Engineering Economics Lecture (# 1 - 6)

Welcome to Heartiest Students
By
Meseret G [Lecturer]
Engineering Economics [CENG 5011]
Department of Civil Engineering
School of Engineering
Adama Science and Technology University
Semester II
2013 4/10/2013
2
Chapter 1
General Introduction
Lecture # 1
Meseret G. [CENG 5011] 4/10/2013

Introduction
3
Definition of Engineering and Science
 Engineering a profession in which knowledge of the

mathematical and natural sciences gained by study,
experience and practice is applied with judgment to
develop ways to
a. Utilize the material and forces economically and
b. For the benefit of mankind

Meseret G. [CENG 5011] 4/10/2013
Introduction
4
Definition Cont’d
 Engineering is not a science but application of science
 Nature of engineering: application
 Role of scientist: Discover the universal laws of nature/

behavior
 Role of engineer: Apply knowledge to particular situation

to produce products and services
Meseret G. [CENG 5011] 4/10/2013
Introduction
5
What is Economy?
 It is the management of household or private expenses.
[From Greek]
 It is the careful management of material resources.

[Dictionary Definition]
Meseret G. [CENG 5011] 4/10/2013

Introduction
6
What is Economics?
 It is the branch of social science that deals with the production
& distribution and consumption of goods and services and
their management.
 It is the study of production and distribution of wealth.
 It is the study of choice and decision-making in the world with
limited resources.
 It is the study of how individuals, businesses & governments use
their limited resources and satisfy unlimited wants. [Others]
Meseret G. [CENG 5011] 4/10/2013
Introduction
7
Engineering and Economics

 Engineering activities are means of satisfying human wants
and requirements
 Concerns – material / forces and needs
 Because of resource constraints, engineering is closely
associated with economics
 Its essential that an engineering proposal is evaluated in
terms of economics [worth & cost] before it is undertaken
 Essential pre-requisite of successful engineering application is
economic feasibility Meseret G. [CENG 5011] 4/10/2013
Introduction
8
Engineering Economics [EEs]

 Engineering economy is a collection of mathematical /
analytical techniques that simplify economic comparison
 Engineering economy: formulation, estimation and

evaluation of the economic outcomes out of various available
alternatives to accomplish a defined purpose
 Discipline that involves the systematic evaluation of the cost

and benefit [economic merits] of proposed technical projects
Meseret G. [CENG 5011] 4/10/2013
Introduction
9
Engineering Economics [EEs] Cont’d

 To be economically acceptable [i.e., affordable], solutions to
engineering problems must demonstrate a positive balance of
long-term benefits over long-term costs.
 The mission of EEs is to balance different types of costs and

the performance [response time, safety, reliability, etc.] in the
most economical manner.
Meseret G. [CENG 5011] 4/10/2013

Introduction
10
Principles of EEs
Principle 1: Develop the alternatives
 The alternatives need to be identified and then defined for

subsequent analysis.
 Since the choice [decision] is among alternatives, developing

and defining the comprehensive list of alternatives for
detailed evaluation is important.
 Creativity and innovation are essential. 4/10/2013

Meseret G. [CENG 5011]
Introduction
11
Principles Cont’d
Principle 2: Focus on the differences
 Only the differences in expected future outcomes among the

alternatives are relevant to their comparison and should be
considered in the decision.
 Outcomes that are common to all alternatives can be

disregarded in the process of comparison and decision.
Meseret G. [CENG 5011] 4/10/2013
Introduction
12
Principles Cont’d
Principle 3: Use a consistent viewpoint
 The prospective outcomes of the alternatives, selection of the
criteria and other, should be consistently developed from a
defined viewpoint [perspective.
 Usually, the viewpoint of the decision maker would be used.
 For example, the perspective of the employees is used for

the problem of designing the employee benefit package.
Meseret G. [CENG 5011] 4/10/2013

Introduction
Principles Cont’d
13
Principle 4: Use a common unit of measure

 Using a common unit of measurement to enumerate as many
of the prospective outcomes as possible will simplify the
analysis of the alternatives.
 For economic outcomes, a monetary unit such as
“birr/dollars” is the common unit of measure.
 If the outcomes cannot be quantified, describe these
consequences explicitly so that the information is useful to the
decision maker in the comparison of the alternatives.
Meseret G. [CENG 5011] 4/10/2013
Introduction
14
Principles Cont’d
Principle 5: Consider all relevant criteria
Principle 6: Make uncertainty explicit
 Risk and uncertainty are inherent in estimating the future

outcomes of the alternatives and should be recognized in the
analysis and comparison.
Meseret G. [CENG 5011] 4/10/2013

Introduction
15
Principles Cont’d
Principle 7: Revisit your decisions
 Improved decision making results from an adaptive process.

The initial projected outcomes of the selected alternative
should be subsequently compared with the actual results
achieved.
Meseret G. [CENG 5011] 4/10/2013

Introduction
16
Procedures of Engineering Economic Analysis

 Problem recognition, formulation, and evaluation.
 Development of the feasible alternatives. [Principle 1]
 Development of the outcomes and cash flows for each
alternative. [Principles 2, 3 and 4]
 Selection of a criterion [or criteria]. [Principles 3 and 5]
 Analysis and comparison of the alternatives.[Principle 5]
 Selection of the preferred alternative. [Principle 6]
 Performance monitoring and post-evaluation of results.
[Principle 7]
Meseret G. [CENG 5011] 4/10/2013
Introduction
17
Example: Say you and two friends find yourself strong

desire for a pizza. Having it delivered is the only option
since you are busy.
a. Pizza House 1 [square pizza] has 0.8in thick and 15in
wide and costs 40 Birr + 5% sales tax + 2.5 Birr delivery
charge.
b. Pizza House 2 [round pizza] has 0.9in thick and 5in
diameter and costs 45 Birr + 5% sales tax.
Applying the seven-step procedure.
Meseret G. [CENG 5011] 4/10/2013
Introduction
18
Example Cont’d
Step 1: Define the Problem
 Hunger satisfaction with a pizza.
Step 2: Develop Alternatives [Principle 1]
 Alternative A: Order pizza 1 [square]
 Alternative B: Order pizza 2 [round]
Meseret G. [CENG 5011] 4/10/2013

Introduction
19
Example Cont’d
Step 3: Outcomes and cash flows for each alternative
 Outcomes should be developed from the consistent viewpoint:
You and your friends’ perspective [Principle 3]
 Compare the differences between Alternatives A and B.
[Principle 2]
 Alternative A varies from B in the aspects of the size of the
pizza, total cash flow and may also delivery time.
Meseret G. [CENG 5011] 4/10/2013
Introduction
20
Example: Step 3 Cont’d

 Common unit of measure: Dollars or cost per unit volume.
[Principle 4]
 Alternative A: Cash outflow = 42 Birr+ [42 * 5%] Birr + 2.5

Birr =46.60 Birr for a square pizza.
 Alternative B: Cash outflow = 45 Birr + [45 * 5%] = 47.25

Birr for a round pizza.
Meseret G. [CENG 5011] 4/10/2013
Introduction
21
Example Cont’d
Step 4: Selection of criteria
 It is important to use consistent viewpoint [Principle 3] to

define the selection criteria.
 Criteria: [Principle 5]: Cost per unit volume, Variety and

quality, Delivery time, Customer service etc
Meseret G. [CENG 5011] 4/10/2013

Introduction
22
Example Cont’d
Step 5: Analysis and comparison of the alternatives
[Principle 5]
 Alternative A: Cost per unit volume = 0.259 Birr per in3.
 Alternative B: Cost per unit volume = 0.297 Birr per in3.
 Further information is required to compare both alternatives

under the other criteria.
Meseret G. [CENG 5011] 4/10/2013
Introduction
23
Example Cont’d
Step 6: Selection of best alternative
 When performing this step, the uncertainly and risk should be
stated explicitly [Principle 6].
 Uncertainties which relevant to the decision includes Quality,
Delivery time etc
 Based on the criteria of the smallest cost per unit volume,
Alternative B is the best choice.
Meseret G. [CENG 5011] 4/10/2013
Introduction
24
Example Cont’d
Step 7: Performance monitoring and post-evaluation of

results [Principle 7]
 Tasted good?
 On-time delivery?
Meseret G. [CENG 5011] 4/10/2013

