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INFO 630 Evaluation of Information SystemsDr. Jennifer BookerWeek 8 Chapters 10-121INFO630 Week 8www.ischool.drexel.eduwww.ischool.drexel.eduFor-Profit Business DecisionsChapter 10INFO630 Week 82www.ischool.drexel.eduwww.ischool.drexel.edu2Minimum Attractive Rate of Return (MARR)Basic for-profit decision processIncremental vs. total cash-flow analysisRank on rate of return

For-Profit Business DecisionsOutline

3INFO630 Week 8www.ischool.drexel.edu3Where are we?Prior chaptersMutually exclusive alternativesWhich is best to carry outCan be based onTotal cash flowDifferential cash flowBasis of comparisonMinimum Attractive Rate of Return (MARR) etc.4INFO630 Week 8www.ischool.drexel.eduMinimum Attractive Rate of Return (MARR)A statement that the organization is confident it can achieve at least that rate of returnA.k.a. Opportunity costBy investing here, you forego the opportunity to invest thereIf youre confident you can get X% there, all other alternatives should be evaluated against that X%5INFO630 Week 8www.ischool.drexel.edu5Significance of the MARRThe MARR is used as the interest rate in for-profit business decisionsPW(MARR) = how much more, or less, valuable that alternative is than investing the same $ in an investment that returns the MARRSo PW(MARR) = $1000 doesnt mean youll gain just $1000, it means that the cash-flow stream is equivalent to $1000 more today than investing those same resources in something that returns the MARR6INFO630 Week 8www.ischool.drexel.edu6Remember the ATE project from earlier? Its PW(10%) is $260K. If MARR=10%, it means that this project will return the equivalent of $260K more than investing in something that returns the MARR.

Since investing at MARR has same PW(i) as investing at MARR, PW(MARR) = $0Since FW(i) = PW(i) * (F/P,I,n), FW(MARR) = 0Since AE(i) = PW(i) * (A/P,i,n), AE(MARR) = 0

Significance of the MARRNote: Usually MARR is often set by policy decision from an organizations management teamToo high or too low?How set MARR? What impact does that have?7INFO630 Week 8www.ischool.drexel.edu7Remember the ATE project from earlier? Its PW(10%) is $260K. If MARR=10%, it means that this project will return the equivalent of $260K more than investing in something that returns the MARR.

Since investing at MARR has same PW(i) as investing at MARR, PW(MARR) = $0Since FW(i) = PW(i) * (F/P,I,n), FW(MARR) = 0Since AE(i) = PW(i) * (A/P,i,n), AE(MARR) = 0

Factors Influencing the MARRType of organizationFor-profit vs. regulated public utilityPrevailing interest rate for typical investmentsAvailable fundsSource of fundsEquity vs. borrowed fundsNumber of competing proposalsEssential vs. electiveAccounting for inflation or notBefore- or after-tax8INFO630 Week 8www.ischool.drexel.edu8If a MARR is defined, it will probably have been defined via a policy statement from the upper-management of the organization

Regulated utilityset by PUCIf borrowed funds, need to consider cost of capitalBefore- and After-tax MARROne way or another, a for-profit organization needs to address income taxes (see Chapter 16)Before-tax MARRUse on pre-income tax cash-flow streamsApproximationAfter-tax MARRUse on post-income tax cash flow streams (Ch 16)More accurateRelationship between themAfter-tax MARR = (Before-tax MARR) * (1-Eff Tax Rate)E.g. Before-tax MARR = 21%, Eff tax rate = 38%After-tax MARR = 0.21 * (1-0.38) = 0.13 = 13%9INFO630 Week 8www.ischool.drexel.edu9Work outand interpreta couple more examples

Basic For-profit Decision ProcessAssume the first alternative is the current bestfor j = 2 to the number of alternativesConsider the jth alternative to be the candidateCompare the candidate to the current bestif the candidate is better than the current best then make the jth alternative the current best{* on ending, the current best is the best alternative *} Find maximum valueSuccessive comparison in pair-wise basisCurrent bestCandidateAlgorithm10INFO630 Week 8www.ischool.drexel.edu10Just like a MAX() function

Basic For-profit Decision Process (cont)All other things equal, an alternative with a smaller initial investment is preferredAlso, if using IRR, order is importantLeads to a small, but important changeSort the alternatives in order of increasing investment, andif the candidate is strictly better than the current best 11INFO630 Week 8www.ischool.drexel.edu11Explain why IRR and Differential CFSs are important (helps differential cash flow stream meet three IRR criteria)

