CE533 Chp4 PW Analysis

7/28/2019 CE533 Chp4 PW Analysis

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ByAssoc. Prof. Dr. Ahmet ZTA

Gaziantep University

Department of Civil Engineering

CHP IV-PRESENT WORTH ANALYSIS

CE 533 - ECONOMIC DECISIONANALYSIS IN CONSTRUCTION


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2 / 68CE533 - PW Analysis

TOPICS

Formulating Alternatives

PW of Equal-Life AlternativesPW of Different-Life alternatives

Future Worth Analysis

Capitalized Cost AnalysisIndependent projects

Payback Period

CHP IV-PRESENT WORTH ANALYSIS


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4.1 FORMULATING MUTUALLY EXCLUSIVEALTERNATIVES

Viable firms/organizations have thecapability to generate potentialbeneficial projects for potential

investmentTwo types of investment categories

Mutually Exclusive Set

Independent Project Set


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Mutually Exclusive set is where acandidate set of alternatives exist(more than one)

Objective: Pick one and only one fromthe set.

Once selected, the remainingalternatives are excluded.


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Mutually exclusive alternatives competewith each other.

Independent alternatives may or may

not compete with each otherThe independent project selectionproblem deals with constraints and mayrequire a mathematical programming orbundling technique to evaluate.


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4.1 Type of Alternatives

Alternatives CF are classified asrevenue-based or cost-based

Revenue/Cost the alternatives consist

of cash inflow and cash outflows Select the alternative with the maximum

economic value

Service the alternatives consist mainlyof cost elements

Select the alternative with the minimumeconomic value (min. cost alternative)


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4.1 Evaluating Alternatives

Part of Engineering Economy is theselection and execution of the bestalternative from among a set of

feasible alternatives

Alternatives must be generatedfrom within the organization

One of the roles of engineers!


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4.1 Evaluating Alternatives

In part, the role of the engineer toproperly evaluate alternatives froma technical and economic view

Must generate a set of feasiblealternatives to solve a specificproblem/concern


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4.1 Alternatives

Problem

DoNothing

Alt.1

Alt.2

Alt.m

Analysis

Selection

Execution

If there are m investment proposals,

we can form up to 2m mutually exclusive alternatives.This includes DN option.


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4.1 Alternatives: The Selected Alternative

Problem Alt.Selected

Execution

Audit and Track

Selection is dependent upon the data, life,discount rate, and assumptions made.


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4.2 Present Worth Approach Equal-Lifes

Simple Transform all of the current

and future estimated cash flow back to

a point in time (time t = 0)

Have to have a discount rate before the

analysis in started

Result is in equivalent dollars now!


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4.2 THE PRESENT WORTH METHOD

At an interest rate usually equal to or

greaterthan the Organizationsestablished MARR.

A process of obtaining theequivalent worth of future cashflows to some point in time

called the Present Worth


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P(i%) = P(+) P(-).

P(i%) = P( + cash flows) +

P( - cash flows)

OR, . . .


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If P(i%) > 0 then the project isdeemed acceptable.

If P(i%) < 0 the project is usuallyrejected.

If P(i%) = 0 Present worth of costs = Presentworth of revenues Indifferent!


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If the present worth of a project turns out to =0, that means the project earned exactly thediscount rate that was used to discount the cashflows!

The interest rate that causes a cash flows NPVto equal 0 is called the Rate of Return of the

cash flow!


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A positive present worth is a dollaramount of "profit" over the minimumamount required by the investors(owners).

For P(i%) > 0, the following holds true:


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Depends upon the Discount Rate Used

The present worth is purely a

function of the MARR (thediscount rate one uses).

If one changes the discount rate, adifferent present worth will result.


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For P(i%) > 0, the followingholds true:

Acceptance or rejection of aproject is a function of the timingand magnitude of the project'scash flows, and the choice of thediscount rate.


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4.2 PRESENT WORTH: SpecialApplications

Present Worth of Equal LivedAlternatives

Alternatives with unequal lives: Beware

Capitalized Cost Analysis

Require knowledge of the discount rate

before we conduct the analysis


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4.2 PRESENT WORTH: Equal Lives

Present Worth of Equal LivedAlternatives straightforward

Compute the Present Worth of eachalternative and select the best, i.e.,smallest if cost and largest if profit.


