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• To learn how to properly apply the profitability
measures described in Chapter 4 to select the best
alternative out of a set of mutually exclusive
alternatives (MEA)
• The cash-flow analysis methods (previously
described) used in this process:
Present Worth ( PW )
Annual Worth ( AW )
Future Worth ( FW )
Internal Rate of Return ( IRR )
• Organizations have the capability to generate
potential beneficial projects for potential
• The alternatives being considered may require
different amounts of capital investment
• The alternatives may have different useful lives
• The subject of this section will help:
– analyze and compare feasible alternatives
– select the preferred alternative
Feasible Design Alternatives
• Three types of investment categories
– Mutually Exclusive Set
– Independent Project Set
– Contingent
• Mutually exclusive set
– The selection of one alternative excludes the consideration of
any other alternative
– Once selected, the remaining alternatives are excluded
• Independent project set
– Selecting the best possible combination of projects from the
set that will optimize a given criteria
– Subjects to constraints
• Contingent
– The choice of the project is conditional on the choice of one or
more other projects
Section 5.2
Fundamental Purpose of Capital Investment
• To obtain at least the MARR for every dollar invested.
• Basic Rule:
• Spend the least amount of capital possible unless the
extra capital can be justified by the extra savings or
• In other words, any increment of capital spent (above the
minimum) must be able to pay its own way.
Section 5.2 Two Types of Decisions
1. Investment Alternatives - each alternative has an initial
investment producing positive cash flows resulting from
increased revenues, reduced costs, or both.
– "Do nothing" (DN) is usually an implicit investment
– If positive cash flows > negative cash flows, then IRR>0.
– If EW(MARR)>0, investment is profitable, or if
EW(MARR)<0, do nothing (DN) is better, where EW refers to
an equivalent worth method (e.g. PW)
2. Cost Alternatives - have all negative cash flows except
for the salvage value (if applicable). These alternatives
represent “must do” situations, and DN is not an option
– IRR not defined for cost alternatives. Can you explain why?
Section 5.3. The Study Period
• Must be appropriate for the decision being made
• Study Period: The time interval over which service is
needed to fulfill a specified function
• Useful Life: The period over time during which an
asset is kept in productive operation
• Case 1: Study period = Useful life
• Case 2: Study period Useful life
• Fundamental Principle: Compare MEAs over the
same period of time
Section 5.4.1
(EW) Methods: PW, AW, FW (Case 1)
• Procedure for Selecting the Best MEA using the EW
– 1. Compute the equivalent worth of each alternative, using the
MARR as the interest rate.
– 2. Investment Alternatives: Select the alternative having the
greatest equivalent worth.
• Note: If all equivalent worths are < 0 for investment alternatives, then
"do nothing" is the best alternative.
– 3. Cost Alternatives: Select the alternative having the
smallest equivalent cost (the one that is least negative).
• All three equivalent worth methods (PW, AW, FW) will
identify the same "best" alternative.
Study Period =Useful Life
Investment cost (I) $100,000
Net Annual receipts
Salvage value (SV)
Useful life
• If the MARR is 12%, use the PW method to select the best
• PW(12%) = -I + A(P|A, 12%, 10) + SV(P|F, 12%, 10)
• Solution
• PWDN (12%) = 0, PWI (12%) = -10,897, PWII (12%) =
+28.241, PWIII (12%) = +23,672, PWIV (12%)=+20,923
• Select Alternative ___ to maximize PW.
Example – Problem 5-12a (p. 234)
• Cost Alternatives
• Study Period = Useful Life
Initial cost (I)
Annual expenses
years 1-7
• Use the AW method to choose the best alternative
(MARR = 12%)
– AWA = -26,155
– AWB = -25,947
– AWC = -25,781
• Assuming one must be chosen (i.e., DN is not an option),
select alternative _ to minimize AW costs.
Rate-of-Return Analysis: Multiple Alt.s
1. Assume we have two or more mutually exclusive Alt.
2. Objective: Which, if any of the alternatives is preferred?
3. Two Investments A and B, Discount rate = 10%, Each
investment requires $100 at t = 0, A is a 1-year investment,
B is a 5- year investment.
4. i*A = 0.20 = 20%, i*B = 0.15 = 15%, PWA(10%) = +$9.09,
PWB(10%) = +24.89
1. Using ROR Ranking _ is superior to _
Using a PW(10%) approach _ is superior to _
• The two methods do not rank the same?
Using the IRR Method: Another Example
• Why not select the investment opportunity that
maximizes IRR?
• Consider 2 alternatives:
• Investment
• Lump-Sum Receipt
• If MARR = 20%, would you rather have A or B if
comparable risk is involved?
• If MARR = 20%, PWA = $733 and PWB = $2,500
Is It Worth It?
• Now the question is….
• Is it worth spending an additional $9,900 to move from
investment A to investment B?
• NEVER simply select the MEA that MAXIMIZES the
• Never compare the IRR to anything except the MARR.
• We don't maximize rate of return. Look at the increment
• Answer: Compute the ROR or PW of the incremental
investment to see!
• IRR A-B : PW A-B = 0 = -9,900 + 14,000(P|F, i'%, 1)
i' A-B = 41.4% > MARR
Calculations of Incremental Cash
Flows for ROR Analysis
• Given two or more alternatives
• Rank the investments based upon their initial time t = 0
investment requirements
• Summarize the investments in a tabular format
• Select the first investment to be the one with the lowest
time t = 0 investment amount.
