Chapter 5 Present Worth (PW) Analysis

7/29/2019 Chapter 5 Present Worth (PW) Analysis

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GENG 360:EngineeringEconomics

Chapter 5. Present Worth

(PW) Analysis

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Contents

Formulating Alternatives PW of Equal-Life Alternatives PW of Different-Life alternatives Future Worth Analysis Capitalized Cost Analysis Payback Period

Life-Cycle Costs PW of Bonds Spreadsheet Applications


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5.1. Formulating Mutually Exclusive

Alternatives

Firms have the capability to generatepotential projects for investment

Two types of investment categories

Mutually Exclusive Set

Independent Project Set

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Mutually Exclusive set

It is a set where a candidate set ofalternatives exist (more than one)

Objective: Pick one and only one from theset.

Once selected, the remaining alternativesare excluded.


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Independent Project Set

Given a set of alternatives (more than one)

Objective is to:

Select the best possible combination of projects fromthe set that will optimize a given criteria.

Subjects to constraints

More difficult problem than the mutually exclusive

approach

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Mutually Exclusive Vs.Independent Alternatives

Mutually exclusive alternatives compete witheach other.

Independent alternatives may or may notcompete with each other

The independent project selection problem dealswith constraints and may require a mathematicalprogramming or bundling technique to evaluate


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Type of Alternatives

(Revenue based or service-based)Cash Flows determine whether the alternativesare revenue based or service based. Revenue/Cost the alternatives consist of cash

inflow and cash outflows

Select the alternative with the maximumeconomic value

Service the alternatives consist mainly of costelements

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

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Problem

DoNothing

Alt.1

Alt.2

Alt.m

Analysis

Selection

Execution

Selection depends upon data, life, discount rate, and assumptions made.


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5.2 Present Worth Approach with Equal-Lives

Simple: Transform all of the current and future

estimated cash flow back to a point in time (time t = 0)

Result is in equivalent dollars now!

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

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If P(%) is:

> 0 then the project is deemed acceptable.

< 0 the project is usually rejected= 0 (Present worth of costs = Present

worth of revenues) Indifferent!If P(%) = 0 that meansthe project earned exactlythe discount rate that wasused to discount the cash

flows!

The interest rate that causesa cash flows NPV to equal 0is called the Rate of Returnof the cash flow!

The net present worth is purely a function of theMARR (the discount rate one uses).

If one changes the discount rate, a different NPV willresult.


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

A positive present worth is a dollar amount

of "profit" over the minimum amount

required by the investors (owners).

Acceptance or rejection of a project is afunction of the timing and magnitude of theproject's cash flows, and the choice of thediscount rate.

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PRESENT WORTH: Special Applications

Present Worth of Equal Lived Alternatives

Alternatives with unequal lives: Beware

Capitalized Cost Analysis

Both Calculations Require knowledge of the discount rate before

we conduct the analysis


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Present Worth of Equal Lived

Alternatives Straightforward: Compute the PW of each

alternative and select the best,

i.e., smallest if cost and largest if profit

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Example:

Consider: Machine A Machine B

First Cost $2,500 $3,500

Annual Operating Cost 900 700

Salvage Value 200 350

Life 5 years 5 years

i = 10% per year

Which alternative should we select?


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Solution: Cash Flow Diagrams:

0 1 2 3 4 5

$2,500A = $900

F5=$200

MA

0 1 2 3 4 5

$3,500

F5=$350

A = $700

MB

PA = 2,500 + 900 (P|A, 0.10, 5) 200 (P|F,0.10, 5)= 2,500 + 900 (3.7908) - 200 (0.6209)= 2,500 + 3,411.72 - 124.18 = $5,788

PB = 3,500 + 700 (P|A, 0.10, 5) 350 (P|F,0.10, 5)= 3,500 + 2,653.56 - 217.31 = $5,936

Which alternativeshould we select?

SELECTMACHINE A:Lower PW cost!

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Present Worth of Alternativeswith Different Lives

BASIC RULE: In an analysis one cannot effectivelycompare the PW of one alternative with a study period

different from another alternative that does not have the

same study period.

Comparison must be made over equal time periods

Compare over the Least Common Multiple (LCM) for their lives

Example:You have different projects over{3,4, and 6}yearsThe LCM life is 12 years (6x2) Evaluate all over 12 years for aPW analysis.


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Example:Machine A Machine B

First Cost $11,000 $18,000

Annual Operating Cost 3,500 3,100

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

A commonmistake isto computethe presentworth of

the 6-yearproject andcompare it

to thepresentworth of

the 9-yearproject.

i = 15% per year

0 1 2 3 4 5 6

$11,000

F6=$1,000

A 1-6

=$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


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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 years

Machine B

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First, Calculate the PW of Machine A:

Machine A

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

6 years 6 years 6 years

PA6 = 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

0 1 2 3 4 5 6

$11,000

F6=$1,000

A 1-6 =$3,500

Machine A


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0 6 12 18

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

Machine A

PA18=23,813 +23,813 (P|F, 0.15, 6) +23,813 (P|F, 0.15, 12)

=23,813 +10,294 +4,451 =$38,558

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Calculate the Present Worth of a 9-year cycle for B

