1Engineering Economy

Chapter 5Present Worth

Analysis

Background An engineering project or alternative is formulated to make or purchase a product,

to develop a process, or to provide a servicewith specified results

An engineering economic analysis evaluates cash flow estimates for parameterssuch as initial cost, annual costs and revenues, etc., over an estimated useful life ofthe product; process, or service

We have learned some basic tools in pervious chapters

In this chapter (and next few chapters.Stage 2) we are going to use the basic tools(we learnt already) with some more techniques to evaluate one or morealternatives using the factors and formulas learned in Stage 1

After completing these chapters, you will be able to evaluate most engineeringproject proposals using awell-accepted economic analysis technique, such aspresent worth, future worth, capitalized cost, life-cycle costing, annual worth, rateof return, or benefit /cost analysis

Why we need this Chapter ?(Present Worth Analysis)

A future amount of money converted to its equivalent value now has apresent worth (PW) that is always less than that of the future cash flow,because all P/F factors have a value less than 1.0 for any interest rategreater than zero

For this reason, present worth values are often referred to asdiscounted cash flows (DCF) , and the interest rate is referred to as thediscount rate

Up to this point, present worth computations have been made for oneproject or alternative

In this chapter, techniques for comparing two or more mutuallyexclusive alternatives by the present worth method. Full details ofcontents of the chapter is on next slide

Content of the Chapter1. Formulate Alternatives

2. Present Worth of equal-life alternatives

3. Present Worth of different-life alternatives

4. Future Worth analysis

5. Capitalized Cost analysis

From Chapter 1:Steps in an Engineering Economy Study

Problem descriptionObjective statement

Available dataAlternatives for solution

Cash flows and otherestimates

Engineering EconomicAnalysis

Measure of worth criterion(PW, B/C, IRR etc)

Best alternative Selection

New Problem description

Step 1 inStudy

Step 2

Step 3

Step 5

Step 4

Step 6

Step 7

One or more approaches tomeet objectives

Expected life Revenues Costs Taxes Project Financing

Implementation andMonitoring

New engineering economicstudy beginsStep 1 inStudy

Time Passes

Tools u will be learning inthis course are used here

Two types of the Events: basicconcept from probability theory

Mutually Exclusive events If two events cannot occur at the

same time, it is called mutuallyexclusive events

Mutually exclusive means twooutcomes cannot happensimultaneously

An example is tossing a coinonce, which can result in eitherheads or tails, but not both.

Independent Events One event is independent of

the other event

In this case the occurrence ofone event is totallyindependent of anotherevent

E.g., it rained today. And achair broke down in officetoday. These are twoindependent events

2Formulating AlternativesTwo types of economic proposals:1. Mutually Exclusive (ME) Alternatives: Only one proposal can be

selected; Compete against each other and are compared pairwise. Theseproposals are normally called alternativese.g., A selection of Best diesel powered engine among the available models

2. Independent Projects: More than one can be selected , these proposals arecalled projects; it competes only against DN

Do Nothing (DN) An ME alternative or independent project tomaintain the current approach; no new costs, revenues or savings

In this chapter, we will learn Present Worth Method toevaluate either type of proposal ..in next chapters we will learnsome more such techniques.

Project or alternatives typesbased on Cash flows

There are two types of alternatives based on Cash flows1. Revenue each alternative project being evaluated generate

costs and revenues over life period of alternative. E.g., new systems, products/services that involve capital costs

Criteria of selection is to maximize the economic measure (e.g.,profit in case of introducing new product).

2. Cost ( or service based) Each alternative has only cost cashflow estimates (revenues are same for all alternatives)

E.g., which 100-seat plane to buy? Criteria of selection is to minimize the economic measures (e.g.,

in this case cost of buying 100-seat plane)

FormulatingAlternatives

Mandates, Ideas, identifications,experience, Plans, Estimates

A BC

Not viable

D E

Not viable21

3

viable

Selectonlyone

123...m

MutuallyExclusiveAlternatives

Selectall

justified

123...

