INFO 630
Evaluation of Information Systems
Dr. Jennifer Booker
Week 8 – Chapters 10-12
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For-Profit Business Decisions
Chapter 10
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For-Profit Business Decisions
Outline
• Minimum Attractive Rate of Return
(MARR)
• Basic for-profit decision process
• Incremental vs. total cash-flow analysis
• Rank on rate of return
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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



Find maximum value
Successive comparison in pair-wise basis
 Current best
 Candidate
Algorithm
Assume the first alternative is the current best
for j = 2 to the number of alternatives
Consider the jth alternative to be the candidate
Compare the candidate to the current best
if the candidate is better than the current best
then make the jth alternative the current best
{* on ending, the current best is the best alternative *}
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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
Start
Arrange the alternatives
in order of increasing
initial investment
Assume the first alternative
is the current best
Are there
more alternatives
to compare
?
Yes
Compare the next candidate
to the current best
No
Is the next
candidate strictly
better than the
current best
?
The current best is
the best overall
No
Yes
Make that next candidate
be the new current best
Stop
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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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Present Worth on Incremental Investment
• Differential cash-flow stream (B-Do
Nothing)
– Initial investment = $6200 - $0 = $6200
– Annual income = $7329 - $0 = $7329
– Annual expenses = $5908 - $0 = $5908
– Salvage value = $340 - $0 = $340
P/A,12%,8
P/F,12%,8
PW(12%) = -$6200 + ($7329 - $5908) ( 4.9676 ) + $340 (0.4039 )
= -$6200 + $1421 * 4.9676 + $340 * 0.4039
= -$6200 + $7059 + $137 = $996
PW of differential >= $0, therefore B is better
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Present Worth on Incremental Investment
• Differential cash-flow stream (C-B)
– Initial investment = $6890 - $6200 = $690
– Annual income = $6601 - $7329 = -$728
– Annual expenses = $5335 - $5908 = -$573
– Salvage value = $190 - $340 = -$150
P/A,12%,8
P/F,12%,8
PW(12%) = -$690 + (-$728 - -$573) ( 4.9676 ) + -$150 (0.4039 )
= -$690 + -$155 * 4.9676 + -$150 * 0.4039
= -$690 + -$770 + -$61 = -$1521
PW of differential <= $0, so B is still better end of alternatives,
and B is best overall
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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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IRR on Incremental Investment
• Differential cash-flow stream (B-Do
Nothing)
– Initial investment = $6200 - $0 = $6200
– Annual income = $7329 - $0 = $7329
– Annual expenses = $5908 - $0 = $5908
– Salvage value = $340 - $0 = $340
IRR = 16.37%
IRR of differential >= MARR, so B is better
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IRR on Incremental Investment
• Differential cash-flow stream (C-B)
– Initial investment = $6890 - $6200 = $690
– Annual income = $6601 - $7329 = -$728
– Annual expenses = $5335 - $5908 = -$573
– Salvage value = $190 - $340 = -$150
PW(0%) < $0, can’t compute IRR (it’s negative)
IRR of differential <= MARR, B is still better end of alternatives, B is best overall
– NOTE: Same conclusion as before, B is best overall
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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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Present Worth on Total Investment
• Cash-flow stream Do Nothing
– Initial investment = $0
– Annual income = $0
– Annual expenses = $0
– Salvage value = $0
PW(12%) = $0
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Present Worth on Total Investment
• Cash-flow stream A
– Initial investment = $5300
– Annual income = $4142
– Annual expenses = $3144
– Salvage value = $210
P/A,12%,8
P/F,12%,8
PW(12%) = -$5300 + ($4142 - $3144) ( 4.9676 ) + $210 (0.4039 )
= -$5300 + $998 * 4.9676 + $210 * 0.4039
= -$5300 + $4958 + $85 = -$257
PW(A) <= PW(Do Nothing), so Do Nothing is better
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Present Worth on Total Investment
• Cash-flow stream B
– Initial investment = $6200
– Annual income = $7329
– Annual expenses = $5908
– Salvage value = $340
P/A,12%,8
P/F,12%,8
PW(12%) = -$6200 + ($7329 - $5908) ( 4.9676 ) + $340 (0.4039 )
= -$6200 + $1421 * 4.9676 + $340 * 0.4039
= -$6200 + $7059 + $137 = $996
PW(B) >= PW(Do Nothing), so B is better
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Present Worth on Total Investment
• Cash-flow stream C
– Initial investment = $6890
– Annual income = $6601
– Annual expenses = $5335
– Salvage value = $190
P/A,12%,8
P/F,12%,8
PW(12%) = -$6890 + ($6601 - $5335) ( 4.9676 ) + $190 (0.4039 )
= -$6890 + $1266 * 4.9676 + $190 * 0.4039
= -$6890 + $6289 + $77 = -$524
PW(B) >= PW(C), B is still better end of alternatives,
So B is best overall – NOTE: Same conclusion…
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Rank on Rate of Return
•
A well-known approach
1. 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
• Alternative A2
– Initial investment = $25,120
– Year 1-10 annual net income = $5966
• Alternative A3
– Initial investment = $31,400
– Year 1-10 annual net income = $7850
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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
MARR
$
A y
PW(MARR) of A
y
A x
IRR of A
x
i
(See the notes below for this slide)
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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 Horizons and Economic Life
Outline
•
•
•
•
•
Planning horizon
Capital recovery with return
Economic life
Finding the economic life
Economic life and planning horizons
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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
}
CR(i) = AE(i) of the loss in value ...