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4.2 Equal Lives Straightforward!

Given two or more alternatives withequal lives.

Alt. 1

Alt. 2

Alt. N

N = for allalternatives

Find PW(i%) for each alternative then compare


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4.2 PRESENT WORTH: Example

Consider: Machine A Machine B

First Cost $2,500 $3,500

Annual Operating Cost 900 700

Salvage Value 200 350Life 5 years 5 years

i = 10% per year

Which alternative should we select?


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4.2 PRESENT WORTH: Cash Flow Diagram

Which alternative should we select?

0 1 2 3 4 5

$2,500A = $900

F5=$200MA

0 1 2 3 4 5

$3,500

F5=$350

A = $700

MB


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4.2 PRESENT WORTH: Solving

PA= 2,500 + 900 (P|A, .10, 5)200 (P|F, .01, 5)

= 2,500 + 900 (3.7908) - 200 (.6209)

= 2,500 + 3,411.72 - 124.18 = $5,788

PB = 3,500 + 700 (P|A, .10, 5)

350 (P|F, .10, 5)= 3,500 + 2,653.56 - 217.31 = $5,936

SELECT MACHINE A: Lower PW cost!


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4.2 Present Worth of Bonds

Often corporations or government obtaininvestment capital for projects by sellingbonds.

A good application of PW method is theevaluation of a bond purchase alternative.

If PW < 0 at MARR, do-nothing alternativeis selected.

A bond is similar to an IOU for time periodssuch as 5, 10, 20 or more years.


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4.2 Present Worth of Bonds

Each bond has a face value V of $100,$1000, $5000 or more that is fullyreturned to the purchaser when the

bond maturity is reached.

Additionaly, bond provide thepurchaser with periodic interest

payments I (bond dividends) using thebond coupon (interest) b, and c, thenumber of payment periods per year.


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4.2 Bonds Notation and Example

(Bond face value)(bond coupon rate) V.b

I = ------------------------------------------ = ----

number of payments per year c

At the time of purchase, the bond may sell formore or less than face value.

Example: V = $5,000 (face value) b = 4.5% per year paid semiannually

c = 10 years


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4.2 PW Bonds Example Continued

The interest the firm would pay to thecurrent bondholder is calculated as:

0.045$5,000( ) $5,000(0.0225)2

$112.50 every 6 months

I

I

The bondholder, buys the bond and will receive$112.50 every 6 months for the life of the bond


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4.2 Example 4.2

Aye has some extra money, requires safeinvestment. Her employer is offering toemployees a generous 5% discount for 10-year 5000 YTL bonds tat carry a coupon rate of

6% paid semiannually.The expectation is to match her return onother safe investments, which have averaged6.7% per year compounded semiannually.(This is an effective rate of 6.81% per year).

Should Aye buy the bond?


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4.2 Example 4.2 Cash-Flow Diagram

A = 150

0 1 2 3 4 . .. 19 20

P=??

$5,000

i=3.35%

Find the PW(3.35%) of the future cashflows to the potential bond buyer


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4.2 Example 4.2 Solving

I = (5000)(0.06)/2 = 150 YTL every 6months for a total of n=20 dividendpayments.The semiannual MARR is 6.7/2 =

3.35%, and the purchase price now is5000(0.95)= -4750 YTL.Using PW evaluation:PW = -4750 + 150(P/A, 3.35%, 20) +

5000(P/F, 3.35%, 20) = - 2.13 YTL


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4.3 Present Worth Analysis of Different-LifeAlternatives

In an analysis one cannot effectivelycompare the PW of one alternative witha study period different from another

alternative that does not have the samestudy period.

This is a basic rule!


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4.3 PRESENT WORTH: Unequal Lives

If the alternatives have different lifes,there are 2 ways to use PW analysis tocompare alternatives:

A) The lowest common Multiple (LCM)

B) Study period (planning horizon)


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4.3 PRESENT WORTH: Lowest CommonMultiple (LCM) of Lives

If the alternatives have different studyperiods, you find the lowest commonlife for all of the alternatives in

question.Example: {3,4, and 6} years. Thelowest common life (LCM) is 12 years.

Evaluate all over 12 years for a PWanalysis.


