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INFO 631 Prof . Glenn Booker

INFO 631 Prof . Glenn Booker. Week 3 – Chapters 7-9. Equivalence. Chapter 7. Equivalence Outline. Simple comparison of two proposals Equivalence, defined Simple equivalence Equivalence with varying cash-flow instances Equivalence with varying interest rates.

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INFO 631 Prof . Glenn Booker

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  1. INFO 631Prof. Glenn Booker Week 3 – Chapters 7-9 INFO631 Week 3

  2. Equivalence Chapter 7 INFO631 Week 3

  3. EquivalenceOutline • Simple comparison of two proposals • Equivalence, defined • Simple equivalence • Equivalence with varying cash-flow instances • Equivalence with varying interest rates INFO631 Week 3

  4. Simple Comparison of Two Proposals • Your company sells a product for $20,000 • A customer offers to pay $2500 at the end of each of the next 10 years instead. Is this a good deal? End of Year Pay now Pay later 0 $20,000 $0 1 $0 $2500 2 $0 $2500 3 $0 $2500 4 $0 $2500 5 $0 $2500 6 $0 $2500 7 $0 $2500 8 $0 $2500 9 $0 $2500 10 $0 $2500 Total $20,000 $25,000 INFO631 Week 3

  5. Simple Comparison of Two Proposals (cont) • How do I evaluate? • Impact of time? • Interest • What does 0% interest mean? • Is this realistic? INFO631 Week 3

  6. Simple Comparison of Two Proposals (cont) • That analysis assumed 0% interest • The interest rate is unlikely to be 0% • What if we use a more reasonable interest rate, say 9%? P/A, 9%, 10 P = $2500 ( 6.4177 ) = $16,044 A/P, 9%, 10 A = $20,000 ( 0.1558 ) = $3116 P/A = equal-payment-series present-worth A/P = equal-payment capital-recovery INFO631 Week 3

  7. Recall - Naming Conventions in Interest Formulas • P • “Principal Amount”—how much is the money worth right now? • Also known as “present value” or “present worth” • F • “Final Amount”—how much will the money be worth at a later time? • Also known as the “future value” or “future worth” • i • Interest rate per period • Assumed to be an annual rate unless stated otherwise • n • Number of interest periods between the two points in time • A • “Annuity”—a stream of recurring, equal payments that would be due at the end of each interest period INFO631 Week 3

  8. Equivalence “Two or more different cash-flow instances (or cash-flow streams) are equivalent at a given interest rate only when they equal the same amount of money at a common point in time. More specifically, comparing two different cash flows makes sense only when they are expressed in the same time frame” INFO631 Week 3

  9. Equivalence (cont) • Equivalence at one time means equivalence at all other times • Equivalence (or more appropriately the lack of it) can be used as a basis of choice • Basis of decision making • If both proposal are equivalent, doesn't matter which one we choose • If different, one is better than the other • Economic comparisons need to be made on an equivalent basis • Or you could make the wrong decision INFO631 Week 3

  10. Simple Equivalence • The compound interest formulas are statements of simple equivalence (single payment compound amount (F/P)) • If i% interest is fair, you would be indifferent to getting $P now compared to getting $F after n interest periods • Note: Fair is important word. Why? • Simple equivalence in action • Fast food joint pays its contest winner $2 million, as $200k annually for 10 years • Using an interest rate of 7%, that’s really P/A, 7%, 10 P = $200k ( 7.0236 ) = $1.4 million INFO631 Week 3

  11. Equivalence with Varying Cash-Flow Instances • Equivalence applied to entire cash flow stream • Each instance translated into common reference time frame, then add them up • Two approaches • Elegant Approach • Better when done by hand • Hard to automate • Brute Force Approach • Easy to automate • A lot of computations if done by hand INFO631 Week 3

  12. Equivalence With Varying Cash-Flow Instances (Elegant Approach) Steps (pg 101) • Choose the reference time frame • Break cash flow stream into segments • For each segment, apply appropriate formula to translate it into the reference time frame • Sum up all the results • Represents the net equivalent value of chase flow stream in terms of reference time frame INFO631 Week 3

