INFO 630 Evaluation of Information Systems Dr. Jennifer Booker

Week 7 – Chapters 7-9 INFO630 Week 7 1 www.ischool.drexel.edu

Equivalence

Chapter 7 INFO630 Week 7 2 www.ischool.drexel.edu

Equivalence

Outline

• Simple comparison of two proposals • Equivalence, defined • Simple equivalence • Equivalence with varying cash-flow instances • Equivalence with varying interest rates INFO630 Week 7 3 www.ischool.drexel.edu

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 INFO630 Week 7 4 www.ischool.drexel.edu

Simple Comparison of Two Proposals (cont)

• How do I evaluate?

• Impact of time?

– Interest – What does 0% interest mean?

– Is this realistic? INFO630 Week 7 5 www.ischool.drexel.edu

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 INFO630 Week 7 6 www.ischool.drexel.edu

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 INFO630 Week 7 7 www.ischool.drexel.edu

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”

INFO630 Week 7 8 www.ischool.drexel.edu

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 INFO630 Week 7 9 www.ischool.drexel.edu

Simple Equivalence

• The compound interest formulas are statements of simple equivalence (single payment compound amount (F/P)) F  P  1  i 

n

– 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 INFO630 Week 7 10 www.ischool.drexel.edu

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 INFO630 Week 7 11 www.ischool.drexel.edu

Equivalence With Varying Cash-Flow Instances (Elegant Approach) Steps (pg 101) 1. Choose the reference time frame 2. Break cash flow stream into segments 3. For each segment, apply appropriate formula to translate it into the reference time frame 4. Sum up all the results  Represents the net equivalent value of chase flow stream in terms of reference time frame INFO630 Week 7 12 www.ischool.drexel.edu

Equivalence With Varying Cash-Flow Instances (Elegant Approach) End of Year 0 $2657.70 $878.36 + $1386.76 - $156.11 + $548.69 1 2 3 $1234 $1234 ( 0.7118 ) 4 5 $678 ( 3.6048 ) = $2444.05 $2444.05 ( 0.5674 ) 6 $678 7 $678 8 $678 9 $678 10 $678 11 -$543 -$543 ( 0.2875 ) 12 13 $890 14 $890 P/F,12,3 P/A,12,5 P/F,12,11 F/A,12,3 Partial present equivalent amounts P/F,12,5 P/F,12,15 15 $890 $890 ( 3.3744 ) = $3003.21 $3003.21 ( 0.1827 ) Assume Interest = 12%, 15 Year, Using Single Payment Present Worth Value = P/F, i, n P = F ( ) INFO630 Week 7 13 www.ischool.drexel.edu

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) INFO630 Week 7 14 www.ischool.drexel.edu

Equivalence With Varying Cash-Flow Instances (Brute Force) Translate each cash flow into reference time frame (now) using Single Payment Compound Interest 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

INFO630 Week 7 15 www.ischool.drexel.edu

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?

INFO630 Week 7 16 www.ischool.drexel.edu

Equivalence With Varying Interest Rates (Elegant Approach) End of Year Partial present equivalent amounts 0 $2706.90 $878.36 + $1168.44 + $222.26 - $161.82 + $599.66 1 2 P/F,12,3 3 $1234 $1234 ( 0.7118 ) 4 P/A,12,4 P/F,12,5 5 $678 ( 3.0373 ) = $2059.29 $2059.29 ( 0.5674 ) 6 $678 7 $678 8 $678 P/F,12,9 P/F,12,9 9 $678 12 13 $890 14 $890 P/F,10,1 $616.37 ( 0.3606 ) -$448.74 ( 0.3606 ) $1662.96 ( 0.3606 ) 10 $678 $678 ( 0.9091 ) F/A,10,3 P/F,10,2 11 -$543 -$543 ( 0.8264 ) P/F,10,6 15 $890 $890 ( 3.3100 ) = $2945.90 $2945.90 ( 0.5645 ) 12% P/F,12,9 10% Notice the interest rate is now 12% above the red line, 10% below it INFO630 Week 7 17 www.ischool.drexel.edu

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

INFO630 Week 7 18 www.ischool.drexel.edu

Equivalence With Varying Interest Rates (more) (Brute Force) • In the 10% region 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

• Translating that to the beginning of the 12% region P/F, 12%, 9 P = $1830.51 ( 0.3603 ) = $660.08

• Adding that to the previous sum for the 12% region $2046.82 + $660.08 = $2706.90