Introduction
25
Generally
Engineering economy is an answer to following questions
 Which engineering projects are worthwhile? [project
worthiness]
 Which engineering projects should have a higher priority?
[priority for available alternatives]
 How should the engineering project be designed?
[economic design]
Meseret G. [CENG 5011] 4/10/2013
Introduction
26
Role of Engineering Economy in Decision Making

 Assist people in making decisions
 Timeframe: future
 Actual value may differ from estimated one
 Sensitivity analysis: changes in decisions with varying
estimates
 Analysis of present and past situations based on observed
data to predict the future
Meseret G. [CENG 5011] 4/10/2013
Introduction
27
Rational Decision Making Process

 Recognize a decision problem
 Define the goals and/or objectives
 Collect the relevant information
 Identify a set of feasible decision alternatives
 Select the decision criteria to use
 Select the best alternatives

Meseret G. [CENG 5011] 4/10/2013
Introduction
28
Types of Strategic Economic Decisions
 Service and quality improvement
 New product and product expansion
 Equipment and equipment selection
 Equipment Replacement
 Cost reduction etc.
Meseret G. [CENG 5011] 4/10/2013

Introduction
29
Fundamental Principles in Engineering Economics

 Principle 1: Money now is worth more than money at later
time.
 Principle 2: All that counts is the differences among
alternatives.
 Principle 3: Marginal revenue must exceed marginal cost.
 Principle 4: Additional risk is not taken without the expected
additional return.
Meseret G. [CENG 5011] 4/10/2013
Introduction
30
Bi-Environmental Nature of Engineering

 Two environments are needed for the work of an engineer:
 Physical and Economical environments
 An engineer needs to establish efficiency in both environments
that are not independent
Engineering Production/consumption Need satisfaction

proposal
Physical Economic
Environment Environment
Meseret G. [CENG 5011] 4/10/2013
Introduction
31
Bi-Environmental Nature of Engineering Cont’d

 Physical environment is based on physical laws [formulas,
maths, calculations], therefore, it is certain
 Economic environment is affected by the behavior of

people, therefore, its not certain and economic laws are not
exact and certain
 The function of engineering is to manipulate the physical

environment to create value in economic environment
Meseret G. [CENG 5011] 4/10/2013
Introduction
32
Physical and Economic Efficiency
 Physical efficiency = output / input
 Physical efficiency is always less than unity or 100%
 Economic efficiency = worth / cost
 Economic efficiency can 100% or even more
 Overall economic efficiency= economic/physical
= physical x worth / cost

Meseret G. [CENG 5011] 4/10/2013
Introduction
33
Engineering Process
 Engineering activities dealing with elements of the physical

environment take place to meet human needs that arise in
an economic setting.
Step 1: Determination of objectives
 Finding out what people need and want that can be supplied
by engineering  Assistance in decision making
Meseret G. [CENG 5011] 4/10/2013
Introduction
34
Engineering Process Cont’d
Step 2: Identification of strategic factors
 The factors that stand in the way of attaining objectives are

known as limiting factors.
 Limiting factors are examined to locate strategic factors 

which can be altered to remove limitations restricting the
success of an undertaking.
Meseret G. [CENG 5011] 4/10/2013
Introduction
35
Step 3: Determination of means [Engineering proposals]
 Discovering what means exist to alter strategic factors in

order to overcome limiting factors [To solve problems from
scratch]
Meseret G. [CENG 5011] 4/10/2013

Introduction
36

Step 4: Evaluation of engineering proposals
 Existence of same result with a variety of means.
 Thus, determine the best means for solving the problem at
hand.
Step 4: Evaluation of engineering proposals
 The decision-maker may not be sufficiently knowledgeable
about the technical aspect.
 Thus, engineer can help to bridge this gap.
Meseret G. [CENG 5011] 4/10/2013
Introduction
37
Engineering Economic Studies

 Four key steps in planning an economic study are:
 Step 1: Creative Step : People with vision and initiative adopt the
premise that better opportunities exist than are known to them. This
leads to exploration, and investigation of potential opportunities.
 Step 2: Definition Step : System alternatives are synthesized with

economic requirements and physical requirements, and
enumerated with respect to inputs/outputs.
Meseret G. [CENG 5011] 4/10/2013
Introduction
38
Engineering Economic Studies Cont’d

 Step 3: Conversion: System alternatives are converted to a
common measure/scale so that systems can be compared.
 Step 4: Decision : Qualitative and quantitative inputs and
outputs to/from each system form the basis for system
comparison and decision making.
 Decisions among system alternatives should be made on the
basis of their differences.
Meseret G. [CENG 5011] 4/10/2013
Introduction
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Examples
 Infrastructure expenditure decision
 Replace versus repair decisions
 Selection of inspection method
 Selection of a replacement for an equipment
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End of Chapter 1
General Introduction
Lecture # 1
Thank You!!!
Meseret G. [CENG 5011] 4/10/2013

1
By
Semester II
2013 4/10/2013
2
Chapter 2
Fundamental Economic and Cost Concepts
Lecture # 2
Meseret G. [CENG 5011] 4/10/2013

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Important Terminology
Utility is the level of satisfaction i.e. obtained by consuming
a commodity [good or service]
Value designate the worth that a person attaches to an
object or service
 Value is a measure or appraisal of utility in some medium
of exchange.
 It is not the same as cost or price.
Meseret G. [CENG 5011] 4/10/2013
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Consumer goods are the goods and services that directly

satisfy human wants. Example: TV, shoes, houses, roads etc.
The goal of the consumer is to maximize his utility.
Producer goods are the goods and services that satisfy

human wants indirectly as a part of the production or
construction process. Example: factory equipment, industrial
chemicals ands materials.
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The Utilities of Goods

 Consumer goods: The utility in this case is considered
objectively and/or subjectively.
 Producer goods: The utility stems for their means to get to

an end. The utility in this case is considered objectively.
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Economy of exchange
 It occurs when utilities are exchanged by two or more

people.
 It is possible because consumer utilities are evaluated

subjectively.
 Mutual benefit in exchange
 Persuasion in exchange. Salesperson

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Economy of Organization
Through organizations, ends can be attained or attained

more economically by:
1. Labor saving
2. Efficiency in manufacturing or capital use
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Cost [Classification]
First [or initial] cost: Cost to get activity started such as

property improvement, transportation, installation, and initial
expenditures
Operation and maintenance cost: Experienced continually

over the useful life of the activity
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Fixed cost: It does not depend on output volume.
 It remain constant over a specific range of operating

conditions.
 It is subject to change when larger changes in the operating

conditions involved such as plant expansion or shutdown.
Variable cost: Vary in total with the output units, e.g.

material cost. [It is per unit of output]
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Incremental or marginal cost: the additional costs that result

from increasing the output level by one or more units. It is
determined from the variable cost.
Sunk cost: It cannot be recovered or altered by future

actions. Usually this cost is not a part of engineering
economic analysis.
Life-cycle cost: Feasibility, design, construction, operation

and disposal costs Meseret G. [CENG 5011] 4/10/2013
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Recurring Costs: Repetitive and occur when an organization

produces similar goods or services on a continuing basis, e.g.,
variable cost and a periodic fixed cost like office rent.
Nonrecurring Costs: Not repetitive, e.g., construction cost of

the manufacturing plant.
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Example
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Direct costs: Attributed to a specific activity or output, e.g.,

labor and material costs directly associated with the a
product.
Indirect costs: Cannot be attributed to a specific activity or

output.
 Allocated through a selected formula, such as by proportion

to the outputs or work activities. E.g. Profit cost, overhead cost
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Standard costs: Planned costs per unit of output that are

established in advance of actual production.
 Play an important role in cost control and other

management functions.
 One typical use is to measure operating performance by

comparing actual cost per unit with the standard unit cost
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Cash Costs: involve payment of cash and would result in a

cash flow.
Book Costs: do not involve cash transaction and is reflected

in the accounting system as a noncash cost, e.g.,
depreciation.
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Opportunity cost is incurred because there are only limited

resources available.
 The opportunity cost of the selected alternative is the value

of the next best alternative opportunity that is forgone
[given up].
 The value of the scarified alternative [Something is what you

sacrifice to get it]
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Example [Opportunity Cost]
 Suppose you have $2,000 in your bank account, you can

either keep it in the bank or invest it in stock market.
 What is the opportunity cost of investing in stock market?
 You will earn the accrued interest of $2,000 if you choose to
keep it in the bank instead of investing it in stock market.
 Therefore, The opportunity cost of investing in stock market is
the accrued interest of $2,000.
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Option 1:
 To keep $2,000 in the bank and earn a guaranteed interest
of $100 at the end of 1 year.
Option 2:
 To invest $2,000 in stock market and earn the expected
profit of $200 at the end of 1 year.
A. Opportunity cost of Option 1 = $200
B. Opportunity cost of Option 2 = $100
 B< A, could we conclude that we SHOULD choose option 2?
 No! We have to take the uncertainly of $200 in Option 2
into account.
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General Economic Concepts