  13. End of Year Partial present equivalent amounts 0 $2657.70 $878.36 + $1386.76 - $156.11 + $548.69 1 2 P/F,12,3 3 $1234 $1234 ( 0.7118 ) 4 P/A,12,5 P/F,12,5 5 $678 ( 3.6048 ) = $2444.05 $2444.05 ( 0.5674 ) 6 $678 7 $678 8 $678 9 $678 10 $678 P/F,12,11 11 -$543 -$543 ( 0.2875 ) 12 13 $890 14 $890 F/A,12,3 P/F,12,15 15 $890 $890 ( 3.3744 ) = $3003.21 $3003.21 ( 0.1827 ) Equivalence With Varying Cash-Flow Instances (Elegant Approach) Assume Interest = 12%, 15 Year, Using Single Payment Present Worth Value = P/F, i, n P = F ( ) INFO631 Week 3

  14. Recall - Compound Interest Formulas • Six different compound interest formulas • Single-payment compound-amount (F/P) • Single-payment present-worth (P/F) • Equal-payment-series compound-amount (F/A) • Equal-payment-series sinking-fund (A/F) • Equal-payment-series capital-recovery (A/P) • Equal-payment-series present-worth (P/A) INFO631 Week 3

  15. Equivalence With Varying Cash-Flow Instances (Brute Force) Year Net cash-flow Present-worth factor Equivalent value n at end of year (P/F,12%,n) at end of year 0 1 $0 0.8929 $0 2 $0 0.7972 $0 3 $1234 0.7118 $878.36 4 $0 0.6355 $0 5 $0 0.5674 $0 6 $678 0.5066 $343.47 7 $678 0.4523 $306.66 8 $678 0.4039 $273.84 9 $678 0.3606 $244.49 10 $678 0.3220 $218.32 11 -$543 0.2875 -$156.11 12 $0 0.2567 $0 13 $890 0.2292 $203.99 14 $890 0.2046 $182.09 15 $890 0.1827 $162.60 Total $2657.71 Translate each cash flow into reference time frame (now) using Single Payment Compound Interest INFO631 Week 3

  16. Equivalence With Varying Cash-Flow Instances • Last scenario assumed a single interest rate • Is that always the correct assumption? • In general yes. • Interest rates do usually change over time • Most business decision are based on “nominal” interest rate • What happens if interest rate varies? INFO631 Week 3

  17. Equivalence With Varying Interest Rates (Elegant Approach) Notice the interest rate is now 12% above the red line, 10% below it INFO631 Week 3

  18. Equivalence With Varying Interest Rates (cont) (Brute Force) • In the 12% region Year Net cash-flow Present-worth factor Equivalent value n at end of year (P/F,12%,n) at end of year 0 1 $0 0.8929 $0 2 $0 0.7972 $0 3 $1234 0.7118 $878.36 4 $0 0.6355 $0 5 $0 0.5674 $0 6 $678 0.5066 $343.47 7 $678 0.4523 $306.66 8 $678 0.4039 $273.84 9 $678 0.3606 $244.49 Total $2046.82 INFO631 Week 3

  19. Year Net cash-flow Present-worth factor Equivalent value n at end of year (P/F,10%,n) at end of year 0 1 $678 0.9091 $616.37 2 -$543 0.8264 -$448.74 3 $0 0.7513 $0 4 $890 0.6830 $607.87 5 $890 0.6209 $552.60 6 $890 0.5645 $502.41 Total $1830.51 Equivalence With Varying Interest Rates (more) (Brute Force) • In the 10% region • Translating that to the beginning of the 12% region • Adding that to the previous sum for the 12% region P/F, 12%, 9 P = $1830.51 ( 0.3603 ) = $660.08 $2046.82 + $660.08 = $2706.90 INFO631 Week 3

  20. Key Points • Comparisons must be made on an equivalent basis • Interest formulas are statements of simple equivalence • Different cash-flow instances can be translated into an equivalent basis • This can be done across different interest rates INFO631 Week 3