DN

Independentprojects

Either ofthese

Type of Cash flow Estimates- Revenue Alternatives

Revenues and costs- Cost AlternativesCosts only

Performance evaluation and make selection

Project ID Present Worth

A $ 30,000B $ 12,500C $ 4,000D $ 2,000

Solution: (a) Select numerically largest PW; alternative A(b) Select all with PW > 0; projects A, B & D

Example 1: Selection of Alternativesby Present Worth Criteria

For the alternatives shown below, which should be selected if theyare (a) mutually exclusive; (b) independent?

Present Worth (PW) analysis

This is the process of obtaining the equivalent worth of futurecash flows at present time

That is, finding PW of cash flows

We say that future cash flows are discounted to time 0

The higher the PW, the better PW is evaluate based on aninterest rate, which is equal to the organizations MARR

Present Worth AnalysisEvaluation

Mutually exclusive projects For one project, it is financially viable if PW 0. For 2 or more alternatives, select the one with the

(numerically) largest PW value.

Independent Projects Select all projects with PW 0 However, in practice a budget limit exists (details in chapter 12)

REMEMBER: This Evaluation is for the case whenalternatives have equal life

3 The present worth method is quite popular in industrybecause all future costs and revenues are transformedto equivalent monetary units NOW

This Criteria work as follows;

Convert all cash flows to Present Worth (same aspresent value) using MARR

Precede costs by minus sign; receipts by plus sign

PW Analysis Procedure Example 2: PW evaluation of equal-lifeMutually Exclusive alternatives

Solution:

PWX = 20,000 9000(P/A,12%,5) + 5000(P/F,12%,5)= $49,606

PWY = 35,000 4000(P/A,12%,5) + 7000(P/F,12%,5)= $45,447

Select alternative Y

Alternative X has a first cost of $20,000, an operating cost of $9,000 per year, and a$5,000 salvage value after 5 years.

Alternative Y will cost $35,000 with an operating cost of $4,000 per year and a salvagevalue of $7,000 after 5 years.

At an MARR of 12% per year, which should be selected?

Convert all cash flows toPresent Worth (sameas present value) usingMARR

Precede costs byminus sign; receipts byplus signPractice: Example 5.1

Find PW at MARR and select numerically larger PW value

Cash flow diagram ? Any one ?

Class Practice: 5 minutesA university lab is a research contractor to NASA for in-spacefuel cell systems that are hydrogen and methanol-based. Duringlab research, three equal-service machines need to beevaluated economically. Perform the present worth analysiswith the costs shown below. The MARR is 10% per year.

Electric powered Gas powered Solar poweredFirst Cost($) 4500 3500 600Annual Operating Costs ($/year) 900 700 50Salvage value S ($) 200 350 100Life years 8 8 810% Single Payments Uniform Series Factors

n CompoundAmount(F/P)

PresentWorth(P/F)

Sinking Fund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

8 2.1436 0.4665 0.08744 11.4359 0.18744 5.3349

Class PracticeElectric powered Gas powered Solar powered

First Cost($) 4500 3500 600Annual Operating Costs ($/year) 900 700 50Salvage value S ($) 200 350 100Life years 8 8 8

PWE = 4500 900( P/A ,10%,8) + 200( P/F ,10%,8) = $9208PWG = 3500 700( P/A ,10%,8) + 350( P/F ,10%,8) =$7071PWS = 6000 50( P/A ,10%,8) + 100( P/F ,10%,8)= $6220

$6220 Solar powered alternative should beselected

10% Single Payments Uniform Series Factorsn Compound

Amount (F/P)PresentWorth (P/F)

Sinking Fund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

8 2.1436 0.4665 0.08744 11.4359 0.18744 5.3349

PW of Different-Life Alternatives For alternatives with unequal lives the rule is:PW must be compared over the same number of years

This is called equal service alternatives requirement (i.e.,alternatives must end at the same time) Why itsimportant ?