Price-demand relationships
 1. Model 1: Selling price per unit [p] is a linear function of
demand [D]
p = a - bD for 0 < D < a/b , and a > 0; b > 0
 2. Model 2: Selling price per unit is independent of demand
Maximizing total revenue [TR]: [Use Model 1]
TR = p * D = [a – bD] * D = aD - bD2
4/10/2013
Price-demand relationships Cont’d

20
 TR is a concave function. [Check the second derivative:

d2TR = - 2b < 0
dD2
 Find the maximizer D^ from the first derivative:
dTR = a - 2bD = 0  D^ = a /ab
dD
 Maximizing profit: [Use Model 1]
 Profit = TR - TC, what is Total Cost [TC]? Assumption:
TC = CF + Cv D
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 Where CF is the fixed cost and Cv is the variable cost per

unit. Thus,
Profit = aD - bD2 – [CF + CvD]
 Maximizing profit:
Profit = aD - bD2 – [CF + CvD] = - bD2 + [a – Cv]D -CF
 In order for a positive profit to occur and to avoid negative
demand, a - Cv > 0 or a > Cv.
 From the first derivative, we have
dProfit = a - Cv - 2bD = 0  D* = a – Cv
dD 2b
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 Since d2Profit = -2b < 0 for all D,

dD2
D* = a – Cv  is the maximizer of the Profit.
2b
 Remark: Companies may also gain insights from Breakeven
points other than the profit-maximizing points.
 At a breakeven point, total revenue is equal to total cost.
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Breakeven Analysis
 Finding the breakeven points: [Model 1]
 Solve, - bD2 + [a – Cv]D - CF = 0. We obtain two roots:
D’ = - [a – Cv] {[a – Cv]2 - 4bCF}1/2 [See Scenario 1]
- 2b
 Finding the breakeven points: [Model 2]
 p is independent of D.
TR = p * D
TC = CF + CvD
TR = TC  D’ = CF [See Scenario 2]
P - Cv Meseret G. [CENG 5011] 4/10/2013
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Figure 2-1 Combined cost and revenue functions and

breakeven, as functions of volume, and their effect on typical
profit [Scenario 1]
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Figure 2-2 Typical breakeven chart with price [p] a constant

[Scenario 2]
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Example 2
 A small-medium enterprise produces half concrete ditch used
to construct the drainage for the anticipated road project.
 The fixed cost is 50,000 birr per month, and the variable
cost is 30 birr per ditch.
 The selling price per unit is p = 150 – 0.02D.
 Maximum output of the enterprise is 2,000 units per month.
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Example 2 Cont’d
a. Determine the optimum demand that maximizes profit?
b. What is the maximum profit per month?
c. What is the breakeven point?
d. What is the enterprise's range of profitable demand?
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Given:
 p = 150 – 0.02D, where a = 150 and b = 0.02; CF = 50, 000;
Cv = 30.
Solution
a. D* = [a – Cv] / 2b = [150 – 30] / 2*0.02 = 3,000 Blocks.
b. Profit = aD - bD2 – [CF + CvD] = [150*3,000] – 0.02[3,000]2 –
[50,000 + [30*3,000]]
= 130,000 birr.
c. D’ = - [150 – 30] +/- {[150 – 30]2 – 4[0.02][50,000]}1/2
- 2[0.02]
D1’ = 450; D2’ = 5550
d. D is profitable in the range 450 - 5550.
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Cost Optimization
 It refers to how to design the product in order to minimize the
cost.
 General approach:
1. Identify the design variable that is the primary cost driver
[e.g., the thickness of the material].
2. Write an expression for the cost model in terms of the
design variable.
3. Find the optimum value of the design variable to have the
minimum cost value.
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Example 3
 Block production machine type 1 and 2 are being
considered for the production of a part of a product.
 The capital investment associated with the machine are about
the same and can be ignored. The important differences
between the machines are:
1. their production capacities [production rate x available
production hours];
2. their reject rates [% of block produced that cannot achieve
the required quality and not to be sold.
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Type 1 Type 2
Production Rate 100 block/hr 130 block/hr
Hours available for 7 hrs/day 6hrs/day

production
Reject rate 3% 10%
 The material cost is 6.00 birr per parts, and all defect-free
parts produced can be sold for 12 birr each. The rejected
block have no value.
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 For either machine, the operator cost is 15.00 birr per hour
and the variable overhead rate for traceable costs is 5.00
birr per hour.
a. Assume that the daily demand for this part is large enough
that all defect-free parts can be sold. Which machine
should be selected?
b. What would be the reject rate have to be for Machine
type 2 to be as profitable as Machine type 1?
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Solution
 Since both machine types have different total daily revenue and
total daily cost, we should base on the daily profit to select the
machine.
 Profit per day = Revenue per day - Cost per day
= [Production rate x Production hrs x 12

birr/block x [1- Reject rate]] – [[Production rate x Production
hrs x 6 birr/block] – [Production hrs x [15 birr/hr + 5 birr/hr]]]
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Solution Cont’d
 Machine type 1: Profit per day = 3,808 birr per day.
 Machine type 2: Profit per day = 3,624 birr per day.
 Therefore, select Machine type 1.
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Solution Cont’d
 The breakeven reject rate, X, for Machine B, can be obtained

by equating the profits from Machine A and Machine B.
3808 = [130][6][12][1 – X] – [130][6][6] – [6][15 + 5]
X = 0.08
 So, the reject rate for Machine B can be no higher than 8%

for it to be as profitable as Machine A.
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The earning power of money
 This power is there because money can be exchanged by

production tools
The purchasing power of money
 The prices of goods and services can go upward or

downward, therefore the purchasing power of money can
change with time
Meseret G. [CENG 5011] 4/10/2013
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End of Chapter 2
Lecture # 2
Thank You!!!
Meseret G. [CENG 5011] 4/10/2013

1
By
Semester II
2013 4/10/2013
2
Chapter 4
Time Value of Money
Lecture # 4
Meseret G. [CENG 5011] 4/10/2013

Time Value of Money
3
Interest Rate [i]
 Called also the rate of capital growth, it is the rate of gain

received from an investment.
 It is usually expressed on an annual basis.
 For the lender, it consists, for convenience, of [1] risk of loss, [2]
administrative expenses, and [3] profit or pure gain.
 For the borrower, it is the cost of using a capital for immediately

meeting his or her needs.
Meseret G. [CENG 5011] 4/10/2013
Time Value of Money
4
Time Value of Money [TVM]  Money- Time Relationships

 As a result of the earning power of money [through interest], time
increases the purchasing power of money by its increase through
earning
 Money has a time value because it can earn interest [or profit]
over time.
 One birr today is worth more than one birr tomorrow
 Failure to pay the bills results in additional charge termed
interest
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Time Value of Money
5
Interest [ I ]
 Interest is usually expressed as a percentage of the amount owed
 It is due and payable at the close of each period of time involved

in the agreed transaction [usually every year or month].
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Interest [ I ] [Simple]
 Total interest is linearly proportional to the amount of loan
[principal], the interest rate, and the number of interest periods
I = [P] [N] [i]
 I : total interest
 P : principal
 N : number of interest periods
 i : interest rate per interest period.
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Time Value of Money
7
Simple Interest [i] :- Example:

 If 1,000.00 birr is borrowed at 14% interest, then interest on the
principal of 1,000.00 birr after one year is 0.14 x 1, 000, or
140.00 birr.
 If the borrower pays back the total amount owed after one year,
she/he will pay 1,140.00 birr.
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Time Value of Money
8
Interest [i] [Compounded] Cont’d
 If someone does not pay back any of the amount owed after one
year, then normally the interest owed, but not paid, is considered
now to be additional principal, and thus the interest is
compounded
 After two years she/he will owen 1,140.00 birr + 0.14 X

1,140.00, or 1,299.60.
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Time Value of Money
9
Economic Equivalency
 The banker in the previous example normally does not care
whether you pay him 1,140.00 birr after one year or 1,299.60
birr after two years.
 To him, the three values [1,000, 1,140, and 1,299.60 birr] are
equivalent.
 1,000 Birr today is equivalent to 1,140 birr one year from today
and 1,000 Birr today is equivalent to 1,299.60 Birr two years
from today.
 NB: The three values are not equal but equivalent
Meseret G. [CENG 5011] 4/10/2013
Time Value of Money
10
Economic Equivalency Cont’d

 It is to be noted that:
1. The concept of equivalence involves timing of money, amount of
money receipt/expenses and a specified rate of interest.
 The three preceding values are only equivalent for an interest
rate of 14%, and then only at the specified times.
2. Equivalence means that one sum or series differs from another
only by the accumulated interest at rate i for n periods of time.