  21. Sample Exercise $100 $80 $50 0 1 2 3 4 5 6 7 8 9 -$20 -$100 INFO631 Week 3

  22. Exercise Answer P/F, 6,1 P/A,6,4P/F,6,3 P/F,6,8 P/F,6,9 -$100 + -$20 (0.9434) + $50 (3.4651)(0.8396) + $80 (0.6247) + $100 (0.5919) -$100 - $19 + $145 + $50 + $59 = $135 INFO631 Week 3

  23. Bases for Comparison Chapter 8 INFO631 Week 3

  24. Bases for ComparisonOutline • Basis for comparison defined • An example • Present worth • Future worth • Annual equivalent • Internal rate of return • Payback period • Discounted payback period • Project balance • Capitalized equivalent amount INFO631 Week 3

  25. Basis for Comparison • Last chapter discusses how cash-flow instances can be added, subtracted, compare at the same time frame • Will expand to different cash-flow streams: • A common frame of reference for comparing two or more cash-flow streams in a consistent way • Basically, all streams are converted into the same basis, such as Present Worth • Then compared INFO631 Week 3

  26. Basis for Comparison con’t • Six Bases • Present worth • Future worth • Annual equivalent • Internal rate of return • Payback period • Capitalized equivalent amount INFO631 Week 3

  27. Comparing Cash-Flow Streams • Need to be converted into same basis • After all proposal expressed in same basis for comparison • Best one obvious • Mechanics of actual choice in Chapter 9 • Caution • Always use the same • Interest (i) • Study Period (n) INFO631 Week 3

  28. An Example - Project: Automated Test Equipment (ATE) Taken from Lecture Ch 3 • One person-year = $125k • Initial investment • $300k for test hardware and development equipment (Year 0) • 20 person-years of software development staff (Year 1) • 10 person-years of software development staff (Year 2) • Operating and maintenance costs • $30k per year for test hardware and dev equipment (Years 1-10) • 5 person-years of software maintenance staff (Years 3-10) • Sales income • None • Cost avoidance • $1.3 million in reduced factory staffing (Years 2-10) • Salvage value • Negligible INFO631 Week 3

  29. Automated Test Equipment (ATE) Simple Example Year Dev Staff Equipment O & M Savings Total 0 0 -$300K 0 0 -$300K 1 -$2.5M 0 -$30K 0 -$2.53M 2 -$1.25M 0 -$30K $1.3M $20K 3 -$625K 0 -$30K $1.3M $645K 4 -$625K 0 -$30K $1.3M $645K 5 -$625K 0 -$30K $1.3M $645K 6 -$625K 0 -$30K $1.3M $645K 7 -$625K 0 -$30K $1.3M $645K 8 -$625K 0 -$30K $1.3M $645K 9 -$625K 0 -$30K $1.3M $645K 10 -$625K 0 -$30K $1.3M $645K Example from Ch 3 Lecture Slides INFO631 Week 3

  30. The ATE Example – Cash Flow Stream $645K $20K 0 1 2 3 4 5 6 7 8 9 10 -$300K -$2.53M Example from Ch 3 Lecture Slides INFO631 Week 3

  31. Present Worth, PW(i) • How much is the future cash-flow stream worth (equivalent to) right now at interest rate, i? • Reference time for PW(i) = • Beginning of first period (end of period 0) • Also called Net Present Value (NPV) • How much is the cash-flow stream worth today? • NOTE: “Present” - can be any arbitrary point in time as appropriate for decision INFO631 Week 3

  32. Present Worth, PW(i) • Formula • Uses single-payment present-worth (P/F,i,n) to translate each individual net-cash flow • Then sum all amounts Ft = net-cash flow instance in period t Notes:Except for Year 0, PW values are always < original cash flow. Process of translating cash-flow backwards is referred to as “discounting” INFO631 Week 3

  33. Present Worth, PW(i) (cont) • Manual calculation of PW(10%) for ATE Year Net cash-flow Present-worth factor Equivalent value n at end of year (P/F,10%,n) at end of year 0 0 -$300K 1.0000 -$300K 1 -$2,530K 0.9091 -$2,300K 2 $20K 0.8264 $17K 3 $645K 0.7513 $485K 4 $645K 0.6830 $441K 5 $645K 0.6209 $400K 6 $645K 0.5645 $364K 7 $645K 0.5132 $331K 8 $645K 0.4665 $301K 9 $645K 0.4241 $274K 10 $645K 0.3855 $249K PW(10%) $260K INFO631 Week 3