Because if this condition is not meet, For COSTALTERNATIVES (which involves only cost) will always favorthe shorter-lived mutually exclusive alternative, even if it isnot the more economical choice, because fewer periods ofcosts are involved

PW of Different-Life AlternativesThe following are two equal ways of meeting the equalservice requirements:1. Least Common Multiple (LCM) of alternative livesCompare the PW of alternatives over a period of timeequal to the least common multiple (LCM) of theirestimated lives2. Study Period ApproachCompare the PW of alternatives using a specified studyperiod of n years. This approach does not necessarilyconsider the useful life of an alternative. The study periodis also called the planning horizon.

4LCM of Alternative LivesApproach

This approach compare the PW of alternatives over aperiod of time equal to the least common multiple(LCM) of their estimated lives

Three assumptions of LCM Approach1. The service provided is needed for LCM years or

more.2. The selected alternative is repeated over each life

cycle of the LCM in exactly the same manner.3. Cash flow estimates are the same in every life cycle

(i.e., change are exactly by the inflation or deflation rate only)

Evaluation of Present WorthUsing a LCM Approach

First, find the LCM for the life of alternatives

Expand the cash flows for each alternatives till theLCM period thus meeting the equal servicerequirement

Calculate the present worth for all the alternatives

Use the criteria used for Equal Life Alternatives toevaluate the alternatives

Study Period Approach Compare the PW of alternatives using a specifiedstudy period of n years. This approach does notnecessarily consider the useful life of analternative. The study period is also called theplanning horizon.

A study period analysis is necessary if the firstassumption of LCM approach (i.e., The serviceprovided is needed for LCM years or more ) cannot bemade.

Evaluation of Present WorthUsing a Study Period

For the study period approach, a time horizon is chosen overwhich the economic analysis is conducted, and only thosecash flows which occur during that time period areconsidered relevant to the analysis

Once a study period is specified, all cash flows after this timeare ignored

Salvage value is the estimated market value at the end ofstudy period (at this stage we will just use Salvage value as it is)

Short study periods are often defined by management whenbusiness goals are short-term

Example: Different-Life Alternatives

National Homebuilders, Inc., plans to purchase new cut-and-finishequipment. Two manufacturers offered the estimates below.

Vendor A Vendor BFirst cost, $Annual cost, $/yearSalvage value, $Life, years

15,000 18,0003,500 3,1001,000 2,000

6 9

(a) Determine which vendor should be selected on the basis of a presentworth comparison, if the MARR is 15% per year.

(b) National Homebuilders has a standard practice of evaluating all options overa 5-year period. If a study period of 5 years is used and the salvage valuesare not expected to change, which vendor should be selected?

Solution: LCM = 18 years; We draw its cash flows to make things easy

(a) Determine which vendor should be selected on the basis of a present worthcomparison, if the MARR is 15% per year.

To meet the criteria of equalservice alternatives we extendedthe project life from 6 years to 18years for the first alternative (& 9to 18 for 2nd alternative)

NOW You have equal life twoalternatives(equal servicecondition meet) with cashflowsyou can use standardprocedure to obtain the presentvalue of both and compare it.

5Example 5.3: Different-Life Alternatives

Solution:

PWA = -15,000 15,000(P/F,15%,6) +1000(P/F,15%,6) 15,000(P/F,15%,12)+1000(P/F,15%,12) + 1000(P/F,15%,18) 3,500(P/A,15%,18)

= $ 45,036PWB = -18,000 18,000(P/F,15%,9)+ 2000(P/F,15%,9)+ 2000(P/F,15%,18)

3100(P/A,15%,18)= $ 41,384

Select vender BWhich one to select ? A or B ?

Example 5.3: Different-Life Alternatives

Solution (b): Nowcomparison is required for5 years. Since cash flowsare of 6 years no cyclerepeat is required

PWA = -15,000 3,500(P/A,15%,5) +1000(P/F,15%,5)= $ 26, 236

PWB = -18,000 3100(PA,15%,5) + 2000(P/F,15%,5)= $ 27,397

Select vender AWhich one to select ? A or B ?

