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KP6443 RISKS ANALYSIS &
ec urer: ro . a ya a a a eros
. .Faculty of Engineering
. .E-mail: [email protected]
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CHAPTER I - FOUNDATIONS OF
LEARNING OBJECTIVES:
ues ons n ers an e ypes o ques ons eng neer ng
economy can answer. Decision Making Determine the role of engineering economy
in the decision making process. Study Approach Identify what is needed to successfully
perform an engineering study economy study.
Interest Rate Perform calculations about interest rates andrates of return. Equivalence Understand what equivalence means in economic
terms. Simple and Compound Interest Calculate simple interest
. Symbols Identify and use engineering economy terminology
and symbols. Minimum Attractive Rate of Return
2
as ows Doubling Time
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Questions
Why Engineering Economy is Important to Engineers (and
other professionals)
mpor ance - ng neers es gn
Engineers must be concerned with the economic aspects of
designs and projects they recommend and perform
(1) Analysis (2) Design (3) Synthesis
Engineers must work within the realm of economics and
Work with limited funds (capital)
Capital is not unlimited rationed
Capital does not belong to the firm Belongs to the Owners of the firm
Capital is not freeit has a cost
3
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ENGINEERING ECONOMY IS INVOLVED WITH THE
FORMULATION, ESTIMATION, AND EVALUATION OF
ECONOMIC OUTCOMES WHEN ALTERNATIVES TO
ACCOMPLISH A DEFINED PURPOSE ARE
AVAILABLE.
ENGINEERING ECONOMY IS INVOLVED WITH THE
APPLICATION OF DEFINED MATHEMATICAL
RELATIONSHIPS THAT AID IN THE COMPARISON OF
4
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Know ledge of Engineering Economy w ill have asignificant impact on you, personally.
a e proper econom c compar sons
In your profession
Public sector
In our ersonal life
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ROLE OF ENGINEERING ECONOMY
Remember: People make decisions not tools
Engineering Economy is a set of tools that aid in decisionma ng u w no ma e e ec s on or you
Engineering economy is based mainly on estimates offuture events must deal w ith the future and risk anduncertainty
The parameters w ithin an engineering economy problem
Parameters that can vary w il l dictate a numerical outcome apply and understand
Sensitivity Analysis plays a major role in the assessment ofmost, if not all, engineering economy problems
6need to master th is valuable tool as an analysis aid
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-
.
2. Col lect al l relevant data/ information
.
4. Evaluate each alternative
5. Select the best al ternative
6. Implement and monitor
Major Role ofEngineeringEconom 7
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-
1. Understand the Prob lem
2. Col lect al l relevant data/ information3. Define the feas ible alternatives
4. Evaluate each alternative
5. Select the best al ternative6. Implement and monitor
One of the moredifficult tasks
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-
.
2. Col lect al l relevant data/ information
.
4. Evaluate each alternative
5. Select the best al ternative
6. Implement and monitorWhere the major
tools of En ineerin
Economy are applied
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-
.
2. Col lect al l relevant data/ information
.
4. Evaluate each alternative
5. Select the best al ternative
6. Implement and monitor ToolsPresent Worth, Future Worth
Annual W orth, Rate of Return
Benefit/ Cost, Payback,Capitalized Cost, Value Added
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Money can make money if Invested
Centers around an interest rate
The change in the amount of money over agiven time period is called the time value of
,in engineering economy
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,
more ways to solve a problem)Alternative w ays to solve a problem must first be identified
Estimate t e cas ows or t e a ternatives
Analyze the cash flows for each alternative
To anal ze, one must have:
Concept of the time value of $$
An Interest Rate
Evaluate and weigh
Factor in non-economic parameters12 Select, implement, and monitor
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Needed ParametersFirst cost (investment amounts)
Estimates of useful or project life
Estimated future cash flows (revenues and expenses and
salvage values)
Interest rate
Inflation and tax effects
Estimate flows of money coming into the firm revenuessalva e values etc. ma nitude and timin ositive cash
flows
Estimates of investment costs, operating costs, taxes paid 13
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AlternativesEach problem w ill have at least one alternative DONOTHING
May not be free and may have future costs associated
Do not overlook this option!
Goal: Define Evaluate Select and Execute
DoNothing Alt . 1
The Question:
Which One do weaccept?