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4.3 PRESENT WORTH: Example UnequalLives

EXAMPLE

Machine A Machine B

First Cost $11,000 $18,000

Annual Operating Cost 3,500 3,100Salvage Value 1,000 2,000

Life 6 years 9 years

i = 15% per year

Note: Where costs dominate a problem it is customary to assign apositive value to cost and negative to inflows


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4.3 PRESENT WORTH: Example UnequalLives

A common mistake is tocompute the present

worth of the 6-yearproject and compare it tothe present worth of the

9-year project.NO! NO! NO!


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4.3 PRESENT WORTH: Unequal Lives

i = 15% per year

0 1 2 3 4 5 6

$11,000

F6=$1,000

A1-6

=$3,500

Machine A

0 1 2 3 4 5 6 7 8 9

F6=$2,000

A 1-9

=$3,100

$18,000

Machine B

LCM(6,9) = 18 year study period will apply for present worth


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4.3 Unequal Lives: 2 Alternatives

i = 15% per year

Machine A

LCM(6,9) = 18 year study period will apply for present worth

Cycle 1 for A Cycle 2 for A Cycle 3 for A

6 years 6 years 6 years

Cycle 1 for B Cycle 2 for B

18 years

9 years 9 yearsMachine B


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4.3 Example: Unequal Lives Solving

LCM = 18 years

Calculate the present worth of a 6-year cyclefor A

PA = 11,000 + 3,500 (P|A, .15, 6)

1,000 (P|F, .15, 6)

= 11,000 + 3,500 (3.7845) 1,000 (.4323)

= $23,813, which occurs at time 0, 6 and 12


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4.3 Example: Unequal Lives

PA= 23,813+23,813 (P|F, .15, 6)+

23,813 (P|F, .15, 12)= 23,813 + 10,294 + 4,451 = 38,558

0 6 12 18

$23,813 $23,813 $23,813

Machine A


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4.3 Unequal Lives Example: Machine B

Calculate the Present Worth of a 9-yearcycle for B

0 1 2 3 4 5 6 7 8 9

F6=$2,000

A 1-9

=$3,100

$18,000


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4.3 9-Year Cycle for B

Calculate the Present Worth of a 9-year cyclefor B

PB = 18,000+3,100(P|A, .15, 9)

1,000(P|F, .15, 9)= 18,000 + 3,100(4.7716) - 1,000(.2843)

= $32,508 which occurs at time 0 and 9


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4.3 AlternativeB 2 Cycles

PB = 32,508 + 32,508 (P|F, .15, 9)

= 32,508 + 32,508(.2843)

PB = $41,750

Choose Machine A

0 9 18

$32,508 $32,508

Machine A: PW =$38,558


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4.3 Unequal Lives Assumed StudyPeriod

Study Period Approach

Assume alternative: 1 with a 5-year life

Alternative: 2 with a 7-year life

Alt-1: N = 5 yrs

Alt-2: N= 7 yrs

LCM = 35 yrs

Could assume a study period of, say, 5 years.


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4.3 Unequal Lives Assumed StudyPeriod

Assume a 5-yr. Study period

Estimate a salvage value for the 7-yearproject at the end of t = 5

Truncate the 7-yr project to 5 years

Alt-1: N = 5 yrs

Alt-2: N= 7 yrs

Now, evaluate bothover 5 years using

the PW method!


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FUTURE WORTH APPROACH

FW(i%) is an extension of the presentworth method

Compound all cash flows forward in

time to some specified time periodusing (F/P), (F/A), factors or,

Given P, the F = P(1+i)N


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Applications of Future Worth

Projects that do not come on line untilthe end of the investment period

Commercial Buildings

Marine Vessels Power Generation Facilities

Public Works Projects

Key long time periods involving

construction activities


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Life Cycle Costs (LCC)

Extension of the Present Worth method

Used for projects over their entire lifespan where cost estimates are

employedUsed for:

Buildings (new construction or purchase)

New Product Lines

Commercial aircraft

New automobile models

Defense systems


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Life Cycle: Two General Phases

TIME

Cost-$

Acquisition Phase Operation Phase

Cumulative Life CycleCosts


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4.4 CAPITALIZED COST ANALYSIS

CAPITALIZED COST- the present worthof a project that lasts forever.

