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Www.ischool.drexel.edu INFO 630 Evaluation of Information Systems Prof. Glenn Booker Week 8 – Chapters 10-12 1INFO630 Week 8.

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Presentation on theme: "Www.ischool.drexel.edu INFO 630 Evaluation of Information Systems Prof. Glenn Booker Week 8 – Chapters 10-12 1INFO630 Week 8."— Presentation transcript:

1 INFO 630 Evaluation of Information Systems Prof. Glenn Booker Week 8 – Chapters INFO630 Week 8

2 For-Profit Business Decisions Chapter 10 INFO630 Week 82

3 Minimum Attractive Rate of Return (MARR) Basic for-profit decision process Incremental vs. total cash-flow analysis Rank on rate of return For-Profit Business Decisions Outline 3INFO630 Week 8

4 Where are we? Prior chapters –Mutually exclusive alternatives Which is best to carry out –Can be based on Total cash flow Differential cash flow Basis of comparison Minimum Attractive Rate of Return (MARR) etc. 4INFO630 Week 8

5 Minimum Attractive Rate of Return (MARR) A statement that the organization is confident it can achieve at least that rate of return A.k.a. “Opportunity cost” –By investing here, you forego the opportunity to invest there –If you’re confident you can get X% there, all other alternatives should be evaluated against that X% 5INFO630 Week 8

6 Significance of the MARR The MARR is used as the interest rate in for-profit business decisions –PW(MARR) = how much more, or less, valuable that alternative is than investing the same $ in an investment that returns the MARR –So PW(MARR) = $1000 doesn’t mean you’ll 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 MARR 6INFO630 Week 8

7 Significance of the MARR –Note: Usually MARR is often set by policy decision from an organization’s management team Too high or too low? How set MARR? What impact does that have? 7INFO630 Week 8

8 Factors Influencing the MARR Type of organization –For-profit vs. regulated public utility Prevailing interest rate for typical investments Available funds Source of funds –Equity vs. borrowed funds Number of competing proposals Essential vs. elective Accounting for inflation or not Before- or after-tax 8INFO630 Week 8

9 Before- and After-tax MARR One way or another, a for-profit organization needs to address income taxes (see Chapter 16) –Before-tax MARR Use on pre-income tax cash-flow streams Approximation –After-tax MARR Use on post-income tax cash flow streams (Ch 16) More accurate Relationship between them –After-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 8

10 Basic For-profit Decision Process Assume the first alternative is the current best for j = 2 to the number of alternatives Consider the j th alternative to be the candidate Compare the candidate to the current best if the candidate is better than the current best then make the j th alternative the current best {* on ending, the current best is the best alternative *}  Find maximum value  Successive comparison in pair-wise basis  Current best  Candidate  Algorithm 10INFO630 Week 8

11 Basic For-profit Decision Process (cont) All other things equal, an alternative with a smaller initial investment is preferred –Also, if using IRR, order is important –Leads to a small, but important change Sort the alternatives in order of increasing investment, and… … if the candidate is strictly better than the current best … 11INFO630 Week 8

12 Basic For-profit Decision Process 12INFO630 Week 8

13 Incremental vs. Total Cash-Flow Analysis Total cash-flow analysis –Comparing on entire cash-flow stream basis Incremental cash-flow analysis –Comparing on difference between cash-flow streams If PW(MARR) of CFS2-CFS1 >0, then CFS2 is better than CFS1 WARNING: if using IRR as the basis of comparison, cash-flow analysis must be done incrementally 13INFO630 Week 8

14 Computing Differential Cash-Flow Streams Cash-flow stream A –Initial investment = $5300 –Annual income = $4142 –Annual expenses = $3144 –Salvage value = $210 Cash-flow stream B –Initial investment = $6200 –Annual income = $7329 –Annual expenses = $5908 –Salvage value = $340 Differential cash-flow stream (B-A) –Initial investment = $ $5300 = $900 –Annual income = $ $4142 = $3187 –Annual expenses = $ $3144 = $2764 –Salvage value = $340 - $210 = $130 14INFO630 Week 8

15 An Example MARR = 12%, 8 year planning horizon Alternatives –Do Nothing –Cash-flow stream A Initial investment = $5300 Annual income = $4142 Annual expenses = $3144 Salvage value = $210 –Cash-flow stream B Initial investment = $6200 Annual income = $7329 Annual expenses = $5908 Salvage value = $340 –Cash-flow stream C Initial investment = $6890 Annual income = $6601 Annual expenses = $5335 Salvage value = $190 15INFO630 Week 8

