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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INFO 630 Evaluation of Information Systems

Prof. Glenn Booker

Week 8 – Chapters 10-12

1INFO630 Week 8


For-Profit Business Decisions

Chapter 10

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• 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

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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.

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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%

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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

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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?

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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

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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%

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Basic For-profit Decision Process

Assume the first alternative is the current bestfor j = 2 to the number of alternatives

Consider the jth alternative to be the candidateCompare the candidate to the current bestif 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 *}

Find maximum value Successive comparison in pair-wise basis

Current best Candidate

Algorithm

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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…

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Basic For-profit Decision Process

Arrange the alternatives in order of increasing

initial investment

Assume the first alternative is the current best

Compare the next candidate to the current best

Is the next candidate strictly

better than the current best

?

Make that next candidate be the new current best

The current best is the best overall

Are there more alternatives

to compare ?

Start

Stop

Yes

No

Yes

No

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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

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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

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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

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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

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• 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

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• 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

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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

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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

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• 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

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• 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

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Present Worth on Total Investment

• Preferred method– Comparison on incremental investment

• Other options– Total Investment

• Can not use IRR• Must use

– PW() – FW()– AE()

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• Cash-flow stream Do Nothing– Initial investment = $0– Annual income = $0– Annual expenses = $0– Salvage value = $0

PW(12%) = $0

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• 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

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• 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

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• 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…

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Rank on Rate of Return

• A well-known approach1. 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)

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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



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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)

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Rank on Rate of Return Flaw—Explained

A x

A y

MARR

i

$

PW(MARR) of A y

IRR of A x

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(See the notes below for this slide)


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

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Planning Horizons and Economic Life

Chapter 11

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• Planning horizon• Capital recovery with return• Economic life• Finding the economic life• Economic life and planning horizons

Planning Horizons and Economic Life

Outline

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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

*

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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)

Acquisition Cost

Salvage Value

CR(i) = AE(i) of the loss in value ...

... plus i times the salvage value

}0 1 2 3 n-1 n

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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

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Capital Recovery with Return, CR(i) (cont)

• Generalizing

CR(i) = (P - F) (A/P, i, n) + F*iAs 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

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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

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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

Total cost

Operating and maintenance cost

Capital recovery with return, CR(i)

AE(i) Costs

Number of years the asset is kept

Economic life

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Finding the Economic Life

(3) (1) (2) Operating (4) (6) (7) Salvage AE(i) cost and PW(i) (5) AE(i) TotalEnd value if if retired maintenance O&M for Sum of year 0 cost of AE(i) ifof retired at in year n costs for year n in O&Ms through operating retired atYear 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

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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

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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

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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

CR(i)

Salvage Value

0 1 2 3 Economic life

5

PW(i) of salvage value at early retirement

64

CR(i) costs avoided by early retirement

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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

* *

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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

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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

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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

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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

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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

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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

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Replacement and Retirement Decisions

Chapter 12

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• 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

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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?

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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

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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

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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

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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

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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) ( 3.1272 ) + $25k ( 0.4371 ) = -$230k + $100k * 3.1272 + $25k * 0.4371 = -$230k + $313k + $11k = $94k

PW of differential >0, so Challenger 1 is better than Defender

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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) ( 3.1272 ) + $5k ( 0.4371 ) = -$50k + $20k * 3.1272 + $5k * 0.4371 = -$50k + $63k + $2k = $15k

PW of differential >0, so Challenger 2 is betterEnd of alternatives, should retire in-house SW & go with COTS solution

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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

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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• …

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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

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Example Retirement Decision

• MARR = 16%

End of Salvage year Revenue O & M cost value 0 - - $1200 1 $775 $300 $900 2 $775 $400 $700 3 $775 $500 $550 4 $775 $600 $450 5 $775 $700 $375 6 $775 $800 $325

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Example Retirement Decision (cont)

• PW(16%) of immediate retirement

• PW(16%) of retiring after 1 year

• PW(16%) of retiring after 2 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) = $1217

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Example Retirement Decision (cont)




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)

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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, …

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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