Select One and only one from a set of feasible alternatives
,are excluded at that point.
I f all of the proposed alternatives are not economically
14
es ra e en
One usually defaults to the DO-NOTHING alternative
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TaxesTaxes represent a significant negative cash flow to the for-profit firm.
A realistic economic ana lysis must assess the impact oftaxes.
Called andAFTER-TAX cash flow analysis
Not consider in taxes is cal led a BEFORE- AX Cash Flowanalysis.
A Before-Tax cash flow analysis (while not as accurate) is.
A final, more complete analysis should be performed usingan After-Tax analysis
Both are valuable analysis approaches
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INTEREST RATES AND RETURNS
Interest can be viewed from two perspectives:
Lending situation
Investing situation
You borrow money (renting someone else's money)
he lender ex ects a return on the mone lent
The return is measured by application of an interest rate
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. Example
ou orrow , or one u year
Must pay back $10 ,700 at the end of one year
, - ,
Interest Amount = $700 for the year
= =
Expressed as a per cent per year
Notation
I = the interest amount is $
i = the interest rate (% / interest period)17
N = No. of interest periods (1 for this problem)
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The interest rate (i) is 7% per year
The interest amount is $700 over one year
The $700 represents the return to the lender for this use of
7% is the interest rate charged to the borrower
7% is the return earned b the lender
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. -
Borrow $20,000 for 1 year at 9% interest per year
i = 0.09 per year and N = 1 Year
Pay $20,000 + (0.09)($20,000) at end of 1 year Interest (I) = (0.09)($20 ,000) = $1,800
Total amt. paid one year hence
$20,000 + $1,800 = $21,800
Note the follow ing Total Amount Due one year hence is
($20,000) + 0.09($20,000)
=$20,000(1.09) = $21,800
The (1.09) factor accounts for the repayment of the$20,000 and the interest amount
This w ill be one of the important interest factors to be19seen later
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Inflation EffectsA social-economic occurrence in which there is morecurrency competing for constrained goods and services
Where a countr s currenc becomes worth less over t ime,thus requiring more of the currency to purchase the sameamount of goods or services in a time period
In ation Rate s
Inflation impacts: Purchasing Power (reduces)
Operating Costs (increases)
Rate of Return on Investments (reduces) Specifically covered in Chapter 14
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EQUIVALENCE
You travel at 68 m iles per hour
Thus:
68 m h is e uivalent to 110 k h
Using two measuring scales
Miles and Kilometers
Is 68 equal to 110?
No, not in terms of absolute numbers
But they are equivalent in terms of the two measuringscales
22
Kilometer [email protected]
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Economic Equivalence
Two sums of money at tw o different points in time can bemade economically equivalent if:
,
No. of time periods between the tw o sums
Return to Example 2$20,000 now iseconomicall
Diagram the loan (Cash Flow Diagram)
The company s perspective is show n
20 000 is
equivalent to$21,800 oneyear from nowIF the interest
T=0 t = 1 Yr
received here rate is set toequal 9% / year
23$21,800 paid
back [email protected]
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Equivalence I llustrated$20,000 now is not equal in magnitude to $21,800 1 yearfrom now
But, $20,000 now is economicall e u ivalent to $21,800one year from now if the interest rate is 9% per year.
Another way to put it is ..
....
Which is worth more?
$20,000 now or $21,800 one year from now?
The tw o sums are economically equivalent but notnumerically equal!
To have economic equivalence you must specify:
Timing of the cash flows
24 Number of interest periods (N)
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Simple and Compound InterestTwo types of interest calculations
Simple Interest
Compound Interest
Compound Interest is more common w orldwide and appliesto most anal s is s ituations
Simple Interest
Calculated on the principal amount only
Easy (simple) to calculate
Simple Interest is:
(principal)(interest rate)(time)
$I = (P)(i)(n)
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Simple and Compound InterestExample 1.7
Borrow $1,000 for 3 years at 5% per year
Let P = the principal sum
i = the interest rate (5% / year)
e = num er o years
Simple Interest
I = P(i)(N)
For Exam le 1.7:
I = $1,000(0.05)(3) = $150.00
Total Interest over 3 Years26
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Simple and Compound InterestYear-by-Year Analysis: Simple Interest
Year 1
I1 = $1,000(0.05) = $50.00
Year 2
2 = , . = .