Government Projects

Roads, Dams, Bridges (projects thatpossess perpetual life)

Infinite analysis period


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4.4 Derivation for Capitalized Cost

Start with the closed form for the P/A factor

Next, let N approach infinity and divide thenumerator and denominator by (1+i)N

(1 ) 1

(1 )

N

N

iP A

i i


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4.4 Derivation - Continued

Dividing by (1+i)N yields

Now, let n approach infinity and the

right hand side reduces to.

11

(1 )NiP A

i


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4.4 Derivation - Continued

1 AP A

i i

Or,

CC(i%) = A/i


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4.4 CAPITALIZED COST

Assume you are called on to maintain acemetery site forever if the interest rate= 4% and $50/year is required to

maintain the site.

Find the PW of an infinite annuity flow

1 2 3 4 5 .. N=inf.

A=$50/yr

P = ?

..


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P0 = $50[1/0.04]

P0 = $50[25] = $1,250.00

Invest $1,250 into an account thatearns 4% per year will yield $50 of

interest forever if the fund is not

touched and the i-rate stays constant.


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4.4 CAPITALIZED COST: Endowments

Assume a wealthy donor wants to endow a

chair in an engineering department.

The fund should supply the department with

$200,000 per year for a deserving faculty

member.

How much will the donor have to come up

with to fund this chair if the interest rate =

8%/yr.


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4.4 CAPITALIZED COST: Endowed Chair

The department needs $200,000 per year.

P = $200,000/0.08 = $2,500,000

If $2,500,000 is invested at 8% then theinterest per year = $200,000

The $200,000 is transferred to the department,but the principal sum stays in the investment tocontinue to generate the required $200,000


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EXAMPLE

Calculate the Capitalized Cost of aproject that has an initial cost of

$150,000. The annual operating cost is$8,000 for the first 4 years and $5000thereafter. There is an recurring$15,000 maintenance cost each 15

years. Interest is 15% per year.

4.4 Capitalized Cost Example

4 4 C h Fl Di


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$4,000

0 1 2 3 4 5 6 7 15 30

$150,000

$8,000

$15,000 $15,000 $15,000 $15,000

i=15%/YR

N=

How much $$ at t = 0 is required to fund thisproject?

The capitalized cost is the total amount of $ at t

= 0, when invested at the interest rate, willprovide annual interest that covers the futureneeds of the project.

4.4 Cash Flow Diagram


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4.4 CAPITALIZED COST - ExampleContinued

1. Consider $4,000 of the $8,000 costfor the first four years to be a one-timecost, leaving a $4,000 annual operating

cost forever.P0= 150,000 + 4,000 (P|A, .15, 4) =

$161,420

2.855


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4.4 CAPITALIZED COST - Continued

Recurring annual cost is $4,000 plus theequivalent annual of the 15,000 end-of-cycle cost.

.0 15 30 45 60 ..

Take any 15-year period and find the equivalentannuity for that period using the F/A factor.


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4.4 CAPITALIZED COST: One Cycle

Take any 15-year period and find the equivalentannuity for that period using the F/A factor

$15,000

A for a 15-year period

0 15 30 45 60 ...


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CE533 - PW Analysis


2. Recurring annual cost is $4,000 plus

the equivalent annual of the 15,000

end-of-cycle cost.

A= 4,000 + 15,000 (A|F, .15, 15)

= 4,000 + 15000 (.0210) = $5,315

Recurring costs = $5,315/i =

5,315/0.15 =$3,443/yr


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CE533 - PW Analysis


Capitalized Cost = 161,420 + 5315/.15

= $196,853

Thus, if one invests $196,853 at time t

= 0, then the interest at 15% willsupply the end-of-year cash flow tofund the project so long as the principalsum is not reduced or the interest rate

changes (drops).


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CE533 - PW Analysis

4.6 USING EXCEL FOR PW ANALYSIS

General format to determine the PW is;

PW = P PV (i%,n,A,F)

When different-life alternatives areevaluated using LCM; develop NPV function

PW = P +

NPV(i%,year_1_CF_cell:last_year_CF_cell)


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CE533 - PW Analysis

Summary: Present Worth

PW represents a family of methods

Annual worth

Future Worth Capitalized Cost

Life-cycle cost analysis application

Bond Problems application


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End of Chapter 4

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CE533 Chp4 PW Analysis