the length of the loan period.
4. Interest Rate. It is the amount a lender charges for the use of assets expressed as a
percentage of the principal.
5. Compounding. It is the process in which an asset's earnings, from either capital gains or
interest, are reinvested to generate additional earnings over time.
Essential Knowledge
MONEY-TIME RELATIONSHIPS AND EQUIVALENCE
The term capital refers to wealth in the form of money or property that can be used to
produce more wealth. Majority of engineering economic studies involve commitment of capital
for extended periods of time, so the effect of time must be considered. In this regard, it is
recognized that a peso today is worth more than a peso years from now because of the interest
(or profit) it can earn. Money, therefore, has a time value.
Capital in the form of money for the people, machines, materials, energy, and other things
needed in the operation of an organization may be classified into two basic categories. Equity
capital is that owned by individuals who have invested their money or property in a business
project or venture in the hope of receiving a profit. Debt capital, often called borrowed capital,
is obtained from lenders for investment. In return, the lenders receive interest from the
borrowers.
There are fundamental reasons why return to capital in the form of interest and profit is
an essential ingredient of engineering economic studies. First, interest and profit pay the
providers of capital for forgoing its use during the time the capital is being used. The fact that the
supplier can realize a return on capital acts as an incentive to accumulate capital by savings, thus
postponing immediate consumption in favor of creating wealth in the future. Second, interest
and profit are payments for the risk the investor takes in permitting another person, or an
organization, to use his or her capital.
Interest and the Time Value of Money
The terms interest, interest period, and interest rate are useful in calculating equivalent
sums of money for one interest period in the past and one period in the future.
However, for more than one interest period, the terms simple interest and compound
interest become important
Simple Interest
When the total interest earned or charged is linearly proportional to the initial amount of
the loan (principal), the interest rate, and the number of interest periods for which the principal
is committed, the interest and the interest rate are said to be simple.
Simple Interest (denoted as I) is defined as the interest on a loan or principal that is based
only on the original amount of the loan or principal. This is usually used for short-term loans
where the period of the loan is measured in days rather than years. Simple interest is not used
frequently in modern commercial practice. It can be calculated using the formula:
I = Pin
Equation 1
where:
P = principal/loan
I = interest
i = interest rate
n = period
The future amount of the principal may be calculated by adding the interest (I) to the
principal (P).
𝐹 = 𝑃 + 𝐼
𝐹 = 𝑃 + 𝑃𝑖𝑛
𝑭 = 𝑷(𝟏 + 𝒊𝒏)
Equation 2
There are two types of simple interest namely, ordinary simple interest and exact simple interest.
Ordinary simple interest is based on one banker’s year. A banker year is composed of 12 months
of 30 days each which is equivalent to a total of 360 days in a year. The value of n that is used in
Equations 1 and 2 may be calculated as
𝑑
𝑛 =
360
Exact simple interest is based on the exact number of days in a given year. A normal year
has 365 days while a leap year has 366 days. Unlike the ordinary simple interest where each
month has 30 days, in this type of simple interest, the number of days in a month is based on the
actual number of days each month contains following the Gregorian calendar. The value of n that
is used in Equations 1 and 2 may be calculated as
𝑑
𝑛 =
for ordinary year, and
365
𝑑
𝑛 =
for leap year
360
Example 1:
Problem: ECE Board April 2001
if Php 1,000 accumulates to Php 1,500 when invested at a simple interest for three
years, what is the rate of interest?
Given:
P = Php 1,000
F = Php 1,500
n=3
Required:
i=?
Solution:
F = P(1 + in)
1500 = 1000[1 + i(3)]
i = 0.1667
i = 16.67%
Answer
Example 2:
Problem: ECE Board November 1999
You loaned from a loan firm an amount of Php 100,000 with a rate of simple interest of
20% but the interest was deducted from the loan at the time the money was borrowed. If at the
end of one year, you have to pay the full amount of Php 100,000, what is the actual rate of
interest?
Given:
P = Php 100,000
n=1
Required:
i=?
Solution:
I = 0.20(100,000)
I = 20,000
Amount Received = 100,000 – 20,000
Amount Received = 80,000
I = Pin
20,000 = 80,000(i)(1)
i = 0.25
i = 25%
Answer
Example 3:
Problem: ECE Board November 1999
Mr. J. Reyes borrowed money from the bank. He received from the bank Php 1,842 and
promise to repay Php 2,000 at the end of 10 months. Determine the simple interest.
Given:
P = Php 1,842
F = Php 2,000
Required:
i=?
Solution:
𝑛=
10
12
F = P(1 + in)
2,000 = 1,842[1 + i(10/12)]
i = 0.1029
i = 10.29%
Answer
Example 4:
Problem:
On her recent birthday, April 22, 2019, Nicole was given by her mother a certain sum of
money as birthday present. She decided to invest the said amount on 20% exact simple interest.
If the account will mature on Christmas day at an amount of Php 10,000, how much did Nicole
receive from her mother on her birthday?
Given:
F = Php 10,000
i = 20%
Required:
P=?
Solution:
Solving for the total number of days the money was invested:
April 22 – 30
=8
May
= 31
June
= 30
July
= 31
August
= 31
September
= 30
October
= 31
November
= 30
December 1 – 25
= 25
247 days
F = P(1 + in)
10,000 = P[1 + (0.20)(247/365)]
P = Php 8,807.92
Answer
SOLVE IT!
Solve the following:
1. A deposit of Php 110,000 was made for 31 days. The net interest after deducting 20%
withholding tax is Php 890.36. Find the rate of return annually.
2. It is the practice of almost all banks in the Philippines that when they grant a loan, the
interest for one year is automatically deducted from the principal amount upon release
of money to the borrower. Let us therefore assume that you applied for a loan with a
bank and the Php 80,000 was approved at an interest rate of 14% of which Php 11,200
was deducted and you were given a check of Php 68,800. Since you have to pay the
amount of Php 80,000 one year after, what then will be the effective rate of interest?
3. If you borrow money from your friend with simple interest of 12%, find the present worth
of Php 20,000, which is due at the end of nine months.
4. Determine the exact simple interest on Php 5,000 invested for the period from January
15, 2016 to October 12, 2016, if the rate of interest is 18%.
Compound Interest
Compound interest is defined as the interest of loan or principal which is based not only
on the original amount of the loan or principal but the amount of loan or principal plus the
previous accumulated interest. This means that aside from the principal, the interest now earns
interest as well. Thus, the interest charges grow exponentially over a period of time.
Compound interest is frequently used in commercial practice than simple interest, more
especially if it is a longer period which spans for more than a year.
The future amount of the principal may be derived by the following tabulation:
Period
1
2
3
n
Principal
P
P(1 + i)
P(1 + i)2
Interest
Pi
P(1 + i)i
P(1 + i)2i
Total Amount
P + Pi = P(1 + i)
P(1 + i)(1 + i) = P(1 + i)2
P(1 + i)2(1 + i) = P(1 + i)3
P(1 + i)n
Table 4. Compound Interest
The tabulation above shows that the future amount (total amount) is just the value P(1 +
i) with an exponent which is numerically equal to the period.
It is also observed that compound interest is based on the principles of geometric
progression and using such method, the total amount after each period are as follows:
First term,
a1 = P(1 + i)
Second term, a2 = P(1 + i)2
Third term, a3 = P(1 + i)3 and so on.
Solving for the common ratio,
𝑟=
𝑟=
𝑎2
𝑎1
𝑃(1+𝑖)2
𝑃(1+𝑖)
𝑟 =1+𝑖
Using the formula for nth term of geometric progression:
an = a1rn-1
an = P(1 + i)(1 + i)n-1
an = P(1 + i)n
a. FUTURE AMOUNT, F:
𝑭 = 𝑷(𝟏 + 𝒊)𝒏
where:
P = principal
i = interest per period(in decimal)
n = number of interest periods
(1 + i)n = single payment compound amount factor
0
1
2
3
P
4
n
F
Cash Flow
Equation 3
b. PRESENT WORTH, P:
𝑭
𝑷=
Equation 4
(𝟏+𝒊)𝒏
where:
1
(1+𝑖)𝑛
0
= single payment present worth factor
1
2
P
3
Cash Flow
4
n
F
Continuous Compounding
The concept of continuous compounding is based on the assumption that cash payments
occur once per year but compounding is continuous throughout the year.
The basic equation for future worth of compound interest is
𝐹 = 𝑃(1 + 𝑖)𝑛 .
For m periods per year,
𝑁𝑅
𝐹 = 𝑃(1 + 𝑚 )𝑚𝑁 .
𝑚
Let x = 𝑁𝑅 ,
1
𝐹 = 𝑃(1 + 𝑥)𝑥(𝑁𝑅)(𝑁)
1
𝐹 = 𝑃[(1 + 𝑥)𝑥 ](𝑁𝑅)(𝑁) .
But,
1 𝑥
lim (1 + 𝑥) = 𝑒.
𝑥→∞
Therefore,
𝐹 = 𝑃𝑒 (𝑁𝑅)(𝑁)
where:
P = principal
𝑒 = 2.71828…
NR = nominal rate
N = number of years
(𝑁𝑅)(𝑁)
𝑒
= continuous compounding amount factor
The present worth of continuous compounding is
𝑃=
𝐹
𝑒 𝑁𝑅(𝑁)
Nominal and Effective Rates of Interest
Rate of interest is the cost of borrowing money. It also refers to the amount earned by a
unit principal per unit time.
There are two types of rates of interest, namely the nominal rate of interest and the
effective rate of interest.
Nominal rate of interest is defined as the basic annual rate of interest while effective rate
of interest is defined as the actual or the exact rate of interest earned on the principal during a
one-year period.
For example: A principal is invested at 5% compounded quarterly. In this statement, the
nominal rate is 5% while the effective rate is greater than 5% because of the compounding which
occurs four times a year. The following formula is used to determine the effective rate of interest:
𝐸𝑅 = [1 + 𝑖]𝑚 − 1
Equation 5
where:
m = number of interest period per year
i
= interest per period =
𝑁𝑅
𝑚
NR = nominal rate of interest
or,
𝐸𝑅 = [1 +
𝑁𝑅 𝑚
]
𝑚
−1
If the mode of compounding is annual, then i = NR.
Substituting the values of m and i:
𝐸𝑅 = [1 +
0.05 4
]
4
−1
𝐸𝑅 = 0.0509
𝐸𝑅 = 5.09%
Example 1:
Problem: ECE Board November 2000, ECE Board April 1999
The amount of Php 50,000 was deposited in the bank earning an interest of 7.5% per
annum. Determine the total amount at the end of 5 years if the principal and interest were not
withdrawn during the period.
0
1
2
3
50,000
0
Given:
4
5
F
P = Php 50,000
i = 7.5%
n=5
Required:
F=?
Solution:
F = P(1 + i)n
F = 50,000(1 + 0.075)5
F = Php 71,781.47
Answer
Example 2:
Problem: ME Board October 1999
If Php 5,000 shall accumulate for 10 years at 8% compounded quarterly, then what is the
compound interest at the end of 10 years?
Given:
P = Php 5,000
m = 4 (Note: there are 4 quarters in 1 year)
N = 10
i = 8%
Required:
I=?
Solution:
F = P(1 + i)n
F = 5,000(1 +
0.08 4(10)
)
4
F = Php 11,040.20
I=F–P
I = 11,040.20 – 5,000
I = Php 6,040.20
Answer
Example 3:
Problem: CE Board May 1998
Compute the equivalent rate of 6% compounded semi-annually to a rate compounded
quarterly.
Given:
i = 6%
Quarterly: m = 4
Semi-annually: m = 2
Required:
i =? when m = 4
Solution:
ERquarterly = ERsemi-annually
𝑖
(1 + 4)4 − 1 = (1 +
i = 0.0596
i = 5.96%
0.06 2
2
) −1
Answer
Example 4:
Problem: ME Board April 1998
Php 1,500 was deposited in a bank account 20 years ago. Today, it is worth Php 3,000.
Interest is paid semi-annually. Determine the interest rate paid on this account.
Given:
P = Php 1,500
F = Php 3,000
N = 20
Required:
NR = ?
Solution:
0
1,500
F = P(1 +
𝑁𝑅 mN
)
𝑚
1
2
3
………
20
3,000
3,000 = 1,500(1 +
NR = 0.035
NR = 3.5%
𝑁𝑅 2(20)
)
2
Answer
Example 4:
Problem: ME Board October 1997
When will an amount be tripled with an interest of 11.56%?
Given:
i = 11.56%
Required:
n= ?, when P becomes F=3P
Solution:
F = P(1 = i)n
3P = P(1 + 0.1156)n
3 = 1.1156n
Take log on both sides:
log 3 = n log 1.1156
log 3
n = log 1.1156
n = 10.04 years
Answer
Example 5:
Problem: ME Board October 1997
A nominal interest of 3% compounded continuously is given on the account. What is the
accumulated amount of P 10,000 after 10 years?
Given:
NR = 3%
P = Php 10,000
N = 10
Required:
F=?
Solution:
F = Pe(NR)N
F = 10,000(0.03)(10)
F = Php 13,498.59
Answer
Example 6:
Problem: ME Board April 1996
A firm borrows Php 2,000 for 6 years at 8%. At the end of 6 years, it renews the loan for
the amount due plus Php 2,000 more for 2 years at 8%. What is the lump sum due?
Given:
P = Php 2,000
i = 8%
n=8
Required:
Plump sum due = ?
Solution:
0
1
2
3
………
6
2,000
2,000
F2
F1
Solving for F1:
F = P(1 + i)n
F1 = 2,000(1 + 0.08)8
F1 = 3,701.86
Solving for F2:
F2 = 2,000(1 + 0.08)2
F2 = 2,332.80
Total Future Worth = F1 + F2
Total Future Worth = 6,034.66
Solving for the present worth of the total future worth:
𝑃=
𝑃=
𝐹
(1+𝑖)𝑛
6,034.66
(1+0.08)8
P = Php 3,260.34
Answer
SOLVE IT!
Solve the following:
1. What is the effective rate corresponding to 18% compounded daily? Take 1 year is equal
to 360 days.
2. Which of these gives the lowest effective rate of interest?
a. 12.35% compounded annually
b. 11.90% compounded semi-annually
c. 12.20% compounded quarterly
d. 11.60% compounded monthly
3. How many years will Php 100,000 earn a compound interest of Php 50,000 if the interest
rate is 9% compounded quarterly?
4. Mandarin Bank advertises 9.5% account that yields 9.84% annually. Find how often the
interest is compounded.
5. Alexander Michael owes Php 25,000 due in one year and Php 75,000 due in four years.
He agrees to pay Php 50,000 today and the balance in two years. How much must he pay
at the end of two years if money is worth 5% compounded semi-annually?
The Concept of Equivalence
Alternative should be compared as far as possible when they produce similar results,
serve the same purpose, or accomplish the same function, though this is not always possible in
some types of economy studies.
How can alternatives for providing the same service or accomplishing the same function
be compared when interest is involved over extended periods of time? Thus, we should consider
the comparison of alternative options, or proposals, by reducing them to an equivalent basis that
is dependent on (1) the interest rate, (2) the amounts of money involved, (3) the timing of
monetary receipts and/or expenses, and (4) the manner in which the interest, or profit, on
invested capital is paid and the initial capital recovered.
To better understand the mechanics of interest and to expand on the notion of economic
equivalence, consider a situation in which we borrow Php 8,000 and agree to repay it in four
years at an interest rate of 10% per year. There are many plans by which the principal of this loan
and the interest on it can be repaid. For simplicity, we have selected four plans to demonstrate
the idea of economic equivalence. Here equivalence means that all four plans are equally
desirable to the borrower. In each that interest rate is 10% per year and the original amount
borrowed is Php 8,000; thus, differences among the plans rest with items (3) and (4) above. The
four plans are shown in Table 5.
(1)
(2)
(3) = 10% x (2) (4) = (2) + (3)
(5)
Amount Owed
Interest
Total Money
at Beginning
Accrued for
Owed at
Principal
Year
of Year
Year
End of Year
Payment
Plan 1: At End of Each Year Pay Php 2,000 Principal Plus Interest Due
(6) = (3) + (5)
Total
End-of-Year
Payment (Cash Flow)
1
2
3
4
Php 2,800
2,600
2,400
2,200
Php 10,000
(total amount repaid)
Php 8,000
6,000
4,000
2,000
20,000 Php-yr
Php 800
600
400
200
Php 2,000
(total interest)
Php 8,800
6,600
4,400
2,200
Php 2,000
2,000
2,000
2,000
Php 8,000
Plan 2: Pay Interest Due at End of Each Year and Principal at End of Four Years
1
2
3
4
Php 8,000
8,000
8,000
8,000
32,000 Php-yr
Php 800
800
800
800
Php 3,200
(total Interest)
Php 8,800
8,800
8,800
8,800
Php
0
0
0
8,000
Php 8,000
Php
800
800
800
8,800
Php 11,200
(total amount repaid)
Php 1,724
1,896
2,086
2,294
Php 8,000
Php
Plan 3: Pay in Four Equal End-of-Year Payments
1
2
3
4
Php 8,000
6,276
4,380
2,294
20,960 Php-yr
Php
800
628
438
230
Php 2,096
(total interest)
Php 8,800
6,904
4,818
2,524
2,524
2,524
2,524
2,524
Php 10,096
(total amount repaid)
Plan 4: Pay Principal and Interest in One Payment at End of Four Years
1
2
3
4
Php 8,000
8,800
9,680
10,648
Php 37,130 Php-yr
Php
800
880
968
1,065
Php 3,713
(total interest)
Php 8,800
9,680
10,648
11,713
Php
0
0
0
8,000
Php 8,000
Php
0
0
0
11,713
Php 11,713
(total amount repaid)
Table 5. Four Plans for Repayment of Php 8,000 in Four Years with Interest at 10% Per Year
In Plan 1, Php 2,000 of the loan principal is repaid at the end of each years one through
four. As a result, the interest we repay at the end of a particular year is affected by how much
we syill owe on the loan at the beginning of that year. Our end-of-year payment is just the sum
of Php 2,000 and interest computed on the beginning-of-year amount owed.