• The next investment is to be the one with the largest
investment at time t = 0
Example – Problem 5-2 on page 232
• Given three MEAs and MARR = 15% per year
Investment (FC) -28,000
Net Cash Flow/year 5,500
Salvage Value
Useful Life
10 yrs
10 yrs
10 yrs
• Use the Incremental IRR procedure to choose the best
Incremental Investment Analysis Procedure
1. Order the feasible alternatives
2. Establish a base alternative
a. Cost alternatives -- The first alternative is the base
b. Investment alternatives - If the first alternative is acceptable,
select as base. If the first alternative is not acceptable, choose
the next alternative
3. Use iteration to evaluate differences (incremental cash
flows) between alternatives until no more alternatives
a. If incremental cash flow between next alternative and current
alternative is acceptable, choose the next
b. Repeat, and select as the preferred alternative the last one for
which the incremental cash flow was acceptable
To Summarize
1. Each increment of capital must justify itself by
producing a sufficient rate of return on that increment.
2. Compare a higher investment alternative against a
lower investment alternative only when the latter is
3. Select the alternative that requires the largest
investment of capital as long as the incremental
investment is justified by benefits that earn at least the
MARR. This maximizes equivalent worth on total
investment at i = MARR.
Section 5.5 Case 2: Study PeriodUseful Life
• Up until now, study periods and useful lives have been
the same length
• The study period is frequently taken to be a common
multiple of the alternatives’ lives when study period 
useful life
• Repeatability Assumption (page 211)
• Conditions:
1. Study period is either indefinitely long or equal to a common
multiple of the lives of the alternative.
2. The cash flows associated with an alternative's initial life span
are representative of what will happen in succeeding life spans.
Example: Problem 5-24a (pp. 236)
• Cost Alternatives; Study Period > Useful Life; MARR =
Investment cost
Annual costs
Useful life
SV (MKT value)
• If the study period = 20 years, which alternative is
Different Lives
• Comparison must be made over equal time
– Compare over the least common multiple,
LCM, for their lives
– Remember – if the lives of the alternatives are
not equal, one must create or force a study
period where the life is the same for all of the
AW for Unequal Lives
• Consider the AW over the useful life of Alternative A:
AWA = -14,000(A|P, 15, 5)- 14,000 + 8,000(A|F, 15, 5)= -16,990
– Life 1: AW 1-5 = -16,990
– Life 2: AW 6-10 = -14,000(A|P, 15, 5)- 14,000 + 8,000(A|F, 15,
5) = -16,990
– Life 3: AW 11-15 = -16,990
– Life 4: AW 16-20 = -16,990
• Shortcut: If the study period equals a common multiple
of the alternatives' lives, simply compare AW computed
over the respective useful lives (assuming repeatability is
• What if the study period is not a common multiple of
the alternatives' lives or repeatability is not applicable?
• A finite and identical study period is used for all
• This planning horizon, combined with appropriate
adjustments to the estimated cash flows, puts the
alternatives on a common and comparable basis
• Used when repeatability assumption is not applicable
• Frequently used in engineering practice
Use the Cotermination Assumption
• Procedure: The cash flows of the alternatives need to
be adjusted to terminate at the end of the study period.
• Cost alternatives: Assuming repeatability, repeat part
of the useful life of the original alternative, and then use
an estimated MV to truncate it at the end of the study
– Without repeatability, we must purchase/lease the service/asset
for the remaining years.
• Investment alternatives: Assume all cash flows will be
reinvested at the MARR to the end of the study period
(i.e., calculate FW at end of useful life and move this to
the end of the study period using the MARR).
Study Period < Useful Life
• When the study period is explicitly stated to be shorter
than the useful life, use the cotermination assumption
• Procedure: The cash flows of the alternatives need to be
adjusted to terminate at the end of the study period.
• Truncate the alternative at the end of the study period
using an estimated Market Value.
Alt-1: N = 5 yrs
Alt-2: N= 7 yrs
Example of Cotermination
• Suppose the study period had been stated to be 20 years.
Which boiler would you recommend?
Boiler A
Investment cost $
Useful life
20 yrs.
[email protected] of useful life
Annual costs
Useful life of A = 20 years = study period
Useful life of B = 40 years > study period
Assume MV B @ EOY 20 = $50,000
The MARR is 10% per year.
Boiler B
40 yrs.
• AWA (10%) = -14,700, AWB (10%) = -19,225
• Select _
• What would the market value of Boiler B @ EOY 20
have to be in order to select Boiler B instead of A?
• Set AWA = AWB
-14,700 = -6,000 - {120,000(A|P, 10, 20) - X(A|F,10, 20)}
• X = $308,571 therefore, MVB > $308,571 to favor B
• Such a value is very unlikely because X is more than
the initial cost of Boiler B.
Problem 5-24a Revisited with Cotermination
• If the study period = 10 years and the estimated market
value for alternate B = $25,000 @ EOY 10, which
alternative is preferred?
Comparing Alternatives Using the
Capitalized Worth Method
• Capitalized Worth (CW) method -- Determining the
present worth of all revenues and / or expenses over an
infinite length of time
• Capitalized cost -- Determining the present worth of
expenses only over an infinite length of time
• Capitalized worth or capitalized cost is a convenient
basis for comparing mutually exclusive alternatives
when a period of needed services is indefinitely long
and the repeatability assumption is applicable
Capitalized Cost
• CAPITALIZED COST- the present worth of a project
which lasts forever.
• Government Projects
• Roads, Dams, Bridges, project that possess perpetual life
• Infinite analysis period
• Start with the closed form for the P/A factor
 (1  i ) N  1 
P  A
• Next, let N approach infinity
1 
P  A  
i 