16 Present Worth on Incremental Investment Differential cash-flow stream (A-Do Nothing) –Initial investment = $ $0 = $5300 –Annual income = $ $0 = $4142 –Annual expenses = $ $0 = $3144 –Salvage value = $210 - $0 = $210 P/A,12%,8 P/F,12%,8 PW(12%) = -$ ($ $3144) ( ) + $210 (0.4039) = -$ $998 * $210 * = -$ $ $85 = -$257 PW of differential <= $0, therefore Do Nothing is better 16INFO630 Week 8

17 Present Worth on Incremental Investment Differential cash-flow stream (B-Do Nothing) –Initial investment = $ $0 = $6200 –Annual income = $ $0 = $7329 –Annual expenses = $ $0 = $5908 –Salvage value = $340 - $0 = $340 P/A,12%,8 P/F,12%,8 PW(12%) = -$ ($ $5908) ( ) + $340 ( ) = -$ $1421 * $340 * = -$ $ $137 = $996 PW of differential >= $0, therefore B is better 17INFO630 Week 8

18 Present Worth on Incremental Investment Differential cash-flow stream (C-B) –Initial investment = $ $6200 = $690 –Annual income = $ $7329 = -$728 –Annual expenses = $ $5908 = -$573 –Salvage value = $190 - $340 = -$150 P/A,12%,8 P/F,12%,8 PW(12%) = -$690 + (-$ $573) ( ) + -$150 ( ) = -$ $155 * $150 * = -$ $ $61 = -$1521 PW of differential <= $0, so B is still better end of alternatives, and B is best overall 18INFO630 Week 8

19 Using IRR on Incremental Investment Can be used for comparisons Incremental investment to carry out candidate is desirable if –IRR cash flow stream > MARR –Recall IRR is the critical i, for which PW(i=IRR) = 0 19INFO630 Week 8

20 IRR on Incremental Investment Differential cash-flow stream (A-Do Nothing) –Initial investment = $ $0 = $5300 –Annual income = $ $0 = $4142 –Annual expenses = $ $0 = $3144 –Salvage value = $210 - $0 = $210 IRR = 10.62% IRR of differential <= MARR, so Do Nothing is better 20INFO630 Week 8

21 IRR on Incremental Investment Differential cash-flow stream (B-Do Nothing) –Initial investment = $ $0 = $6200 –Annual income = $ $0 = $7329 –Annual expenses = $ $0 = $5908 –Salvage value = $340 - $0 = $340 IRR = 16.37% IRR of differential >= MARR, so B is better 21INFO630 Week 8

22 IRR on Incremental Investment Differential cash-flow stream (C-B) –Initial investment = $ $6200 = $690 –Annual income = $ $7329 = -$728 –Annual expenses = $ $5908 = -$573 –Salvage value = $190 - $340 = -$150 PW(0%) < $0, can’t compute IRR (it’s negative) IRR of differential <= MARR, B is still better end of alternatives, B is best overall – NOTE: Same conclusion as before, B is best overall 22INFO630 Week 8

23 Present Worth on Total Investment Preferred method –Comparison on incremental investment Other options –Total Investment Can not use IRR Must use –PW() –FW() –AE() 23INFO630 Week 8

24 Present Worth on Total Investment Cash-flow stream Do Nothing –Initial investment = $0 –Annual income = $0 –Annual expenses = $0 –Salvage value = $0 PW(12%) = $0 24INFO630 Week 8

25 Present Worth on Total Investment Cash-flow stream A –Initial investment = $5300 –Annual income = $4142 –Annual expenses = $3144 –Salvage value = $210 P/A,12%,8 P/F,12%,8 PW(12%) = -$ ($ $3144) ( ) + $210 ( ) = -$ $998 * $210 * = -$ $ $85 = -$257 PW(A) <= PW(Do Nothing), so Do Nothing is better 25INFO630 Week 8

26 Present Worth on Total Investment Cash-flow stream B –Initial investment = $6200 –Annual income = $7329 –Annual expenses = $5908 –Salvage value = $340 P/A,12%,8 P/F,12%,8 PW(12%) = -$ ($ $5908) ( ) + $340 ( ) = -$ $1421 * $340 * = -$ $ $137 = $996 PW(B) >= PW(Do Nothing), so B is better 26INFO630 Week 8

