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INFO 630 Evaluation of Information Systems Prof. Glenn Booker

INFO 630 Evaluation of Information Systems Prof. Glenn Booker. Week 8 – Chapters 10-12. For-Profit Business Decisions. Chapter 10. For-Profit Business Decisions Outline. Minimum Attractive Rate of Return (MARR) Basic for-profit decision process Incremental vs. total cash-flow analysis

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INFO 630 Evaluation of Information Systems Prof. Glenn Booker

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  1. INFO 630Evaluation of Information SystemsProf. Glenn Booker Week 8 – Chapters 10-12 INFO630 Week 8

  2. For-Profit Business Decisions Chapter 10 INFO630 Week 8

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

  10. Basic For-profit Decision Process • Find maximum value • Successive comparison in pair-wise basis • Current best • Candidate • Algorithm Assume the first alternative is the current best for j = 2 to the number of alternatives Consider the jth alternative to be the candidate Compare the candidate to the current best if 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 *} INFO630 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 … INFO630 Week 8

  12. Basic For-profit Decision Process INFO630 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 INFO630 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 = $6200 - $5300 = $900 • Annual income = $7329 - $4142 = $3187 • Annual expenses = $5908 - $3144 = $2764 • Salvage value = $340 - $210 = $130 INFO630 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 INFO630 Week 8

  16. Present Worth on Incremental Investment • Differential cash-flow stream (A-Do Nothing) • Initial investment = $5300 - $0 = $5300 • Annual income = $4142 - $0 = $4142 • Annual expenses = $3144 - $0 = $3144 • Salvage value = $210 - $0 = $210 P/A,12%,8 P/F,12%,8 PW(12%) = -$5300 + ($4142 - $3144) ( 4.9676 ) + $210 (0.4039) = -$5300 + $998 * 4.9676 + $210 * 0.4039 = -$5300 + $4958 + $85 = -$257 PW of differential <= $0, therefore Do Nothing is better INFO630 Week 8

  17. Present Worth on Incremental Investment • Differential cash-flow stream (B-Do Nothing) • Initial investment = $6200 - $0 = $6200 • Annual income = $7329 - $0 = $7329 • Annual expenses = $5908 - $0 = $5908 • Salvage value = $340 - $0 = $340 P/A,12%,8 P/F,12%,8 PW(12%) = -$6200 + ($7329 - $5908) ( 4.9676 ) + $340 (0.4039 ) = -$6200 + $1421 * 4.9676 + $340 * 0.4039 = -$6200 + $7059 + $137 = $996 PW of differential >= $0, therefore B is better INFO630 Week 8

  18. Present Worth on Incremental Investment • Differential cash-flow stream (C-B) • Initial investment = $6890 - $6200 = $690 • Annual income = $6601 - $7329 = -$728 • Annual expenses = $5335 - $5908 = -$573 • Salvage value = $190 - $340 = -$150 P/A,12%,8 P/F,12%,8 PW(12%) = -$690 + (-$728 - -$573) ( 4.9676 ) + -$150 (0.4039 ) = -$690 + -$155 * 4.9676 + -$150 * 0.4039 = -$690 + -$770 + -$61= -$1521 PW of differential <= $0, so B is still better end of alternatives, and B is best overall INFO630 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 whichPW(i=IRR) = 0 INFO630 Week 8

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

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

  22. IRR on Incremental Investment • Differential cash-flow stream (C-B) • Initial investment = $6890 - $6200 = $690 • Annual income = $6601 - $7329 = -$728 • Annual expenses = $5335 - $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 INFO630 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() INFO630 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 INFO630 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%) = -$5300 + ($4142 - $3144) ( 4.9676 ) + $210 (0.4039 ) = -$5300 + $998 * 4.9676 + $210 * 0.4039 = -$5300 + $4958 + $85 = -$257 PW(A) <= PW(Do Nothing), so Do Nothing is better INFO630 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%) = -$6200 + ($7329 - $5908) ( 4.9676 ) + $340 (0.4039 ) = -$6200 + $1421 * 4.9676 + $340 * 0.4039 = -$6200 + $7059 + $137 = $996 PW(B) >= PW(Do Nothing), so B is better INFO630 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%) = -$6890 + ($6601 - $5335) ( 4.9676 ) + $190 (0.4039 ) = -$6890 + $1266 * 4.9676 + $190 * 0.4039 = -$6890 + $6289 + $77 = -$524 PW(B) >= PW(C), B is still better end of alternatives, So B is best overall – NOTE: Same conclusion… INFO630 Week 8

  28. Rank on Rate of Return • A well-known approach • Calculate IRR for each proposal • Sort proposals in order of decreasing IRR • 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) INFO630 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 = $7850 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) INFO630 Week 8

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

  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 INFO630 Week 8

  33. Planning Horizons and Economic Life Chapter 11 INFO630 Week 8

  34. Planning Horizons and Economic LifeOutline • Planning horizon • Capital recovery with return • Economic life • Finding the economic life • Economic life and planning horizons INFO630 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 * INFO630 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) INFO630 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 * 0.2502 = $3423 • Both are annual costs so they can be added: cost of ownership = $3423 + $171 = $3494/year INFO630 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 INFO630 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 $6950 2 $6667 $5862 3 $4444 $5048 4 $2963 $4432 5 $1975 $3960 6 $1317 $3594 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 INFO630 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 $8950 2 $6667 $5862 $3400 $2663 $4433 $2657 $8519 3 $4444 $5048 $4800 $3327 $7759 $3286 $8335 4 $2963 $4432 $6200 $3803 $11,562 $3887 $8319 5 $1975 $3960 $7600 $4125 $15,687 $4460 $8420 6 $1317 $3594 $9000 $4323 $20,010 $5005 $8600 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 INFO630 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 INFO630 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 INFO630 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 INFO630 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 INFO630 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 INFO630 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 INFO630 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 INFO630 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 INFO630 Week 8

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