Year 3
3 , . .
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Accrued Interest: Year 1
Accrued means owed but not yet paid
First Year:
P=$1,000
1 2 3
I1=$50.00
$50.00 interest accrues but is not paid
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Accrued Interest: Year 2
Year 2
= ,
I1=$50.00 I2=$50.00
$50.00 interest accrues but is not paid
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$150 of interest has accrued
P=$1,000
1 2 3
I2=$50.00I1=$50.00 I3=$50.00
Pay back $1 ,000
interestThe unpaid interest did notearn interest over the 3-year
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mp e n eres : ummaryIn a multi-period situation w ith simple interest:
The accrued interest does not earn interestduring the succeeding time period.
Normally, the total sum borrowed (lent) is paidback at the end of the agreed time period PLUS
.
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COMPOUND INTEREST
Compound Interest is much different
Compound means to stop and compute
In this application, compounding means to compute the
interest owed at the end of the period and then add it to theunpaid balance of the loan
Interest then earns interest
To COMPOUND stop and compute the associated interest.
When interest is compounded, the interest that is accruedat the end of a given time period is added in to form a NEW
principal balance. That new balance then earns or is charged interest in thesucceedin t ime eriod 32
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EXAMPLE 1.8 - COMPOUND INTEREST
ssume:
P = $1,000
i = 5% per year compounded annually (C.A.)
N = 3 years
For compound interest, 3 years, we have:
=
1 2 3
Owe at t = 3years:
I1=$50.00
I =$52.50
$1,000 + 50.00 +52.50 + 55.13 =
33I3=$55.13
$1,157.63
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COMPOUND INTEREST: CALCULATED
or e examp e:
P0 = +$1,000
1 , . .
Owe P1 = $1,000 + 50 = $1,050 (but w e dont pay yet!)
= = , .
Compound Interest: t = 2Principal and end of year 1: $1,050.00 I2 = $1,050(0.05) = $52.50 (owed but not paid)
o e curren unpa a ance y e s: $1,050 + 52 .50 = $1,102.50
34
Now, go to year 3 . [email protected]
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COMPOUND INTEREST: T = 3
ew r nc pa sum: , .
I3 = $1102.50(0.05) = $55.125 = $55.13
$1102.50 + 55.13 = $1157.63
Note how the interest amounts were added to form a new
principal sum w ith interest calculated on that new amount
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EXAMPLE 1.9
Five lans are shown that w ill a off a loan of 5 000 over5 years w ith interest at 8% per year.
Plan1 . Simple I nterest, pay all at the end
Plan 2 . Compound Interest, pay all at the end
Plan 3. Simple interest, pay interest at end of each year.=
Plan 4 . Compound Interest and part of the principal each
year (pay 20% of the Principal Amount)Plan 5 . Equal Payments of the compound interest andprincipal reduction over 5 years with end-of- year payments.
Note: The follow ing tables w ill show the five approaches.For now, do not try to understand how all of the numbers aredetermined (that w ill come later!). Focus on the methods
36an ow ese a es us ra e econom c equ va ence.
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PLAN 1: @ 8% SIMPLE INTEREST
,
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PLAN 3: SIMPLE INTEREST PAIDANNUALLY
Principal Paid at the End (balloon Note)
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PLAN 4: COMPOUND INTEREST
20% of Principal Paid back annually
hjbaba
eng.ukm.my
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PLAN 5: EQUAL REPAYMENT PLAN
Equal Annual Payments (Part P rincipal andPart Interest
hjbaba
eng.ukm.my
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COMPARISONS 5 PLANS
Plan 1 Simple interest = (original principal)(0.08)
Plan 2 Compound interest = (total owed previous year)(0.08)
Plan 3 Simple interest = (original principal)(0.08)
Plan 4 Compound interest = (total owed previous year)(0.08)
an ompoun nterest = tota ow e prev ous year .
AnalysisNote that the amounts of the annual payments are different foreach repayment schedule and that the total amounts repaid formost plans are different, even though each repayment planrequires exact y 5 years.
The difference in the total amounts repaid can be explained (1)by the time value of money, (2) by simple or compound interest,
an y e par a repaymen o pr nc pa pr or o year .