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Time Value of Money
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Cash Flow Diagrams
 P = a present single amount of money

 F = a future single amount of money, after n periods of time
 A = end-of-period cash flows in a uniform series for a specified
number of periods, starting at the end of first period and
continuing through the last period.
 i = the rate of interest per interest period [usually one year]
 n = the number of periods of time [usually years]
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Cash Flows Over Time: Example
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Time Value of Money
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In a cash-flow diagram:
 Horizontal line represents time scale,
 Arrows represent cash flows. Downward arrows represent

expenses [negative cash flows or cash outflows] and upward
arrows represent receipts [positive cash flows or cash inflows].
 The CFD is dependent on the point of view. In the course, without

explicitly mention, the company’s [investor’s] point of view will be
taken.
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Time Value of Money
14
Cash-Flow Diagram Cont’d
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Time Value of Money
15
Example [CFD]
 You are analyzing a project with five-year life. The project

requires a capital investment of $50,000 now, and it will
generate uniform annual revenue of $6,000. Further, the project
will have a salvage value of $4,500 at the end of the fifth year
and it will require $3,000 each year for the operations.
 Develop the cash-flow diagram for this project from the investor’s
viewpoint.
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Time Value of Money
16
Example [CFD]: Solution
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Time Value of Money
17
Rules for performing arithmetic calculations with cash flows

[Three Rules]
1. Cash flows cannot be added or subtracted unless they occur at

the same point in time.
2. To move a cash flow forward in time by one time unit, multiply the
magnitude of the cash flow by [1 + i].
3. To move a cash flow backward in time by one time unit, divide the
magnitude of the cash flow by [1 + i].
Meseret G. [CENG 5011] 4/10/2013
Time Value of Money
18
Financial Engineering Analysis [Single Payment]

1. Single Payment Compound-Amount Factor [SPCAF]: Find F
OR
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Time Value of Money
19
Financial Engineering Analysis [Single Payment]

2. Single Payment Present-Worth Factor [SPPWF]: Find P when F is
given;
OR
 Notation: P = F [P/F, i%, N] where the factor in the parentheses
is read "find P given F at i% interest per period for N interest
periods.
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Time Value of Money
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Single Payment Analysis
 To calculate the future value F of a single payment P after n

periods at an interest rate i, we make the following calculation:
 At the end of the first period: F1 = P + Pi
 At the end of the second period: F2 = P + Pi + [P + Pi]i = P[1 + i]2
 At the end of the nth period: Fn = P[1 + i]n
 The future single amount of a present single amount is F = P[1 + i]n
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21

 Note: F is related to P by a factor which depends only on i and n.
This factor, termed the single payment compound amount factor
[SPCAF], makes F equivalent to P. SPCAF may be expressed in a
functional form:
 The present single amount of a future single amount is
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Time Value of Money
22
 Note: The factor 1/[1+i]n is called the present worth compound

amount factor [PWCAF]
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Time Value of Money
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E.G 1: [Single Payment Analysis]

 A contractor wishes to set up a revolving line of credit at the bank to
handle her cash flow during the construction of a project. She believes
that she needs to borrow 12,000 Birr with which to set up the account,
and that she can obtain the money at 1.45% per month.
 If she pays back the loan and accumulated interest after 8 months, how
much will she have to pay back?
 F = 12,000[1 + 0.0145]8 = 12,000[1.122061]= 13,464.73 = 13,465

Birr. The amount of interest will be: 13,465 - 12,000 = 1,465 Birr.
Meseret G. [CENG 5011] 4/10/2013
Time Value of Money
24
E.G 2: [Single Payment Analysis]

 A construction company wants to set aside enough money today in
 an interest-bearing account in order to have 100,000 Birr five

years from now for the purchase of a replacement piece of
equipment.
 If the company can receive 8% interest on its investment, how

much should be set aside now to collect the 100,000 Birr five
years from now?
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Time Value of Money
25
 E.G 2: Solution
 P = 100,000/[I + 0.08]5 =100,000/[1.46933] = 68,058.32 Birr
= 68,060 Birr
 To solve this problem you can also use the interest tables.
 P = 100,000 [P/F, 8%, 5] = 100,000[0.6805832] = 68,058.32

= 68,060 Birr
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Time Value of Money
26
Uniform Series of Payments Analysis

 Often payments or receipts occur at regular intervals, and such
uniform values can be handled by the use of additional functions.
 Another symbol: A = uniform end-of-period payments or

receipts continuing for a duration of n periods
 If a uniform amount A is invested at the end of each period for n

periods at a rate of interest i per period, then the total equivalent
amount F at the end of the n periods will be:
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Time Value of Money
27
 By multiplying both sides of above equation by [1+i] and

subtracting from the original equation, the following expression is
obtained:
 Which can be rearrange to give
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Time Value of Money
28
3. Uniform [Equal payment] Series Compound-Amount Factor

[USCAF]: Find F when A is given;
OR
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Time Value of Money
29
4. Uniform [Equal payment] Series Sinking-Fund Factor [USSFF]: Find

A when F is given;
OR
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Time Value of Money
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5. Uniform [Equal payment] Series Capital-Recovery Factor [USCRF]:

Find A when P is known;
OR
Note: This is the case of loans [mortgages]
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Time Value of Money
31
6. Uniform [Equal payment] Series Present-Worth Factor [USPWF]:

Find P when A is given;
OR
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32
Cash Flow Diagram for Single Payment
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33
E.G 3: You plan to deposit $2,000 to your savings account at the

end of every month for the next 15 months starting from the next
month. If the interest rate you can earn is 2% per month how much
money will accumulate immediately after your last deposit at the
end of the 15th month?
 Solution: A = $2,000, i = 2% per month, N = 15 months.
 F =?
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Time Value of Money
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E.G. 4: What uniform monthly amount should you deposit in your

savings account at the end of each month for the following 10
months in order to accumulate $75,000 at the time of the 10th
deposit? Assume that the interest rate you can earn is 4% per
month and the first deposit will be made next month.
 Solution: F = 75; 000, i = 4% per month, N = 10 months.
 A =?
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Time Value of Money
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E.G 5: How much should you deposit to your savings account now
at an annual interest rate of 10% to provide for 5 end-of-year
withdrawals of $15,000 each?
 Solution: A = 15; 000, i = 10% per year, N = 5 years.
 P =?
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Time Value of Money
36
E.G 6: You plan to borrow a loan of $100,000 which you will

repay with equal annual payments for the next 5 years. Suppose
the interest rate you are charged is 8% per year and you will
make the first payment one year after receiving the loan. How
much is your annual payment?
 Solution: P = 100; 000, i = 8% per year, N = 5 years.
 A =?
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Time Value of Money
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E.G. 7: A machine cost $45,000 to purchase. Fuel, oil, grease
[FOG], and minor maintenance are estimated to cost $12.34
per operating hour [those hours when the engine is
operating and the machine is doing work]. A set of tires
cost $3,200 to replace, and their estimated life is 2,800
use hours. A $6,000 major repair will probably be required
after 4,200 hr of use.
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Time Value of Money
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E.G. 7 Cont’d
 The machine is expected to last for 8,400 hr, after which it

will be sold at a price [salvage value] equal to 10% of the
original purchase price. A final set of new tires will not be
purchased before the sale. How much should the owner of
the machine charge per hour of use, if it is expected that the
machine will operate 1,400 hr per year? The company's cost
of capital rate is 15%.
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Time Value of Money
39
 First solve for n, the life;
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Solution
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Meseret G. [CENG 5011] 4/10/2013