Plan 2 indicates that none of the loan principal is repaid until the end of the fourth year.
Our interest cost each year is Php 800, and it is repaid at the end of years one through four.
Because interest does not accumulate in either Plan 1 or Plan 2, compounding of interest is not
present. Notice that Php 3,200 in interest paid in Plan 2, whereas only Php 2,000 is paid in Plan
1. We had the use of the Php 8,000 principal for four years in Plan 2 but, on average, had the use
of much less than Php 8,000 in Plan 1.
Plan 3 requires that we repay equal end-of-year amounts of Php 2,524 each. Observe that
the four end-of-year payments in Plan 3 completely repay the Php 8,000 loan principal with
interest at 10% per year.
Finally, Plan 4 shows that no interest and no principal are repaid for the first three years
of the loan period. Then at the end of the fourth year, the original loan principal plus the
accumulated interest for four years is repaid in a single lump-sum amount of Php 11,712.80
(rounded in the table). Plan 4 involves compound interest. The total amount of interest repaid in
Plan 4 is highest of all the plans considered. Not only was the principal repayment in Plan 4
deferred until the end of year four, but we also deferred all interest payment until that time.
Going back to the notion of economic equivalence, if interest rates remain constant at
10% for the plans shown in Table 5, all four plans are equivalent. This assumes that one can freely
borrow and lend at the 10% rate. Hence, we would be indifferent about whether the principal is
repaid early in the loan’s life (e.g. Plans 1 and 3) or repaid at the end of year four (e.g. Plans 2 and
4). Economic equivalence is established, in general, when we are indifferent between a future
payment, or series of future payments, and a present sum of money.
When total peso-years are calculated for each plan and divided into total interest paid
over the four years (the sum of column 3), the ratio is found to be constant:
Plan
1
2
3
4
Area Under Curve
(Peso-Years)
(Sum of Column 2 in
Table 5)
Php 20,000
32,000
20,960
37,130
Total
Interest Paid
(Sum of Column 3 in
Table 5)
Php 2,000
3,200
2,096
3,713
Ratio of
Total Interest
to Peso-Years
0.10
0.10
0.10
0.10
Table 6. Total interest to Peso-Years Ratio for All Four Plans
Because the ratio is constant at 0.10 for all plans, we can deduce that all repayment
methods considered in Table 5 are equivalent, even though each involves a different total endof-year payment in column 6. Dissimilar peso-years of borrowing, by itself does not necessarily
mean that different loan repayment plans are or are not equivalent. In summary, equivalence is
established when total interest paid, divided by peso-years of borrowing, is a constant ratio
among financing plans.
Notation and Cash Flow Diagram/Tables
The following notation is utilized in formulas for compound interest calculations:
i =
N =
P =
F
=
A =
effective interest rate per interest period
number of compounding periods
present sum of money; the equivalent value of one of more cash flows at a
reference point in time called the present
future sum of money; the equivalent value of one or more cash flows at a
reference point in time called the future
end-of-period cash flows (or equivalent end-of-period values) in a uniform series
continuing for a specified number of periods, starting at the end of the first period
and continuing through the last period
Cash flows are important in engineering economy because they form the basis for
evaluating alternatives. The use of cash flow (time) diagrams and/or tables is strongly
recommended for situations in which the analyst needs to clarify or visualize what is involved
when flows of money occur at various times. In addition, viewpoint (Principle 3) is an essential
feature of cash flow diagrams.
The difference between total cash inflows (receipts) and cash outflows (expenditures) for
a specified period of time is the net cash flow for the period.
Figure 3 shows a cash flow diagram for Plan 4 of Table 5, and Figure 4 depicts the net cash
flow of Plan 3. These two figures also illustrate the definition of the preceding symbols and their
placement on a cash flow diagram. Notice that all cash flows have been placed at the end of the
year to correspond with the convention used in Table 5. In addition, a viewpoint has been
specified.
Figure 3. Cash Flow Diagram for Plan 4 of Table 5
Figure 4. Cash Flow Diagram for Plan 3 of Table 5
The cash flow diagram employs several conventions:
1. The horizontal line is a time scale, with progression of time moving from left to right.
The period (e.g. year, quarter, month) labels can be applied to intervals of time rather
than to points on the time scale.
2. The arrows signify cash flows and are placed at the end of the period. If a distinction
needs to be made, downward arrows represent expenses (negative cash flows or cash
outflows) and upward arrows represent receipts (positive cash flows or cash inflows).
3. The cash flow diagram is dependent on the point of view. For example, the situations
shown in Figures 3 and 4 were based on cash flow as seen by the lender. If directions
of all arrows have been reversed, the problem would have been diagrammed from
the borrower’s point of view.
QUIZ 1
Before evaluating the economic merits of a proposed investment, the XYZ Corporation
insists that its engineers develop a cash flow diagram of the proposal. An investment of Php
10,000 can be made that will produce uniform annual revenue of Php 5,310 for five years and
then have a market value of Php 2,000 at the end of year five. Annual expenses will be Php 3,000
at the end of each year for operating and maintaining the project. Draw a cash flow diagram for
the five-year life of the project. Use the corporation’s viewpoint.
Key Points
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Engineering economy involves the systematic evaluation of the economic
merits of proposed solution to engineering problems.
The foundation of the discipline of Engineering Economy can be seen in terms
of seven principles which include (1) develop the alternatives, (2) focus on
the differences, (3) use a consistent viewpoint, (4) use a common unit of
measure, (5) consider all relevant criteria, (6) make risk and uncertainty
explicit, and (7) revisit your decisions.
The engineering economic analysis procedure involves seven steps: (1)
problem recognition, definition, and evaluation, (2) development of the
feasible alternatives, (3) development of the outcomes and cash flows for
each alternative, (4) selection of a criteria, (5) analysis and comparison of the
alternatives, (6) selection of the preferred alternative, and (7) performance
monitoring and post-evaluation results.
Fixed costs are those unaffected by changes in activity level over a feasible
range of operations for the capacity or capability available.
Variable costs are those associated with an operation that vary in total with
the quantity of output or other measures of activity level.
An incremental cost, or incremental revenue, is the additional cost, or
revenue, that results from increasing the output of a system by one (or more)
units.
Recurring costs are those that are repetitive and occur when an organization
produces similar goods or services on a continuing basis.
Nonrecurring costs are those that are not repetitive, even though the total
expenditure may be cumulative over a relatively short period of time.
Direct costs are those that can be reasonably measured and allocated to a
specific output or work activity.
Indirect costs are those that are difficult to attribute or allocate to a specific
output or work activity.
Overhead costs consist of plant operating costs that are not direct labor or
direct material costs.
Standard costs are representative costs per unit of output that are
established in advance of actual production or service delivery.
A cost that involves payment of cash is called a cash cost.
A book cost is one that does not involve a cash transaction and is reflected in
the accounting system as a noncash cost.
A sunk cost is one that has occurred in the past and has no relevance to
estimates of future costs and revenues related to an alternative course of
action.
An opportunity cost is incurred because of the use of limited resources, such
that the opportunity to use those resources to monetary advantage in an
alternative use is foregone.
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Life-cycle cost refers to a summation of all costs, both recurring and
nonrecurring, related to a product, structure, system, or service during its life
span.
When the influence of time on money is not a significant consideration, cost
analyses are usually called present economy studies.
The term capital refers to wealth in the form of money or property that can
be used to produce more wealth.
Equity capital is that owned by individuals who have invested their money or
property in a business project or venture in the hope of receiving a profit.
Debt capital, often called borrowed capital, is obtained from lenders for
investment. In return, the lenders receive interest from the borrowers.
When the total interest earned or charged is linearly proportional to the
initial amount of the loan (principal), the interest rate, and the number of
interest periods for which the principal is committed, the interest and the
interest rate are said to be simple.
There are two types of simple interest namely, ordinary simple interest and
exact simple interest.
Compound interest is defined as the interest of loan or principal which is
based not only on the original amount of the loan or principal but the amount
of loan or principal plus the previous accumulated interest.
The concept of continuous compounding is based on the assumption that
cash payments occur once per year but compounding is continuous
throughout the year.
Rate of interest is the cost of borrowing money.
There are two types of rates of interest, namely the nominal rate of interest
and the effective rate of interest.
Nominal rate of interest is defined as the basic annual rate of interest while
effective rate of interest is defined as the actual or the exact rate of interest
earned on the principal during a one-year period.
Economic equivalence is established, in general, when we are indifferent
between a future payment, or series of future payments, and a present sum
of money.
The difference between total cash inflows (receipts) and cash outflows
(expenditures) for a specified period of time is the net cash flow for the
period.
BIG PICTURE
Week 4 – 5: Unit Learning Outcomes (ULO)
At the end of the unit, you are expected to:
a. solve engineering economy problems involving annuity; and
b. apply money-time relationships.
Big Picture in Focus: ULOa. Solve engineering economy problems involving annuity.
Essential Knowledge
ANNUITY
Annuity is defined as a series of equal payments occurring at equal interval of time. When
an annuity has a fixed time span, it is known as annuity certain. The following are annuity certain:
1. Ordinary Annuity.
It is a type of annuity where the payments are made at the end of each period
beginning from the first period.
Derivation of formula for the sum of ordinary annuity:
Let A be the periodic or uniform payment and assuming only four payments:
Annuity is based on the principles of compound interest. Hence, computation of
the sum of annuity may be done using the formulas for geometric progression.
Solving for common ratio:
𝑎2
𝑟=
𝑎1
𝐴(1 + 𝑖)
𝑟=
𝐴
𝑟 =1+𝑖
Solving for the sum:
𝑎1 (𝑟 𝑛 − 1)
𝑆=
𝑟−1
𝐴[(1 + 𝑖 )4 − 1]
𝑆=
1+𝑖−1
𝐴[(1 + 𝑖 )4 − 1]
𝑆=
𝑖
A. SUM OF ORDINARY ANNUITY:
0
1
2
3
A
A
A
…
n
A
F
𝐹=
𝐴[(1+𝑖)𝑛 −1]
𝑖
where:
i
n
A
𝑛
𝐴[(1+𝑖) −1
𝑖
= interest period
= number of periods
= uniform payment
= uniform series compound factor
Equation 6
B. PRESENT WORTH OF ORDINARY ANNUITY:
0
1
2
3
A
A
A
…
n
A
F
P
Using compound interest formula:
𝐹
𝑃=
(1 + 𝑖)𝑛
But:
𝐹=
𝐴[(1 + 𝑖 )𝑛 − 1]
𝑖
Substituting the value of F:
𝑃=
𝐴[(1+𝑖)𝑛 −1]
𝑖(1+𝑖)𝑛
Equation 7
where:
[(1+𝑖)𝑛 −1]
𝑖(1+𝑖)𝑛
= uniform series present worth factor
2. Annuity Due
It is a type of annuity where the payments are made at the beginning of each
period starting from the first period.
1
0
3
2
… n-1
n
Cash Flow of Annuity Due
A
A
A
A
A
3. Deferred Annuity
It is a type of annuity where the first payment does not begin until some later
date in the cash flow.
0
1
2
3
4
5
6
A
A
A
A
…
n
A
Cash Flow of Deferred Annuity
When an annuity does not have a fixed time span but continues indefinitely, then it is
referred to as a perpetuity. The sum of a perpetuity is an infinite value.
PRESENT WORTH OF PERPETUITY:
𝑃=
where:
i = interest per period
A = uniform payment
𝐴
𝑖
Equation 8
0
1
2
3
4
A
A
A
A
…
∞
A
P
Cash Flow of Perpetuity
Example 1:
Problem: ECE Board April 2001
What is the present worth of a Php 500 annuity starting at the end of the third year and
continuing to the end of the fourth year, if the annual interest rate is 10%?
Given (Cash Flow Diagram):
0
1
P
2
P1
Required:
P=?
Solution:
Solving for P1, the present worth of ordinary annuity:
𝐴[(1 + 𝑖 )𝑛 − 1]
𝑃1 =
𝑖(1 + 𝑖)𝑛
500[(1 + 0.10)2 − 1]
0.10(1 + 0.10)2
𝑃1 = 𝑃ℎ𝑝 867.77
𝑃1 =
3
4
500
500
Solving for P:
P1 is the future worth of P with n=2.
𝐹
𝑃=
(1 + 𝑖)𝑛
𝑃1
(1 + 𝑖)𝑛
867.77
𝑃=
(1 + 0.10)2
𝑷 = 𝑷𝒉𝒑 𝟕𝟏𝟕. 𝟏𝟕
𝑃=
Answer
Another Solution:
Cash Flow Diagram:
0
P1
P2
Present worth = P1 + P2
𝑃=
𝐹
(1 + 𝑖)𝑛
Solving for P1:
𝑃1 =
500
(1 + 0.10)3
𝑃1 = 𝑃ℎ𝑝 375.66
Solving for P2:
500
(1 + 0.10)4
𝑃2 = 𝑃ℎ𝑝 341.51
Solving for the present worth:
Present Worth = P1 + P2
𝑃2 =
1
2
3
4
500
500
Present Worth = 375.66 + 341.51
Present Worth = Php 717.17
Answer
Example 2:
Problem: EE Board October 1999
A manufacturer desires to set aside a certain sum of money to provide funds to cover the
yearly operating expenses and the cost of replacing every year the dyes of a stamping machine
used in making radio chassis as model changes for a period of 10 years.
Operating cost per year
=
Php 500.00
Cost of dye
=
Php 1,200.00
Salvage value of dye
=
Php 600.00
The money will be deposited in a saving account which earns 6% interest. Determine the
sum of money that must be provided, including the cost of the initial dye.
Given (Cash Flow Diagram):
C
0
A
C
C
2
C
C
4
C
C
C
C
8
6
10
A
A
A
A
A
A
A
A
A
B
B
B
B
B
B
B
B
B
where:
A = Php 1,200
B = Php 500
C = Php 600
Required:
Amount to set aside = ?
C
B
Solution:
From the cash flow diagram:
Let D = A + B – C
D = 1,200 + 500 – 600
D = 1,100
Reduced Cash Flow Diagram (Equivalent Cash Flow Diagram):
P2
100
0
2
D
D
4
D
A
P1
Using the equivalent (reduced) cash flow:
Amount to set aside = A + P1 – P2
Solving for P1:
𝐷[(1 + 𝑖 )𝑛 − 1]
𝑃1 =
𝑖(1 + 𝑖)𝑛
1100[(1 + 0.06)9 − 1]
𝑃1 =
0.06(1 + 0.06)9
𝑃1 = 𝑃ℎ𝑝 7,481.86
Solving for P2:
𝐹
𝑃2 =
(1 + 𝑖)𝑛
𝑃2 =
100
(1 + 0.06)10
𝑃2 = 𝑃ℎ𝑝 55.84
Therefore,
D
8
6
D
D
D
D
10
D
Amount to set aside = A + P1 – P2
Amount to set aside = 1200 + 7481.86 – 55.84
Amount to set aside = Php 8,626.02
Answer
Example 3:
Problem: ME Board October 1997
Maintenance cost of an equipment is Php 20,000 for 2 years, Php 40,000 at the end of 4
years and Php 80,000 at the end of 8 years. Compute the semi-annual amount that will be set
aside for this equipment. Money is worth 10% compounded annually.
Given (Cash Flow Diagram):
0
1
P1
3
2
4
5
80,000
at i = 10% compounded annually
Required:
Asemi-annual to set aside = ?
Solution:
Solving for the present worth of maintenance cost:
P = P1 + P2 + P3
EQ.1
𝐹
𝑃=
(1 + 𝑖)𝑛
Solving for P2:
40000
𝑃2 =
(1 + 0.10)4
8
40,000
P3
𝑃1 = 𝑃ℎ𝑝 16,528.92
7
20,000
P2
Solving for P1:
20000
𝑃1 =
(1 + 0.10)2
6
𝑃2 = 𝑃ℎ𝑝 27,320.54
Solving for P3:
80000
𝑃3 =
(1 + 0.10)8
𝑃3 = 𝑃ℎ𝑝 37,320.59
Substitute P1, P2, and P3 in EQ. 1:
P = 16,528.93 + 27,320.54 + 37,320.59
P = Php 81,170.06
Solving for the semi-annual amount to set aside for the equipment, A:
0
A
1
2
A
A A
3
A
A
4
A
P
𝑃=
𝐴[(1 + 𝑖 )𝑛 − 1]
𝑖(1 + 𝑖)𝑛
Solving for the semi-annual interest:
ERsemi-annual = ERannual
(1 + i)2 – 1 = (1 + 0.10)1 – 1
1 + i = 1.0488
i = 0.0488
i = 4.88%
Substituting the values:
𝐴[(1 + 0.0488)16 − 1]
81,170.06 =
0.0488(1 + 0.0488)16
𝑨 = 𝑷𝒉𝒑 𝟕, 𝟒𝟐𝟓. 𝟕𝟐
Answer
A
5
A
A
A
6
7
A
A A
8
A
A
Example 4:
Problem: ME Board April 1997
Mr. Ramirez borrowed Php 15,000 two years ago. The terms of the loan are 10% interest
for 10 years with uniform payments. He just made his second annual payment. How much
principal does he still owe?
