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MS-291: Engineering Economy(3 Credit Hours)

Mid-Term

Section A-Total Students: 164- Average Score: 34.00- S.D: 6.60- Highest Score: 50- Lowest Score: 15

Section B-Total Students:163- Average Score:36.07- S. D: 7.37- Highest Score: 56- Lowest Score: 18

HOW was it ???

Low Score!!!!Approach to Attempt MCQs paper is different from Subject nature paper

You can check your Mid-Papers with TA.We will let you Know in next class for timing in which you can check your Mid-Papers if you want so.

Summary so for…

Present WorthConvert all cash flows to a single sum equivalent at timezero using the MARRFuture WorthConvert all cash flows to a single sum equivalent at theend of the planning horizon using the MARRCapitalized Worth MethodDetermine the single sum at time zero that is equivalentat i=MARR to a cash flow pattern that continuesindefinitely

Engineering Economy

Chapter 6Annual Worth

Analysis

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Content of this Chapter

1) Annual Worth Basics

2) Advantages and Uses of AW

3) Capital Recovery and AW values

4) AW analysis

5) Perpetual life

6) Life-Cycle Cost analysis

Annual Worth(AW): Basics

• Annual worth is an equal periodic series of dollar amounts thatis equivalent to the cash flows (inflows & outflows), at MARR (orinterest rate)

• It can be interchangeably use for Annuity “A” we studied earlier

• AW, PW and FW has the following relationship:AW = PW (A/P, i, n) = FW(A/F, i, n)

• The n in the factors is the number of years for equal-servicecomparison. This is the LCM or the stated study period of thePW or FW analysis.

Example: Different-Life Alternatives PW(Example 5.3 used in Example 6.1)

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 present worth

comparison, if the MARR is 15% per year.

(b) National Homebuilders has a standard practice of evaluating all options over a 5-year period. If a study period of 5 years is used and the salvage values are notexpected to change, which vendor should be selected?

We did this example in Last chapter.

National Homebuilders, Inc. evaluated cut-and-finish equipmentfrom vendor A (6-year life) and vendor B (9-year life) (In Example5.3). The PW analysis used the LCM of 18 years.

Consider only the vendor A option now. Consider the cash flows forall three life cycles, 18 years… (first cost $15,000; annual M&Ocosts $3500; salvage value $1000).

Demonstrate the equivalence of PW over three life cycles andAW over one cycle at i 15% .

In Example 5.3, present worth for vendor A was calculated asPW $45,036.

Example 6.1

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Example 6.1Awfirst cycle = ― 15000(A/P, i, n)+1000(A/F, i, n)

― 35000AW = ― 15000(A/P, 15%, 6)+1000(A/F, 15%, 6)

― 35000AW = ― $7349

But we know that PW of alternative A for 18years is given as: ― $45,036

AW = ― 45,036(A/P, 15%, 18)AW = ― $7349

The one-life-cycle AW value andthe AW value based on 18 years (3life-cycle) are equal.

AW = ― 45,036(A/P, i, n)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

PW = $ 45036

3 life cycles

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20….

AW = ‒$ 7349

$15,000

0 1 2 3 4 5 6

$3500

Life cycle 1

$1,000

$15,000

0 1 2 3 4 5 6

$3500

Life cycle 2

$1,000

$15,000

0 1 2 3 4 5 6

$3500

Life cycle 3

$1,000

i=15%

Important!!!!!

• The AW value determined over one life cycle isthe AW for all future life cycles.

• Therefore, it is not necessary to use the LCM oflives to satisfy the equal-service requirement

• But its important to remember, that AW methodis also based on assumption of equal servicerequirement or equal life assumption.

• Other names for AW are equivalent annual worth(EAW), equivalent annual cost (EAC), annualequivalent (AE), and equivalent uniform annualcost (EUAC)

• AW and PW (and all other alternative evaluationmethods such as, FW, BCR, etc also) select samealternatives , provided they are performed correctly

Annual Worth(AW): Basics AW Method: Assumption1. The services provided are needed for at least the LCM of the

lives of the alternatives.2. The selected alternative will be repeated for succeeding life

cycles in exactly the same manner as for the first life cycle.3. All cash flows will have the same estimated values in every life

cycle.

• If first two assumptions are not reasonable, a study period mustbe used

• for third assumption, if cash flows are expected to change exactlywith the inflation (or deflation) rate its Okay….if not…., new cashflow estimates must be made for each life cycle, and again astudy period must be used

Exactly the same as for PW method!!!!