INFO630 Week 7 19 www.ischool.drexel.edu

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 INFO630 Week 7 20 www.ischool.drexel.edu

Sample Exercise $100 $50 $80 0 1 -$100 -$20 2 3 4 5 6 7 8 9 INFO630 Week 7 21 www.ischool.drexel.edu

Exercise Answer

P/F, 6,1 P/A,6,4 P/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 INFO630 Week 7 22 www.ischool.drexel.edu

Bases for Comparison

Chapter 8 INFO630 Week 7 23 www.ischool.drexel.edu

Bases for Comparison

Outline

• 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 INFO630 Week 7 24 www.ischool.drexel.edu

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 INFO630 Week 7 25 www.ischool.drexel.edu

Basis for Comparison con’t

• Six Bases – Present worth – Future worth – Annual equivalent – Internal rate of return – Payback period – Capitalized equivalent amount INFO630 Week 7 26 www.ischool.drexel.edu

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) INFO630 Week 7 27 www.ischool.drexel.edu

• 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 INFO630 Week 7 28 www.ischool.drexel.edu

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 INFO630 Week 7 29 www.ischool.drexel.edu

The ATE Example – Cash Flow Stream $645K 0 -$300K 1 $20K 2 3 4 5 6 7 8 9 10 -$2.53M

Example from Ch 3 Lecture Slides INFO630 Week 7 30 www.ischool.drexel.edu

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 INFO630 Week 7 31 www.ischool.drexel.edu

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 PW(i) 

t n

  0 F t  1 

i

 

t

F t = 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

” INFO630 Week 7 32 www.ischool.drexel.edu

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 INFO630 Week 7 33 www.ischool.drexel.edu

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) INFO630 Week 7 34 www.ischool.drexel.edu

Comments on PW(i) (cont)

• 2 nd most widely used basis for comparison – Future Value is 1 st • PW(i) over -1 < i < oo – Only 0 < i < oo is meaningful is 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 INFO630 Week 7 35 www.ischool.drexel.edu

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?

INFO630 Week 7 36 www.ischool.drexel.edu

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 FW(i) 

t n

  0 F t  1 

i



n



t

INFO630 Week 7 37 www.ischool.drexel.edu

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 INFO630 Week 7 38 www.ischool.drexel.edu

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 F/P, i, n FW(i) = PW(i) ( ) F/P, 10%, 10 FW(10%) = $260K ( 2.5937 ) = $675K • 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 INFO630 Week 7 39 www.ischool.drexel.edu

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) ( ) INFO630 Week 7 40 www.ischool.drexel.edu

Annual Equivalent, AE(i) (cont)

• Formula AE(i)   

t n

  0 F t  1 

i

 

t

     (

i

( 1 1  

i

)

n i

)

n

 1   • 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 INFO630 Week 7 41 www.ischool.drexel.edu

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 INFO630 Week 7 42 www.ischool.drexel.edu

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 0  PW(i * ) 

t n

  0 F t  1 

i

 

t

INFO630 Week 7 43 www.ischool.drexel.edu

PW(i) = 0, Critical i at IRR

Critical i, where PW(i) = 0, discussed later IRR Yup, the same figure from slide 34 INFO630 Week 7 44 www.ischool.drexel.edu

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 INFO630 Week 7 45 www.ischool.drexel.edu

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)

INFO630 Week 7 46 www.ischool.drexel.edu

Computing IRR (cont)

7.

8.

9.

1.

2.

3.

4.

5.

6.

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.

INFO630 Week 7 47 www.ischool.drexel.edu

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  0

t n

  0 F t F t = net-cash flow instance in period t INFO630 Week 7 48 www.ischool.drexel.edu

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 INFO630 Week 7 49 www.ischool.drexel.edu

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 INFO630 Week 7 50 www.ischool.drexel.edu

Discounted Payback Period, DPP(i) • Payback period

doesn’t address interest

– Discounted payback period does – So DPP is a much more realistic measure!

• Formula – Smallest n where

t n

  0 F t  1 

i

 

t

 0 NOTE: DPP for next slide is before end of 9 th year. Use linear interpolation technique in Appendix C to find precise DPP.