Given (Cash Flow Diagram):
15,000
2
4
8
6
10
0
A
A
A
A
A
A
A
A
A
A
P
at i = 10%
Required:
Balance = ?, after paying 2 annual payments
Solution:
Solving for the annual amortization, A:
𝐴[(1 + 𝑖 )𝑛 − 1]
𝑃=
𝑖(1 + 𝑖)𝑛
But P = 15,000
𝐴[(1 + 0.10)10 − 1]
15,000 =
0.10(1 + 0.10)10
A = Php 2,441.18
After paying two annual payments,
15,000
F1
2
4
8
6
10
0
A
A
F2
A
A
A
A
A
A
A
A
F = P(1 + i)n
Solving for F1:
F1 = 15,000(1 + 0.10)2
F1 = Php 18,150
Solving for F2:
𝐴[(1 + 𝑖 )𝑛 − 1]
𝐹2 =
𝑖
2,441.18[(1 + 0.10)2 − 1]
𝐹2 =
0.10
F2 = Php 5,126.48
Solving for the Balance:
Balance = F1 – F2
Balance = 18,150 – 5,126.48
Balance = Php 13,023.52
Answer
Example 5:
Problem: ME Board June 1990
A machinery supplier is offering a certain machinery on a 10% downpayment and the
balance payable in equal end of year payments without interest for 2 years. Under this
arrangement, the price is pegged to be Php 250,000. However, for cash purchase, the machine
would only cost Php 195,000. What is the equivalent interest rate that is being charged on the 2year payment plan if interest is compounded quarterly?
Given:
If payable for two years,
Cost of Machine = Php 250,000
Downpayment = 10% = 0.10(250,000) = Php 25,000
If paid in cash,
Cost of Machine = Php 195,000
Required:
NR = ?, compounded quarterly for the 2-year payment plan
Solution:
For the 2-year payment plan, since no interest is considered for the balance, the annual payment,
A, is:
𝐴 =
𝐵𝑎𝑙𝑎𝑛𝑐𝑒
2
250,000 − 25,000
𝐴 =
2
A = Php 112,500
For cash payment, solving for interest, i:
Cash Price
1
2
A
A
0
25,000
A
P
P + 25,000 = Cash Price
P + 25,000 = 195,000
P = Php 170,000
𝐴[(1 + 𝑖 )𝑛 − 1]
𝑃=
𝑖(1 + 𝑖)𝑛
170,000 =
112,500[(1 + 𝑖 )2 − 1]
𝑖(1 + 𝑖)2
By trial and error:
i = 0.209
i = 20.9%
NR = 20.9% compounded annually
Solving for NR compounded quarterly:
ERquarterly = ERannually
𝑁𝑅
(1 + 4 )4 − 1 = (1 + 0.209)1 − 1
(1 +
1+
𝑁𝑅 4
)
4
𝑁𝑅
4
= 1.209
= 1.0486
NR = 0.1947
NR = 19.47% compounded quarterly
Answer
SOLVE IT!
Solve the following:
1. If Php 10,000 is deposited each year for 9 years, how much annuity can a person get
annually from the bank every year for 8 years starting 1 year after the 9th deposit is made?
Cost of money is 14%
2. A man inherited a regular endowment of Php 100,000 every end of 3 months for n years.
However, he may choose to get a single lump sum of Php 3,702,939.80 at the end of 4
years. If the rate of interest was 14% compounded quarterly, what is the value of n?
3. You need Php 4,000 per year for four years to go to college. Your father invested Php
5,000 in a 7% account for your education when you were born. If you withdraw Php 4,000
at the end of your 17th, 18th, 19th and 20th birthday, how much will be left in the account
at the end of the 21st year?
4. A service car whose cash price was Php 540,000 was bought with a downpayment of Php
162,000 and monthly installment of Php 10,874.29 for 5 years. What was the rate of
interest if compounded monthly?
5. A company has approved a car plan for its six senior officers in which the company will
shoulder 25% of the cost and the difference payable by each officer to a financing
company in 48 equal end of the month installments at an interest rate of 1.5% per month.
if the cost of each car is Php 350,000, determine the following:
A. The total amount that each officer will have to shoulder.
B. The amount each officer has to pay the financing company per month.
Big Picture in Focus: ULOb. Apply money-time relationships.
Metalanguage
1. Disbursement. It is the payment of money from a fund.
Essential Knowledge
THE BASIC ECONOMY STUDY METHODS
All engineering economy studies of capital projects should consider the return that a given
project will or should produce. The basic question is whether a proposed capital investment and
its associated expenditures can be recovered by revenue (or savings) over time in addition to a
return on the capital that is sufficiently attractive in view of the risks involved and the potential
alternative uses. The interest and money-time relationships discussed in the previous unit
emerge as essential ingredients in answering that question.
Because patterns of capital investment, revenue (or savings) cash flows, and
disbursement cash flows can be quite different in various projects, there is no single method for
performing engineering economic analyses that is ideal for all cases. Consequently, several
methods are commonly used.
We will now concentrate on the correct use of five methods for evaluating the economic
profitability of a single proposed problem solution (i.e., alternative). The five methods are
present worth (PW), annual worth (AW), future worth (FW), internal rate of return (IRR), and
external rate of return (ERR). The first three methods convert cash flows resulting from a
proposed solution into their equivalent worth at some point (or points) in time by using an
interest rate known as the minimum attractive rate of return (MARR). The IRR and ERR methods
produce annual rates of profit, or returns, resulting from an investment, and are then compared
against the MARR.
A sixth method, the payback period, is a measure of the speed with which an investment
is recovered by the cash flows it produces. This measure, in its most common form, ignores time
value of money principles. For this reason, the payback method is often used to supplement
information produced by the five primary methods.
The Minimum Attractive Rate of Return
Business managers constantly consider investments for new products and capital
expenditures. But they need to have a measure that helps them determine if these new projects
are a worthwhile use of the company's funds. Managers evaluate capital expenditure projects by
calculating the internal rate of return (IRR) and comparing the results to the minimum acceptable
rate of return (MARR), also known as the hurdle rate. If the IRR exceeds the hurdle rate, it gets
approved. If not, management is likely to reject the project. However, these decision rules are
not rigid; other considerations might change the MARR. For example, management might decide
to use a lower MARR, say 10 percent, to approve new plants, but require a 20 percent MARR for
expansions to existing facilities. This is because all projects have different characteristics; some
have more uncertainty of future cash flows while others have shorter or longer time spans for
realizing the return on investment.
For most corporations, the MARR is the company's weighted average cost of capital
(WACC). The WACC is determined by the cost of obtaining the funds needed to pay for a project.
A company has access to funds by taking on additional debt, increasing equity capital or using
retained earnings. Each source of funds has a different cost. Interest rates on debt vary
depending on current economic conditions and the credit rating of the business. The cost of
equity capital is the return that stockholders demand for investing their money in the business.
The WACC is calculated by multiplying the proportions of debt and equity by their respective
costs to arrive at a weighted average.
While the WACC is the benchmark most commonly used as the MARR, it is not the only
one. If a company has an unlimited budget and access to capital, it can invest in any project that
meets the MARR. But with a limited budget, the opportunity cost of other projects becomes a
factor. Suppose the WACC of a company is 12 percent, and it has two projects: one has an IRR of
15 percent, and the other has an IRR of 18 percent. The IRR of both projects exceeds the MARR,
as defined by the WACC, and on this basis, management could authorize both projects.
In this case, the MARR becomes the highest IRR, 18 percent, of the available projects
under consideration. This IRR represents the "opportunity cost" that all other projects must be
compared to.
The Present Worth Method
The present worth (PW) method is based on the concept of equivalent worth of all cash
flows relative to some base or beginning point in time called the present. That is, all cash inflows
and outflows are discounted to the present point in time at an interest rate that is generally the
MARR.
The PW of an investment alternative is a measure of how much money an individual or a
firm could afford to pay for the investment in excess of its cost. Or, stated differently, a positive
PW or an investment project is a peso amount of profit over the minimum amount required by
investors. It is assumed that cash generated by the alternative is available for other uses that
earn interest at a rate equal to the MARR.
To find the PW as a function of i% (per interest period) of a series of cash inflows and
outflows, it is necessary to discount future amounts to the present by using the interest rate over
the appropriate study period (years, for example) in the following manner:
𝑃𝑊(𝑖%) = 𝐹0 (1 + 𝑖)0 + 𝐹1 (1 + 𝑖)−1 + 𝐹2 (1 + 𝑖)−2 + ⋯ + 𝐹𝑘 (1 + 𝑖 )−𝑘 + ⋯ + 𝐹𝑁 (1 + 𝑖)−𝑁
−𝑘
𝑃𝑊(𝑖%) = ∑𝑁
𝑘=0 𝐹𝑘 (1 + 𝑖)
Equation 9
where:
i
k
Fk
N
=
=
=
=
effective interest rate, or MARR, per compounding period
index for each compounding period (0 ≤ 𝑘 ≤ 𝑁)
future cash flow at the end of period k
number of compounding periods in the planning horizon
The relationship given in Equation 9 is based on the assumption of a constant interest
rate throughout the life of a particular project. The higher the interest rate and the further into
the future of a cash flow occurs, the lower its PW is. As long as the PW is greater than or equal
to zero, the project is economically justified; otherwise, it is not acceptable.
Example:
A piece of new equipment has been proposed by engineers to increase the productivity
of a certain manual welding operation. The investment cost is Php 25,000 and the equipment will
have a market value of Php 5,000 at the end of the study period of five years. Increased
productivity attributable to the equipment will amount to Php 8,000 per year after extra
operating costs have been subtracted from the revenue generated by the additional production.
If the firm’s MARR (before income taxes) is 20% per year, is this proposal a sound one? Use the
PW method.
Given (Cash Flow Diagram):
5,000
0
8,000
8,000
8,000
8,000
8,000
1
2
3
4
5
@ i = 20% per year
25,000
Required:
PW = ?, to justify the proposed equipment
Solution:
PW = PW of cash inflows – PW of cash outlflows
PW (20%) = 8,000(P/A, 20%, 5) + 5,000(P/F, 20%, 5) – 25,000
Note:
* 8,000(P/A, 20%, 5) is read as solving for the present value (P) given A(8,000) at 20%
interest per year for 5 years.
PW (20%) =
8,000[(1+0.20)5 −1]
0.20(1+0.20)5
+
5,000
(1+0.20)5
− 25,000
PW (20%) = 23,924.90 + 2,009.39 – 25,000
PW (20%) = Php 934.29
Therefore, because PW (20%) > 0, this equipment is economically justified.
The Future Worth Method
Because a primary objective of all time value of money methods is to maximize the future
wealth of the owners of a firm, the economic information provided by the future worth (FW)
method is very useful in capital investment decision situations. The future worth is based on the
equivalent worth of all cash inflows and outflows at the end of the planning horizon (study
period) at an interest rate that is generally the MARR. Also, the FW of a project is equivalent to
its PW; that is, FW = PW(F/P, i%, N). If FW≥0 for a project, it would be economically justified.
Equation 10 summarizes the general calculations necessary to determine a project’s future
worth:
𝐹𝑊(𝑖%) = 𝐹0 (1 + 𝑖)𝑁 + 𝐹1 (1 + 𝑖)𝑁−1 + ⋯ + 𝐹𝑁 (1 + 𝑖 )0
𝑁−𝑘
𝐹𝑊(𝑖%) = ∑𝑁
𝑘=0 𝐹𝑘 (1 + 𝑖)
Equation 10
Example:
Evaluate the FW of the potential improvement project described in the previous example.
Show the relationship between FW and PW for this example.
Solution:
FW (20%) = -25,000(F/P, 20%, 5) + 8,000(F/P, 20%, 4) + 8,000(F/P, 20%, 3) + 8,000(F/P,
20%, 2) + 8,000(F/P, 20%, 1) + 13,000
𝐹𝑊 (20%) = −25,000(1 + 0.20)5 + 8,000(1 + 0.20)4 + 8,000(1 + 0.20)3
+ 8,000(1 + 0.20)2 + 8,000(1 + 0.20)1 + 13,000
𝐹𝑊 (20%) = −62,208 + 16,588.80 + 13,824 + 11,520 + 9,600 + 13,000
𝐹𝑊 (20%) = 𝑃ℎ𝑝 2,324.80
Again, since FW > 0, the project is shown to be a good investment. The FW is a multiple
of the equivalent PW value:
PW (20%) = Php 2,324.80(P/F, 20%, 5) = Php 934.29
The Annual Worth Method
The annual worth (AW) method of a project is an equal annual series of peso amounts for
a stated study period, that is equivalent to the cash inflows and outflows at an interest rate that
is generally the MARR. Hence, the AW of a project is annual equivalent revenues or savings (R)
minus annual equivalent expenses (E), less its annual equivalent capital recovery (CR) amount,
which is defined in Equation 11. An annual equivalent value of R, E, and CR is computed for the
study period, N, which is usually in years. In equation form, the AW, which is a function of i%, is
𝐴𝑊 (𝑖%) = 𝑅 − 𝐸 − 𝐶𝑅(𝑖%)
Equation 10
Also, we need to notice that the AW of a project is equivalent to its PW and FW. That is,
AW = PW(A/P, i%, N), and AW = FW(A/F, i%, N). Hence, it can be easily computed for a project
from these other equivalent values.
As long as the AW is greater than or equal to zero, the project is economically attractive;
otherwise, it is not. An AW of zero means that an annual return exactly equal to the MARR has
been earned.
The capital recovery (CR) amount for a project is the equivalent uniform annual cost of
the capital invested. It is an annual amount that covers the following two items:
1. Loss in value of the asset
2. Interest on invested capital (i.e., at the MARR)
The easiest formula in computing the capital recovery amount involves finding the annual
equivalent of the capital investment and then subtracting the annual equivalent of the salvage
value. Thus,
CR(i%) = I(A/P, i%, N) – S(A/F, i%, N)
Equation 11
where:
I =
S =
N =
initial investment for the project
salvage (market) value at the end of the study period
project study period
Example:
By using the AW method and Equation 10, determine whether the equipment described
in the previous examples should be recommended.
Solution:
AW (20%) = 8,000 – [25,000(A/P, 20%, 5) – 5,000(A/F, 20%, 5)]
R-E
𝐴𝑊 (20%) = 8,000 – [
CR Amount (Equation 11)
25,000[0.20(1 + 0.20) 5]
(1+0.20)5 −1
−
5,000(0.20)
]
(1+0.20) 5 −1
𝐴𝑊 (20%) = 8,000 – (8,359.49 – 671.90)
𝐴𝑊 (20%) = 𝑃ℎ𝑝 312.41
Therefore, because its AW is positive, the equipment more than pays for itself over
a period of five years while earning a 20% return per year on the unrecovered
investment. In fact, the annual “surplus” is Php 312.41, which means that the
equipment provided more than 20% return on beginning-of-year unrecovered
investment. This piece of equipment should be recommended as an attractive
investment opportunity.
Also, we can verify that:
AW (20%) = PW(A/P, 20%, 5) = Php 934.29(A/P, 20%, 5) = Php 312.41
AW (20%) = FW(A/F, 20%, 5) = Php 2,324.80(A/F, 20%, 5) = Php 312.41
SOLVE IT!
Solve the following:
1. A capital investment of Php 10,000 can be made in a project that will produce a uniform
annual revenue of Php 5,310 for five years and then have a salvage value of Php 2,000.
Annual expenses will be Php 3,000. The company is willing to accept any project that will
earn at least 10% per year, before income taxes, on all invested capital. Show whether
this is a desirable investment by using the PW method.
2. An investment company is considering building a 25-unit apartment complex in a growing
town. Because of the long-term growth potential of the town, it is felt that the company
could average 90% full occupancy for the complex each year. If the following items are
reasonably accurate estimates, what is the minimum monthly rent that should be charged
if a 12% MARR (per year) is desired? Use the AW method.
Land investment cost
Php 50,000
Building investment cost
Php 225,000
Study period, N
20
Rent per unit per month
?
Upkeep expense per unit per month
Php 35
Property taxes and insurance per year
10% of the total initial investment
The Internal Rate of Return Method
The internal rate of return (IRR) method is the most widely used rate of return method
for performing engineering economic analyses. It is sometimes called by several other names,
such as investor’s method, discounted cash flow method, and profitability index.
This method solves for the interest rate that equates the equivalent worth of an
alternative’s cash inflows (receipts or savings) to the equivalent worth of cash outflows
(expenditures, including investment costs). Equivalent worth may be computed with any of the
three methods discussed earlier. The resultant interest rate is termed the internal rate of return
(IRR).
For a single alternative, the IRR is not positive unless (1) both receipts and expenses are
present in the cash flow pattern and (2) the sum of receipts exceeds the sum of all cash outflows.
Be sure to check both of these conditions in order to avoid the unnecessary work involved with
finding that the IRR is negative. (Visual inspection of the total net cash flow will determine
whether the IRR is zero or less.)
By using a PW formulation, the IRR is the i’% at which
𝑁
′
′
∑𝑁
𝑘=0 𝑅𝑘 (𝑃⁄𝐹 , 𝑖 %, 𝑘 ) = ∑𝑘=0 𝐸𝑘 (𝑃⁄𝐹 , 𝑖 %, 𝑘)
Equation 12
where:
Rk =
Ek =
N =
net revenues or savings for the kth year
net expenditures including any investment costs for the kth year
project life (or study period)
Once i' is calculated, it is compared with the MARR to assess whether the alternative in
question is acceptable. If i' ≥ MARR, the alternative is acceptable; otherwise, it is not.
A popular variation of Equation 12 for computing the IRR for an alternative is to determine
the i' at which its net PW is zero. In equation form, the IRR is the value of i' at which
𝑁
′
′
𝑃𝑊 = ∑𝑁
𝑘=0 𝑅𝑘 (𝑃⁄𝐹 , 𝑖 %, 𝑘 ) − ∑𝑘=0 𝐸𝐾 (𝑃⁄𝐹 , 𝑖 %, 𝑘 ) = 0
Equation 13
The value of i'% can also be determined as the interest rate at which FW = 0 or AW = 0.