42
Meseret G. [CENG 5011] 4/10/2013

43
End of Chapter 4
Time Value of Money
Lecture # 4
Thank You All!!!
Meseret G. [CENG 5011] 4/10/2013

1
By
Semester II
2013 4/10/2013
2
Chapter 5 and 6
Evaluating Investment Alternative/s
Lecture # 5
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Evaluating Alternative/s
3
Time Value of Money: Applications
 We will learn how to evaluate the profitability and liquidity of a

single problem solution [or alternative]. Minimum Attractive Rate
of Return [MARR] is useful for this analysis. MARR ["hurdle rate“] is
usually organization-specific and determined based on the
following:
1. Cost of money available for investment
2. Number of good projects available for investment
3. Risks involved in investment opportunities 4/10/2013
4
How to use MARR?
 Use it as an interest rate to convert cash flows into equivalent

worth at some point in time.
 The proposed problem solution [project or alternative] is

profitable if it generates sufficient cash flow to recover the initial
investment and earn an interest rate that is at least as high as
MARR.
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5
Quantitative Methods to evaluate profitability

 Net Present Worth [NPW]
 Incremental Net Present Value [INPV]
 Future Worth [FW]
 Annual Worth [AW]
 Rate of Return [ROR]: Internal Rate of Return [IRR] and External
Rate of Return [ERR]
 Incremental Rate of Return
 Payback Period with and without interest
 Project Balance [PB]
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1. Net Present Value or Present Worth[NPV]

 It compares alternatives based on their present values at the time
of the initial investment at the MARR.
 If NPV is positive, the alternative produces a return greater than
the MARR [Accepted].
 If NPV is zero, the alternative produces a return equal to the
MARR.
 If NPV is negative, the alternative produces a return less than the
MARR and, if possible, the investment should be rejected.
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E.G. 1: Your company is looking at purchasing a front-end

loader at a cost of $120,000. The loader would have a useful
life of five years with a salvage value of $12,000 at the end of
the fifth year. The loader can be billed out at $95.00 per hour.
It costs $30.00 per hour to operate the frontend loader and
$25.00 per hour for the operator. Using 1,200 billable hours per
year determine the net present value for the purchase of the
loader using a MARR of 20%. Should your company purchase
the loader? Meseret G. [CENG 5011] 4/10/2013
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Solution: The hourly profit [HP] on the loader equals the billing
rate less the operation cost and the cost of the operator.
 HP = $95.00 – [$30.00 + $25.00] = $40.00 per hr
 Annual Profit [AP] = $40.00/hr x [1,200 hr/yr] = $48,000/yr
Cash Flow
Diagram
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9
 The present value of the annual profits [PAP] by using USPWF:
PAP = $48,000 [(1+0.20)5-1] / [0.20 (1+0.20)5] = $143,549
 The PAP is positive because it is a cash receipt.
 The present value of the salvage value [PSV] by using SPPWF
PSV = $12,000/(1+0.20)5 = $4,823
 The PSV is positive because it is a cash receipt.
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 The present value purchase price [PPP] of the loader =

purchase price. Because the net present value is measured at
the time of the initial investment. The PPP is negative because it
is a cash disbursement.
NPV = PAP + PSV + PPP = $143,549 + $4,823 - 120,000 =

$28,372
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11
 Because the NPV is greater than zero, the purchase of the front-
end loader will produce a return greater than the MARR and your
company should invest in the front-end loader.
 NB: When comparing two alternatives with positive net present

values, the alternative with the largest net present value
produces the most profit in excess of the MARR.
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E.G. 2: Your company needs to purchase a dump truck and

has narrowed the selection down to two alternatives. The first
alternative is to purchase a new dump truck for $65,000. At
the end of the seventh year the salvage value of the new
dump truck is estimated to be $15,000. The second
alternative is to purchase a used dump truck for $50,000. At
the end of the fourth year the salvage value of the used
dump truck is estimated to be $5,000. The annual profits,
revenues less operation costs, are $17,000 per year for
either truck. Using a MARR of 18% and a twenty-eight year
study period, calculate the net present value for each of the
dump trucks. Which truck should your company purchase?
4/10/2013
13
Solution: The present value of the annual profits for either

truck is determined by using USPWF:
 PAP = $17,0003 [(1+0.18)28–1]/0.18[1+0.18]28] =
$93,527
1. Alternative 1[New]: The PSV for the new dump truck is
determined by summing the PSVs occurring in years 7, 14,
21, and 28. The present value for each salvage value is
calculated using SPPWF as follows:
Meseret G. [CENG 5011] 4/10/2013

14
 The PPP for the new dump truck is determined by summing the
present value of purchase prices occurring in years 0, 7, 14, and
21. The present value for each purchase price is calculated using
SPPWF as follows:
Meseret G. [CENG 5011] 4/10/2013

15
 The NPV for the purchase of the new dump truck is calculated as
follows:
 NPV [New] = $93,527 + $6,797 + [ -$93,822] = +$6,502
Meseret G. [CENG 5011] 4/10/2013

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2. Alternative 2[Used]: The PSVs for the used dump truck is

determined by summing the present value of salvage values
occurring in years 4, 8, 12, 16, 20, 24, and 28. The present
value for each salvage value is calculated using SPPWF as
follows:
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17
 The PPPs for the used dump truck is determined by summing

the present value of purchase prices occurring in years 0, 4,
8, 12, 16, 20, and 24. The present value for each purchase
price is calculated using SPPWF as follows:
Meseret G. [CENG 5011] 4/10/2013

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 The net present value [NPV] for the purchase of the used dump
truck is calculated as follows:
NPV = $93,527 + $5,275 + [ -$102,257] = -$3,455

 The new truck has the highest NPV; therefore, your company
should purchase the new truck.
Meseret G. [CENG 5011] 4/10/2013

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Incremental Net Present Value [INPV]
 Step 1: Order the alternatives by increasing initial capital

investment.
 Step 2: Find a base alternative [current best alternative]: Cost

alternatives: the first alternative in the ordered list [the one with
the least capital investment].
 Step 3: Evaluate the difference between the next alternative and

the current best alternative.
Meseret G. [CENG 5011] 4/10/2013
20
 Step 3 [Cont’d]: If the incremental cash flow is positive, choose the

next alternative as the current best alternative. Otherwise, keep
the current best alternative [i.e. negative] and drop the next
alternative from further consideration.
 Step 4: Repeat Step 3 until the last alternative is considered.

Select the current best alternative as the preferred one.
Meseret G. [CENG 5011] 4/10/2013

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E.G. 3: Your company is looking at purchasing a new front-end

loader and has narrowed the choice down to four loaders. The
purchase price, annual profit, and salvage value at the end of
five years for each of the loaders is found in figure below.
Which front-end loader should your company purchase based on

the incremental net present values using a MARR of 20% and a
useful life of five years?
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Given:
Cash Flow Loader A [$] Loader B [$] Loader C [$] Loader D [$]
Purchase Price 110,000 127,000 120,000 130,000
Annual Profit 37,000 43,000 40,000 44,000
Salvage Value 10,000 13,000 12,000 13,000
Meseret G. [CENG 5011] 4/10/2013

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Solution:
Step 1: Rank the alternative in order of initial cost [purchase price].
Loader A  Loader C  Loader B  Loader D. Because Loader
A has the lowest initial cost [current best alternative].
Step 2: Compare Loader A to Loader C.
 Difference in purchase price is $10,000 [$120,000 - $110,000].
 Difference in annual profit is $3,000 [$40,000 - $37,000].
 Difference in salvage value is $2,000 [$12,000 - $10,000].