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TERMINOLOGY AND SYMBOLS
i = interest rate or rate of return per time period; percentper year, percent per month
=
The symbols P and F represent one-time occurrences: Specifically:
$F
0 1 2 n-1 n
t = n
$P
It shou ld be clear that a present value P represents a44single sum of money at some time prior to a future value F
This is an important basic point to remember
A A
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ANNUALAMOUNTS
represents a uniform amount (i.e., the same amount eachperiod) that extends through consecutive interest periods.
Cash Flow diagram for annual amounts might look l ike
the follow ing:
$A $A $A $A $A
0 1 2 3 .. N-1 n
A = equal, end of period cash flow amounts
I R %
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INTEREST RATE I% PER PERIOD
e n eres ra e s assume o e a compoun ra e,unless specifically stated as simple interest
example, 12% per year.
For many engineering economy problems:
nvo ve e mens on o me
At least 4 of the symbols { P, F, A, i% and n }
known w ith certainty.
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MINIMUMATTRACTIVE RATE OF RETURN (MARR)
An investment is a commitment of funds and resources in aproject w ith the expectation of earning a return over and
Economic Efficiency means that the returns should exceedthe inputs.
In the for-profit enterprise, economic efficiencies greaterthan 100% are required!
that all accepted projects must meet or exceed.
The rate, once established by the firm is termed the.
The MARR is expressed as a percent per year.
47estimating w hat this rate should be in a given time period.
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MARR HURDLE RATE
In some circles, the MARR is termed the Hurdle Rate.
Capital (investment funds) is not free.
It costs the firm money to raise capital or to use the owners
of the firms capital.
.
Cost of Capital: Personal Example .
Assume you have a charge card that carries an 18% peryear interest rate.
If you charge the purchase, YOUR cost of capital is the18% interest rate.
48 ery [email protected]
COST TO A FIRM
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COST TO AFIRM Firms raise ca ital from the follow in sources:
Equity using the owners funds (retained earnings,cash on handbelongs to the ow ners)
Owners expect a return on their money and hence, there
is a cost to the firm
interest rate on the borrowed funds
Costing Capital Financial models exist that w il l approximate the firmsweighted average cost of capital for a given time period.
,
for funding MUST return at least the cost of the funds usedin the project PLUS some additional percent return.
49 e cos s expresse as a per year us e an n eresrate.
S MARR S I
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SETTING THE MARR: SAFE INVESTMENT
First, start w ith a safe investment possibil ity
A firm could always invest in a short-term CD paying around-
But investors w il l expect more that that! The firm should compute its current weighted average costof capital (See Chapter 10)
This cost w il l almost always exceed a safe external
Assume the weighted average cost of capital (WACC) is,say, 10.25% (for the sake of presentation)
Certainly, the MARR must be greater than the firms costof capital in order to earn a profit or return thatsatisfies the owners!
50 Thus, some addit ional buffer must be provided toaccount for risk and uncertainty!
SETTING A MARR
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SETTING AMARR
ar w e
Add a buffer percent (?? Varies from firm to firm)
This becomes the Hurdle Rate that all prospectiveprojects should earn in order to be considered for funding.
G P MARR
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GRAPHICAL PRESENTATION: MARR
RoR - %
Acceptable range for newro ects
-
0%
Safe Investment WACC - %
OPPORTUNITY FORGONE
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OPPORTUNITY FORGONEAssume a firms MARR = 12%
Two projects, A and B
A costs $400,000 and presents an estimated 13% peryear.
B cost $100,000 w ith an estimated return of 14.5%
What if the firm has a budget of, say, $150,000?
A cannot be funded not sufficient funds! B is funded and earns 14.5% return or more
A is not funded, hence the firm loses the OPPORTUNITY
This often happens!
CASH FLOW DIAGRAMMING
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CASH FLOWDIAGRAMMING
ng neer ng conomy as eve ope a grap catechnique for presenting a problem dealing w ith cash flowsand their timing.
It s called a CASH FLOW DIAGRAM
It s similar to a free-body diagram in statics
First, some important TERMS . . . .
Important TERMS CASH INFLOWS
Money flow ing INTO the firm from outside
, , , .