For example, by setting the net FW equal to zero, Equation 14 would result:
𝑁
′
′
𝐹𝑊 = ∑𝑁
𝑘=0 𝑅𝑘 (𝐹 ⁄𝑃 , 𝑖 %, 𝑁 − 𝑘 ) − ∑𝑘=0 𝐸𝐾 (𝐹 ⁄𝑃 , 𝑖 %, 𝑁 − 𝑘 ) = 0
Equation 14
The method of solving Equations 12 through 14 normally involves trial-and-error
calculations until the i'% is converged upon or can be interpolated. The example that follows
presents a typical solution using the common convention of “+” signs for cash inflows and “-“
signs for cash outflows.
Example:
A capital investment of Php 10,000 can be made in a project that will produce a uniform
annual revenue of Php 5,310 for five years and then have a salvage value of Php 2,000. Annual
expenses will be Php 3,000. The company is willing to accept any project that will earn at least
10% per year, before income taxes, on all invested capital. Determine whether it is acceptable by
using the IRR method.
Given (Cash Flow Diagram):
0
10,000
2,000
5,310
5,310
5,310
5,310
5,310
1
2
3
4
5
3,000
3,000
3,000
3,000
3,000
Required:
i'% = ?, to determine whether the investment is acceptable or not
Solution:
Equivalent cash flow diagram:
2,000
0
2,310
2,310
2,310
2,310
2,310
1
2
3
4
5
10,000
Test if the sum of receipts exceeds the sum of all cash outflows:
2,310 + 2,310 + 2,310 + 2,310 + 2,310 + 2,000 = 13,550
Since 13,550 > 10,000, thus, it is likely that a positive-valued i'% can be determined.
Using Equation 13:
PW = 0 = -10,000 + 2,310(P/A, i'%, 5) + 2,000(P/F, i'%, 5)
PW = 0 = -10,000 +
2,310[(1+𝑖′)5 −1]
𝑖′(1+𝑖′)5
2,000
+ (1+𝑖′)5
Using linear interpolation to find the approximation of IRR:
At i' = 5%:
PW = -10,000 + 2,310(4.3295) + 2,000(0.7835) = +1,568
At i' = 15%:
PW = -10,000 + 2,310(3.3522) + 2,000(0.4972) = -1,262
Since we both have a positive and negative PW, the answer is bracketed, the dashed curve
in the figure below is what we are linearly approximating. The answer i'%, can be determined by
using the similar triangles dashed in the figure.
𝑙𝑖𝑛𝑒 𝐴𝐵
𝑙𝑖𝑛𝑒 𝐵𝐶
=
𝑙𝑖𝑛𝑒 𝐴𝑑
𝑙𝑖𝑛𝑒 𝑑𝑒
15%−5%
1,568−(−1,262)
𝑖 ′ % = 5% +
=
𝑖 ′ %−5%
1,568−0
1,568
1,568−(−1,262)
(15% − 5%)
𝑖 ′ % = 5% + 5.5% = 10.5%
This approximate solution illustrates the trial-and-error process, together with linear
interpolation. The error in this answer is due to nonlinearity of the PW function and would be
less if the range of interest rates used in the interpolation had been smaller.
From the result, we know that the project is minimally acceptable and that i' = MARR =
10% per year. This can be confirmed by substituting i =10% in the PW equation as follows:
PW (10%) = -10,000 + 2,310(P/A, 10%, 5) + 2,000(P/F, 10%, 5) = 0
Difficulties Associated with the IRR Method
The PW, FW, and AW methods assume that net receipts less expenses (positive recovered
funds) each time period are reinvested at the MARR during the study period, N. However, the
IRR method is based on the assumption that recovered funds, if not consumed in each time
period, are reinvested at i'% rather than at MARR. This latter assumption may not mirror reality
in some problems, thus making IRR an unacceptable method for analyzing engineering
alternatives.
Other difficulties with the IRR method include its computational intractability and the
occurrence of multiple IRRs in some types of problems. Generally speaking, multiple rates are
not meaningful for decision-making purposes, and another method of evaluation (e.g., PW)
should be utilized.
Another drawback to the IRR method is that it must be carefully applied and interpreted
in the analysis of two or more alternatives when only one of them is to be selected (i.e., mutually
exclusive alternatives). The key advantage of the method is its widespread acceptance by
industry, where various types of rates of returns and ratios are routinely used in making project
selections.
SOLVE IT!
Solve:
A small company needs to borrow Php 160,000. The local (and only) banker makes this
statement: “We can loan you Php 160,000 at a very favorable rate of 12% per year for a five-year
loan. However, to secure this loan you must agree to establish a checking account (with no
interest) in which the minimum average balance is Php 32,000. In addition, your interest
payments are due at the end of each year and the principal will be repaid in a lump-sum amount
at the end of year five”. What is the true effective annual interest rate being charged?
The External Rate of Return Method
The reinvestment assumption of the IRR method noted previously may not be valid in
engineering economy study. For instance, if a firm’s MARR is 20% per year and the IRR for a
project is 42.4%, it may not be possible for the firm to reinvest net cash proceeds from the project
at much more than 20%. This situation, coupled with the computational demands and possible
multiple interest rates associated with the IRR method, has given rise to other rate of return
methods that can remedy some of these weaknesses.
One such method is the external rate of return (ERR) method. It directly takes into account
the interest rate (є) external to the project at which net cash flows generated (or required) by
the project over its life can be reinvested (or borrowed). If this external reinvestment rate, which
is usually the firm’s MARR, happens to equal the project’s IRR, then the ERR method produces
identical to those of the IRR method.
In general, three steps are used in the calculating procedure. First, all net cash outflows
are discounted to time 0 (the present) at є% per compounding period. Second, all net cash inflows
are compounded to period N at є%. Third, the external rate of return, which is the interest rate
that establishes equivalence between the two quantities, is determined. The absolute value of
the present equivalent worth of the net cash outflows at є% (first step) is used in this last step.
In equation form, the ERR is the i'% at which
𝑁
′
∑𝑁
𝑘=0 𝐸𝑘 (𝑃 ⁄𝐹 , є%, 𝑘 )(𝐹 ⁄𝑃 , 𝑖 %, 𝑁 ) = ∑𝑘=0 𝑅𝑘 (𝐹 ⁄𝑃 , є%, N − k)
Equation 15
where:
Rk
Ek
N
є
=
=
=
=
excess of receipts over expenses in period k
excess of expenditures over receipts in period k
project life or number of periods for the study
external reinvestment rate per period
A project is acceptable when i'% of the ERR method is greater than or equal to the firm’s
MARR.
The external rate of return method has two basic advantages over the IRR method:
1. It can usually be solved for directly rather than by trial and error.
2. It is not subject to the possibility of multiple rates of return.
Example:
When є = 15% and MARR = 20%, determine whether the project whose total cash flow
diagram appears below is acceptable. Notice in this example that the use of an є% different from
MARR is illustrated. This might occur if, for some reason, part or all of the funds related to a
project are “handled” outside the firm’s normal capital structure.
6,000
1
6,000
3
2
1,000
6,000
4
1,000
6,000
5
1,000
6,000
6
1,000
5,000
10,000
Solution:
E0 =
E1 =
Rk =
10,000 (k = 0)
5,000 (k = 1)
5,000 for k = 2, 3, …. , 6
Using Equation 15:
|-10,000 – 5,000(P/F, 15%, 1)|(F/P, i'%, 6) = 5,000(F/A, 15%, 5)
1,000
|-10,000 –
5,000
(1+0.15)1
|[(1 + 𝑖′)6 ] =
5,000[(1+0.15)5 −1]
0.15
6
14,347.83(1 + i') = 33,711.91
33,711.91
(1 + i'%)6 = 14,347.83
(1 + i'%)6 = 2. 35
Taking the 6th root on both sides:
1+ i'% = 1.1530
i'% = 1.1530 – 1
i'% = 0.1530
i'% = 15.30%
Thus, since the i'% is less than the MARR = 20%, this project would be unacceptable according to
the ERR method.
SOLVE IT!
Solve:
Refer to the IRR example. Suppose that є = MARR = 20% per year. What is the alternative’s
external rate of return, and is the alternative acceptable?
The Payback (Payout) Period Method
All methods presented thus far reflect the profitability of a proposed alternative for a
study period of N. The payback method, which is often called the simple payout method, mainly
indicates a project’s liquidity rather than its profitability. Historically, the payback method has
been used as a measure of a project’s riskiness, since liquidity deals with how fast an investment
can be recovered. A low-valued payback period is considered desirable. Quite simply, the
payback method calculates the number of years required for cash inflows to just equal to cash
outflows. Hence the simple payback period is the smallest value of 𝜃 (𝜃 ≤ 𝑁) for which this
relationship is satisfied under our normal end-of-year cash flow convention. For a project where
all capital investment occurs at time 0, we have:
∑𝜃𝑘=1(𝑅𝑘 − 𝐸𝑘 ) − 𝐼 ≥ 0
Equation 16
The simple payback period, 𝜃, ignores the time value of money and all cash flows that
occur after 𝜃. The payback period can produce misleading results, and it is recommended as
supplemental information only in conjunction with one or more of the five methods previously
discussed.
Sometimes the discounted payback period, 𝜃 ′ (𝜃 ′ ≤ 𝑁), is calculated so that the time value of
money is considered:
∑𝜃′
𝑘=1(𝑅𝑘 − 𝐸𝑘 )(𝑃⁄𝐹 , 𝑖%, 𝑘) − 𝐼 ≥ 0
Equation 17
where i% is the MARR, I is the capital investment made at the present time (k = 0), and 𝜃′ is the
smallest value that satisfies Equation 17. Because 𝜃 ′ ≥ 𝜃, we can reduce the number of
iterations required by Equation 17 to determine 𝜃 ′ .
This variation of the simple payback period produces the breakeven life of a project in
view of the time value of money. However, neither payback period calculation includes cash flows
occurring after 𝜃 (𝜃 ′ ). This means that they do not take into consideration the economic life of
physical assets. Thus, these methods will be misleading if one alternative that has a longer (less
desirable) payout period than another that produces a higher rate of return (or PW) on the
invested capital.
Example:
Considering the alternative presented in the PW, FW, and AW example that has end-ofyear cash flows, calculate the simple payback period and the discounted payback period. In view
of the minimum acceptable payback period of five years, is this alternative attractive?
Recall:
Given (Cash Flow Diagram):
5,000
0
25,000
8,000
8,000
8,000
8,000
8,000
1
2
3
4
5
@ i = 20% per year
Solution:
Payback Period:
[∑𝜃=3.125
(8,0000)𝑘 ] − 25,000 ≥ 0
𝑘=1
Note:
𝜃=
𝐼
(𝑅𝑘 − 𝐸𝑘 )
=
25,000
8,000
= 3.125
Thus, 𝜽 ≅ 𝟒 years, and the alternative is acceptable.
Notice that the Php 5,000 salvage value was not considered in Equation 16 to determine the
simple payback period.
Discounted Payback Period:
′ =5
(8,000)𝑘 (𝑃⁄𝐹 , 20%, 𝑘)] + 5,000(𝑃⁄𝐹, 20%, 5) − 25,000 > 0
[∑𝜃𝑘=1
or
8,000(𝑃⁄𝐴 , 20%, 5) + 5,000(𝑃⁄𝐹 , 20%, 5) − 25,000 > 0
at the project’s expected life of N = 5 years. The alternative is minimally acceptable with the
discounted payback method since 𝜃 ′ = 5 years. In this case, the salvage value was considered in
determining the value of 𝜃 ′ . Observe that when k = 4, Equation 17 is not satisfied.
The Benefit/Cost Ratio Method
As the name implies, the benefit/cost ratio method involves the calculation of a ratio of
benefits to costs. Whether evaluating a project in the private sector, or in the public sector, the
time value of money must be considered to account for the timing of cash flows (or benefits)
occurring after the inception of the project. Thus the B/C ratio is actually a ratio of discounted
benefits to discounted costs.
Any method for formally evaluating projects in the public sector must consider the
worthiness of allocating resources to achieve social goals. For nearly 60 years, the B/C ratio
method has been the accepted procedure for making a go/no-go decisions on independent
projects and for comparing alternative projects in the public sector, even though the other
methods previously discussed will lead to identical recommendations, assuming all these
procedures are applied properly.
The B/C ratio is defined as the ratio of the equivalent worth of benefits to the equivalent
worth of costs. The equivalent worth measure applied can be present worth, annual worth, or
future worth, but customarily, either PW or AW is used. The benefit/cost ratio is also known as
the savings-investment ratio (SIR) by some governmental agencies.
Several different formulations of the B/C ratio have been developed. Two of the more
commonly used formulations are presented here, illustrating the use of both present worth and
annual worth.
Conventional B/C ratio with PW:
𝑃𝑊 (𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑠𝑒𝑑 𝑝𝑟𝑜𝑗𝑒𝑐𝑡)
𝑃𝑊(𝐵)
𝐵 ⁄𝐶 =
=
𝑃𝑊 (𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑠𝑒𝑑 𝑝𝑟𝑜𝑗𝑒𝑐𝑡) 𝐼 + 𝑃𝑊 (𝑂&𝑀)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 18
where:
PW(•)
B
I
O&M
=
=
=
=
present worth of (•)
benefits of the proposed project
initial investment in the proposed project
operating and maintenance costs of the proposed project
Modified B/C ratio with PW:
𝐵 ⁄𝐶 =
𝑃𝑊(𝐵)−𝑃𝑊(𝑂&𝑀)
𝐼
Equation 19
The numerator of the modified benefit/cost ratio expresses the equivalent worth of the
benefits minus the equivalent worth of the O&M costs, and the denominator includes only the
initial investment costs. A project is acceptable when the B/C ratio, as defined in either Equation
18 or 19, is greater than or equal to 1.0.
Equations 18 and 19 can be rewritten in terms of equivalent annual worth as follows:
Conventional B/C ratio with AW:
𝐴𝑊 (𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑠𝑒𝑑 𝑝𝑟𝑜𝑗𝑒𝑐𝑡)
𝐴𝑊(𝐵)
𝐵 ⁄𝐶 =
=
𝐴𝑊 (𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑠𝑒𝑑 𝑝𝑟𝑜𝑗𝑒𝑐𝑡) 𝐶𝑅 + 𝐴𝑊 (𝑂&𝑀)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 20
where:
AW(•) =
B
=
CR
=
O&M =
annual worth of (•)
benefits of the proposed project
capital recovery amount (i.e., the equivalent annual cost of the
initial investment, I, including an allowance for salvage value, if any)
operating and maintenance costs of the proposed project
Modified B/C ratio with AW:
𝐵 ⁄𝐶 =
𝐴𝑊(𝐵)−𝐴𝑊(𝑂&𝑀)
𝐶𝑅
Equation 21
Note that when using the annual worth approach, the annualized equivalent of any
salvage value associated with the investment is effectively subtracted from the denominator in
the calculation of the capital recovery amount (CR) in Equations 20 and 21. Similarly, when using
the present worth approach to calculate a benefits/cost ratio, it is customary to reduce the
investment in the denominator by the discounted equivalent of any salvage value. Equations 18
and 19 are rewritten to incorporate the salvage value of an investment below:
Conventional B/C ratio with PW, Salvage Value included:
𝐵 ⁄𝐶 =
𝑃𝑊 (𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑠𝑒𝑑 𝑝𝑟𝑜𝑗𝑒𝑐𝑡)
𝑃𝑊 (𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑠𝑒𝑑 𝑝𝑟𝑜𝑗𝑒𝑐𝑡)
𝑃𝑊(𝐵)
𝐵⁄𝐶 = 𝐼−𝑃𝑊 (𝑆)+𝑃𝑊 (𝑂&𝑀)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 22
where:
PW(•)
B
I
S
O&M
=
=
=
=
=
present worth of (•)
benefits of the proposed project
initial investment in the proposed project
salvage value of investment
operating and maintenance costs of the proposed project
Modified B/C ratio with PW, Salvage Value included:
𝐵 ⁄𝐶 =
𝑃𝑊(𝐵)−𝑃𝑊(𝑂&𝑀)
𝐼−𝑃𝑊(𝑆)
Equation 23
The resulting B/C ratios for all above formulations will give consistent results in
determining the acceptability of a project (i.e., either B/C > 1.0 or B/C < 1.0 or B/C = 1.0). The
conventional B/C ratio will give identical results for both PW and AW formulations; similarly, the
modified B/C ratio gives identical numerical results whether PW or AW is used.
Example:
The city of Bugtussle is considering extending the runways of its Municipal Airport so that
commercial jets can use the facility. The land necessary for the runway extension is currently
farmland, which can be purchased for Php 350,000. Construction costs for the runway extension
are projected to be Php 600,000, and the additional annual maintenance costs for the extension
are estimated to be Php 22,500. If the runways are extended, a small terminal will be constructed
at a cost of Php 250,000. The annual operating and maintenance costs for the terminal are
estimated at Php 75,000. Finally, the projected increase in flights will require the addition of two
air traffic controllers, at an annual cost of Php 100,000. Annual benefits of the runway extension
have been estimated as follows:
Php 325,000 rental receipts from airlines leasing space at the facility
Php 65,000 airport tax charged to passengers
Php 50,000 convenience benefit for residents of Bugtussle
Php 50,000 additional tourism peso for Bugtussle
Apply the B/C ratio method with a study period of 20 years and an interest rate of 10% to
determine whether the runways at Bugtussle Municipal Airport should be extended.