Meseret G. [CENG 5011] 4/10/2013
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 The difference in the cash flows for these two alternatives is shown
in Figure below.
Meseret G. [CENG 5011] 4/10/2013

25
 The present value of the difference in annual profits is

determined by using USPWF as follows:
PAP = $3,000[(1 0.20)5 -1]/ [0.20(1+0.20)5] = $8,972
 The present value of the difference in salvage values is
determined by using SPPWF as follows:
PSV = $2,000[1+0.20]5 = $804
 The incremental net present value for the purchase of Loader C in
lieu of Loader A is calculated as follows:
INPV = $8,972 + $804 + [ -$10,000] = -$224 4/10/2013
26
 Because the incremental net present value is negative, Loader A

continues to be the current best alternative.
Next, we compare Loader A to Loader B, the loader with the next

lowest initial cost.
 Difference in purchase price is $17,000 [$127,000 - $110,000].
 Difference in annual profit is $6,000 [$43,000 - $37,000].
 Difference in salvage value is $3,000 [$13,000 - $10,000].
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27
 The difference in the cash flows for these two alternatives is shown
in Figure below.
 The present value of the difference in annual profits is determined

by using USPWF as follows:
 PAP = $6,0003 [(1+0.20)5 -1] / 0.20(1+0.20)5 = $17,944
4/10/2013
28
 The present value of the difference in salvage values is

determined by using SPPWF as follows:
PSV = $3,000 [1+ 0.20]5 = $1,206
 The incremental net present value for the purchase of Loader B in

lieu of Loader A is calculated as follows:
INPV = $17,944 + $1,206 + [-$17,000] = $2,150
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29
 Because the incremental net present value is positive, Loader B

becomes the new current best alternative and Loader A is
eliminated from comparison.
Next, Compare Loader B to Loader D
 Difference in PP is $3,000 [$130,000 - $127,000].
 Difference in AP is $1,000 [$44,000 - $43,000].
 Difference in SV is zero [$13,000 - $13,000].
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 The present value of the difference in annual profits is determined

by using USPWF as follows:
PAP = $1,000 [(1 0.20)5 -1] / [0.20 (1+0.20)5] = $2,991
 The incremental net present value for the purchase of Loader B in
lieu of Loader D is calculated as follows:
INPV = $2,991 + $0 + [- $3,000] = - $8
 Because the INPV is negative, Loader B continues to be the current
best alternative. Therefore, your company should purchase
Loader B. Meseret G. [CENG 5011] 4/10/2013
31
3. Future Value [FW]: Net Future Value

 It compares alternatives based on their future values at the end of
the study period.
 If the FW is positive, the alternative produces a return greater
than the MARR.
 If the FW is zero, the alternative produces a return equal to the
MARR.
 If the FW is negative, the alternative produces a return less than
the MARR and, if possible, the investment should be rejected.
Meseret G. [CENG 5011] 4/10/2013
32
E.G. 4: Your company is looking at purchasing the front-end

loader at cost of $120,000. The loader would have a useful life
of five years with a salvage value of $12,000 at the end of the
fifth year. The annual profit of loader [revenue less operation
cost] is $48,000. Determine the future worth for the purchase of
the loader using a MARR of 20%. Should your company purchase
the loader?
Meseret G. [CENG 5011] 4/10/2013

33
Solution: The future value of the purchase price is determined by

using SPCAF as follows:
FPP = $120,000 [1+0.20)5 = - $298,598
 The future value of the purchase price is negative because it is a
cash disbursement.
 The FW of the annual profits is determined by using USCAF as
follows:
FAP = $48,0003 [(1+0.20)5 -1] / 0.20 = $357,197
 The FW of the annual profits is positive because it is a cash
receipt.
Meseret G. [CENG 5011] 4/10/2013
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 The future value of the salvage value is equal to the salvage

value because the future value is measured at the end of the
study period. The FW of the salvage value is positive because it is
a cash receipt.
 The future worth for purchasing the loader equals the sum of the
future values of the individual cash flows and is calculated as
follows:
 FW = - $298,598 + $357,197 + $12,000 = $70,599 > MARR
 So, it is attractive for the company to purchase 4/10/2013
35
4. Annual Equivalent [AE]

 It compares alternatives based on their equivalent annual receipts
less the equivalent annual disbursements.
 The AE is calculated by converting the cash receipts and

disbursements into a uniform series of annual cash flows occurring
over the study period using the equations.
 If the AE is positive, the alternative produces a return greater than

the MARR.
Meseret G. [CENG 5011] 4/10/2013
36
 If the AE is zero, the alternative produces a return equal to the

MARR.
 If the AE is negative, the alternative produces a return less than

the MARR and, if possible, the investment should be rejected.
 Because any PV can be converted to a uniform series by USSFF.

The AE produces the same result as the net present value.
 Similarly, because any FW can be converted to a uniform series

by USCRF, the AE produces the same result as the future value.
Meseret G. [CENG 5011] 4/10/2013
37
E.G 5: Your company needs to purchase a dump truck and

has narrowed the selection down to two alternatives. The first
alternative is to purchase a new dump truck for $65,000. At
the end of the seventh year the salvage value of the new
dump truck is estimated to be $15,000.
The second alternative is to purchase a used dump truck for
$50,000. At the end of the fourth year the salvage value of
the used dump truck is estimated to be $5,000.
The annual profits, revenues less operation costs, are
$17,000 per year for either truck. Using a MARR of 18%
calculate the annual worth for each of the dump trucks.
Which truck should your company purchase? 4/10/2013
38
Solution: Alternative 1[New]  The useful life of the new truck is

seven years, which is used as the study period for the new truck.
 The purchase price for the new truck is converted to a uniform

series of annual cash flows by USCRF as follows:
APP = - $65,000 [0.18(1+0.18)7] / [(1 0.18)7 -1] = - $17,054
 The salvage value for the new truck is converted to a uniform

series of annual cash flows by USSFF as follows:
ASV = $15,000(0.18) / [ (1+0.18)7 -1] = $1,235

Meseret G. [CENG 5011] 4/10/2013
39
 The annual profits for the new truck are already a uniform series.
The annual equivalent for purchasing new loader is;
AE [New] = - $17,054 + $1,235 + $17,000 = $1,181
Alternative 2 [Used]
APP = $50,000 [0.18(1+0.18)4] / [(1+0.18)4 1] = - $18,587
ASV = $5,000 [0.18] / [(1+0.18)4 -1] = $959
The annual equivalent for purchasing used loader is;
AE = - $18,587 + $959 + $17,000 = - $628 4/10/2013

40
 Decision: The new truck has the highest annual equivalent;

therefore, your company should purchase the new truck.
Meseret G. [CENG 5011] 4/10/2013

41
Meseret G. [CENG 5011] 4/10/2013

42
Meseret G. [CENG 5011] 4/10/2013

43
Meseret G. [CENG 5011] 4/10/2013

44
Meseret G. [CENG 5011] 4/10/2013

Engineering Economics
45
End of Chapter 5 and 6
Evaluating Investment Alternative/s
Lecture # 5
Thank You All!!!
Meseret G. [CENG 5011] 4/10/2013

1
By
Semester II
2013 4/10/2013
2
Chapter 7
Depreciation Cost
Lecture # 6
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
3
Introduction
 Depreciation is the loss in value of a piece of equipment over

time, generally caused by wear and tear from use, deterioration,
obsolescence, or reduced need.
 The profitable owner of equipment must recover this loss during

its useful life.
 Depreciation accounting is the systematic allocation of the costs

of a capital investment over some specific number of years.
Meseret G. [CENG 5011] 4/10/2013
Depreciation Cost
4
Reasons for calculating the depreciation accounting value [usually

termed book value] of a piece of equipment:
1. To provide the construction owner and project manager with an

easily calculated estimate of the current market value of the
property [equipment etc].
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
5
Reasons Cont’d
2. To provide a systematic method for allocating the depreciation

portion of equipment ownership costs over a period of time and
to a specific productivity rate.
3. To allocate the depreciation portion of ownership costs in such a

manner that the greatest tax benefits will accrue.
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
6
Information needed for depreciation accounting:
 The purchase price of the piece of equipment, P
 The optimum period of time to keep the equipment or the

recovery period allowed for income tax purposes, N
 The estimated resale value at the close of the optimum period of

time, F [Salvage Value]
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
7
Depreciation accounting Method [Most Common]
1. Straight-line [SL] Method
2. Sum-of-the-years [SOY] Method
3. Declining-balance [DB] Method
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
8
1. Straight-line [SL] Method: Easiest to calculate and most

widely used in construction.
 The annual amount of depreciation Dm, for any year m, is a
constant value, and thus the book value BV m decreases at a
uniform rate over the useful life of the equipment.
 Depreciation rate: R m = 1/N
 Annual depreciation amount: Dm = R m [P – F] = [P – F]/N
 Book value at year m: BV m = P – [m*Dm 4/10/2013