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Advantages and Uses ofAnnual Worth

• AW method offers a prime computational and interpretationadvantage because the AW value needs to be calculated for onlyone life cycle

• AW is not necessary to use the LCM of lives to satisfy the equal-service requirement because AW for one life cycle is the AW for allfuture life cycles

• AW more desirable compared to Present worth because:i. AW is easy to calculate;ii. AW is easy to understand (express in $ per year)iii. Its assumptions are identical to those of the PW

method(such as equal life services or cash flow same in each life cycle)

An alternative have the following cash flow estimates:• Initial investment P. This is the total first cost of all assets and

services required to initiate the alternative (some time this investmenttake place overall several years)

• Salvage value S. This is the terminal estimated value of assets at theend of their useful life.

• The S is zero if no salvage is anticipated; S is negative when it will costmoney to dispose of the assets.

• For study periods shorter than the useful life, S is the estimated marketvalue or trade-in value at the end of the study period.

• Annual amount A. This is the equivalent annual amount (costsonly for cost alternatives; costs and receipts for revenuealternatives). Often this is the annual operating cost (AOC) orM&O cost, so the estimate is already an equivalent A value.

Calculation of Annual Worth (AW)

Calculation of Annual Worth (AW)

AW = CR + A

Annual Worth is composed of two components1. Capital Recovery for initial investment (at MARR)2. Equivalent Annual Amount, A

Capital Recovery Amount

• Capital recovery (CR) is the equivalent annualamount that the asset, process, or system mustearn (new revenue) each year to just recover theinitial investment plus a stated rate of return overits expected life

• Any expected salvage value is considered in thecomputation of CR

CR = ― P (A/P,i,n ) + S (A/F,i,n )

CR = ―(P ―S)[A/P, i, n) ― S(i)]or

There is fullderivation how toget the secondformula…but wedo not need that.

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Example: AW, CR

Lockheed Martin is increasing its booster thrust power in order towin more satellite launch contracts from European companiesinterested in opening up new global communications markets. Apiece of earth-based tracking equipment is expected to require aninvestment of $13 million, with $8 million committed now andthe remaining $5 million expended at the end of year 1 of theproject. Annual operating costs for the system are expected to startthe first year and continue at $0.9 million per year. The useful life ofthe tracker is 8 years with a salvage value of $0.5 million.Calculate the CR and AW values for the system, if the corporateMARR is 12% per year.

Example: AW, CR

Capital Recovery in millions is given as

P = 8 + 5(P/F, i, n) = 8+5(P/F, 12%, 1) = $12.46

Now CR = ― P(A/P, i, n) + S(A/F, i, n)

CR = ― $2.47CR = ― 12.46(0.20130) + 0.5(0.08130)

CR = ― P (A/P,i,n ) + S (A/F,i,n )

CR = ― 12.46(A/P, 12%, 8) + 0.5(A/F, 12%, 8)

Interpretation: It means that each and every year for 8 years, the equivalenttotal net revenue from the tracker must be at least $2,470,000 just to recover theinitial present worth investment plus the required return of 12% per year. Thisdoes not include the AOC of $0.9 million each year.

Example: AW, CR

Annual worth: To determine AW, the cash flows must beconverted to an equivalent AW series over 8 years

Now, AW = ? CR = -$2.47 and A = -$0.9

AW = -2.47 + (- 0.9)

AW = CR + A

AW = ― $3.37 million per year

Interpretation : $3.37 million is the AW for all future life cycles of 8 years,provided the costs rise at the same rate as inflation, and the same costs andservices are expected to apply for each succeeding life cycle.

Annual Worth Analysis:Mutually Exclusive Events

• Convert all cash flows to equivalent uniform annual amounts overthe planning horizon using the MARR

For mutually exclusive alternatives, whether cost- or revenue-based, the guidelines are as follows:

• One alternative: If AW ≥ 0, the alternative is economicallyjustified

• Two or more alternatives: Select the alternative with the AWthat is numerically largest, that is, less negative or morepositive.

• This indicates a lower AW of cost for cost alternatives or alarger AW of net cash flows for revenue alternatives.

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• If the projects are independent, the AW at the MARRis calculated.

• All projects with AW ≥ 0 are acceptable.