Solution:
Using Equation 18 – Conventional B/C:
𝐵 ⁄𝐶 =
𝐵 ⁄𝐶 =
𝑃𝑊(𝐵)
𝐼+𝑃𝑊 (𝑂&𝑀)
(325,000+65,000+50,000+50,000)(𝑃 ⁄𝐴,10%,20)
(350,000+600,000+250,000)+(22,500+75,000+100,000)(𝑃 ⁄𝐴,10%,20)
𝐵 ⁄𝐶 =
490,000(𝑃 ⁄𝐴,10%,20)
1,200,000+197,500(𝑃 ⁄𝐴,10%,20)
𝑩⁄𝑪 = 1.448
Using Equation 19 – Modified B/C:
𝐵 ⁄𝐶 =
𝐵 ⁄𝐶 =
𝑃𝑊(𝐵)−𝑃𝑊(𝑂&𝑀)
𝐼
490,000(𝑃 ⁄𝐴,10%,20)−197,500(𝑃 ⁄𝐴,10%,20)
1,200,000
𝑩⁄𝑪 = 𝟐. 𝟎𝟕𝟓
Using Equation 20 – Conventional B/C:
𝐵 ⁄𝐶 =
𝐵 ⁄𝐶 =
𝐴𝑊(𝐵)
𝐶𝑅+𝐴𝑊 (𝑂&𝑀)
490,000
1,200,000(𝐴⁄𝑃,10%,20)+197,500
𝑩⁄𝑪 = 𝟏. 𝟒𝟒𝟖
Using Equation 21 – Modified B/C:
𝐵 ⁄𝐶 =
𝐵 ⁄𝐶 =
𝐴𝑊(𝐵)−𝐴𝑊(𝑂&𝑀)
𝐶𝑅
490,000−197,500
1,200,000(𝐴⁄𝑃,10%,20)
𝑩⁄𝑪 = 2.075
Key Points







Annuity is a series of equal payments occurring at equal interval of time.
Ordinary annuity is an annuity certain where the payments are made at the
end of each period beginning from the first period.
Annuity due is an annuity certain where the payments are made at the
beginning of each period starting from the first period.
Deferred annuity is an annuity certain where the first payment does not
begin until some later date in the cash flow.
When an annuity does not have a fixed time span but continues indefinitely,
then it is referred to as a perpetuity whose sum is an infinite value.
The five basic methods for evaluating the financial profitability of a single
project are present worth, future worth, annual worth, internal rate of
return, and external rate of return.
The two supplemental methods for assessing a project’s liquidity are the
simple payback period and the discounted payback period.
BIG PICTURE
Week 6 – 7: Unit Learning Outcomes (ULO)
At the end of the unit, you are expected to:
a. determine mutually exclusive investment combination;
b. associate depreciation concept to after-tax analysis;
c. determine whether an asset should be kept or replaced using the
considerations involved in replacement studies;
d. determine the breakeven point with respect to production and sales
volumes; and
e. use decision tree analysis in evaluating alternatives.
Big Picture in Focus: ULOa. Determine mutually exclusive investment combination.
Metalanguage
1. Mutually Exclusive. Two events are mutually exclusive if they cannot occur both at the
same time.
Essential Knowledge
DECISIONS UNDER CERTAINTY
Mutually Exclusive Investment Alternatives
It is helpful to categorize investment opportunities (projects) into three major groups as
follows:
1. Mutually Exclusive: at most one project out of the group can be chosen.
2. Independent: the choice of a project is independent of the choice of any other project
in the group, so that all or none of the projects may be selected or some number in
between.
3. Contingent: the choice of a project is conditional on the choice of one or more other
projects.
It is common for decision makers to be faced with sets of mutually exclusive,
independent, and/or contingent investment projects. For example, a construction contractor
might be considering investing in a dump truck, and/or a backhoe, and/or expansion of the
headquarters office building. For each of these investment projects, there may be two or more
mutually exclusive alternatives (i.e., brands of dump trucks, types of backhoes, and designs for
expansion of the office building). While the choice of a design for the office building is probably
independent of that of either dump trucks or backhoes, the choice of any type of backhoe may
be contingent (conditional) on the decision to purchase a dump truck.
A general approach, then, requires that all projects be listed and that all the feasible
combinations of projects be enumerated. Such combinations of projects will then be mutually
exclusive. Each combination of projects is mutually exclusive because each is unique and the
acceptance of one combination of investment projects precludes the acceptance of any of the
other combinations. The total net cash flow of each combination is determined simply by adding,
period by period, the cash flows of each project included in the mutually exclusive combination
being considered.
For example, suppose that we have three projects: A, B, and C. Each project can be
selected once or not at all (i.e., multiple project As are not possible). If the projects themselves
are all mutually exclusive, then the four mutually exclusive combinations are shown in binary
form in Table 7. If, by chance, the firm feels that one of the projects must be chosen (i.e., it is not
permissible to turn down all projects), then mutually exclusive combination one would be
eliminated from consideration.
Mutually
Project
Exclusive
Explanation
XA
XB
XC
Combination
1
0
0
0
Accept None
2
1
0
0
Accept A
3
0
1
0
Accept B
4
0
0
1
Accept C
Table 7. Combinations of Three Mutually Exclusive Projects
If the three projects are independent, there are eight mutually exclusive combinations, as
shown in Table 8.
Mutually
Exclusive
Combination
1
2
3
4
5
6
7
8
Project
XA
XB
XC
0
1
0
0
1
1
0
1
0
0
1
0
1
0
1
1
0
0
0
1
0
1
1
1
Explanation
Accept None
Accept A
Accept B
Accept C
Accept A and B
Accept A and C
Accept B and C
Accept A, B, and
C
Table 8. Mutually Exclusive Combinations of Three Independent Projects
To illustrate one of the many possible instances of contingent projects, suppose that A is
contingent on the acceptance of B. Now there are four mutually exclusive combinations: (1) do
nothing, (2) B only, (3) B and C, and (4) A, B, and C.
Suppose that a company is considering two independent sets of mutually exclusive
projects. That is, projects A1 and A2 are mutually exclusive, as are B1 and B2. However, the
selection of any project from the set of projects A1 and A2 is independent of the selection of any
project from the set of projects B1 and B2. Independent means that the choice of a project from
the A set does not affect the choice from the B set. For example, the decision problem may be to
select at most one dump truck out of two models being considered and to select at most one
design for expansion of the office building out of two designs being considered. Table 9 shows all
mutually exclusive combinations for this situation.
Mutually
Exclusive
Combination
Project
Explanation
XA1
XA2
XB1
XB1
1
0
0
0
0
Accept None
2
1
0
0
0
Accept A1
3
0
1
0
0
Accept A2
4
0
0
1
0
Accept B1
5
0
0
0
1
Accept B2
Accept A1 and
B1
Accept A1 and
7
1
0
0
1
B2
Accept A2 and
8
0
1
1
0
B1
Accept A2 and
9
0
1
0
1
B2
Table 9. Mutually Exclusive Combinations for Two Independent Sets of Mutually Exclusive
Projects
6
1
0
1
0
Example:
The following are five proposed projects being considered by an engineer in an integrated
transportation company for upgrading an intermodal shipment transfer facility for less than
truckload lots of consumer goods. The interrelationships among the projects, and their respective
cash flows for the coming budgeting period, are as shown. Some of the projects are mutually
exclusive, as noted, and B1 and B2 are independent of C1 and C2. Also, certain projects are
dependent on others that may be included in the final portfolio. Using the PW method and MARR
= 10% per year, determine what combination of projects is best if the capital invested is (a)
unlimited, and (b) limited to Php 48,000.
Given:
Project B1
Project B2
Project C1
Project C2
Project D
mutually exclusive and independent of C set
mutually exclusive and dependent (contingent) on the acceptance of B2
contingent on the acceptance of C1
Cash Flow (Php 000s) for End of Year
Project
B1
B2
C1
C2
D
0
-50
-30
-14
-15
-10
1
2
3
4
20
20
20
20
12
12
12
12
4
4
4
4
5
5
5
5
6
6
6
6
Table 10. Project Cash Flows and PWs (Example)
PW
(Php 000s)
at MARR =
10% per
year
13.4
8.0
-1.3
0.8
9.0
Required:
Best combination of projects if the capital to be invested is (a) unlimited, and (b) limited
to Php 48,000.
Solution:
Note:
The PW for each project by itself is shown in the right-most column of Table 10. Sample
calculation for the PW of project B1:
PWB1 = -50,000 + 20,000(P/A, 10%, 4) = Php 13,400
The mutually exclusive project combinations are shown in Table 11. Project C1, with a
negative PW, has not been eliminated from further consideration because project D is contingent
on it.
Mutually
Project
Exclusive
B1
B2
C1
C2
D
Combination
1
0
0
0
0
0
2
1
0
0
0
0
3
0
1
0
0
0
4
0
1
1
0
0
5
0
1
0
1
0
6
0
1
1
0
1
Table 11. Mutually Exclusive Project Combinations (Example)
The combined cash flows and the PW for each mutually exclusive combination are shown
in Table 12.
PW
Cash Flow (Php 000s) for End of
Mutually
Invested
(Php 000s)
Year
Capital
at MARR =
Exclusive
(Php 000s)
10% per
Combination
0
1
2
3
4
year
1
2
3
4
5
6
0
0
0
0
0
0
-50
20
20
20
20
50
-30
12
12
12
12
30
-44
16
16
16
16
44
-45
17
17
17
17
45
-54
22
22
22
22
54
Table 12. Combined Project Cash Flows and PWs (Example)
0
13.4
8.0
6.7
8.9
15.7
Examination of the right-most column reveals that mutually exclusive combination 6 has
the highest PW if capital available (in year 0) is unlimited. If, however, capital available is limited
to Php 48,000, mutually exclusive combinations 2 and 6 are not feasible. Of the remaining
mutually exclusive combinations, 5 is best, which means that a portfolio consisting of projects B2
and C2 would be selected with a PW = Php 8,888.
SOLVE IT!
A large corporation is considering the funding of three independent, nonrepeating
projects for enlarging freshwater harbors supporting its operations in three areas of the country.
Its available capital investment budget this year for such projects is Php 200 million, and the
firm’s MARR is 10% per year. In view of the following data, which project(s), if any, should be
funded?
Capital
Net Annual
PW = I +
Project
Investment, I
Benefits, A
Useful Life, N
A(P/A, 10%, N)
(in Php)
(in Php)
(in Php)
H1
-93,000,000
13,000,000
15 years
5,879,300
H2
-55,000,000
9,500,000
10 years
3,373,700
H3
-71,000,000
10,400,000
30 years
27,039,760
Big Picture in Focus: ULOb. Associate depreciation concept to after-tax analysis.
Metalanguage
1. Useful Life. The expected (estimated) period of time that a property will be used in a trade
or business or to produce income. It is not how long the property will last but how long
the owner expects to productively use it. It is also sometimes referred to as depreciable
life, although the actual useful life of an asset may be different from its depreciable life.
2. First Cost. It is the sum of the initial expenditures involved in capitalizing a property or
building a project; includes items such as transportation, installation, preparation for
service, as well as other related costs.
3. Book Value. It represents the value of a particular asset on the company's balance sheet
after taking accumulated depreciation into account.
4. Market Value. It is the highest estimated price that a buyer would pay and a seller would
accept for an item in an open and competitive market.
Essential Knowledge
Depreciation and After-Tax Economic Analysis
Taxes have been paid since the dawn of civilization. Most organizations consider the
effect of income taxes on the financial results of a proposed engineering project because income
taxes usually represent a significant cash outflow that cannot be ignored in decision making.
Depreciation
Depreciation is the decrease in value of physical properties with the passage of time and
use. The actual amount of depreciation can never be established until the asset is retired from
service. Because depreciation is noncash cost that affects income taxes, we must consider it
properly when making after-tax engineering economy studies.
Depreciable property is property for which depreciation is allowed under income tax laws
and regulations. To determine if depreciation deductions can be taken, the classification of
various types of property must be understood. In general, property is depreciable if it meets the
following basic requirements:
1. It must be used in business or held to produce income.
2. It must have a determinable useful life, and the life must be longer than one year.
3. It must be something that wears out, decays, gets used up, becomes obsolete, or loses
value from natural causes.
4. It is not inventory, stock in trade, or investment property.
Depreciable property is classified as either tangible or intangible. Tangible property can
be seen or touched, and it includes two main types called personal property and real property.
Personal property includes assets such as machinery, vehicles, equipment, furniture, and similar
items. In contrast, real property is land and generally anything that is erected on, growing on, or
attached to a land. Land itself, however, is not depreciable because it does not have a
determinable life.
Intangible property is personal property such as copyright, patent, or franchise.
A company can begin to depreciate property it owns when the property is placed in
service for use in the business or for production of income. Property is considered to be placed
in service when it is ready and available for a specific use, even if it is not actually used yet.
Depreciation stops either when the cost of placing it in service has been recovered or it is retired
from service.
Depreciation has two types:
1. Physical Depreciation. This is due to the reduction of the physical ability of an
equipment or asset to produce results.
2. Functional Depreciation. This is due to the reduction in the demand for the function
that the equipment or asset was designed to render. This type of depreciation is often
called obsolescence.
Methods of Computing Depreciation
Straight Line Method
In this method of computing depreciation, it is assumed that the loss in value is directly
proportional to the age of the equipment or asset.
The formula in computing for the annual depreciation charge, d, is given by:
𝑑=
𝐶0 − 𝐶𝑛
𝑛
Equation 24
where:
C0 =
Cn =
n =
first cost
cost after “n” years (salvage/scrap value)
life of the property
The book value at the end of “m” years, C m, of using the equipment or asset is computed
using:
𝐶𝑚 = 𝐶0 − 𝐷𝑚
Equation 25
where:
Dm =
Dm =
total depreciation after “m” years
dm
Sinking Fund Method
In this method of computing depreciation, it is assumed that a sinking fund is established
in which funds will accumulate for replacement purposes.
The formula in computing for the annual depreciation charge, d, is given by:
𝑑=
where:
C0 =
first cost
(𝐶0 − 𝐶𝑛 )(𝑖)
(1+𝑖)𝑛 −1
Equation 26
Cn = cost after “n” years (salvage/scrap value)
n = life of the property
The book value at the end of “m” years, C m, of using the equipment or asset is computed
using:
𝐶𝑚 = 𝐶0 − 𝐷𝑚
Equation 27
where:
Dm =
Dm =
total depreciation after “m” years
𝑑[1+𝑖)𝑚 −1
𝑖
Declining Balance Method
In this method of computing depreciation, it is assumed that the annual cost of
depreciation is a fixed percentage of the book value at the beginning of the year. This method is
sometimes known as constant percentage method or the Matheson Formula.
The Matheson Formula is given by:
𝑛
𝐶
𝑘 = 1 − √ 𝐶𝑛
or
0
𝑚
𝐶
𝑘 = 1 − √ 𝐶𝑚
0
Equation 28
The value k is the constant percentage. Hence, k must be decimal and a value less than 1. In this
method, the salvage or scrap value must not be zero.
Sum-of-Years’ Digit (SYD) Method
In this method, the following are the respective depreciation charges:
First Year:
𝑛
𝑑1 = (𝐶0 − 𝐶𝑛 ) ∑
𝑦𝑒𝑎𝑟𝑠
Second Year:
𝑛−1
𝑑2 = (𝐶0 − 𝐶𝑛 ) ∑ 𝑦𝑒𝑎𝑟𝑠
Third Year:
𝑛−2
𝑑3 = (𝐶0 − 𝐶𝑛 ) ∑ 𝑦𝑒𝑎𝑟𝑠
or
𝑑𝑁 = (𝐶0 − 𝐶𝑛 )
𝑛−(𝑁−1)
∑ 𝑦𝑒𝑎𝑟𝑠
Equation 29
where
N = specific year
The book value at the end of “m” years, C m, of using the equipment or asset is computed
using:
𝐶𝑚 = 𝐶0 − (𝑑1 + 𝑑2 + ⋯ + 𝑑𝑚
The sum of years’ digit, ∑ 𝑦𝑒𝑎𝑟𝑠, is computed using:
Equation 30
∑ 𝑦𝑒𝑎𝑟𝑠 =
𝑛(𝑛+1)
Equation 31
2
Comparison Between the Different Methods of Depreciation
In order to establish the comparison between the depreciation methods mentioned
above, let us consider the following data:
First Cost, C0
=
Php 10,000
Salvage value, Cn
=
Php 500
Life of property, n
=
5 years
A. Using the Straight Line Method:
𝐶 −𝐶
𝑑 = 0𝑛 𝑛
𝑑=
10,000 − 500
5
𝑑 = 𝑃ℎ𝑝 1,900
Tabulation of book value:
Period Depreciation
(in Php)
0
1
1,900
2
1,900
3
1,900
4
1,900
5
1,900
Book Value
(in Php)
10,000
8,100
6,200
4,300
2,400
500
B. Using the Sinking Fund Method:
Assume 10% interest rate
𝑑=
𝑑=
(𝐶0 − 𝐶𝑛 )(𝑖)
(1+𝑖)𝑛 −1
(10,000− 500)(0.10)
(1+0.10)5−1
𝑑 = 𝑃ℎ𝑝 1,556.076
𝐷1 = 𝑑 = 1,556.076
𝐷2 =
𝐷3 =
𝐷4 =
𝑑[(1+𝑖)2 −1
𝑖
𝑑[(1+𝑖)3 −1
𝑖
𝑑[(1+𝑖)4 −1
𝑖
=
=
=
1,556.076[(1+0.10) 2 −1
0.10
1,556.076[(1+0.10) 3 −1
0.10
1,556.076[(1+0.10) 4 −1
0.10
= 3,267.7596
= 5,150.6116
= 7,221.7487
𝐷5 =
𝑑[(1+𝑖)5 −1
𝑖
=
1,556.076[(1+0.10) 5 −1
0.10
Tabulation of book value:
Period Depreciation
(in Php)
0
1
1,556.08
2
3,267.76
3
5,150.61
4
7,221.75
5
9,500
= 9,500.00
Book Value
(in Php)
10,000
8,443.92
6,732.24
4,849.39
2,778.29
500
C. Using the Declining Balance Method:
𝑛
𝐶
𝑘 = 1 − √ 𝐶𝑛
0
5
500
𝑘 = 1 − √ 10,000
𝑘 = 1 − 0.5493
𝑘 = 0.4507
Tabulation of book value:
Period
Depreciation
Book Value
(in Php)
(in Php)
0
10,000
1
0.4507(10,000) = 4,507.00
5,493.00
2
0.4507(5,493) = 2,475.70
3,017.30
3
0.4507(3,017.30) = 1,359.90
1,657.40
4
0.4507(1,657.40) = 746.99
910.41
5
0.4507(910.41) = 410.32
500.09
Note: Slight difference is a result of rounding off of values.