Salvage value isthe estimatedmarket value at theend of study period

LCM or Study PeriodApproach

In Previous Example LCM suggest to select Vendor B

Study Period approach suggest to select Vendor A

In such situations, the standard practice of using a fixedstudy period should be carefully examined

It should be ensured that the appropriate approach,that is, LCM or fixed study period, is used to satisfy theequal-service requirement

PW of Different-Life Alternatives forindependent alternatives

For independent projects , use of the LCMapproach is unnecessary since each project iscompared to the do-nothing alternative, not toeach other

Equal-service requirement is not a problem

Use the MARR to determine the PW over therespective life of each project, and select allprojects with a PW 0

Future Worth Analysis Future Worth is exactly like PW analysis, except

Future Worth Must compare alternatives for equalservice (i.e. alternatives must end at the same time)

The selection guidelines for FW analysis are the sameas for PW analysis; FW 0 means the MARR is met orexceeded

For two or more mutually exclusive alternatives, selectthe one with the numerically largest FW value.

Future Worth Analysis If life of two alternatives are not equal, one need to fulfill the

equal service requirement for using FW criteria.

Two ways to compare equal service:1. Least common multiple (LCM) of lives2. Specified study period

Same way as used for Present Worth Analysis expect oncelife of alternatives are equal for cash flows, one need tocompare the Future Worth instead of Present Worth

6Try: Future Worth Analysis(Problem 5.26)

An industrial engineer is considering two robotsfor purchase by a fiber-optic manufacturingcompany. Robot X will have a first cost of$80,000, an annual maintenance and operation(M&O) cost of $30,000, and a $40,000 salvagevalue. Robot Y will have a first cost of $97,000, anannual M&O cost of $27,000, and a $50,000salvage value. Which should be selected on thebasis of a future worth comparison at an interestrate of 15% per year? Use a 3-year study period.

Future Worth Analysis(Problem 5.26)

F = ?

FWX = -80,000(F/P,15%,3) 30,000(F/A,15%,3) + 40,000= -80,000(1.5209) 30,000(3.4725) + 40,000= $-185,847

FWY = -97,000(F/P,15%,3) 27,000(F/A,15%,3) + 50,000= -97,000(1.5209) 27,000(3.4725) + 50,000= $-191,285

A = $30,000

i = 15%

0 1 2 3

$40,000

$80,000

A = $27,000

i = 15%

0 1 2 3

$50,000

$97,000

F = ?

Robot X CF Robot Y CF

Select robot X

Capitalized Worth Analysis

Capitalized worth is the present worth of all revenuesor expenses over an infinite length of time

If only expenses are considered this is referred ascapitalized cost

The capitalized worth method is especially useful inproblems involving public projects with indefinite lives,or permanent endowments(or donations) for charitableorganizations and universities

The Capitalized Worth Analysis The Capitalized Worth of a series of end-of-period uniform

payments A, with interest at i% per period, is

CW (or CC) = A(P/A, i%, n) where n

As N becomes very large (if the A are perpetual payments)

We already know that P/A is given as

So, the above equations become as:

P/A= ( ) The term in the bracket becomes as n tends to infinityCapitaized Worth (or Capitalized Costs) = 1CW or CC = or

Example: Capitalized Worth(Costs) Problems

Capitalized worth (or costs) type problems vary fromvery simple to somewhat complex

Consider a simple Capitalized Cost type problem

A person want to donate $100, 000 for scholarshipsin a university. Consider, 20% per year interest rate;How much money can be withdrawn forever fromthis account?

Example: Capitalized Worth(Costs) Problem

Draw a Cash Flow Diagram

$100,000

1 2 3 - - - - - -

$ A per year = ?