Not necessary to use LCM for different life alternatives whileusing AW (does not matter if its mutually exclusive events orindependent events)

Annual Worth Analysis:Independent Events

Class Practice: Comparing AlternativesUsing AW criteria (3 Minutes)

Machine R Machine SFirst Cost, $ ‒250,000 ‒ 370,000Annual operating cost, $/year ‒ 40,000 ‒ 50,000Salvage value, $ 20,000 30,000Life, years 3 5

TT Racing and Performance Motor Corporation wishes to evaluate twoalternative CNC machines for NHRA engine building. Use the AW method at10% per year to select the better alternative.

10% Single Payments Uniform Series Factors

n CompoundAmount (F/P)

Present Worth(P/F)

Sinking Fund(A/F)

Capital Recovery(A/P)

3 1.3310 0.7513 0.30211 0.402115 1.6105 0.6209 0.16380 0.2638015 4.1772 0.2394 0.03147 0.13147

Solution

Solution: What if we want to compare on PW criteria ?

Select alternative R

Machine R Machine SFirst Cost, $ ‒ 250,000 ‒ 370,000Annual operating cost, $/year ‒ 40,000 ‒ 50,000Salvage value, $ 20,000 30,000Life, years 3 5

AWR = – 250,000(A/P,10%,3) – 40,000 + 20,000(A/F,10%,3)= – 250,000(0.40211) – 40,000 + 20,000(0.30211)= $ – 134,485

AWS = – 370,500(A/P,10%,5) – 50,000 + 30,000(A/F,10%,5)= – 370,500(0.26380) – 50,000 + 30,000(0.16380)= $ – 142,824

TT Racing and Performance Motor Corporation wishes to evaluate twoalternative CNC machines for NHRA engine building. Use the AW method at10% per year to select the better alternative.

Annual Worth Analysis forPermanent Investment

• We may have a permanent Investment (such as publicsector projects) for which we need evaluation based onAnnual Worth

• In such case; convert cash flows (recurring & non-recurring) to equivalent uniform annual amounts Afor one cycle

• This automatically annualizes them for eachsucceeding life cycle.

• Add all the A values to the CR amount to find total AW(as AW = CR + A, where CR = ― P (A/P,i,n ) + S (A/F,i,n ))

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Question 6.26: Annual Worth Analysisfor Permanent Investment

Compare two alternatives for a security system surrounding apower distribution substation using annual worth analysis and aninterest rate of 10% per year.

C TFirst Cost, $ ― 25,000 ― 130,000Annual operating cost, $/year ― 9,000 ― 2,500Salvage value, $ 3,000 150,000Life, years 3 ∞

AWC = ―25,000(A/P,10%,3) – 9000 + 3000(A/F,10%,3)= ― 25,000(0.40211) – 9000 + 3000(0.30211)= $ ― 18,146 per year

AWT = ― 130,000(0.10) – 2500 = $ ― 15,500 per yearSelect the T system

How to findpresent worth forundefined lifealterative ?

( CC = A / i)( AW = CC x i)

Life Cycle Cost Analysis

Life Cycle Cost …SoundFamiliar ?

Life-cycle Costs (Old Slide)

• Life-cycle - all the time from the initial conception ofan idea to the death of a product (process)

• Life-cycle costs - sum total of all the costs incurredduring the life cycle

• Life-cycle costing - designing a product with anunderstanding of all the costs associated with aproduct during it’s life-cycle

Product Life-cycle (Old Slide)

1-28

Begin EndTime

Needsassessment

andjustification

Conceptual orpreliminary

design phase

Conceptual orpreliminary

design phase

Detaileddesignphase

Production orConstruction

Phase

OperationalPhase

Decline andretirement

phase

Requirements

OverallFeasibility

ConceptualDesignPlanning

Impact Analysis

Proof ofconcept

PrototypeDevelopmentand testing

Detailed designplanning

Allocation ofresources

Detailedspecification

Componentand supplierselection

Production orconstructionphase

Product,goods andservicebuilt

Allsupportingfacilitiesbuilt

Operational useplanning

Operational Use

Use by ultimatecustomer

Maintenance andsupport

Process,materials andmethods use

Declined andretirementplanning

DecalingUse

Phase out

Retirement

Responsibledisposal

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Life-Cycle Cost Analysis

• Life-cycle cost (LCC) analysis use Annual Worth or PresentWorth methods to evaluate cost estimates for the entire life cycleof one or more projects

• Estimates will cover the entire life span from the earlyconceptual stage, through the design & developmentstages, throughout the operating stage, and even the phaseout and disposal stages.

• Both direct and indirect costs are included to the extentpossible, and differences in revenue and savings projectionsbetween alternatives are included.