... plus i times the salvage value
0
1
2
3
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n-1
Salvage
Value
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*i
As in
CR(i) = ($15,000 - $1317) (A/P, 13%, 6) + $1317 (0.13)
= $3494
Where P = acquisition cost, F = salvage cost, n = length of time
asset kept, i = interest rate
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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
Year
n
1
2
3
4
5
6
Salvage value
at end of year
$10,000
$6667
$4444
$2963
$1975
$1317
CR(13%) if kept
n years
$6950
$5862
$5048
$4432
$3960
$3594
• O&M tends to increase over time
– Why?
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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
Economic life
Total cost
AE(i)
Costs
Operating and
maintenance cost
Capital recovery
with return, CR(i)
Number of years the asset is kept
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Finding the Economic Life
(1)
Salvage
End value if
of
retired at
Year year n
1
$10,000
2
$6667
3
$4444
4
$2963
5
$1975
6
$1317
(3)
(2)
Operating
AE(i) cost
and
if retired maintenance
in year n
costs for
[CR(i)]
year n
$6950
$2000
$5862
$3400
$5048
$4800
$4432
$6200
$3960
$7600
$3594
$9000
(4)
PW(i)
(5)
O&M for Sum of year 0
year n in O&Ms through
year 0
year n
$1770
$1770
$2663
$4433
$3327
$7759
$3803
$11,562
$4125
$15,687
$4323
$20,010
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(6)
(7)
AE(i)
Total
cost of
AE(i) if
operating retired at
for n years year n
$2000
$8950
$2657
$8519
$3286
$8335
$3887
$8319
$4460
$8420
$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
PW(i) of salvage value
at early retirement
CR(i) costs avoided
by early retirement
CR(i)
Salvage
Value
0
1
2
3
4
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Economic
life
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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
$8319 (P/A, 13%, 2) = $8319 (1.6681) = $13,877
– Owner also has access to estimated salvage value 2 years earlier
$2963 (P/F, 13%, 2) = $2963 (0.7831) = $2320
– Implied salvage vale is sum of those
$13,877 + $2320 = $16,197
• Generalizing
P/A, i, n-n *
F *n = CR(i) (
P/F, i, n-n *
)+F(
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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 and Retirement Decisions
Outline
•
•
•
•
•
•
Replacement decisions, defined
Special issues in replacement decisions
Outsider’s viewpoint
Example of replacement analysis
Retirement decisions, defined
Example of retirement analysis
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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
–
–
–
–
•
Challenger 1: redevelop software in-house
–
–
–
–
•
Initial investment = $30,000
Annual income = $130,000
Annual operating and maintenance = $100,000
Salvage value = $30,000
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 better
End 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
year
0
1
2
3
4
5
6
Revenue
$775
$775
$775
$775
$775
$775
O & M cost
$300
$400
$500
$600
$700
$800
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Salvage
value
$1200
$900
$700
$550
$450
$375
$325
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Example Retirement Decision (cont)
• PW(16%) of immediate retirement
Salvage value today = $1200
• PW(16%) of retiring after 1 year
P/A,16%,1
P/F,16%,1
P/F,16%,1
$775 * (0.8621) - $300 * (0.8621) + $900 * (0.8621) = $1185
• PW(16%) of retiring after 2 years
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
• PW(16%) of retiring after 3 years
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)
• PW(16%) of retiring after 4 years
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
• PW(16%) of retiring after 5 years
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
• PW(16%) of retiring after 6 years
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 forprofit 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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