CC= A (or AW)i

Or AW= CC (i)A = $100,000(0.20) = $20,000 per period

Solution:

7The Capitalized Worth Analysis

The equation can be understand by thinking of .. What present amount invested today at i will enable aninvestor to periodically withdraw an amount A forever

If investor withdraw more than amount A each period, he/shewill be withdrawing a portion of the initial principle andeventually it will exhausted

If amount being withdrawn each period is equals the interestearned on the principal for that period, the principal remainsintact, thus series of withdraw will continue forever

Capitalized Worth (Costs) Recurringand Non-recurring

More complex problems will have two types of costs associated;1. Recurring Periodic and repeat2. Non-recurring One time present or future cash flows

For more complex CC problems one must separate therecurring from the non-recurring

You will not just face problems to calculate Capital Costsof a single amount (like the previous example) but youconfronts situations in which you have to make selectionamong alternatives using CC criteria

Capitalized Cost Analysis For the comparison of two alternatives on the basis of

capitalized cost, you will use formula (CC = A/i)

So find the A value (& CCT (the sum of recurring and non-recurring costs) for each alternative and select the one whichhas lowest present worth of costs (or equal to say Lowestcapitalized costs).

Alternatives are automatically compared for same life periodbecause CCT represents the total present worth of financing andmaintaining a given alternative forever (i.e., infinity).

Summary: How to calculate CC and A and how to useCC criteria to select an alternative

Cost (cash flows)(Step 1)

Non-recurringOne time present or future cashflows (e.g., first cost, cost oncein 25th year etc)

RecurringPeriodic and repeated

Non-recurringConvert it to PW (will be PW ofall non-recurring costs forwhole life)

(Step 2)

Add values of step 4 and Step 2to obtain CC of overall cashflows

(Step 5)

Select the alternative withlowest capitalized costs

A in a only one costCycle: e.g., cash flowevery 5th year or every20th year

Uniform EqualRecurring Amounts:e.g., Annuity Series(say A2)

Convert this to a UniformSeries (say A1)

Add A1 and A2 toget one UniformSeries (AnnuitySeries) startingfrom time 0 andcontinue till infinity

(Step 3)

Divide the value of UniformAnnuity Series by i" (usingCC= A/i) to get the value ofCapitalized worth foruniform series

(Step 4)Step 3 can be skipped if you convert both recurring costsdirectly to present worth

Example: The Haverty County Transportation Authority (HCTA) has just installed

new software to charge and track toll fees. The director wants to knowthe total equivalent cost of all future costs incurred to purchase thesoftware system. If the new system will be used for the indefinite future,(a) find the equivalent cost now: i.e., a CC value. (b) for each yearhereafter, an AW value.

The system has an installed cost of $150,000 and an additional cost of$50,000 after 10 years. The annual software maintenance contract costis $5000 for the first 4 years and $8000 thereafter. In addition, there isexpected to be a recurring major upgrade cost of $15,000 every 13 years.Assume that i is 5% per year for county funds.

Can you try to draw its cash flow with you for 2 cycle of recurring costs?

Step 1: Draw cash flow Diagram(for at least 2 recurring cost cycles)

The system has an installed cost of $150,000 and an additional cost of $50,000 after 10years. The annual software maintenance contract cost is $5000 for the first 4 years and$8000 thereafter. In addition, there is expected to be a recurring major upgrade cost of$15,000 every 13 years. Assume that i is 5% per year for county funds.

8Non- Recurring Costs: $150,000 and $50,000Recurring Costs (A in a life cycle): $15000

Recurring Costs (uniform A series): $5000, $8000

Step 1: Draw cash flow Diagram(for at least 2 recurring cost cycles)

Step 2: Convert Non-Recurring costs into PW

CC1 = 150,000 50,000(P/F, 5%, 10)

CC1 = $-180,695

Non- Recurring Costs:$150,000 (Initial Costs) and$50,000 (in year 10th)

Step 3 & 4: Convert Recurring Costs (A in one lifecycle) into Uniform Costs and add it with Uniform

Recurring Costs

CC3 = 5000 (P/A, 5%, )= 5000/i or 5000/0.05

CC3 = = $ 100,000

Recurring Costs (A in one life cycle): $15000A1 = 15000(A/F, 5%, 13)