INFO630 Week 7 51 www.ischool.drexel.edu

Discounted Payback Period, DPP(i) (cont) • Manual calculation of DPP(10%) for ATE Year Net cash-flow Present-worth factor Equivalent value Running sum n at end of year (P/F,10%,n) at end of year 0 through year n 0 -$300K 1.0000 -$300K -$300K 1 -$2,530K 0.9091 -$2,300K -$2,600K 2 $20K 0.8264 $17K -$2,583K 3 $645K 0.7513 $485K -$2,099K 4 $645K 0.6830 $441K -$1,658K 5 $645K 0.6209 $400K -$1,258K 6 $645K 0.5645 $364K -$894K 7 $645K 0.5132 $331K -$563K 8 $645K 0.4665 $301K -$262K 9 $645K 0.4241 $274K $12K n = 9 INFO630 Week 7 52 www.ischool.drexel.edu

Project Balance, PB(i)

• Not really a basis of comparison but closely related to DPP(i) – Simply continues DPP(i) calculations for the life of the cash-flow stream – PB(i) = profile that shows the equivalent amount of dollars invested, or earned from, the proposal at the end of time period over life of cash-flow stream.

• Formula

PB

(

i

)

T



t T

  0 F t  1 

i



T



t

INFO630 Week 7 53 www.ischool.drexel.edu

Project Balance, PB(i) (cont)

• Manual calculation of PB(10%) for ATE Year Net cash-flow Present-worth factor Equivalent value Running sum n at end of year (P/F,10%,n) at end of year 0 through year n 0 -$300K 1.0000 -$300K -$300K 1 -$2,530K 0.9091 -$2,300K -$2,600K 2 $20K 0.8264 $17K -$2,583K 3 $645K 0.7513 $485K -$2,099K 4 $645K 0.6830 $441K -$1,658K 5 $645K 0.6209 $400K -$1,258K 6 $645K 0.5645 $364K -$894K 7 $645K 0.5132 $331K -$563K 8 $645K 0.4665 $301K -$262K 9 $645K 0.4241 $274K $12K 10 $645K 0.3855 $249K $260K INFO630 Week 7 54 www.ischool.drexel.edu

0 -$300K 1 -$2.60M

Graph of PB(10%) for ATE

8 $12K 9 2 3 4 5 6 7 -$262K -$563K -$894K -$1.26M

-$1.67M

-$2.10M

-$2.58M

10 $260K Net Equiv $ Earned Net Equiv $ Exposed Risk INFO630 Week 7 55 www.ischool.drexel.edu

Capitalized Equivalent Amount CE(i) • Formal: – CE(i) = dollar amount now, that at a given interest rate, will be equivalent to the net difference of the income and payments if the cash-flow pattern is repeated indefinitely • Informal – Amount to invest at interest rate i to produce an equivalent cash flow stream on interest alone • Example – Self-supporting endowments • Formula

CE

(

i

) 

AE

(

i

)

i

INFO630 Week 7 56 www.ischool.drexel.edu

Capitalized Equivalent Amount CE(i) for ATE Get AE(i) for project ATE from slide 41

CE

( 10 %)  $ 42 .

3

K

0 .

10  $ 423

K

INFO630 Week 7 57 www.ischool.drexel.edu

Key Points

• A basis for comparison is a common frame of reference – Use of equivalence • Eight different bases were discussed: – Present worth—how much is it worth today?

– Future worth—how much will it be worth later?

– Annual equivalent—how much as a set of equal cash-flow instances?

– Internal rate of return—what’s the equivalent interest rate – Payback period -- how long to recover the investment?

– Discounted payback period—how long to recover the investment with interest?

– Project balance—what is the balance over time?

– Capitalized equivalent amount—how much capital is frozen?

INFO630 Week 7 58 www.ischool.drexel.edu

Developing Mutually Exclusive Alternatives

Chapter 9 INFO630 Week 7 59 www.ischool.drexel.edu

Developing Mutually Exclusive Alternative

Outline

• Independent proposals • Dependent proposals – Co-dependent proposals – Mutual exclusive proposals – Contingent proposals • Developing mutually-exclusive alternatives • “Do-nothing” alternative • Cash-flow streams for alternatives INFO630 Week 7 60 www.ischool.drexel.edu

Independent Proposals

• A set of proposals are

independent

when selecting any one from that set has no effect on accepting any other – Ignoring, for now, resource constraints • Example – A proposal to develop a system that predicts the stock market vs. a proposal to develop a system that plays chess INFO630 Week 7 61 www.ischool.drexel.edu

Dependent Proposals

• A set of proposals are

dependent

when selecting any one from that set can have an effect on accepting any other • Forms of dependency – Co-dependent – Mutually exclusive – Contingent INFO630 Week 7 62 www.ischool.drexel.edu