Basic For-profit Decision Process

12INFO630 Week 8www.ischool.drexel.edu12Incremental vs. Total Cash-Flow AnalysisTotal cash-flow analysisComparing on entire cash-flow stream basisIncremental cash-flow analysisComparing on difference between cash-flow streamsIf PW(MARR) of CFS2-CFS1 >0, then CFS2 is better than CFS1WARNING: if using IRR as the basis of comparison, cash-flow analysis must be done incrementally13INFO630 Week 8www.ischool.drexel.edu13Will show how to calculate differential CFS on next slide

Computing Differential Cash-Flow StreamsCash-flow stream AInitial investment = $5300Annual income = $4142Annual expenses = $3144Salvage value = $210Cash-flow stream BInitial investment = $6200Annual income = $7329Annual expenses = $5908Salvage value = $340Differential cash-flow stream (B-A)Initial investment = $6200 - $5300 = $900Annual income = $7329 - $4142 = $3187Annual expenses = $5908 - $3144 = $2764Salvage value = $340 - $210 = $13014INFO630 Week 8www.ischool.drexel.edu14

An ExampleMARR = 12%, 8 year planning horizonAlternativesDo NothingCash-flow stream AInitial investment = $5300Annual income = $4142Annual expenses = $3144Salvage value = $210Cash-flow stream BInitial investment = $6200Annual income = $7329Annual expenses = $5908Salvage value = $340Cash-flow stream CInitial investment = $6890Annual income = $6601Annual expenses = $5335Salvage value = $19015INFO630 Week 8www.ischool.drexel.edu15Here are the 3 alternatives to considerAlready ordered by increasing initial investmentMARR = 12%, 8 year planning horizon

Present Worth on Incremental InvestmentDifferential cash-flow stream (A-Do Nothing)Initial investment = $5300 - $0 = $5300Annual income = $4142 - $0 = $4142Annual expenses = $3144 - $0 = $3144Salvage value = $210 - $0 = $210 P/A,12%,8 P/F,12%,8PW(12%) = -$5300 + ($4142 - $3144) ( 4.9676 ) + $210 (0.4039) = -$5300 + $998 * 4.9676 + $210 * 0.4039 = -$5300 + $4958 + $85 = -$257PW of differential = $0, B is betterPresent Worth on Incremental InvestmentDifferential cash-flow stream (C-B)Initial investment = $6890 - $6200 = $690Annual income = $6601 - $7329 = -$728Annual expenses = $5335 - $5908 = -$573Salvage value = $190 - $340 = -$150 P/A,12%,8 P/F,12%,8PW(12%) = -$690 + (-$728 - -$573) ( 4.9676 ) + -$150 (0.4039 ) = -$690 + -$155 * 4.9676 + -$150 * 0.4039 = -$690 + -$770 + -$61 = -$1521PW of differential MARR are selected

This method doesnt always lead to the best decisionProposals must be independent with no limit on resourcesAlternative with highest IRR may not maximize PW(MARR)28INFO630 Week 8www.ischool.drexel.edu28Rank on Rate of Return FlawExampleMARR = 15%Alternative A0Initial investment = $0Year 1-10 annual net income = $0Alternative A1Initial investment = $15,700Year 1-10 annual net income = $4396Alternative A2Initial investment = $25,120Year 1-10 annual net income = $5966Alternative A3Initial investment = $31,400Year 1-10 annual net income = $785029INFO630 Week 8www.ischool.drexel.edu29Rank on Rate of Return FlawExample (cont)Alternative A0IRR = 15%PW(MARR) = $0Alternative A1IRR = 24.9%PW(MARR) = $12,512Alternative A2IRR = 19.9%PW(MARR) = $13,168Alternative A3IRR = 21.4%PW(MARR) = $18,979Alternative A1 has highest IRR, but A3 has highest PW(MARR)30INFO630 Week 8www.ischool.drexel.edu30Alternative A1 has highest IRR, but A3 has highest PW(MARR)

Rank on Rate of Return FlawExplainedAxAyMARRi$PW(MARR) of AyIRR of Ax31INFO630 Week 8(See the notes below for this slide)www.ischool.drexel.edu31Remember that mathematically, IRR is the i-value where the PW(i) function crosses zero(given those 3 assumptions)Since PW(i) function is an n-order polynomial it could have any slope it wanted (so long as that slope is negative)The polynomial with farthest zero-intersection point might not have highest PW(i) at MARR.Key PointsMARR is lowest rate of return the organization thinks is a good investmentMARR is the interest rate used in comparisonsDifferential cash-flow analysis is better because it works with all bases of comparisonRank on rate of return doesnt always lead to the right decision

32INFO630 Week 8www.ischool.drexel.edu32MARR = opportunity costMARR is used as interest rate in comparisonsWhen using IRR for comparison, MUST use differential CFSPlanning Horizons and Economic LifeChapter 11INFO630 Week 833www.ischool.drexel.eduwww.ischool.drexel.edu33Planning horizonCapital recovery with returnEconomic lifeFinding the economic lifeEconomic life and planning horizonsPlanning Horizons and Economic LifeOutline