27 Present Worth on Total Investment Cash-flow stream C –Initial investment = $6890 –Annual income = $6601 –Annual expenses = $5335 –Salvage value = $190 P/A,12%,8 P/F,12%,8 PW(12%) = -$ ($ $5335) ( ) + $190 ( ) = -$ $1266 * $190 * = -$ $ $77 = -$524 PW(B) >= PW(C), B is still better end of alternatives, So B is best overall – NOTE: Same conclusion… 27INFO630 Week 8

28 Rank on Rate of Return A well-known approach 1.Calculate IRR for each proposal 2.Sort proposals in order of decreasing IRR 3.All proposals with IRR > MARR are selected This method doesn’t always lead to the best decision –Proposals must be independent with no limit on resources –Alternative with highest IRR may not maximize PW(MARR) 28INFO630 Week 8

29 Rank on Rate of Return Flaw—Example MARR = 15% Alternative A0 –Initial investment = $0 –Year 1-10 annual net income = $0 Alternative A1 –Initial investment = $15,700 –Year 1-10 annual net income = $4396 Alternative A2 –Initial investment = $25,120 –Year 1-10 annual net income = $5966 Alternative A3 –Initial investment = $31,400 –Year 1-10 annual net income = $ INFO630 Week 8

30 Rank on Rate of Return Flaw—Example (cont) Alternative A0 –IRR = 15% –PW(MARR) = $0 Alternative A1 –IRR = 24.9% –PW(MARR) = $12,512 Alternative A2 –IRR = 19.9% –PW(MARR) = $13,168 Alternative A3 –IRR = 21.4% –PW(MARR) = $18,979 Alternative A1 has highest IRR, but A3 has highest PW(MARR) 30INFO630 Week 8

31 Rank on Rate of Return Flaw—Explained A x A y MARR i $ PW(MARR) of A y IRR of A x 31INFO630 Week 8 (See the notes below for this slide)

32 Key Points MARR is lowest rate of return the organization thinks is a good investment MARR is the interest rate used in comparisons Differential cash-flow analysis is better because it works with all bases of comparison Rank on rate of return doesn’t always lead to the right decision 32INFO630 Week 8

33 Planning Horizons and Economic Life Chapter 11 INFO630 Week 833

34 Planning horizon Capital recovery with return Economic life Finding the economic life Economic life and planning horizons Planning Horizons and Economic Life Outline 34INFO630 Week 8

35 Planning Horizon Up to now (prior chapters) assumed same lifespan –Not always the case To compare proposals consistently, they need to have the same time span –That common time span is called the planning horizon (or “study period”, or “n “) Planning horizons can be based on: –Company policy –How far in the future reasonable estimates can be made –Economic life of the shortest-lived asset –Economic life of the longest-lived asset –Best judgment of the person doing the decision analysis * 35INFO630 Week 8

36 Capital Recovery with Return, CR(i) Cost of owning an asset, in equal-payment-series terms over some given time span Determined by –Asset’s loss in value over time –Interest/opportunity cost of frozen capital (invested capital) 36INFO630 Week 8

37 Capital Recovery with Return, CR(i) (cont) Example –Paid $15,000, sold after six years for $1317, MARR = 13% –Interest/opportunity cost of salvage value is 13% of $1317 annually, or $171 –Drop in value is $13,683 over six years, AE(i) is $13,683k (A/P, 13%, 6) = $13,683k * = $3423 –Both are annual costs so they can be added: cost of ownership = $ $171 = $3494/year 37INFO630 Week 8

38 Capital Recovery with Return, CR(i) (cont) Generalizing CR(i) = (P - F) (A/P, i, n) + F*i As in CR(i) = ($15,000 - $1317) (A/P, 13%, 6) + $1317 (0.13) = $3494 Where P = acquisition cost, F = salvage cost, n = length of time asset kept, i = interest rate 38INFO630 Week 8

39 Economic Life Total lifetime costs are driven by –CR(i) –Operation and maintenance (O&M) costs Capital Recovery (CR(i)) tends to decrease over time O&M tends to increase over time –Why? Year Salvage value CR(13%) if kept n at end of year n years 1 $10,000 $ $6667 $ $4444 $ $2963 $ $1975 $ $1317 $ INFO630 Week 8

40 Economic Life (cont) Total cost of ownership, operation, and maintenance is sum –Economic life is when total cost of ownership is lowest A.k.a. minimum-cost life, or optimum replacement interval 40INFO630 Week 8