  34. Comments on PW(i) • There is a single value of PW(i) for any i • Generally, as i increases PW(i) decreases Critical i, where PW(i) = 0, is IRR (slide 43) INFO631 Week 3

  35. Comments on PW(i) (cont) • 2nd most widely used basis for comparison • Future Value is 1st • PW(i) over -1 < i < oois meaningful • Only 0 <i < oois important • Negative interest rates almost impossible • Graph shows several important things • Equivalent profit or loss at any i • What ranges of i would be profitable • The “critical i” where PW(i)=0 INFO631 Week 3

  36. Future Worth, FW(i) • Just like PW(i) except it’s referenced to a future point in time • Reference time for FW(i) = • Usually the end of the cash-flow stream • Answer the question: • How much is this proposal worth in the end-of-the-proposal time frame? INFO631 Week 3

  37. Future Worth, FW(i) Con’t • Formula • Uses single-payment compound-amount (F/P,i,n) to translate each individual net-cash flow instance • Then sum all amounts INFO631 Week 3

  38. Present Worth, FW(i) (cont) • Manual calculation of FW(10%) for ATE Year Net cash-flow Future-worth factor Equivalent value n at end of year (F/P,10%,n) at end of year 0 -$300K 2.5937 -$778K 1 -$2,530K 2.3579 -$5,965K 2 $20K 2.1436 $42K 3 $645K 1.9487 $1256K 4 $645K 1.7716 $1142K 5 $645K 1.6105 $1038K 6 $645K 1.4641 $944K 7 $645K 1.3310 $858K 8 $645K 1.2100 $780K 9 $645K 1.1000 $709K 10 $645K 1.0000 $645K FW(10%) $675K INFO631 Week 3

  39. Comments on FW(i) • The only difference between PW(i) and FW(i) is the time frame • PW(i) and FW(i) are mathematically related • Number from class example above • For fixed i and n, FW(i) = PW(i) times a constant • FW(i) = 0 when PW(i) = 0 • for the same value of “critical i” • Comparing cash-flow streams in FW(i) terms will always lead to the same conclusion as comparing with PW(i) • Assuming used consistently for all cash-flow streams F/P, i, n FW(i) = PW(i) ( ) F/P, 10%, 10 FW(10%) = $260K ( 2.5937 ) = $675K INFO631 Week 3

  40. Annual Equivalent, AE(i) • PW(i) and FW(i) represent the cash-flow stream as an equivalent one-time cash-flow instance • Either: • at the beginning (PW) or • at the end (FW) of the cash-flow stream • AE(i) represents it as a series of equal cash-flow instances over the life of the study • AE(i) relates to PW(i) the same as A relates to P A/P, i, n AE(i) = PW(i) ( ) INFO631 Week 3

  41. Annual Equivalent, AE(i) (cont) • Formula • Manual calculation of AE(10%) • Start with PW(i) and multiple by equal-payment-series capital recovery (A/P,i,n) factor. A/P, 10%, 10 AE(10%) = PW(10%) ( ) = $260K ( 0.1627 ) = $42.3K Cash flow stream equivalent = $42.3 K at the end of each of the next 10 yrs INFO631 Week 3

  42. Comments on AE(i) • For fixed i and n, AE(i) = PW(i) times a constant • AE(i) = 0 when PW(i) = 0 • for the same value of “critical i” • Comparing cash-flow streams in AE(i) terms will always lead to the same conclusion as comparing with PW(i) • Assuming used consistently for all proposals • Advantage • AE(i) form is useful for repeating cash-flow streams • Easy to represent as annual equivalents • If the ATE project can be repeated, AE(i) = $42.3K over 20 years, or over 30 years, … • Example: renewable bond INFO631 Week 3

