INFO 630 Evaluation of Information Systems Prof. Glenn Booker

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INFO 630 Evaluation of Information Systems Prof. Glenn Booker

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INFO 630 Evaluation of Information Systems Prof. Glenn Booker

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INFO 630Evaluation of Information SystemsProf. Glenn Booker

Week 7 – Chapters 7-9

INFO630 Week 7

Equivalence

Chapter 7

INFO630 Week 7

- Simple comparison of two proposals
- Equivalence, defined
- Simple equivalence
- Equivalence with varying cash-flow instances
- Equivalence with varying interest rates

INFO630 Week 7

- 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

- How do I evaluate?
- Impact of time?
- Interest
- What does 0% interest mean?
- Is this realistic?

INFO630 Week 7

- 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

- 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

“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

- 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

- 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

INFO630 Week 7

- 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

- Elegant Approach

INFO630 Week 7

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

INFO630 Week 7

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 )

Assume Interest = 12%, 15 Year,

Using Single Payment Present Worth Value =

P/F, i, n

P = F ( )

INFO630 Week 7

- 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

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

INFO630 Week 7

- 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

- In general yes.
- What happens if interest rate varies?

INFO630 Week 7

Notice the interest rate is now 12% above the red line, 10% below it

INFO630 Week 7

- 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

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

- 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

INFO630 Week 7

- 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

$100

$80

$50

0

1

2

3

4

5

6

7

8

9

-$20

-$100

INFO630 Week 7

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

INFO630 Week 7

Bases for Comparison

Chapter 8

INFO630 Week 7

- 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

- 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

- A common frame of reference for comparing two or more cash-flow streams in a consistent way

INFO630 Week 7

- Six Bases
- Present worth
- Future worth
- Annual equivalent
- Internal rate of return
- Payback period
- Capitalized equivalent amount

INFO630 Week 7

- 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)

- Always use the same

INFO630 Week 7

- 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

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

$645K

$20K

0

1

2

3

4

5

6

7

8

9

10

-$300K

-$2.53M

Example from Ch 3 Lecture Slides

INFO630 Week 7

- 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

- Reference time for PW(i) =

INFO630 Week 7

- 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”

INFO630 Week 7

- 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

- 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

- 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

INFO630 Week 7

- 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?

- Answer the question:

- Usually the end of the cash-flow stream

- Reference time for FW(i) =

INFO630 Week 7

- Formula
- Uses single-payment compound-amount (F/P,i,n) to translate each individual net-cash flow instance
- Then sum all amounts

INFO630 Week 7

- 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

- 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

- FW(i) = 0 when PW(i) = 0

F/P, i, n

FW(i) = PW(i) ( )

F/P, 10%, 10

FW(10%) = $260K ( 2.5937 ) = $675K

INFO630 Week 7

- 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

- Either:
- 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

- 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

INFO630 Week 7

- 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

- AE(i) = 0 when PW(i) = 0
- 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

- If the ATE project can be repeated, AE(i) = $42.3K over 20 years, or over 30 years, …

INFO630 Week 7

- 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)

- IRR expresses the cash-flow stream as an interest rate
- Formula

INFO630 Week 7

Critical i,

where PW(i) = 0, discussed later - IRR

Yup, the same figure from slide 34

INFO630 Week 7

- 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

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

- 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

- 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”

- Payback period expresses the cash-flow stream as a time
- Formula
- Smallest n where

- Ft = net-cash flow instance in period t

INFO630 Week 7

- 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

- 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

- Example

INFO630 Week 7

- Payback period doesn’t address interest
- Discounted payback period does
- So DPP is a much more realistic measure!

- Formula
- Smallest n where

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

INFO630 Week 7

- 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

- 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

INFO630 Week 7

- 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

Net Equiv $ Earned

$260K

$12K

0

8

9

1

2

4

5

6

7

10

3

-$262K

-$300K

-$563K

-$894K

Net Equiv $ Exposed

Risk

-$1.26M

-$1.67M

-$2.10M

-$2.60M

-$2.58M

INFO630 Week 7

- 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

INFO630 Week 7

Get AE(i) for project ATE from slide 41

INFO630 Week 7

- 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

Developing Mutually Exclusive Alternatives

Chapter 9

INFO630 Week 7

- 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

- 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

- 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

- 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

- 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

- 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

- 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

- 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

k-1

INFO630 Week 7

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

- 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

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

- 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)

- You’re required to do something

INFO630 Week 7

- 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

- 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

- 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