41 Finding the Economic Life (3) (1) (2) Operating (4) (6) (7) Salvage AE(i) cost and PW(i) (5) AE(i) Total End value if if retired maintenance O&M for Sum of year 0 cost of AE(i) if of retired at in year n costs for year n in O&Ms through operating retired at Year year n [CR(i)] year n year 0 year n for n years year n 1 $10,000 $6950 $2000 $1770 $1770 $2000 $ $6667 $5862 $3400 $2663 $4433 $2657 $ $4444 $5048 $4800 $3327 $7759 $3286 $ $2963 $4432 $6200 $3803 $11,562 $3887 $ $1975 $3960 $7600 $4125 $15,687 $4460 $ $1317 $3594 $9000 $4323 $20,010 $5005 $ INFO630 Week 8

42 Economic Life – Special Case Annual O&M stays constant over life, and salvage stays constant –Example: Purchased software Salvage value usually 0 –Longer asset in service, means lower AE(i), so economic life = service life Operating upgrade make software unusable Acquisition and salvage cost constant, O&M always increasing –Economic life is shortest possible – example 1 yr 42INFO630 Week 8

43 Economic Lives and Planning Horizons Two situations to address –Economic life longer than planning horizon –Economic life shorter than planning horizon Note: IF economic life = planning horizon – trivial case, not worked out - Uncommon 43INFO630 Week 8

44 Economic Life Longer than the Planning Horizon There will be residual value to consider –Implied salvage value is an estimate of that residual value Avoided AE(i) cost of ownership (CR(i)) Earlier access to salvage value 44INFO630 Week 8

45 Economic Life Longer than the Planning Horizon Economic life is 4 years, planning horizon is 2 years –Owner avoids 2 years cost of ownership at $8319/year –Owner also has access to estimated salvage value 2 years earlier –Implied salvage vale is sum of those Generalizing $8319 (P/A, 13%, 2) = $8319 (1.6681) = $13,877 $2963 (P/F, 13%, 2) = $2963 (0.7831) = $2320 $13,877 + $2320 = $16,197 P/A, i, n-n P/F, i, n-n F = CR(i) ( ) + F ( ) *n*n * * 45INFO630 Week 8

46 Economic Life Shorter than the Planning Horizon Three different methods are available –Method 1: Replacement service –Method 2: Multiple iterations of same asset –Method 3: Invest profits somewhere else Each is valid given its assumptions 46INFO630 Week 8

47 Method 1: Replacement Service Assumes a replacement service is available and cash-flow stream can be estimated e.g., computer with 7 year economic life and planning horizon of 10 years –Lease a computer to fill the 3-year gap –Calculate cash-flow stream for 3 years of lease 47INFO630 Week 8

48 Method 2: Multiple Iterations Assumes asset can be replaced by identical asset e.g., computer with 7 year economic life and planning horizon of 10 years –Buy another computer after 7 years and use its implied salvage value on remaining 4 years of second iteration 48INFO630 Week 8

49 Revenue vs. Service Alternatives Revenue alternative –Described in terms of its total cash-flow stream (the typical case, expense-income CF) Service alternative –Alternatives are assumed to provide equivalent utility so only expense cash-flows are included e.g., buy one of two compilers Which to use: –Method 1, Method 2 valid for both –Method 3 is only valid for revenue alternatives 49INFO630 Week 8

50 Method 3: Invest Elsewhere Assumes funds can be invested at MARR at end of economic life e.g., computer with 7 year economic life and planning horizon of 10 years –Use cash-flow stream over 7 years –Calculate FW(i) at end of year 7 –Use FW(i) at MARR as cash flow in remaining 3 years of planning horizon 50INFO630 Week 8

51 Key Points A planning horizon is a common time frame for comparing proposals Capital recovery with return, CR(i), expresses the cost of ownership as an equivalent annual amount The economic life is the time of ownership with minimum total cost –Driven by CR(i) and the cost of operating and maintaining Economic lives need to be fit into planning horizons 51INFO630 Week 8

52 Replacement and Retirement Decisions Chapter 12 INFO630 Week 852

53 Replacement decisions, defined Special issues in replacement decisions Outsider’s viewpoint Example of replacement analysis Retirement decisions, defined Example of retirement analysis Replacement and Retirement Decisions Outline 53INFO630 Week 8

54 Replacement Decisions An organization will be faced with a replacement decision when: –They are involved in some activity, like sales order fulfillment –That activity depends on one or more assets, like order fulfillment software –The lifetime of the asset(s) is shorter than the duration of the activity, like 7 years vs. 20+ years Reasons for replacement –Deterioration –Obsolescence –What is the difference? 54INFO630 Week 8