Depreciation Cost
9
 Note: The value [P – F] is often referred to as the depreciable

value of the investment.
E.G. 1: A piece of equipment is available for purchase for

$12,000, has an estimated useful life of 5 years, and has an
estimated salvage value of $2,000. Determine the depreciation
and the book value for each of the 5 years using the SL method.
R m = 1/5 = 0.2
D m= 0.2[12,000 - 2,000] = $2,000 per year

Meseret G. [CENG 5011] 4/10/2013
Depreciation Cost
10
 BV2 = $12,000 – 2[2,000] = $8,000

 The table of values is:
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
11
 If the equipment is expected to be used about 1,400 hours per

year then its estimated hourly depreciation portion of the
ownership cost is $2,000/1,400 = $1.428 = $1.43 per hour.
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
12
2. Sum-of-the-years [SOY] Method: SOY is an accelerated

depreciation method [fast write-off], which is a term applied to
accounting methods which permit rates of depreciation faster than
straight line.
 The rate of depreciation is a factor R m [depreciation rate] times

the depreciable value [P – F].
 Dm = R m [P-F]
 SOY = N [N+1] / 2
Meseret G. [CENG 5011] 4/10/2013
Depreciation Cost
13
R m = [N-m+1]/SOY
The annual depreciation Dm for mth year [at any age m] is
 D m = {[N-m+1]/SOY}[P-F]
 The book value at the end of year m is
 BV m = P-[P-F] [m(N-m/2+0.5)/SOY]
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
14
E.G. 2: Using the same values as given in Example 1, calculate the

allowable depreciation and the book value for each of the 5 years
using the SOY method.
 SOY = 1+2+3+4+5 = 15 or = N [N+] / 2 = 5[5+1]/2 = 15
 R m = [5-m+1]/15
 D m = R m [12,000-2,000] = [[5-m+1]*10,000]/15
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
15
E.G. Cont’d: Then tabulate the results as follows:
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
16
Declining-balance [DB] Methods: These also are accelerated

depreciation methods that provide for even larger portions of
the cost of a piece of equipment to be written off in the early
years.
DB method often more nearly approximates the actual loss in

market value with time.
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
17
Declining-balance [DB] Cont’d
Declining methods range from 1.25 times the current book value
divided by the life to 2.00 times the current book value divided
by the life [the latter is termed double declining balance].
Note: Although the estimated salvage value F is not included in

the calculation, the book value cannot go below the salvage
value.
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
18
The following equations are necessary to use the declining-

balance methods.
The symbol R is used for the depreciation rate for the declining-
balance method of depreciation:
1. For 1.25 declining-balance [1.25DB] method, R = 1.25/N
For 1.50 declining-balance [1.5DB] method, R = 1.50/N
For 1.75 declining-balance [1.75DB] method, R = 1.75/N
For double-declining-balance [DDB] method, R = 2.00/N 4/10/2013

Depreciation Cost
19
2. The allowable depreciation Dm, for any year m and any

depreciation rate R is
 Dm = R P[1 – R]m-1 or Dm = [BVm-1]*R
3. The book value for any year m is
 BV m = P[1-R]m or BV m = BV m-1- D m provided that BV m > F
 Since [BV] can never go below [F], the declining balance method
must be forced to intersect the value [F] at time [N].

4/10/2013
Depreciation Cost
20
E.G. 3: A piece of equipment is available for purchase for

[$12000], has an estimated useful life of [5 years], and an
estimated salvage value of [$2000]. Determine the depreciation
and the book value for each of the 5 years using the DDB
method.
 Solution:
 Calculate R: 2/N = 2/5 = 0.4
 Calculate Dm: Dm = 0.4 [BVm-1]
 Calculate BVm: BVm = BVm-1 - Dm
Meseret G. [CENG 5011] 4/10/2013
Depreciation Cost
21
 Then tabulate the results as follows:
Meseret G. [CENG 5011] 4/10/2013

Depreciation Cost
22
 Depreciation Curves for the above Examples
Meseret G. [CENG 5011] 4/10/2013

23
End of Chapter 7
Depreciation Cost
Lecture # 6
Thank You!!!
Meseret G. [CENG 5011] 4/10/2013

1
By
Semester II
2013 4/10/2013
2
Chapter 9
Public Project Evaluation
[Benefit – Cost Ratio Method]
Lecture # 8
Meseret G. [CENG 5011] 4/10/2013

Public Project Evaluation [B-C Ratio Method]
3
Introduction
 The project evaluation and selection methods we have learned so
far are directed towards “private projects".
 The primary goal of a private project is to make profit for the
project owners and investors.
 In these projects, a unified purpose is usually agreed on and the
projects’ evaluation is only based on monetary values.

Meseret G. [CENG 5011] 4/10/2013
Public Project Evaluation [B-C Ratio Method
4
 On the other hand, “public projects" do not aim to generate

profit, but instead have the [social] goal of providing service and
benefit to the general public.
 Difficulties:
a. Quantifying service in monetary terms
b. Multiple purposes and conflicting interests
 Their evaluation requires different methods. One such method is

called Benefit-Cost (B-C) ratio method.
Meseret G. [CENG 5011] 4/10/2013
5
Benefit-Cost [B-C] Ratio Method
 B-C ratio method calculates the ratio of project’s benefits to its

costs. The components are:
a. Benefits: Positive consequences to the public.
b. Costs: Monetary expenditures required for the project [financed

by government often through taxation].
c. Disbenefits: Negative consequences to a segment of the public.
Meseret G. [CENG 5011] 4/10/2013

6
Benefit-Cost [B-C] Ratio Method Cont’d
 As in the private projects, time value of money is important for the

public projects, so an interest rate needs to be determined.
 Several factors are considered in selecting an appropriate interest

rate, such as;
1. interest rate on borrowed capital and
2. opportunity cost of capital both to the taxpayers and the

governmental agency.
Meseret G. [CENG 5011] 4/10/2013
7
Notation:
 B: Benefits of the project
 I: Initial investment
 O & M: Operating and Maintenance cost of the project
 CR: Capital Recovery amount
 D: Disbenefits of the project
 MV: Market Value of investment.

Meseret G. [CENG 5011] 4/10/2013
8
B-C Ratio Cont’d
Meseret G. [CENG 5011] 4/10/2013

9
Decision Criteria
 A project is acceptable if B-C > 1, otherwise not.
 Conventional and modified B-C methods give identical

acceptability results [not necessarily the same numerical results!].
Meseret G. [CENG 5011] 4/10/2013

10
E. G. 1: The project is to extend the runways of an airport and it is

considered by the city municipality.
 Costs have been identified:
 Land: $350,000
 Construction: $600,000
 Annual maintenance: $22,500
 Terminal construction: $250,000
 Annual operating and maintenance for the terminal: $75,000
 Addition of air traffic controllers per year: $100,000
Meseret G. [CENG 5011] 4/10/2013
11
 Estimated Project Benefits:
 Rental receipts: $325,000
 Tax to passengers: $65,000
 Convenience: $50,000
 Tourism: $50,000.
 Apply the B-C ratio method with a study period of 20 years and
a MARR of 10% per year, and determine whether this is an
acceptable project or not. 4/10/2013
12
Solution: [a] Conventional B-C ratio:

 PW version:
 AW version:
CR – Capital Recovery
[b] Modified B-C ratio:
 PW version:
Meseret G. [CENG 5011] 4/10/2013

13
 AW version:
Extend runways
 How to include disbenefits?
 Two alternatives: reduce benefits or increase costs.
 Conventional B-C ratio [with disbenefits]:
Meseret G. [CENG 5011] 4/10/2013

14
Suppose the increased noise level caused by the project will bring
a disbenefit of $100,000 per year to the neighborhood. Still
extend the runway?
Meseret G. [CENG 5011] 4/10/2013