= $ 847 cost of one cycleCC2 = 847 / 0.05

= $ 16,940 (for all cycles)

Recurring Costs (uniform):$5000, $8000

Alternative:A1 = $ 847A2= $ 5000A = A1+A2 = $ 5847CCA = $5847(P/A, 5%, )

= 5487/0.05 = $116940(same as CC2 + CC3 above)

Step 3 & 4: Convert Recurring Costs (A in onelife cycle) into Uniform Costs(for whole life)and add it with Uniform Recurring Costs

CC4 = (A/i)(P/F, 5%, 4)3000 (1/0.05) (P/F, 5%, 4)

CC4 = = 49,362

Recurring Costs (uniform):If we calculate the present worth of A = $3000 (which is startingfrom year 5)it will be on 4th year. Need to multiply it with singlefactor for four year to bring it to time 0.

Step 5: Add values of Step 2 and Step 4 toObtain CC

CCT = CC1 + CC2 + CC3 + CC4= 180,695 16,940 100,000 49,362= 346,997

b) CC = A/iAW= CCT (i)

= 346.997 (0.05)= $17,350

Interpretation: This means Haverty County officials have committed theequivalent of $17,350 forever to operate and maintain the tollmanagement software

Interpretation: The $-346,997represents the one-time t = 0amount that ifinvested at 5%/year would fund thefuture cash flows asshown on the cash flow diagramfrom now to infinity!

Capitalized Cost Analysis If you havemore than one alternative and youhave to chose one out of it on Capitalized CostAnalysis.

You need to calculate the CC for everyalternative as explained in 5 step procedureand then you select the alternative whichvalue is lowest

9Capitalized cost analysis for afinite-life alternative

If a finite-life alternative is compared to one with an infinitelife on the basis of their capitalized costs, proceed asfollows:

To determine capitalized cost for the alternative with afinite life, calculate the equivalent A value for one life cycleand divide by the interest rate.

It maybe noted that CC/CW method can be applied foralternatives whose lives are not necessary infinite.

ExampleGiven the following two mutually exclusivealternatives, use the Capitalized Worth method todetermine which project you should invest in (MARR= 15%):

A BFirst Cost, $ $12000 $40000Annual Operating Cost, $ per year 2200 1000Salvage value, $ 0 10000Life, years 10 25

Solution

First, find AW of each alternative: (bz CC = AW/i) AWA = $12,000(A/P, 15%, 10) $2,200= $4,592

AWB = $40,000(A/P, 15%, 25) $1,000 +$10,000(A/F, 15%, 25)= $7,141

Solution

Second, find CW of each alternative:CWA = AWA/i = -$4,592/0.15 = -$30,613CWB = AWB/i = -$7,141/0.15 = -$47,607

Since CW of alternative A yields less cost, itshould be selected.

Practice: Finite versus Infinite lifealternatives

Compare the alternatives shown on the basis of their capitalized costsusing an interest rate of 10% per year

Alternative M Alternative NFirst Cost, $ 150000 800000Annual Operating Cost, $ per year 50,000 12000Salvage value, $ 8000 1000000Life, years 5 10% Single Payments Uniform Series Factorsn Compoun

d Amount(F/P)

PresentWorth(P/F)

SinkingFund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

5 1.6105 0.6209 0.16380 6.1051 0.26380 3.7908

Solution!!!!For M, first find AW and then divide by i to find CC.

CCM = AWM/i but we don't know AWMTo calculate Annual worth one need to convert all cashflows into Uniform series!!!!!!

AWM = 150,000(A/P,10%,5) 50,000 + 8000(A/F,10%,5)= 150,000(0.26380) 50,000 + 8000(0.16380)= $ 88,260

CCM = 88,260/0.10= $ 882,600

CCN = 800,000 12,000/0.10= $ 920,000

Select alternative M

Salvage value foralternative withinfinite life isnever realizedbecause n isnever reached.

10

Thank You