CASH OUTFLOWS
54 First costs of assets, labor, salaries, taxes paid,
utilit ies, rents, interest , etc.
CASH FLOWS
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CASH FLOWS
For many practical engineering economy problems the cashflows must be:
Estimated
Generated from an assumed distribution and simulated
Net Cash FlowsANET CASH FLOW is
Cash Inflows Cash Outflows
(for a given time period) We normally assume that all cash flows occur:
55At the END of a given time period
End-of-Period Assumption
END-OF-PERIOD ASSUMPTION
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END-OF-PERIODASSUMPTION
- - onven onALL CASH FLOWS ARE ASSUMED TO OCCUR AT THE ENDOF AN INTEREST PERIOD EVEN IF THE MONEY FLOWS AT
TIM ES WITHIN THE INTEREST PERIOD.
THIS IS FOR SIMPLIFICATION PURPOSES
The Cash Flow Diagram: CFD
Extremel valuable anal sis tool
First step in the solution process
Graphical Representation on a time scale
Does not have to be drawn to exact scale
But, should be neat and properly labeled
56 Required on most in-class exams and part of thegrade for the problem at hand
EXAMPLE CASH FLOW DIAGRAMS
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Assume a 5- ear roblem
The basic time line is shown below
Now is denoted as t = 0
A sign convention is applied
the time l ine
Negative cash flows are normally drawn downward
57rom e me ne
SAMPLE CF DIAGRAM
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SAMPLE CF DIAGRAM
Positive CF at t = 1
Negative CFs at t = 2 & 3
PROBLEM PERSPECTIVES
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PROBLEM PERSPECTIVES
Before solving, one must decide upon the perspective ofthe problem
Assume a borrow ing situation; for example: Perspective 1: From the lenders view
Perspective 2: From the borrow ers view
Impact upon the sing convention
Lending Borrow ing ExampleAssume $5,000 is borrowed and payments are $1,100 per year.
Draw the cash flow diagram for this
First, whose perspective w ill be used?
59 Problem w ill infer or you must decide .
LENDING BORROWING
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From the Lenders Perspective
A = +$1,100/ yr
0 1 2 3 4 5
-$5,000
P = +$5,000
0 1 2 3 4 5
60
A = -$1,100/ yr
EXAMPLE 1.17
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into an investment opportunity 2 years from now that islarge enough to w ithdraw $4,000 per year for state
If the rate of return is estimated to be 15.5% per year,construct the cash flow diagram.
Example 1.17 CF Diagram
RULE OF 72 FOR INTEREST
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RULE OF 72 FOR INTERESTA common question most often asked by investors is:
How long w il l it take for my investment to double in
Must have a known or assumed compound interestrate in advance
Assume a rate of 13% / year to i l lustrate .
The Rule of 72 states: The approximate time for an investment to double invalue given the compound interest rate is:
Estimated time (n) = 72/ i
For i = 13% : 72/ 13 = 5.54 years
RULE OF 72 FOR INTEREST
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Likew ise one can estimate the re uired interest rate for
an investment to double in value over time as:
i approximate = 72/ n
Assume we want an investment to double in, say, 3
years.
Chapter Summary Engineering Economy:
Application of economic factors and criteriao eva ua e a erna ves cons er ng e me
value of money (interest and time)
Chapter Summary
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Chapter Summary
Engineering Economy Study:
Involves modeling the cash flows
ompu ng spec c measures o econom c w or
Using an interest rate(s)
The concept ofequivalencehelps in understanding how
different sums of money at different times are equal ineconomic terms.
Simple and Compound Interest
principal only) and compound interest (based onprincipal and interest upon interest) have beendescribed in formulas tab les and ra hs. 64
Chapter Summary
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C apte Su a y
Compounding of Interest The pow er of compounding is very noticeable, especially over longperiods of time.
o on o compu ng n eres on n eres
The MARR The MARR is a reasonable rate of return established as a hurdle
.
The MARR is always higher than a return from a safe investment.
Attributes of Cash Flows Difficulties w ith their estimation.
Difference between estimated and actual value.
End-of-year convention for cash-flow location.
Attributes of Cash Flows Net cash flow computation.
Different ers ect ives in determin in the cash-flow si n 65
convention.
Construction of acash-flow diagram.