0 1 2 3 4 5 6 7 8 9

F6=$2,000

A 1-9

=$3,100

$18,000

PB9 = 18,000+3,100(P|A, .15, 9) 2,000(P|F, .15, 9)= 18,000 + 3,100(4.7716) - 2,000(.2843)= $32, 223 which occurs at time 0 and 9

0 9 18

$32,508 $32,508

PB18 = 32,508 + 32,508 (P|F, .15, 9)= 32,508 + 32,508(.2843) = $41,750

Choose Machine A because it costs less


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11000

18 years

Alternative (A) cash flow:

3500 35003500

11000 11000

1000 10001000

Alternative (B) cash flow:

18000

3100 3100

18000

20002000

18 years

PWA = - 11,000 3,500 (P/A, 15%, 18) 10,000 (P/F, 15%, 6)-10,000 (P/F, 15%, 12) + 1,000 (P/F, 15%, 18)= - 38,559

PWB = - 18,000 3,100 (P/A, 15%, 18) 16,000 (P/F, 15%, 9)

+ 2,000 (P/F, 15%, 18) = -41,384

We Select (A)

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5.4. FUTURE WORTH APPROACH

Compound all cash flows forward in time to somespecified time period using (F/P), (F/A), factors or,

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

Applications of FW Analysis: Projects that do not come on line until the 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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FW Exercise See Example 5.3

Calculate the Future Worth of determiningthe selling price in order to earn exactly25% on the investment

Draw the cash-flow diagram!!

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5.5 Capitalized Cost

Capitalized Cost: the PW of a project that lastsforever, such as:Government Projects

Roads, Dams, Bridges (projects that possessperpetual life)

Infinite analysis period

Perpetual Investments: Investments that lastfor such a long time (n ~ )


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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

1

1 (1 )NiP A

i

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Now, let n approach infinity and the right hand sidereduces to.

iAPn

1CostdCapitalize

Or,

CC(i%) = A/i


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Example:Calculate the Capitalized Cost of a project

that has an initial cost of $150,000. Theannual operating cost is $8,000 for thefirst 4 years and $4,000 thereafter. Thereis an recurring $15,000 maintenancecost each 15 years. Interest is 15% per

year.

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Cash Flow Diagram:

$4,000

0 1 2 3 4 5 6 7 15 30

$150,000

$8,000

$15,000 $15,000 $15,000 $15,000How much $$ at t = 0is required to fundthis project?

The capitalized cost is thetotal amount of $ at t = 0,when invested at theinterest rate, will provideannual interest that coversthe future needs of theproject.

i=15%/YR

Solution:


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1. Consider $4,000 of the $8,000 cost for thefirst four years to be a one-time cost, leaving

a $4,000 annual operating cost forever:

P0= 150,000 + 4,000 (P|A, .15, 4) = $161,420

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

.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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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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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) = $4,315/yr

Recurring costs = $4,315/i = 4,315/0.15

=$28,767

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Capitalized Cost = 161,420 + 4315/.15

= $190,187

Thus, if one invests $190,187 at time t = 0,then the interest at 15% will supply the end-of-year cash flow to fund the project so longas the principal sum is not reduced or theinterest rate changes (drops).


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Example: Find the capitalized cost for the following project (use i =15%).

Time $

Initial Cost At end of year 0 150,000

Additional Cost At end of year 10 50,000

Annual Cost From end of year 1 to end of year 4 5000

Annual Cost From end of year 5 to 8000

Operation Cost End of every13 years 15000

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Solution:

8 15 ( / ,15%,13)150 50 ( / ,15%,10) 5 8 ( / ,15%,4)

0.15 0.15

15 0.029118150 50 0.1229 3 5.8474

0.15 0.15

$194.847K

K K A FP K K P F K K P A

KKK K K

5000

150000

8000

5000015000 15000 15000

0 1 4 5 10 13 26 39

13 years

8 15150 50 ( / ,15%,10) 5 8 / ,15%,40.15

8 15150 50 0.1229 3 5.8474

0.15 5.1528

$194.847K

K KP K K P F K K P A

K KK

i

K K

OR:

i13 years =(1+0.15)13 1 =5.1528


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Example:

38Which one do you choose?


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Incremental Analysis:The value of the present worth can be

calculated by using either incrementalmethod or individual method.

Both methods should lead to the sameanswer.

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Example:


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5.6 Payback Period Analysis Payback period is the period of time it

takes for the cash flows to recover theinitial investment.

Two forms for this methodDiscounted Payback Period (uses an interest

rate)

Conventional Payback Period (does not use

an interest rate)

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Payback period is only a rough estimator of desirability

Use it as an initial screening method

Avoid using this method as a primary analysis techniquefor selection projects

Totally avoid the no-return payback period

The No-return method (example follows):

Does not employ the time value of money

Disregards all cash flows past the payback time period

If used, can lead to conflicting selections when compared tomore technically correct methods like present worth!