D. Using the SYD Method:
𝑛(𝑛+1)
∑ 𝑦𝑒𝑎𝑟𝑠 =
∑ 𝑦𝑒𝑎𝑟𝑠 =
2
5(5+1)
2
∑ 𝑦𝑒𝑎𝑟𝑠 = 15
𝑛
5
𝑛−1
4
𝑛−2
3
𝑑1 = (𝐶0 − 𝐶𝑛 ) ∑ 𝑦𝑒𝑎𝑟𝑠 = (10,000 − 500) 15 = 3,166.67
𝑑2 = (𝐶0 − 𝐶𝑛 ) ∑ 𝑦𝑒𝑎𝑟𝑠 = (10,000 − 500) 15 = 2,533.33
𝑑3 = (𝐶0 − 𝐶𝑛 ) ∑ 𝑦𝑒𝑎𝑟𝑠 = (10,000 − 500) 15 = 1,900
𝑑4 = (𝐶0 − 𝐶𝑛 ) ∑
𝑛−3
𝑦𝑒𝑎𝑟𝑠
= (10,000 − 500)
𝑛−4
2
15
= 1,266.67
1
𝑑5 = (𝐶0 − 𝐶𝑛 ) ∑ 𝑦𝑒𝑎𝑟𝑠 = (10,000 − 500) 15 = 633.33
Tabulation of book value:
Period Depreciation
(in Php)
0
1
3,166.67
2
2,533.33
3
1,900.00
4
1,266.67
5
633.33
Book Value
(in Php)
10,000
6,833.33
4,300.00
2,400.00
1,133.33
500.00
From the above comparison, the declining balance method has the largest annual
depreciation charge while the sinking fund method has the smallest annual depreciation charges.
Example 1:
Problem: ECE Board April 1999
A telephone company purchased a microwave radio equipment for Php 6 million, freight
and installation charges amounted to 4% of the purchased price. If the equipment will be
depreciated over a period of 10 years with a salvage value of 8%, determine the depreciation
cost during the 5th year using SYD.
Given:
C0 = Php 6,000,000 + 0.04(Php 6,000,000) = Php 6,240,000
Cn = 0.08(Php 6,240,000) = Php 499,200
n = 10
Required:
d5 = ?
Solution:
Solving for depreciation during the 5th year, d5:
𝑛−4
𝑑5 = (𝐶0 − 𝐶𝑛 )
∑ 𝑦𝑒𝑎𝑟𝑠
But
∑ 𝑦𝑒𝑎𝑟𝑠 =
∑ 𝑦𝑒𝑎𝑟𝑠 =
𝑛(𝑛+1)
2
10(10+1)
2
= 55
Substituting,
𝑑5 = (6,240,000 − 499,200)
𝒅𝟓 = 𝑷𝒉𝒑 𝟔𝟐𝟔, 𝟐𝟔𝟗. 𝟎𝟗
10−4
55
Answer
Example 2:
ECE Board November 1998
ABC Corporation makes it a policy that for any new equipment purchased, the annual
depreciation cost should not exceed 20% of the first cost at any time with no salvage value.
Determine the length of service life necessary if the depreciation used is the SYD method.
Given:
d <= 20% of C0
Cn = 0
Required:
n=?
Solution:
In SYD method, the largest depreciation is on the first year. Analyze the first year depreciation,
d1:
𝑛
𝑑1 = (𝐶0 − 𝐶𝑛 )
∑ 𝑦𝑒𝑎𝑟𝑠
But
∑ 𝑦𝑒𝑎𝑟𝑠 =
𝑛(𝑛+1)
2
Substituting ∑ 𝑦𝑒𝑎𝑟𝑠 to d1:
𝑛
𝑑1 = (𝐶0 − 0) 𝑛(𝑛+1)
2
𝑑1 = (𝐶0 )
2
(𝑛+1)
But d1 = 0.20C0
2
0.20𝐶0 = 𝐶0 [𝑛+1]
n + 1 = 10
n = 9 years
Answer
Example 3:
Problem: ME Board October 1996
A machine has an initial cost of Php 50,000 and a salvage value of Php 10,000 after 10
years. What is the straight line method depreciation rate as a percentage of the initial cost?
Given:
C0 = Php 50,000
Cn = Php 10,000
n = 10
Required:
Let x be the percentage of depreciation:
x=?
Solution:
Solving for the annual depreciation, d:
𝑑=
𝑑=
𝐶0 − 𝐶𝑛
𝑛
50,000− 10,000
10
𝑑 = 𝑃ℎ𝑝 4,000
Solving for x with respect to the initial cost:
𝑑
𝑥 = 𝐶 𝑥 100
0
𝑥=
4,000
50,000
𝒙 = 𝟖%
𝑥 100
Answer
Example 4:
ECE Board April 2000, ECE Board November 1999
A VOM has a selling price of Php 400. If its selling price is expected to decline at a rate of
10% per annum due to obsolescence, what will be its selling price after 5 years?
Given:
C0 = Php 400
k = 10%
m=5
Required:
C5 = ?
Solution:
Using Matheson’s Formula:
𝑚
𝐶
𝑘 =1− √ 𝑚
𝐶
0
5
𝐶
5
0.10 = 1 − √400
𝐶5
400
= 0.905
𝑪𝟓 = 𝑷𝒉𝒑 𝟐𝟑𝟔. 𝟐𝟎
Answer
SOLVE IT!
Solve the following:
1. An engineer bought an equipment for Php 500,000. He spent an additional amount of Php
30,000 for installation and other expenses. The estimated useful life of the equipment is
10 years. The salvage value is x% of the first cost. Using the straight line method of
depreciation, the book value at the end of 5 years will be Php 291,500. What is the value
of x?
2. An asset is purchased for Php 9,000. Its estimated economic life is 10 years after which it
will be sold for Php 1,000. Find the depreciation in the first three years using SYD method.
Income Taxes
Up to this point, there has been no consideration of income taxes, except for the influence
of depreciation and other types of deductions. However, there is a wide variety of capital
investment problems in which income taxes do affect the choice among alternatives, and aftertax studies are essential.
Income taxes affect a project’s estimated cash flows. Income taxes resulting from the
profitable operation of a firm are usually taken into account in evaluating engineering projects.
The reason is quite simple: income taxes associated with a proposed project may represent a
major cash outflow that should be considered together with other cash inflows and outflows in
assessing the overall economic profitability of that project.
The mystery behind the sometimes complex computation of income taxes is reduced
when we recognize that income taxes paid are just another type of expense, but income taxes
saved (through depreciation, expenses, and direct tax credits) are identical to other kinds of
reduced expenses (e.g. savings in maintenance expenses).
There is a need to distinguish between income taxes and several other types of taxes.
1. Income taxes are assessed as a function of gross revenues minus allowable
deductions. They are levied by the government.
2. Property taxes are assessed as a function of the value of the property owned, such as
land, buildings, equipment, and so on, and the applicable tax rates. Hence, they are
independent of the income or profit of a firm. They are levied by the government.
3. Sales taxes are assessed on the basis of purchases of goods and/or services, and are
thus independent of gross income or profits. They are levied by the government. Sales
taxes are relevant in engineering economy studies only to the extent that they add to
the cost of the item purchased.
4. Excise taxes are assessed as a function of the sale of certain goods or services often
considered non-necessities, and are hence independent of the income or profit of the
business. Although they are usually charged to the manufacturer or original provider
of the goods or services, the cost is passed on to the purchaser.
Income taxes are usually the most significant type of tax encountered in engineering
economic studies.
The Before-Tax and After-Tax Minimum Attractive Rates of Return
In the previous units, we have taken income taxes, in general, by using a before-tax MARR,
which is larger than the after-tax MARR. An approximation of the before-tax MARR requirement,
which includes the effect of income taxes, for studies involving only before-tax cash flows can be
obtained from the following relationship:
𝐵𝑒𝑓𝑜𝑟𝑒 − 𝑡𝑎𝑥 𝑀𝐴𝑅𝑅 =
𝐴𝑓𝑡𝑒𝑟−𝑡𝑎𝑥 𝑀𝐴𝑅𝑅
(1−𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑡𝑎𝑥 𝑟𝑎𝑡𝑒)
Equation 32
This approximation is exact if the asset is non-depreciable and there are no gains or losses
on disposal, tax credits, or other types of deductions involved.
Taxable Income of Corporations (Business Firms)
At the end of each tax year, a corporation must calculate its net (i.e. taxable) before-tax
income or loss. Several steps are involved in this process, beginning with the calculation of gross
income. Gross income represents the gross profits from operations (revenues from sales minus
the cost of goods sold) plus income from dividends, interest, rent, royalties, and gains (or losses)
on the sale or exchange of capital assets. The corporation may deduct from gross income all
ordinary and necessary operating expenses, including interest, to conduct the business except
capital investments. Deductions for depreciation are permitted each tax period as a means of
consistently and systematically recovering capital investment. Consequently, allowable expenses
and deductions may be used to determine taxable income using Equation 33 below:
𝑇𝑎𝑥𝑎𝑏𝑙𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 = 𝐺𝑟𝑜𝑠𝑠 𝑖𝑛𝑐𝑜𝑚𝑒 − 𝐴𝑙𝑙 𝑒𝑥𝑝𝑒𝑛𝑠𝑒𝑠 𝑒𝑥𝑐𝑒𝑝𝑡 𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡𝑠 − 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 (𝑑𝑒𝑝𝑙𝑒𝑡𝑖𝑜𝑛)𝑑𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛𝑠
This taxable income is also referred to as net income before taxes (NIBT). When income
taxes are subtracted, the remainder is called the net income after taxes (NIAT). In summary,
𝑡𝑎𝑥𝑎𝑏𝑙𝑒 𝑖𝑛𝑐𝑜𝑚𝑒
} − 𝑖𝑛𝑐𝑜𝑚𝑒 𝑡𝑎𝑥𝑒𝑠
𝑁𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 𝑎𝑓𝑡𝑒𝑟 𝑡𝑎𝑥𝑒𝑠 = {
Equation 34
(𝑖. 𝑒. , 𝑁𝐼𝐵𝑇)
Example:
A company generates Php 1,500,000 of gross income during its tax year and incurs
operating expenses of Php 800,000. Interest payments on borrowed capital amount to Php
48,000. The total depreciation deductions for the tax year equal Php 114,000. What is the taxable
income (NIBT) of this firm?
Given:
Gross income = Php 1,500,000
Expenses = Php 800,000 + Php 48,000 = Php 848,000
Depreciation = Php 114,000
Required:
Taxable income (NIBT) = ?
Solution:
𝑇𝑎𝑥𝑎𝑏𝑙𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 = 𝐺𝑟𝑜𝑠𝑠 𝑖𝑛𝑐𝑜𝑚𝑒 − 𝐴𝑙𝑙 𝑒𝑥𝑝𝑒𝑛𝑠𝑒𝑠 𝑒𝑥𝑐𝑒𝑝𝑡 𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡𝑠 − 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 (𝑑𝑒𝑝𝑙𝑒𝑡𝑖𝑜𝑛)𝑑𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛𝑠
𝑇𝑎𝑥𝑎𝑏𝑙𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 = 1,500,000 − 848,000 − 114,000
𝑻𝒂𝒙𝒂𝒃𝒍𝒆 𝒊𝒏𝒄𝒐𝒎𝒆 = 𝑷𝒉𝒑 𝟓𝟑𝟖, 𝟎𝟎𝟎
Answer
Gain (Loss) on the Disposal of an Asset
When a depreciable asset (tangible personal or real property) is sold, the market value is
seldom equal to its book value. In general, the gain (loss) on sale of depreciable property is the
fair market value minus its book value at that time. That is,
[𝑔𝑎𝑖𝑛 (𝑙𝑜𝑠𝑠)𝑜𝑛 𝑑𝑖𝑠𝑝𝑜𝑠𝑎𝑙]𝑁 = 𝑀𝑉𝑁 − 𝐵𝑉𝑁
Equation 35
where:
MVN
BVN
=
=
market value at tax year N
book value at tax year N
When the sale results in a gain, it is often referred to as depreciation recapture. The tax
rate for the gain (loss) on disposal of depreciable personal property is usually the same as for
ordinary income or loss, which is the effective income tax rate.
When a capital asset is sold or exchanged, the gain (loss) is referred to as a capital gain
(loss). Examples of capital assets are stocks, bonds, gold, silver, and other metals, as well as real
property such as a house.
Example:
A corporation sold a piece of equipment during the current tax year for Php 78,600. The
accounting records show that its cost basis, B, is Php 190,000 and the accumulated depreciation
is Php 139,200. Assume that the effective income tax rate is 40%. Based on this information, what
is (a) the gain (loss) on disposal, (b) the tax liability (or credit) resulting from this sale, and (c) the
tax liability (or credit) if the accumulated depreciation was Php 92,400 instead of Php 139,200?
Given:
BVN = Php 190,000 – Php 139,200(depreciation) = Php 50,800,
MVN = Php 78,600
income tax rate = 40%
Required:
(a) gain (loss) on disposal = ?
(b) tax owed from the sale = ?
(c) tax owed from sale if depreciation is Php 92,400 = ?
Solution:
(a)
[𝑔𝑎𝑖𝑛 (𝑙𝑜𝑠𝑠)𝑜𝑛 𝑑𝑖𝑠𝑝𝑜𝑠𝑎𝑙]𝑁 = 𝑀𝑉𝑁 − 𝐵𝑉𝑁
[𝑔𝑎𝑖𝑛 (𝑙𝑜𝑠𝑠)𝑜𝑛 𝑑𝑖𝑠𝑝𝑜𝑠𝑎𝑙]𝑁 = 78,600 − 50,800
[𝒈𝒂𝒊𝒏 (𝒍𝒐𝒔𝒔)𝒐𝒏 𝒅𝒊𝒔𝒑𝒐𝒔𝒂𝒍]𝑵 = 𝑷𝒉𝒑 𝟐𝟕, 𝟖𝟎𝟎
Answer
(b)
𝑡𝑎𝑥 𝑜𝑤𝑒𝑑 𝑓𝑟𝑜𝑚 𝑔𝑎𝑖𝑛 = −𝑖𝑛𝑐𝑜𝑚𝑒 𝑡𝑎𝑥 𝑟𝑎𝑡𝑒 𝑥 𝑔𝑎𝑖𝑛 (𝑙𝑜𝑠𝑠)𝑜𝑛 𝑑𝑖𝑠𝑝𝑜𝑠𝑎𝑙
𝑡𝑎𝑥 𝑜𝑤𝑒𝑑 𝑓𝑟𝑜𝑚 𝑔𝑎𝑖𝑛 = −0.40 (27,800)
𝒕𝒂𝒙 𝒐𝒘𝒆𝒅 𝒇𝒓𝒐𝒎 𝒈𝒂𝒊𝒏 = −𝑷𝒉𝒑 𝟏𝟏, 𝟏𝟐𝟎 Answer
(c)
If depreciation = Php 92,400,
BVN = Php 190,000 – Php 92,400 = Php 97,600
So,
[𝑔𝑎𝑖𝑛 (𝑙𝑜𝑠𝑠)𝑜𝑛 𝑑𝑖𝑠𝑝𝑜𝑠𝑎𝑙]𝑁 = 78,600 − 97,600 = −𝑃ℎ𝑝 19,000
Therefore,
𝑡𝑎𝑥 𝑜𝑤𝑒𝑑 𝑓𝑟𝑜𝑚 𝑙𝑜𝑠𝑠 = −0.40 (−19,000)
𝒕𝒂𝒙 𝒐𝒘𝒆𝒅 𝒇𝒓𝒐𝒎 𝒍𝒐𝒔𝒔 = 𝑷𝒉𝒑 𝟕, 𝟔𝟎𝟎
Answer
General Procedure for Making After-Tax Economic Analyses
After-tax economic analyses can be performed by exactly the same methods as beforetax analyses. The only difference is that ATCFs are used in place of before-tax cash flows (BTCFs)
by including expenses (or savings) due to income taxes and then making equivalent worth
calculations using an after-tax MARR. The tax rates and governing regulations may be complex
and subject to changes, but once those rates and regulations have been translated into their
effect on ATCFs, the remainder of the after-tax analysis is relatively straightforward. To formalize
the procedure, let
Rk
=
revenues (and savings) from the project; this is the cash inflow from the
project during period k
Ek
=
cash outflows during year k for deductible expenses and interest
dk
=
sum of all noncash, or book, costs during year k, such as depreciation and
depletion
t
=
effective income tax on ordinary income; t is assumed to remain constant
during the study period
Tk
=
income taxes paid during year k
ATCFk =
ATCF from the project during year k
Table 13 shows how the computation of after-tax cash flows is facilitated:
Year
(A)
BTCF
k
Rk - Ek
(B)
Depreciation
(C) = (A) – (B)
Taxable
Income
(D) = -t(C)
Cash Flow for
Income Taxes
dk
Rk - Ek - dk
-t(Rk - Ek - dk)
Table 13. After-tax Cash Flow Computation
(E) = (A) + (D)
ATCF
(1 – t)( Rk - Ek) + tdk
Example:
If the revenue from a project is Php 10,000 during a tax year, out-of-pocket expenses are
Php 4,000, and depreciation deductions for income tax purposes are Php 2,000, what is the ATCF
when t = 0.40? What is the NIAT?
Given:
t = 0.40
Rk = Php 10,000
Ek = Php 4,000
dk = Php 2,000
Required:
(a) ATCF = ?
(b) NIAT = ?
