risk analysis and finance

8/4/2019 Risk Analysis and Finance

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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

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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


5. Select the best al ternative6. Implement and monitor

One of the moredifficult tasks

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6. Implement and monitorWhere the major

tools of En ineerin

Economy are applied

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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

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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

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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

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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

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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,

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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

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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.

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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.

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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.

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G P MARR


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GRAPHICAL PRESENTATION: MARR

RoR - %

Acceptable range for newro ects

-

0%

Safe Investment WACC - %

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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!

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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

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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

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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

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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

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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)

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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

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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

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convention.

Construction of acash-flow diagram.

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