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Discounted Payback Approach: Find the value of np such that:

1

0 ( / , , )pt n

tt

P NCF P F i t

NCFt = Net CashFlow at time t

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Example 5.8 (Discounted Analysis)Machine A: N=7 i = 15%

Payback

isbetween6 and 7yeas(6.57yrs)


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Non-Discounted Analysis

At a 0interest ratethe PB timeis seen to

equal 4years!

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Machine B (Discounted):Machine B: N=14 i = 15%

Payback for B isbetween 9 and

10 years!Longer time

period torecover theinvestment.

9.52 years


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Machine B (Undiscounted):

Payback for B at0% is 6 years!

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Example Summary

Discounted

Machine A: 6.57 years

Machine B: 9.52 years

Undiscounted

Machine A: 4.0 years

Machine B: 6.0 years Go with Machine A lower time period payback

to recover the original investment


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5.8 Present Worth of Bonds

Bonds represent a source of funds for thefirm.

Bonds are sold (floated) by investmentbanks for firms in order to raise additionaldebt capital

A bond is similar to an IOU

Bonds are evidence of Debt

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Bond Types: Treasury bonds

Issued by Federal Government

Full backing of the Government

Conservative-type investment

State and Municipal Bonds

Issued by states and local governments

Generally tax-exempt by the Federal Government

Used to finance state and local projects

Backed by future tax and user fees to pay the interestand face value


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Mortgage Bonds Issued by Corporations

Secured by the firms assets Money received by the firm is used to fund projects

Referred to a Debt Capital

Buyers of these bonds are not owners they are lenders to thefirm

Debenture Bond Issued by Corporations

Not backed by specific assets

Backing good faith of the firm

Pays higher interest rates

Higher risks involved Bond interest rate may float

Could be convertible to common stock

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PW of Bonds - Overview

The Firm

Investment Bankers

Proceeds from

The sale

Sell the Bonds toThe lending public

Bondholders


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Basics of Bonds: The bond itself is just a piece of paper Bonds are negotiable instruments

Can be traded by the current bondholder

Source of funds to the firm (Debt capital)

Bondholders are loaning $$ to the firm, they Earn periodicinterest

They can sell the bonds at any time

The bondholders are lenders not owners

The firm pays periodic interest payments to the currentbond holders

At the end of the bonds life, the bonds are redeemed(bought back) from the current bond holder

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Bonds Notations: P0:The time (t = 0) selling price of the bond the cost to

the buyer of the bond

V:The face value of the bond

The value printed on the bond

Face values are usually: $100, $1,000, $5,000, $10,000 increments

N:The life of the bond in years (Defining the Maturity Date)

r: The nominal annual bond interest rate

Given the nominal annual bond interest rate, the payment

frequency of the interest (monthly, quarterly semi-annually, etc.) is

also stated(Facevalue)(Bondinerestrate)

Numberof paymentperiodperyearI I: Amount of interest

paid per period


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Example: Calculate the Bond Semiannual interest if:

V = $5,000 (face value)

r = 4.5% per year paid semiannually

N = 10 years

Solution:

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

2$112.50 every 6 months

I

I

The bondholder,buys the bondand will receive

$112.50 every 6months for thelife of the bond

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Example 5.11

Given V = $5,000 (Face value of the bond)

r = 4.5% paid semiannually

N = 10 years or 20 interest periods

$I per 6 months = $5,000(0.045/2) = $112.50 paid tothe current bondholder

The buyer will consider buying this bond if he canearn a nominal rate of return of 8%/yr c.q.

(compounded quarterly)

Calculate the acceptable purchase price

Bonds are bought and sold in a bond market. Thus the priceof the bond is subject to the pressures of the bond market.


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Solution: What is fixed?

The future interest payments are fixed

The future face value of the bond is fixed

What can vary?

The purchase price such that the buyer can earn at least the 8%/yr c.q.

8% c.q is the same as

0.08/4 =0.02 =2% per quarter.

Bond interest flows every 6 months

Need an effective 6-month rate

The effective 6-month rate is then

(1.02)2 1 =0.0404 =4.04%/6 months

This is the potential buyers required interest rate

The objective is to determine the purchase price of this bonddiscounted at the buyers required rate of 4.04% per 6 months

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Draw the cash-flow diagram

Work the problem with N = 20 (not 10): We have 20 interest payments (every 6 months for 10 years)

A = 112.50/6 months

0 1 2 3 4 . .. 19 20

P=??

$5,000

i=4.04%/6 months

Find the PW(4.04%) of the future cashflows to the potential bond buyer


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P = $112.50(P/A,4.04%,20)+

$5,000(P/F,4.04%,20)= $3,788

IF the buyer can buy this bond for $3,788 orless, he/she will earn at least the 8% c.q. rate.

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Example:

A 15-year bond with a face value of$10,000 is offered for sale.

The rate of interest on the bond is 6%,paid semi-annually.

Assume that MARR = 8% nominal

compounded semi-annually, what would be the selling price of the

bond?


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Solution:( )( ) (10000)(0.06) $300

2Facevalue Bond inerest rateI

Numberof payment period per year

I =$300

Price of bond =?

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Face value =$ 10,000

P =300(P/A, 4%, 30) +10,000(P/F, 4%, 30) =$8,270.6

Chapter 5 Present Worth (PW) Analysis

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