34INFO630 Week 8www.ischool.drexel.edu34Planning HorizonUp to now (prior chapters) assumed same lifespanNot always the caseTo compare proposals consistently, they need to have the same time spanThat common time span is called the planning horizon (or study period, or n )Planning horizons can be based on:Company policyHow far in the future reasonable estimates can be madeEconomic life of the shortest-lived assetEconomic life of the longest-lived assetBest judgment of the person doing the decision analysis*35INFO630 Week 8www.ischool.drexel.edu35Comparing a 5-year proposal to a 7-year proposal begs the question of what about the last 2 years after the 5-year proposal?

0, so Challenger 2 is betterEnd of alternatives, should retire in-house SW & go with COTS solution

60INFO630 Week 8www.ischool.drexel.edu60PW of differential >0, Defender 2 is betterEnd of alternatives, should retire in-house SW & go with COTS solutionReplacement, Economic Life, and Planning HorizonsEconomic lives of assets need to be consideredUse techniques in previous chapter if economic life is different than planning horizonImplied salvage valueReplacement service, multiple iterations, invest elsewhere

If replacement of replacements needs to be considered, you can assume either (long planning horizon):No replacementReplacement by identical asset(s)Replacement by best challengerAll possible combinations61INFO630 Week 8www.ischool.drexel.edu61Retirement Decisions (Asset)Activities dont continue foreverVAX/VMS, Intel 286, Organization needs to decide whether to continue an activity or abandon itStrictly speaking, only the defender is consideredMutually exclusive alternatives are:Retire immediatelyContinue for 1 more yearContinue for 2 more yearsContinue for 3 more years62INFO630 Week 8www.ischool.drexel.edu62This is ASSET retirement, not persons retirement (thats later)Retirement Decisions (cont)Find the alternative that maximizes the PW(MARR) of its net cash-flow streamPW() immediate retirement = salvage valueIn any later year:PW = PW(MARR) of salvage value in that year + PW(MARR) of revenue cash-flow stream thru that year PW(MARR) of O&M cash-flow stream thru that year63INFO630 Week 8www.ischool.drexel.edu63Example Retirement DecisionMARR = 16%

End of Salvage year Revenue O & M cost value 0 - - $1200 1 $775 $300 $900 2 $775 $400 $700 3 $775 $500 $550 4 $775 $600 $450 5 $775 $700 $375 6 $775 $800 $32564INFO630 Week 8www.ischool.drexel.edu64Example Retirement Decision (cont)PW(16%) of immediate retirement

PW(16%) of retiring after 1 year

PW(16%) of retiring after 2 years

PW(16%) of retiring after 3 years

P/A,16%,1 P/F,16%,1 P/F,16%,1$775 * (0.8621) - $300 * (0.8621) + $900 * (0.8621) = $1185 P/A,16%,2 P/F,16%,1 P/F,16%,2 P/F,16%,2$775 * (1.6052) - $300 * (0.8621) - $400 * (0.7432) + $700 * (0.7432) = $1208Salvage value today = $1200 P/A,16%,3 P/F,16%,1 P/F,16%,3 P/F,16%,3$775 * (2.2459) - $300 * (0.8621) - - $500 * (0.6407) + $550 * (0.6407) = $121765INFO630 Week 8www.ischool.drexel.edu65Example Retirement Decision (cont)PW(16%) of retiring after 4 years

PW(16%) of retiring after 5 years

PW(16%) of retiring after 6 years

P/A,16%,4 P/F,16%,1 P/F,16%,4 P/F,16%,4$775 * (2.7982) - $300 * (0.8621) - - $600 * (0.5223) + $450 * (0.5223) = $1210 P/A,16%,5 P/F,16%,1 P/F,16%,5 P/F,16%,5$775 * (3.2743) - $300 * (0.8621) - - $700 * (0.4761) + $375 * (0.4761) = $1175 P/A,16%,6 P/F,16%,1 P/F,16%,6 P/F,16%,6$775 * (3.6847) - $300 * (0.8621) - - $800 * (0.4104) + $325 * (0.4104) = $1120Conclusion: PW is highest for retiring the asset after 3 years (previous slide)66INFO630 Week 8www.ischool.drexel.edu66PW is highest at retiring after 3 years (prev page)

Key PointsReplacement decisions are needed when activities outlast the assets supporting themRetirement decisions are about continuing or abandning an activityReplacement and retirement are special cases of for-profit decisionsOutsiders viewpoint properly accounts for sunk cost and salvage value of defenderRetirement alternatives include now,after 1 year, after 2 years,

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