15
Selection of alternatives by B-C ratio
 Independent projects: Select any project that has a B-C ratio of

greater than or equal to one. [if no budget restriction!]
 Mutually exclusive projects: We must perform incremental B-C

analysis. [Selecting the project with the highest B-C ratio is not
correct!]
Meseret G. [CENG 5011] 4/10/2013

16
Selection of alternatives by B-C Ratio [Incremental analysis for

mutually exclusive projects]:
1. For each alternative, determine its C and B [or [B – D] if

disbenefits are considered] in the B-C ratio.
 Conventional B-C ratio:
PW version: C = I - PW[MV] + PW[O&M]; B = PW[B].
AW version: C = CR + AW[O&M]; B = AW[B].
Meseret G. [CENG 5011] 4/10/2013

17
 Modified B-C ratio:
PW version: C = I - PW[MV]; B = PW[B] - PW[O&M].
AW version: C = CR; B = AW[B] - AW[O&M].
2. Rank the projects in order of increasing C.
3. If there is DN, select DN as the baseline alternative. Otherwise,

select the first alternative in the ordered list of Step 2 as the
baseline alternative.
Meseret G. [CENG 5011] 4/10/2013

18
4. Calculate the incremental B-C ratio ∆B/∆C [Next project -current

baseline project]. Select next project as the new baseline project
if the incremental ratio > 1; otherwise maintain the [last] baseline
project. Iterate until all projects are considered.
Meseret G. [CENG 5011] 4/10/2013

19
E.G. 2: The city of Adama has received two design proposals for
a new wing to the municipality hospital.
Design 1 Design 2
Construction cost [Birr] 10M 15M
Building maintenance cost [Birr/year] 35,000 55,000
Patient benefits [Birr/year] 0.8M 1.05M
 The MARR is 5% per year, and the life of the addition is
estimated at 30 years. Suppose there is DN alternative.
a. Use AW version of conventional B-C ratio method to make the
selection.
Meseret G. [CENG 5011] 4/10/2013
20
b. Once the two designs were publicized, the privately owned

hospital in the adjacent city of ‘Wonji’ a complaint that Design 1
will reduce its own municipal hospital’s income by an estimated
0.6M Birr per year because some of the day-surgery features of
Design 1 duplicate its services.
 Subsequently, the Adama Merchant’ association argued that
Design 2 could reduce its annual revenue by an estimated 0.4M
Birr because it will eliminate an entire parking lot used for short-
term parking. What is the decision now?
Meseret G. [CENG 5011] 4/10/2013
21
Solution:
a. AW1 [C] = 10,000,000 Birr [A/P , 5% , 30] + 35,000 Birr =
685,500 Birr
AW2 [C] = 15,000,000 Birr [A/P , 5% , 30] + 55,000 Birr =
1,030,750 Birr
 Rank order: DN-Design 1-Design 2.
∆B/∆C [1 - DN] = 800,000 = 685,500 = 1.17.

 Design 1 is the base alternative.
∆B [2 -1] =1,050,000 - 800,000 = 250,000 Birr

∆C [2 -1]=1,030,750 - 685,500 = 345,250 Birr
∆B/∆C [2 -1] = 250,000 / 345,250 = 0.72.
 Design 1 is selected.
Meseret G. [CENG 5011] 4/10/2013
22
b. The revenue loss estimates are considered disbenefits.
 Since the disbenefits of Design 2 are 200,000 Birr less than those
of 1, this positive difference is added to the 250,000 Birr benefits
of 2 to give it a total benefit of 450,000 Birr.
∆B/∆C [2 -1] = 450,000 Birr / 345,250 Birr = 1.30
 Design 2 is now favored. The inclusion of disbenefits reversed

the decision.
Meseret G. [CENG 5011] 4/10/2013

23
Incremental B-C analysis with unequal project lives
 If some of the projects in a set of mutually exclusive public-works

projects have different lives, we need to use the techniques
introduced in lecture 5 to make them to have equal lives before
we conduct the incremental B-C analysis.
Meseret G. [CENG 5011] 4/10/2013

24
E.G. 3: Two mutually exclusive alternative public-works projects

are under consideration.
Project Type 1 Project Type 2
Capital investment [Birr] 750,000 625,000
Annual operating & maintenance 120,000 110,000
[Birr/year]
Annual benefit [Birr/year] 245,000 23,000
Useful life [years] 35 25
 The MARR is 9% per year. Assume the assumption of repeatability
is valid.
Meseret G. [CENG 5011] 4/10/2013
25
 Suppose there is DN alternative. Use the AW version of
conventional B-C ratio method to make your selection.

Solution:
AW1 [Costs] = 750,000 Br [A/P, 9%, 35] + 120,000 Br =
190,977 Br
AW2 [Costs] = 625,000 Br [A/P, 9%, 25] + 110,000 Br =
173,629 Br
 Rank order: DN - Project 2 - Project 1.
∆B/∆C [2 - DN) = 230; 000=173; 629 = 1:3247 > 1.0
Meseret G. [CENG 5011] 4/10/2013

Public Project Evaluation [B-C Ratio Method]
26
 Therefore, Project II is the base alternative.
∆B/∆C [1 - 2] = [245,000 – 230,000] = [190,977 - 173,629] =

0:8647 < 1:0.
 Therefore, increment required for Project I is not acceptable.
 Decision: Project II should be selected.
Meseret G. [CENG 5011] 4/10/2013

27
End of Chapter 9
Public Project Evaluation
[Benefit-Cost Ratio Method]
Lecture # 8
Thank You!!!
Meseret G. [CENG 5011] 4/10/2013

Engineering Economics Lecture (# 1 - 6)

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Welcome to Heartiest Students

Engineering Economics [CENG 5011]

Department of Civil Engineering

Adama Science and Technology University

Meseret G. [CENG 5011] 4/10/2013

Definition of Engineering and Science

 Engineering a profession in which knowledge of the

a. Utilize the material and forces economically and

b. For the benefit of mankind

 Engineering is not a science but application of science

 Nature of engineering: application

 Role of scientist: Discover the universal laws of nature/

 Role of engineer: Apply knowledge to particular situation

 It is the careful management of material resources.

Meseret G. [CENG 5011] 4/10/2013

Engineering and Economics

Engineering Economics [EEs]

 Engineering economy: formulation, estimation and

 Discipline that involves the systematic evaluation of the cost

Engineering Economics [EEs] Cont’d

 The mission of EEs is to balance different types of costs and

Meseret G. [CENG 5011] 4/10/2013

 The alternatives need to be identified and then defined for

 Since the choice [decision] is among alternatives, developing

 Creativity and innovation are essential. 4/10/2013

Principle 2: Focus on the differences

 Only the differences in expected future outcomes among the

 Outcomes that are common to all alternatives can be

 For example, the perspective of the employees is used for

Meseret G. [CENG 5011] 4/10/2013

Principle 4: Use a common unit of measure

Principle 5: Consider all relevant criteria

Principle 6: Make uncertainty explicit

 Risk and uncertainty are inherent in estimating the future

Meseret G. [CENG 5011] 4/10/2013

Principle 7: Revisit your decisions

 Improved decision making results from an adaptive process.

Meseret G. [CENG 5011] 4/10/2013

Procedures of Engineering Economic Analysis

Example: Say you and two friends find yourself strong

 Hunger satisfaction with a pizza.

Step 2: Develop Alternatives [Principle 1]

 Alternative A: Order pizza 1 [square]

 Alternative B: Order pizza 2 [round]

Meseret G. [CENG 5011] 4/10/2013

Example: Step 3 Cont’d

 Alternative A: Cash outflow = 42 Birr+ [42 * 5%] Birr + 2.5

 Alternative B: Cash outflow = 45 Birr + [45 * 5%] = 47.25

Step 4: Selection of criteria

 It is important to use consistent viewpoint [Principle 3] to

 Criteria: [Principle 5]: Cost per unit volume, Variety and

Meseret G. [CENG 5011] 4/10/2013

 Alternative A: Cost per unit volume = 0.259 Birr per in3.

 Alternative B: Cost per unit volume = 0.297 Birr per in3.

 Further information is required to compare both alternatives

Step 7: Performance monitoring and post-evaluation of

Meseret G. [CENG 5011] 4/10/2013

Role of Engineering Economy in Decision Making