Life-Cycle Cost Analysis

• Direct costs refers to material, human labor,equipment, supplies, and other costs directlyrelated to a product, process, or system

• Indirect cost refers to taxes, management, legal,warranty, quality, human resources, insurance,software, purchasing, etc.

Life-Cycle Cost Analysis

• Life-cycle cost (LCC) analysis use Annual Worth or Present Worthmethods to evaluate cost estimates for the entire life cycle of one ormore projects

• Most commonly the LCC analysis includes costs (costalternatives), and the AW method is used for the analysis,especially if only one alternative is evaluated

• If there are expected revenue(revenue alternatives) or otherbenefit differences between alternatives, a PW analysis isrecommended.

•Public sector projects are usually evaluated using a benefit/costanalysis (Chapter 9), rather than LCC analysis, because estimatesto the citizenry are difficult to make with much accuracy

Life-Cycle Cost Applications

• Life-Cycle Costs applications are: life-span analysisfor military & commercial aircraft, new manufacturingplants, new automobile models, new and expandedproduct lines, and government systems at federal andstate levels

• For example, the U.S. Department of Defense requiresthat a government contractor include an LCC budgetand analysis in the originating proposal for mostdefense systems

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When to use Life-Cycle Costs ?

• Life-Cycle Costs analysis is more effective when significant percentage of thelife span (post purchase) costs, (relative to the initial investment) is expected tobe spend on operating, maintenance, and similar costs once the system isoperational.

• Example 1: The evaluation of two equipment purchase alternatives withexpected useful lives of 5 years and M&O costs of 5% to 10% of the initialinvestment does not require an LCC analysis.

• Example 2: Exxon-Mobil wants to evaluate the design, construction, operation,and support of a new type and style of tanker that can transport oil over longdistances of ocean. If the initial costs are in the $100 millions with support andoperating costs ranging from 25% to 35% of this amount over a 25-year life, thelogic of an LCC analysis will offer a better understanding of the economic viabilityof the project.

How Life-Cycle Cost AnalysisWorks

• To do LCC analysis one need to identify each stage ofproduct, system or an alternative during entire life

• However, at this stage, we will be given with all stagesof life cycle & we will asked to calculate LCC.

We, then follow the following steps:1. Calculate the PW by phase and stage2. Add all PW values of each stage3. Using PW calculated in Step 2,…Calculate the AW

over entire life-cycle of the alternative

Practice Question: Life-Cycle CostAnalysis (5 Minutes)

A medium-size municipality plans todevelop a software system to assistin project selection during the next 10years. A life-cycle cost approachhas been used to categorize costsinto development, programming,operating, and support costs foreach alternative.There are three alternatives underconsideration, identified as M, N, andO. The costs are summarized below.Use an annual life-cycle cost approachto identify the best alternative at 8%per year.

Practice Question: Life-Cycle CostAnalysis

8% Single Payments

n CompoundAmount (F/P)

Present Worth (P/F)

2 1.1664 0.8573

3 1.2597 0.7938

4 1.3605 0.7350

5 1.4693 0.6806

10 2.1589 0.4632

8% Uniform Series Factorsn Capital

Recovery (A/P)Present Worth(P/A)

2 0.56077 1.78333 0.38803 2.57714 0.30192 3.31215 0.25046 3.992710 0.14903 6.7101

Practice Question: Life-Cycle Cost Analysis

(5 Minutes)

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Question 6.33: Life-Cycle Cost AnalysisPWM = – 250,000 – 150,000(P/A,8%,4) – 45,000 –35,000(P/A,8%,2) – 50,000(P/A,8%,10) –30,000(P/A,8%,5)PWM = – 250,000 – 150,000(3.3121) – 45,000 –35,000(1.7833)-50,000(6.7101) – 30,000(3.9927)PWM = $ – 1,309,517Annual LCCM = – 1,309,517(A/P,8%,10)

= – 1,309,517(0.14903)= $-195,157

PWN = -10,000 – 45,000 - 30,000(P/A,8%,3) –80,000(P/A,8%,10) – 40,000(P/A,8%,10)PWN = – 10,000 – 45,000 - 30,000(2.5771) – 80,000(6.7101) - 40,000(6.7101)PWN = $ – 937,525Annual LCCN = – 937,525(A/P,8%,10)Annual LCCN = – 937,525(0.14903)

Annual LCCO = $ – 175,000

Select Alternative NAnnual LCCN = $ – 139,719

THANK YOU