Solution:
(a)
ATCF = (1 – t)( Rk - Ek) + tdk
ATCF = (1 – 0.40)(10,000 – 4,000) + 0.40(2,000)
ATCF = Php 4,400
Answer
(b)
NIAT = ATCF – dk
NIAT = 4,400 – 2,000
NIAT = Php 2,200
Answer
SOLVE IT!
The XYZ Semiconductor Company is attempting to evaluate the profitability of adding
another integrated circuit production line to its present operations. The company would need to
purchase two or more acres of land for Php 275,000 (total). The facility would cost Php
60,000,000 and have no net MV at the end of five years. The facility could be depreciated using
a recovery period of five years. An increment of working capital would be required, and its
estimated amount is Php 10,000,000. Gross income is expected to increase by Php 30,000,000
per year for five years, and operating expenses are estimated to be Php 8,000,000 per year for
five years. The firm’s effective income tax rate is 40%. (a) Set up a table and determine the ATCF
for this project. (b) What is the NIAT in tear three? (c) Is the investment worthwhile when the
after-tax MARR is 12% per year?
Big Picture in Focus: ULOc. Determine whether an asset should be kept or replaced using
the considerations involved in replacement studies.
Metalanguage
1. Obsolescence. It is the loss of value resulting from external economic factors to an asset
or group of assets.
Essential Knowledge
Replacement Studies
A decision situation often encountered in business firms and government organizations,
as well as by individuals, is whether an existing asset should be retired from use, continued in
service, or replaced with a new asset. As the pressures of worldwide competition continue to
increase, requiring better quality in goods and services, shorter response times, and other
changes, this type of decision is occurring more frequently. Thus, the replacement problem, as it
is commonly called, requires careful engineering economy studies to provide the information
needed for sound decisions that improve the operating efficiency and the competitive position
of an enterprise.
Engineering economy studies of replacement situations are performed using the same
basic methods as other economic studies involving two or more alternatives. The specific
decision situation, however, occurs in different forms. Sometimes it may be whether to retire an
asset without replacement (abandonment) or whether to retain the asset for back-up rather than
primary use. Also, the decision may be whether the changed requirements can be met by
augmenting the capacity or capability of the existing asset(s). Often, however, the decision is
whether to replace an existing (old) asset, descriptively called the defender, with a new asset.
The one or more alternative replacement (new) assets are then called challengers.
Reasons for Replacement Analysis
The need to evaluate the replacement, retirement, or augmentation of assets results from
changes in the economics of their use in an operating environment. Various reasons can underlie
these changes, and unfortunately they are sometimes accompanied by unpleasant financial facts.
The four major reasons that summarize most of the factors involved are:
1. Physical Impairment (Deterioration). These are changes that occur in the physical
condition of the asset. Normally, continuing use (aging) results in the operation of an
asset becoming less efficient. Routine maintenance as well as breakdown repair costs
increase, energy use may increase per unit output, more operator time is required,
and so forth. Or, some unexpected incident such as an accident occurs that affects the
physical condition and the economics of ownership and use of the asset.
2. Altered Requirements. Capital assets are used to produce goods and services that
satisfy human wants. When the demand for a good or service either increases or
decreases or the design of a good or service changes, the related asset(s) may have
the economics of its use affected.
3. Technology. The impact of changes in technology varies among different types of
assets. For example, the relative efficiency of heavy highway construction equipment
is impacted less rapidly by technological changes than automated manufacturing
equipment. In general, the costs per unit of production, as well as the quality and
other factors, are favorably impacted by changes in technology, which results in more
frequent replacement of existing assets with new and better challengers.
4. Financing. Financial factors involve economic opportunity changes external to the
physical operation or use of assets and may involve income tax considerations. For
example, the rental (lease) of assets may become more attractive than ownership.
Regardless of the financial factors involved, the replacement asset does not
necessarily have to be different from the existing asset.
Reason 2 and Reason 3 are sometimes referred to as different categories of obsolescence.
Even Reason 4 could be a form of obsolescence. In any replacement problem, however, factors
from more than one of these four major areas may be involved. Regardless of the specific
considerations and even though there is a tendency to regard it with some apprehension, the
replacement of assets often represents economic opportunity for the firm.
For purposes of discussion of replacement studies, the following is a distinction between
various types of live for typical assets.
Economic life is the period of time (years) that results in the minimum equivalent uniform
annual cost (EUAC) of owning and operating an asset. If we assume good asset management,
economic life should coincide with the period of time extending from the date of acquisition to
the date of abandonment, demotion in use, or replacement from the primary intended service.
Ownership life is the period between the date of acquisition and the date of disposal by a
specific owner. A given asset may have different categories of use by the owner during this
period. For example, a car may serve as the primary family car for several years and then serve
only for local commuting for several more years.
Physical life is the period between original acquisition and final disposal of an asset over
its succession of owners. For example, the car just described may have several owners over its
existence.
Useful life is the time period (years) that an asset is kept in productive service (either
primary or backup). It is an estimate of how long an asset is expected to be used in a trade or
business to produce income.
Factors that Must be Considered in Replacement Studies
There are several factors that must be considered in replacement studies. If they are not
properly included in the analysis, erroneous study results may jeopardize reaching a sound
decision. Once a proper perspective has been established regarding these factors, however, little
difficulty should be experienced in making replacement studies.
1. Recognition and acceptance of past errors. The economic factors in a replacement
study is the future. Any estimation errors made in a previous study related to the
defender are not relevant (unless there are income tax implications). For example,
when an asset’s BV is greater than its current MV, the difference frequently has been
designated as an estimation error. Such “errors” also arise when capacity is
inadequate, maintenance expenses are higher than anticipated, and so forth.
2. Sunk costs. Sunk cost, for the purpose of this unit, is defined as the difference between
an asset’s BV and its MV at a particular point in time. Sunk costs have no relevance to
the replacement decisions that must be made (except to the extent that they affect
income taxes).
3. Existing asset value and the outsider viewpoint. Recognition of the non-relevance of
BVs and sunk costs leads to the proper viewpoint to use in placing value on existing
assets for replacement study purposes. The “outsider viewpoint” is used in
approximating the investment amount of an existing asset (defender). In particular,
the outsider viewpoint is the perspective that would be taken by an impartial third
party to establish the fair MV of a used (secondhand) asset. This viewpoint forces the
analyst to focus on present and future cash flows in a replacement study, thus
avoiding the temptation to dwell on past (sunk) costs.
4. Income tax considerations. The replacement of assets often results in capital gains or
losses, or gains or losses from the sale of land or depreciable property. Consequently,
to obtain an accurate economic analysis in such cases, the studies must be made on
an after-tax basis. It is evident that the existence of a taxable gain or loss, in
connection with replacement, can have a considerable effect on the results of an
engineering economy study.
5. Economic life of the proposed replacement asset (challenger). The economic life of an
asset minimizes the equivalent uniform annual cost of owning and operating an asset,
and it is often shorter than the useful or physical life. It is essential to know a
challenger’s economic life in view of the principle that new and existing assets should
be compared over their economic (optimum) lives.
6. Remaining (economic) life of the old asset (defender). The economic life of a defender
is often one year. Consequently, care must be taken when comparing the defender
with a challenger asset because different lives are involved in the analysis. The
defender should be kept longer than its apparent economic life as long as its marginal
cost is less than the minimum equivalent uniform annual cost of the challenger.
Example:
A construction company is considering the replacement analysis of a pump. The old pump
has an annual operating and maintenance costs of Php 50,000 per year. It can be kept for 5 years
more. It can then be sold to the manufacturer of the new pump for Php 25,000.
The new pump will have a purchase price of Php 120,000. In 5 years, it will have a value
of Php 50,000. The annual operating and maintenance costs for this pump is Php 20,000.
Using a MARR of 20%, evaluate the investment alternative by comparing the present
worths.
Given (Cash Flow Diagram):
Old Pump:
0
1
2
3
50,000
50,000
50,000
@ i = 20% per year
4
50,000
5
50,000
New Pump:
50,000
25,000
0
120,000
1
20,000
2
3
20,000
20,000
4
20,000
5
20,000
@ i = 20% per year
Required:
PWOLD = ? vs. PWNEW = ?
Solution:
PW = PW of cash inflows – PW of cash outlflows
Solving for PWOLD:
PWOLD (20%) = - 50,000(P/A, 20%, 5)
𝑃𝑊𝑂𝐿𝐷 (20%) = −
50,000[(1+0.20)5−1]
5
0.20(1+0.20)
𝑃𝑊𝑂𝐿𝐷 = −𝑃ℎ𝑝 149,530.61
Solving for PWNEW:
𝑃𝑊𝑁𝐸𝑊 (20%) = −120,000 + 25,000 −
20,000[(1+0.20)5−1]
50,000
+ (1+0.20)
5
0.20(1+0.20)5
𝑃𝑊𝑁𝐸𝑊 = −𝑃ℎ𝑝 134,718.36
Taking the absolute value of the present worth, the PWOLD > PWNEW. The recommendation,
therefore, is to replace the old pump.
SOLVE IT!
Solve the following:
1. A petroleum company, whose minimum attractive rate of return is 10%, needs to paint
the vessels and pipes in its refinery periodically to prevent rust. “Tuff-Coat”, a durable
paint, can be purchased for Php 402.50 a gallon while “Quick-Cover”, a less durable paint,
costs Php 162.50 a gallon. The labor cost of applying a gallon of paint is Php 300.00. Both
paints are equally easy to apply and will cover the same area per gallon. Quick-Cover is
expected to last 5 years. How long must TuffCoat promise to last to justify its use? Assume
that the pipes need to be painted indefinitely (i.e., forever) and that the paint and costs
of painting never change (i.e., no inflation or technological improvements affecting the
paint or the cost to produce and sell paint, or to apply the paint).
2. Ten years ago Hyway Robbery, Inc. installed a conveyor system for Php 8,000. The
conveyor system has been fully depreciated to a zero salvage value. The company is
considering replacing the conveyor because maintenance costs have been increasing. The
estimated end-of-year maintenance costs for the next five years are as follow:
Year
Maintenance
1
Php 1,000
2
Php 1,250
3
Php 1,500
4
Php 1,750
5
Php 2,000
At any time the cost of removal just equals the value of the scrap metal recovered from
the system. The replacement the company is considering has an equivalent annual cost
of Php 1,028 at its most economic life. The company has a minimum attractive rate of
return (MARR) of 10%.
a. Should the conveyor be replaced now? Show the basis used for the decision.
b. Now assume the old conveyor could be sold at any time as scrap metal for $500
more than the cost of removal. All other data remain the same. Should the
conveyor be replaced?
Big Picture in Focus: ULOd. Determine the breakeven point with respect to production and
sales volumes.
Essential Knowledge
Breakeven Analysis
Breakeven analysis is concerned with finding the point at which revenues and costs are
exactly equal. This point is known as breakeven point. Thus this is a volume of output at which
neither a profit is made nor a loss is incurred. Therefore, production or sale must not be allowed
to fall beyond this point.
Shown in Figure 5 is the most convenient chart used in breakeven analysis. It is known as
the breakeven chart. It represents the costs, the expected income, etc.
Figure 5. The Breakeven Chart
Example 1:
Problem: CE Board May 1999
A manufacturer produces certain items at a labor cost per unit of Php 315.00, material
cost per unit is Php 100.00, variable cost of Php 3.00 each. If the item has a selling price of Php
995, how many units must be manufactured each month for the manufacturer to breakeven if
the monthly overhead is Php 461,600.00?
Given:
Let x = number of units to be manufactured per month to breakeven,
Labor cost = Php 315x
Material cost = Php 100x
Variable cost = Php 3x
Overhead cost = Php 461,600
Selling price/unit = Php 995
Required:
x=?
Solution:
Solving for the total expenses:
Total expenses = 315x + 100x + 3x + 461,600
Total expenses = 418x + 461,600
The total income is:
Total income = 995x
To breakeven:
Income = Expenses
Thus,
995x = 418x + 461,600
x = 800 units Answer
SOLVE IT!
Solve the following:
1. A telephone switchboard 100-pair cable can be made up with either enameled wire or
tinned wire. There will be 400 soldered connections. The cost of soldering a connection
on the enameled wire will be Php 1.65, on the tinned wire it will be Php 1.15. A 100-pair
cable made up with enameled wire cost Php 0.55 per linear foot and those made up of
tinned wire cost Php 0.75 per linear foot. Determine the length of cable run in feet so that
the cost of each installation would be the same.
2. A local factory assembling calculators produces 400 units per month and sells them at Php
1,800 each. Dividends are 8% on the 8,000 shares with par value of Php 250 per unit. The
fixed operating cost per month is Php 25,000. Other costs are Php 1,000 per unit.
Determine the breakeven point. If 200 units were produced per month, determine the
profit/loss.
3. A factory engaged in the fabrication of an automobile part with a production capacity of
700,000 units per year is only operating at 62% capacity due to unavailability of the
necessary foreign currency to finance the importation of their raw materials. The annual
income is Php 430,000. Annual fixed cost is Php 190,000 and variable costs are Php 0.348
per unit.
a. What is the current profit or loss?
b. What is the breakeven point?
Big Picture in Focus: ULOe. Use decision tree analysis in evaluating alternatives.
Essential Knowledge
DECISIONS RECOGNIZING RISK
Expected Monetary Value of Alternatives
Before we dive into probability and expected monetary value (EMV), we will introduce a
motivational problem from the petroleum industry. You are exploring the possibility of drilling a
potential new oil field. You can either do the drilling yourself, or you can “farm out” the drilling
operation to a partner. The field in which you are proposing to drill may or may not have oil –
you don’t know until you drill and find out. If you drill yourself, you take on the risk if the field is
not a producer, but if the field is a producer, you don’t need to share your profits with anyone. If
you farm out the drilling operation, you are not exposed to any losses if the field is not a producer
but if the field is a producer, the drilling company will take the lion’s share of the profits. Table
14 shows the net present value of the prospective oil field.
Field is Dry
Field is a Producer
Drill Yourself
- Php 250,000
Php 500,000
Farm Out
Php 0
Php 50,000
Table 14. Net Present Values (NPVs) of the Oil Field Problem
Clearly, if you knew the field was a producer you would want to drill yourself (and if you
knew it weren’t, then you would not want to drill at all). But you don’t know this before you make
your drilling decision. What should you do?
Suppose that you had enough information on the productivity of wells drilled in a similar
geology to estimate that the probability of a dry hole was 65% and the probability of a producing
well was 35%. Do these probabilities make your life any easier?
These decision problems can be solved by calculating a quantity known as the “expected
monetary value” (EMV) – basically a probability-weighted average of net present values of
different outcomes. Formally, the EMV is defined by determining probabilities of each distinct or
“mutually exclusive” outcome, determining the NPV under each of the possible outcomes, and
then weighting each possible value of the NPV by its probability. In mathematical terms, if Z is
some alternative; Y1, Y2, …, Yn represent a set of possible outcomes of some uncertain variable;
X1, X2, …, Xn represent the NPVs associated with each of the possible outcomes; and P(Y 1), P(Y2),
…, P(Yn) represent the probabilities of each of the outcomes, then the EMV is defined by:
𝐸𝑀𝑉 (𝑍) = 𝑃(𝑌1 )𝑋1 + 𝑃 (𝑌2 )𝑋2 + ⋯ + 𝑃 (𝑌𝑛 )𝑋𝑛
Equation 36
The alternative with the highest EMV would be the option chosen. A decision-maker who
chooses among alternatives in this way would be called an “expected-value decision-maker.”
Some things to remember about probabilities:
A probability is a number between zero and one. So P(Yj) = 0.05 means that there
is a 5% chance of outcome j occurring. If P(Yj) = 0, it means that outcome j never occurs
and if P(Yj) = 1, it means that outcome j always occurs.
If a set of outcomes is mutually exclusive (meaning that multiple outcomes cannot
occur) and exhaustive (meaning that the set captures all possible outcomes), then the
probabilities of all outcomes in that set would be equal to one. Rolling a six-sided die with
the outcomes {1,2,3,4,5,6} is an example. This set of outcomes is mutually exclusive
because the face of the die cannot show two numbers at once (unless the die is crooked).
The set of outcomes is exhaustive since it contains all possible outcomes of the die being
rolled. In this case P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1. The set of outcomes {1,2,3} is
an example of non-exhaustive outcomes, since the die could show a 4, 5 or 6 upon being
rolled.
Now, back to our oil field problem. We’ll describe the problem again, using the language
of expected monetary value. There are two alternatives – to drill yourself or to farm-out. The
uncertainty is in the outcome of the drilling process. The set of mutually exclusive and exhaustive
outcomes is {dry hole, producer}. These are the only two possible outcomes regardless of
whether you choose to drill yourself or farm-out. The NPVs of each alternative, under each
possible outcome, are shown in Table 14. To decide whether to drill or farm out, you would
calculate the EMV of each option as follows:
𝐸𝑀𝑉 (𝐷𝑟𝑖𝑙𝑙) = 𝑃(𝑑𝑟𝑦 ℎ𝑜𝑙𝑒) 𝑥 − 𝑃ℎ𝑝 250,000 + 𝑃 (𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑟) 𝑥 𝑃ℎ𝑝 500,000
𝐸𝑀𝑉 (𝐷𝑟𝑖𝑙𝑙 ) = 0.65 𝑥 − 𝑃ℎ𝑝 250,000 + 0.35 𝑥 𝑃ℎ𝑝 500,000 = 𝑃ℎ𝑝 12,500
𝐸𝑀𝑉 (𝐹𝑎𝑟𝑚𝑜𝑢𝑡 ) = 𝑃(𝑑𝑟𝑦 ℎ𝑜𝑙𝑒) 𝑥 − 𝑃ℎ𝑝 0 + 𝑃 (𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑟) 𝑥 𝑃ℎ𝑝 50,000
𝐸𝑀𝑉 (𝐹𝑎𝑟𝑚𝑜𝑢𝑡 ) = 0.65 𝑥 − 𝑃ℎ𝑝 0 + 0.35 𝑥 𝑃ℎ𝑝 50,000 = 𝑃ℎ𝑝 17,500
In this case, you should choose to farm out the drilling operation.