55 Special Issues in Replacement Decisions Sunk cost –Defined as acquisition cost minus salvage value –Psychological issues Want to “squeeze every penny out of it” Decision to replace is seen as more significant Salvage value –Holding an asset carries its opportunity cost –That cost needs to be allocated to the existing asset in the analysis 55INFO630 Week 8

56 Outsider’s Viewpoint Addresses sunk cost & salvage value Assume you are an outsider who needs the service provided by either the existing asset or one of the replacements –Outsider would need to buy in at existing asset’s salvage value Proposals that keep any existing assets require salvage values as initial investments 56INFO630 Week 8

57 Replacement Analysis Just like basic for-profit process, except –“Defender” is proposal representing staying with the existing asset(s) Usually low capital cost but relatively high O&M Little capital investment to keep it –“Challenger” is any proposal considered as a replacement Usually high acquisition cost but low O&M Require large capital investment 57INFO630 Week 8

58 Example Replacement Analysis MARR = 18%, 5 year planning horizon Defender: existing in-house software –Initial investment = $30,000 –Annual income = $130,000 –Annual operating and maintenance = $100,000 –Salvage value = $30,000 Challenger 1: redevelop software in-house –Initial investment = $260,000 –Annual income = $210,000 –Annual operating and maintenance = $80,000 –Salvage value = $55,000 Challenger 2: purchase COTS software –Initial investment = $310,000 –Annual income = $200,000 –Annual operating and maintenance = $50,000 –Salvage value = $60,000 58INFO630 Week 8

59 Example Replacement Analysis: Present Worth on Incremental Investment Differential cash-flow stream (Challenger 1-Defender) –Initial investment = $260k - $30k = $230k –Annual income = $210k - $130k = $80k –Annual expenses = $80k - $100k = -$20k –Salvage value = $55k - $30k = $25k P/A,18%,5 P/F,18%,5 PW(18%) = -$230k + ($80k - -$20k) ( ) + $25k ( ) = -$230k + $100k * $25k * = -$230k + $313k + $11k = $94k PW of differential >0, so Challenger 1 is better than Defender 59INFO630 Week 8

60 Present Worth on Incremental Investment (cont) Differential cash-flow stream (Challenger 2 – Challenger 1) –Initial investment = $310k - $260k = $50k –Annual income = $200k - $210k = -$10k –Annual expenses = $50 - $80k = -$30k –Salvage value = $60k - $55k = $5k P/A,18%,5 P/F,18%,5 PW(18%) = -$50k + (-$10k - -$30k) ( ) + $5k ( ) = -$50k + $20k * $5k * = -$50k + $63k + $2k = $15k PW of differential >0, so Challenger 2 is better End of alternatives, should retire in-house SW & go with COTS solution 60INFO630 Week 8

61 Replacement, Economic Life, and Planning Horizons Economic lives of assets need to be considered –Use techniques in previous chapter if economic life is different than planning horizon Implied salvage value Replacement service, multiple iterations, invest elsewhere If replacement of replacements needs to be considered, you can assume either (long planning horizon): –No replacement –Replacement by identical asset(s) –Replacement by best challenger –All possible combinations 61INFO630 Week 8

62 Retirement Decisions (Asset) Activities don’t continue forever –VAX/VMS, Intel 286, … Organization needs to decide whether to continue an activity or abandon it –Strictly speaking, only the defender is considered –Mutually exclusive alternatives are: Retire immediately Continue for 1 more year Continue for 2 more years Continue for 3 more years … 62INFO630 Week 8

63 Retirement Decisions (cont) Find the alternative that maximizes the PW(MARR) of its net cash-flow stream –PW() immediate retirement = salvage value –In 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 year 63INFO630 Week 8

64 Example Retirement Decision MARR = 16% End of Salvage year Revenue O & M cost value $ $775 $300 $900 2 $775 $400 $700 3 $775 $500 $550 4 $775 $600 $450 5 $775 $700 $375 6 $775 $800 $325 64INFO630 Week 8

65 Example 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) = $1208 Salvage 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) = $ INFO630 Week 8

66 Example 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) = $1120 Conclusion: PW is highest for retiring the asset after 3 years (previous slide) 66INFO630 Week 8

67 Key Points Replacement decisions are needed when activities outlast the assets supporting them Retirement decisions are about continuing or abandning an activity Replacement and retirement are special cases of for- profit decisions Outsider’s viewpoint properly accounts for sunk cost and salvage value of defender Retirement alternatives include now, after 1 year, after 2 years, … 67INFO630 Week 8


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