  43. Internal Rate of Return, IRR • PW(i), FW(i), and AE(i) express the cash-flow stream as equivalent dollar amounts • IRR expresses the cash-flow stream as an interest rate • What interest rate would a bank have to pay to match your payments and withdraws and end up with $0 at the end of the cash-flow stream? • Also called Return on Investment (ROI) • Occurs when “critical i” brings PW(i) to zero (next slide) • Formula INFO631 Week 3

  44. PW(i) = 0, Critical i at IRR Critical i, where PW(i) = 0, discussed later - IRR Yup, the same figure from slide 34 INFO631 Week 3

  45. Internal Rate of Return, IRR (cont) • To compute IRR, the cash-flow stream must have these properties: • First nonzero net cash-flow is negative (expense) • That is followed by 0..n further expenses followed by incomes from there on • Only one sign change in the cash-flow stream • The net cash-flow stream is profitable • Sum of all income > sum of all expenses • PW(0%)>$0 • If not met, do not use • Criteria might not have IRR or • Might have more then 1 IRR INFO631 Week 3

  46. Computing IRR Algorithm Given the cash flow stream with the first non-zero cash flow being negative, and only 1 sign change, and PW(0%) > 0 Start with the estimated IRR = 0% Assume we will move IRR in an increasing (+) direction Assume an initial step amount (say, 10%) Calculate PW(i=0%) and save the result Move the IRR in the current direction by the step amount repeat recalculate the PW(i=IRR) if the PW(i=IRR) is closer to $0.00 than before then move the estimated IRR in the same direction by the step amount else switch direction and cut the step amount in half until the PW(i=IRR) is within a pre-determined range of $0.00 (say, 50 cents) INFO631 Week 3

  47. Computing IRR (cont) • Start with IRR = PW(0%) = $2,350K, step = 10%, direction = increasing • Calculate PW(10%), it’s $260K. That’s closer to zero than $2,350K so move the estimated IRR in the same direction (up) by another 10%. It’s now estimated to be 20%. • Calculate PW(20%), it’s -$676K. That’s farther from zero than -$260K so switch direction and cut the step amount in half, to 5%. The estimated IRR is now 15%. • Calculate PW(15%), it’s -$296K. That’s closer to zero than -$676K so move the estimated IRR in the same direction by another 5%. It’s now estimated to be 10%. • Calculate PW(10%), it’s $260K. That’s closer to zero than -$296K so move the estimated IRR in the same direction (down) by another 5%. It’s now estimated to be 5%. • Calculate PW(5%), it’s $1090K. That’s farther from zero than -$260K so switch direction and cut the step amount in half, to 2.5%. The estimated IRR is now 7.5%. • Calculate PW(7.5%), it’s $633K. That’s closer to zero than $1090K so move the estimated IRR in the same direction by another 2.55%. It’s now estimated to be 10%. • Calculate PW(10%), it’s $260K. That’s closer to zero than $633K so move the estimated IRR in the same direction (up) by another 2.5%. It’s now estimated to be 12.5%... • … and so on while the PW(i) at the estimated IRR converges on $0.00. When the PW(i) is within +/- $0.50 of $0, the loop stops and the estimated IRR of 12.1% is returned. INFO631 Week 3

  48. Payback Period, PP • PW(i), FW(i), and AE(i) express the cash-flow stream as equivalent dollar amounts and IRR expresses it as an interest rate • Payback period expresses the cash-flow stream as a time • how long to recover the investment • Like saying “This investment will pay for itself in 5 years” • Formula • Smallest n where • Ft = net-cash flow instance in period t INFO631 Week 3

  49. Payback Period, PP (cont) • Manual calculation of PP for ATE Year Net cash-flow Running sum n at end of year thru year n 0 -$300K -$300K 1 -$2,530K -$2,830K 2 $20K -$2,810K 3 $645K -$2,165K 4 $645K -$1,520K 5 $645K -$875K 6 $645K -$230K 7 $645K $415K n = 7 INFO631 Week 3

  50. Comments on Payback Period • PW(i), FW(i), AE(i), and IRR • Indicators of profitability • Payback Period • Indicator of liquidity • Organization’s exposure to risk of financial loss • Example • If the project starts but gets canceled before the end of the payback period, the organization loses money • Payback = 5 is better then Payback = 10 yrs • Less financial risk INFO631 Week 3

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