The basic idea behind EMV is fairly straightforward, assuming that you can actually
determine the relevant probabilities with some precision. But the meaning of the EMV is a little
bit subtle and requires some degree of care in interpretation. Let’s take a very basic situation – a
coin flip. Suppose that we were to flip a coin. If it shows heads, you must pay me Php 1. If it shows
tails, then I must pay you Php 1. The EMV of this game, assuming that heads and tails have equal
probabilities, is Php 0. (See if you can figure out why, based on the EMV equation and the fact
that P(heads) = 0.5 and P(tails) = 0.5.) But if you think about this for a minute, how useful is the
EMV? If you play the coin-flipping game once, you will never ever have an outcome where the
payoff to you is Php 0. The payoff will either be that you gain or lose one peso.
If you look at the EMVs from the oil-field problem, you will see the same thing. The EMV
of drilling is Php 12,500 but there is no turn of events under which you would wind up earning
Php 12,500 – if you drill, it would either be that you lose Php 250,000 or gain Php 500,000.
Similarly, if you farm out you will never earn exactly Php 17,500. You will either lose nothing
(payoff of Php 0) or you will gain Php 50,000. So what does the EMV mean when it tells you that
farming-out is the better option?
It’s important to remember that the EMV is a type of average. If you were to play the
coin-flip game or the oil-field game a large number of times under identical circumstances, and
make the same decision each time (i.e., to drill or to farm out), then over the long run you would
expect to wind up with Php 12,500 if you choose to drill and Php 17,500 if you choose to farm
out. While the EMV may be useful for gamblers or serial investors, using EMV needs to be done
with some care for stand-alone projects in the face of uncertainty.
Decision Tree Analysis
A decision tree is a decision support tool that uses a tree -like graph or model of decisions
and their possible consequences, including chance event outcomes, resource costs, and utility.
Decision trees provide an alternative and more convenient way of viewing and managing large
sets of rules, especially when these rules are not symmetric.
Decision trees are easy to use once you understand that:
 A condition is declared in its diamond-shaped node.
 The possible values for the condition are represented by branches.
 The actions are declared at the end of each branch.
To help you better understand the basic concepts and how they are applied, this section
is divided into the following sub-sections:
1. Decision tree notation (nodes and branches)
2. Payoff values
3. Outcome probability
4. Expected value
5. Decision tree analysis
Decision Tree Notation (Nodes and Branches)
Any decision includes two or more decision alternatives. Any decision alternative
might lead to multiple possible outcomes. One outcome may depend on another, a situation
called dependent uncertainty. Decisions may also be linked in a sequence, a condition called
sequential decisions. Use decision tree notation to keep these myriad paths and possibilities
easy to understand and compare.
Figure 6. Decision Tree Notations
In the example shown in Figure 6, a company is evaluating whether to invest $1M in a
project immediately or wait for a marketing report that may affect project development. Two
other alternatives are also possible: invest $1M in a fixed yield bond or do nothing. A fixed-yield
investment and doing nothing are examples of baseline alternatives: choices that can be used to
compare the overall merits of the decision alternatives.
A. Decision nodes and the root node: Small squares identify decision nodes. A decision
tree typically begins with a given “first decision.” This first decision is called the root
node. For example, the root node in a medical situation might represent a choice to
perform an operation immediately, try a chemical treatment, or wait for another
opinion.
Draw the root node at the left side of the decision tree.
B. Chance nodes: Small circles identify chance nodes; they represent an event that can
result in two or more outcomes. In this illustration two of the decision alternatives
connect to chance nodes. Chance nodes may lead to two or more decision or chance
nodes.
C. Endpoints: An endpoint, or termination node, indicates a final outcome for that
branch. Small triangles identify endpoints. Show an endpoint by touching one point
of the triangle to the branch it terminates.
D. Branches: Lines that connect nodes are called branches. Branches that emanate from
a decision node (and towards the right) are called decision branches. Similarly,
branches that emanate from a chance node (and towards the right) are called chance
branches. In other words, the node that precedes a branch identifies the branch type.
A branch can lead to any of the three node types: decision node, chance node, or
endpoint.
Tip: Draw branches from the root node with a generous amount of space between the
branches. As branches extend outwards they may spawn any number of additional
nodes and branches. Start with enough room between branches to easily
accommodate the alternatives and outcomes that may result.
Payoff Values
The payoff value is equivalent to the net profit (or net loss) expected at the end of any
outcome. Write payoff values at their respective branch endpoints. Although you can express
payoff in various ways, it is common to use monetary units in most business applications.
Payoff is the difference between investment cost and gross revenue. This section adopts the
convention of indicating investment costs as negative values to simplify calculating payoff
values. Payoff values can be positive or negative. Negative payoff values indicate a net loss.
Figure 7. Payoff Value
Figure 7 shows the expected payoffs at two endpoints. The fixed-yield investment results
in $1,050,000 revenue, and therefore a $50,000 payoff. The payoff for doing nothing is $0. The
other branches lead to chance nodes at this stage of the decision tree. You can assign payoff
values only after these chance nodes lead to endpoints.
Outcome Probability
A chance node leads to two or more outcomes, each outcome represented by a new
branch. As with a game of chance, an outcome has a particular probability of happening. The
total of all outcomes for a given chance node must equal 100% (or 1.0). A standard decision
tree convention expresses probabilities as decimal fractions in parentheses at the chance
branches.
In Figure 8, the decision alternative to develop a project immediately can lead to one
of three outcomes. The company has determined that there is a 20% chance that the project
can meet all the criteria for success in international and domestic markets, but a 50% chance
that the project will only meet the criteria for the domestic market. In addition, there is a 30%
chance that the project will not meet enough criteria for either market as a result of
insufficient information. The company can also wait for a marketing study before developing
the project. The marketing information may help the company create a successful project.
But the information may also suggest unfavorable conditions that the company probably
cannot overcome. The company uses its best judgment and guess, with a 50% favorable and
a 50% unfavorable outcome.
The decision tree shows these probabilities as decimal fractions in parentheses on
their respective chance branches.
Figure 8. Outcome Probabilities
Expected Value
Expected value (EV) is a way to measure the relative merits of decision alternatives.
The expected value term is a mathematical combination of payoffs and probabilities. You
calculate the expected values after all probabilities and payoff values are identified. The goal
of the calculations is to find the EV for each decision alternative emerging from the root node.
For the purposes of this primer, the decision alternative with the highest EV is the best choice.
Although you can apply the formal definition of expected value, in practice you can
calculate EV calculations by applying the following rules. To calculate EV, start from the
endpoints and work back towards the root. An easy way to find expected values is to calculate
an EV for each terminated branch, then each chance node and each decision node.
 For a terminated decision branch, EV is equal to the payoff.
 For a terminated chance branch, EV is the product of its payoff and probability.
 For a chance node, EV is the sum of each chance branch payoff multiplied by
the probability for that payoff.
 For a decision node, EV is the greater EV value of any decision branch. Mark
the lower value EV branches with double-hatch marks to disregard these
branch paths. Since the root node is also the first decision node, the decision
alternative with the greater EV is the overall best decision.
 As the calculations are carried from right to left, use a “resolved” EV at any
node type as the payoff “input” at the node closer to the root.
Tip: Start an EV calculation from the endpoint and then proceed from right to left. The
EV for a node becomes the payoff “input” for the subsequent EV calculation to the left. For
example, use the EV for the topmost decision node (types of markets to enter) as an input to
calculate the chance node (criteria outcomes).
Example, calculate the EV for the decision alternative, as shown in Figure 9, to develop
the project by following the given EV rules:
 Decision node (“international and domestic marketing” vs. “domestic
marketing only”). The EV is the greatest value given by all the decision
branches, $3,000,000. This value then becomes the payoff “input” for the next
node to the left.
 Chance node (“all criteria” vs. “domestic criteria only” vs. “not enough
criteria”). *$3,000,000 x 0.2+ + *$500,000 x 0.5+ + *(-$1,000,000) x 0.3+ = EV
= $550,000. This value becomes the payoff “input” for this alternative when
considering the root node.
 By a similar calculation, the EV for the alternative to wait for the report before
deciding whether or not to develop the project is EV = $750,000. This value
becomes the payoff “input” for this alternative when considering the root
node.
 The EV for the alternative to invest the capital in a fixed-yield investment is
just the payoff value, EV = $50,000.

The EV for doing nothing is EV = $0.
Figure 9. Expected Values
Decision Tree Analysis
The EV at the root node shows that the decision to wait for the marketing report is
the best decision. This result may come as a surprise. You can better understand the result
after calculating expected values.
Before the decision tree is analyzed you may be tempted to assume that the decision
to develop the project immediately is the better choice. After all, the project will only cost
$1,000,000 instead of a rushed cost of $1,500,000. Furthermore, there are fewer
complications to consider, like waiting to determine the potential for international
distribution. The EV for the decision alternative to wait for the report is complicated by two
major chance factors. One factor is that the company knows that waiting makes finding an
international distributor more difficult than if the project begins immediately. The company
has determined that the likelihood of finding an international distributor is less certain (by
50%) if they wait for the report.
The other chance factor is the information in the marketing report. The company
estimates that the marketing report has a 50% chance of delivering favorable data which will
help project development. Two or more chance nodes directly connected in this way indicate
a dependent uncertainty, a condition that can readily be evaluated through decision tree
analysis. Another complication is that rushing development raises development costs by a
third, to ($1,500,000), and this alone reduces the payoff for the international and domestic
marketing by $500,000.
The decision tree method requires probabilities for all chance outcomes. In this
example, the successive chance outcomes of waiting for report results and then securing an
international distributor reduce the EV along the branch path.
But a similar analysis of the competing decision alternative reveals important
information. Without the benefit of the marketing report the chances of “getting it right” for
the international market are fairly low, at 20%. This factor significantly weakens the value
of that branch. This alternative is also risky; there is a 30% chance that the company will lose
all investment costs. The absence of reliable market information means that the project may
not meet criteria for success in any market.
The analyst can sum up the decision tree analysis with the following major
presentation points and with the decision tree:
 The international market potential is $3,000,000 in revenue, while the
domestic market is only $500,000.
 Immediate project development costs only $1,000,000.
 Waiting to develop the project results in rush costs, pushing the total to
$1,500,000.
 Marketing information plays the most important role in the potential success
of this project. In the absence of valid marketing data, the chance for success
in the international market is poor (20%) and the chance for complete failure
is significant (30%). These risk factors significantly reduce the potential for
product success.
 Waiting for the marketing report can complicate project development. There
is a 50% chance that the report will be favorable enough to proceed. Waiting
also reduces the chance (by 50%) for recruiting a distributor in time to capture
the international market. However, these risk factors of waiting do not affect
the chance of success as much as the absence of marketing data.
 Therefore the best decision, given the known assumptions, uncertainties, and
information, is to wait for the results of the marketing report before deciding
to develop the project.
Figure 10. Decision Tree Analysis
SOLVE IT!
Problem:
An investor had Php 2,000,000 in his account and wants to invest it in oil business. He had
an idea of 1 month, 3 months, 6 months and 1 year investment plan. He wishes to invest 27% of
the funds for the 3-month plan, 24% on the 6-month plan, 21% on the 1-year plan and 28% for
the 1-month plan. He wants a return of at least 20% on any of this investment plan. The risks of
achieving this target are: 94% for Plan1, 96% for Plan2, 97% for Plan 3, and 98% for Plan 4. Using
a decision tree analysis, please advise the investor on the appropriate investment plan to invest
in.
Key Points
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Invest opportunities can be categorized as mutually exclusive, independent,
and contingent.
Depreciation is the decrease in value of physical properties with the passage
of time and use and can only be established when asset is retired from
service.
The two types of depreciation are physical depreciation – depreciation due
to the reduction of the physical ability of an equipment or asset to produce
results, and functional depreciation – depreciation due to the reduction in
the demand for the function that the equipment or asset was designed to
render.
The four methods in computing depreciation includes straight line method,
sinking fund method, declining balance method, and sum-of-year’s digit
method.
The declining balance method has the largest annual depreciation charge
while the sinking fund method has the smallest annual depreciation charges
Deductions for depreciation are permitted each tax period as a means of
consistently and systematically recovering capital investment.
The need to evaluate the replacement, retirement, or augmentation of
assets results from changes in the economics of their use in an operating
environment.
Breakeven analysis is concerned with finding the point, called the breakeven
point, at which revenues and costs are exactly equal.
The expected monetary value is basically a probability-weighted average of
net present values of different outcomes.
A decision tree is a decision support tool that uses a tree -like graph or model
of decisions and their possible consequences, including chance event
outcomes, resource costs, and utility.
BIG PICTURE
Week 8 – 9: Unit Learning Outcomes (ULO)
At the end of the unit, you are expected to:
a. perform a single variable sensitivity analysis.
Big Picture in Focus: ULOa. Perform a single variable sensitivity analysis.
Essential Knowledge
DECISIONS ADMITTING UNCERTAINTY
Sensitivity Analysis
In sensitivity analysis, we aim to discover the magnitude of change in one variable (here,
output variables) with respect to change in other variable(s) (here, input parameters). Then, we
can rank the variables based on their sensitivity. It helps the decision maker to understand the
parameters that have the biggest impact on the project.
The following example introduces a single variable sensitivity analysis. Please note that
here we assume variables are independent and have no effect on each other. For example, it is
assumed that magnitude of initial investment doesn’t affect the operating costs.
Example:
For a project, the most expected case includes an initial investment of Php 150,000 at
present time, annual income of Php 40,000 for five years (starting from first year) and salvage
value of Php 80,000. Evaluate the sensitivity of the project ROR to 20% and 40% increase and
decrease in initial investment.
Given (Cash Flow Diagram):
40,000
0
150,000
1
80,000
40,000
2
40,000
3
40,000
4
40,000
5
Required:
Sensitivity analysis
Solution:
Solving for ROR (i) using the PW method:
150,000 = 40,000(P/A, i%, 5) + 80,000(P/F, i%, 5)
i = 20.5%
A. Sensitivity Analysis of Initial Investment
If initial investment is decreased by 40%:
(1 – 0.4)(150,000) = 90,000
Solving for the new ROR (i) using the PW method:
90,000 = 40,000(P/A, i%, 5) + 80,000(P/F, i%, 5)
i = 43.5%
If initial investment is decreased by 20%:
(1 – 0.2)(150,000) = 120,000
Solving for the new ROR (i) using the PW method:
120,000 = 40,000(P/A, i%, 5) + 80,000(P/F, i%, 5)
i = 29.6%
If initial investment is increased by 20%:
(1 + 0.2)(150,000) = 180,000
Solving for the new ROR (i) using the PW method:
180,000 = 40,000(P/A, i%, 5) + 80,000(P/F, i%, 5)
i = 13.8%
If initial investment is increased by 40%:
(1 + 0.4)(150,000) = 210,000
Solving for the new ROR (i) using the PW method:
210,000 = 40,000(P/A, i%, 5) + 80,000(P/F, i%, 5)
i = 8.6%
Summary:
%age Change in
ROR
20.5% ROR
Prediction
90,000
-40%
43.5%
112.7%
120,000
-20%
29.6%
44.8%
150,000
0
20.5%
0%
180,000
+20%
13.8%
-32.6%
210,000
+40%
8.6%
-57.8%
Table 15. Changes in ROR with respect to changes in initial investment
Initial Investment
(in Php)
Change in
Prediction
As you can see, changes in ROR with respect to changes in initial investment are
considerably high. In general, parameters that are close to time zero have a higher impact on
the ROR of the project.
SOLVE IT!
Problem:
For the same project as that in the example, evaluate the sensitivity of the project ROR
to 20% and 40% increase and decrease in annual income, project life and salvage value. Which
among the initial investment, annual income, project life and salvage value has the least effect
on the ROR of the project?
Decision Analysis Models
Decision analysis (DA) is a systematic, quantitative, and visual approach to addressing and
evaluating the important choices that businesses sometimes face. Ronald A. Howard, a professor
of Management Science and Engineering at Stanford University, is credited with originating the
term in 1964. The idea is used by large and small corporations alike when making various types
of decisions, including management, operations, marketing, capital investments, or strategic
choices.
Decision analysis uses a variety of tools to evaluate all relevant information to aid in the
decision-making process and incorporates aspects of psychology, management techniques,
training, and economics. It is often used to assess decisions that are made in the context of
multiple variables and that have many possible outcomes or objectives. The process can be used
by individuals or groups attempting to make a decision related to risk management, capital
investments, and strategic business decisions.
A graphical representation of alternatives and possible solutions, as well as challenges
and uncertainties, can be created on a decision tree or influence diagram. More sophisticated
computer models have also been developed to aid in the decision-analysis process.
The goal behind such tools is to provide decision-makers with alternatives when
attempting to achieve objectives for the business, while also outlining uncertainties involved and
providing measures of how well objectives will be reached if final outcomes are achieved.
Uncertainties are typically expressed as probabilities, while frictions between conflicting
objectives are viewed in terms of trade-offs and utility functions. That is, objectives are viewed
in terms of how much they are worth or, if achieved, their expected value to the organization.
Despite the helpful nature of decision analysis, critics suggest that a major drawback to
the approach is "analysis paralysis," which is the overthinking of a situation to the point that no
decision can be made. In addition, some researchers who study the methodologies used by
decision-makers argue that this type of analysis is not often utilized.
Key Points
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Sensitivity analysis aims to discover the magnitude of change in one variable
with respect to change in other variable(s).
Decision analysis (DA) is a systematic, quantitative, and visual approach to
addressing and evaluating the important choices that businesses sometimes
face.