Co-Dependent Proposals

• A set of proposals are

co-dependent

when selecting any one from that set requires accepting another – These proposals should be combined into one • Example – Upgrade to new operating system and buy more memory INFO630 Week 7 63 www.ischool.drexel.edu

Mutually Exclusive Proposals

• A pair of proposals are

mutually exclusive

when selecting one from that pair negates accepting the other • Example – Get Java compiler from Vendor A vs. Java compiler from Vendor B INFO630 Week 7 64 www.ischool.drexel.edu

Contingent Proposals

• A pair of proposals are

contingent

when selecting one from that pair requires accepting the other, but not the other way • Example – Using the Swing UI toolkit vs. switching to Java INFO630 Week 7 65 www.ischool.drexel.edu

Mutually Exclusive Alternatives

• Mutual exclusion among choices is easiest to work with • In many cases you’ll have resources to do more than one proposal at the same time • A systematic way of turning proposals, along with their dependencies, into a set of mutually exclusive possible courses of action would be handy – An

alternative

is a unique, mutually exclusive course of action consisting of a set of zero or more proposals INFO630 Week 7 66 www.ischool.drexel.edu

Developing Mutually Exclusive Alternatives, Step 1

• Generate the set of all theoretically possible combinations of proposals – Build a matrix with a column for each proposal and a row for each alternative – Fill in the cells to form all potential alternatives • “1” in cell (I,J) means Proposal(I) is in Alternative(J) • “0” in cell (I,J) means it’s not – Notice the binary counting • Under Proposal(1) alternate 0,1,0,1,… • Under Proposal(2) alternate 0,0,1,1,0,0,1,1,… • Under Proposal(k) alternate 2 0’s followed by an equal number of 1’s INFO630 Week 7 67 www.ischool.drexel.edu

Example of Step 1

Alternative P1 P2 P3 Meaning A0 0 0 0 “Do nothing” A1 1 0 0 P1 only A2 0 1 0 P2 only A3 1 1 0 P1 and P2 A4 0 0 1 P3 only A5 1 0 1 P1 and P3 A6 0 1 1 P2 and P3 A7 1 1 1 All INFO630 Week 7 68 www.ischool.drexel.edu

Developing Mutually Exclusive Alternatives, Step 2

• Remove all invalid alternatives – Any alternative containing mutually exclusive proposals – Any alternative containing unsatisfied contingencies – Any alternatives exceeding resource constraints • Example, assume: – P1 and P2 are mutually exclusive – P3 is contingent on P2 – Can’t afford to do all three at same time INFO630 Week 7 69 www.ischool.drexel.edu

Example of Step 2

Alternative P1 P2 P3 Meaning A0 0 0 0 “do nothing” A1 1 0 0 P1 only A2 0 1 0 P2 only A3 1 1 0 P1 and P2 A4 0 0 1 P3 only A5 1 0 1 P1 and P3 A6 0 1 1 P2 and P3 A7 1 1 1 All INFO630 Week 7 70 www.ischool.drexel.edu

The “Do Nothing” Alternative

• Notice alternative A0 is called “do nothing” – Doesn’t really mean doing nothing at all – Only means that none of the proposals in the set being considered are carried out – Instead, money is put into other investments that give a pre determined rate of return • Bonds, interest bearing accounts, a more profitable part of the corporation, etc.

• “Do nothing” should always be considered except when – You’re required to do something • e.g., repair or replace broken equipment – You’re working with “service alternatives” (see Chapter 11) INFO630 Week 7 71 www.ischool.drexel.edu

The “Do Nothing” Alternative (cont) • Sometimes even the best of the proposals is worse than what could be achieved by investing somewhere else – When the “do nothing” alternative comes out the best, it means the organization would be better off not carrying out any of the proposals being considered and should put the money into a more profitable investment elsewhere • A0 is assumed to have – PW(i) = $0 – FW(i) = $0 – AE(i) = $0 INFO630 Week 7 72 www.ischool.drexel.edu

Cash-Flow Stream for Alternatives • The cash-flow stream for any alternative (other than A0) will be the sum of the cash-flow streams of all proposals it contains INFO630 Week 7 73 www.ischool.drexel.edu

Key Points

• There are several forms of dependency between proposals • Decisions are easiest when choices are mutually exclusive • An alternative is a set of zero to many proposals • There is a process for turning proposals with dependencies into valid, mutually exclusive alternatives • The “do nothing” alternative doesn’t really mean do nothing at all, just none of the projects proposed • The cash-flow stream for an alternative is the sum of the cash-flow streams for all its proposals INFO630 Week 7 74 www.ischool.drexel.edu