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COMPARING ALTERNATIVES

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2. Cost Alternatives - have all negative cash flows except for the salvage value (if applicable) ... Salvage value (SV) 10,000 0 15,000 20,000. Useful life 10 10 10 10 ... – PowerPoint PPT presentation

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Title: COMPARING ALTERNATIVES


1
CHAPTER 6
  • COMPARING ALTERNATIVES

2
Objective
  • 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 )

3
Alternatives
  • Organizations have the capability to generate
    potential beneficial projects for potential
    investment
  • 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

4
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

5
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 benefits.
  • In other words, any increment of capital spent
    (above the minimum) must be able to pay its own
    way.

6
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 alternative.
  • If ?positive cash flows gt ?negative cash flows,
    then IRRgt0.
  • If EW(MARR)gt0, investment is profitable, or if
    EW(MARR)lt0, 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?

7
Section 6.3 - 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 ¹ Useful life
  • Fundamental Principle Compare MEAs (mutually
    exclusive alternatives) over the same time period

8
Section 6.4.1 - (EW) Methods PW, AW, FW (Case 1)
  • Procedure for selecting the best MEA using the EW
    method
  • 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 lt 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.

9
Study Period Useful Life
  • I II III IV
  • Investment cost (I) 100,000 152,000
    184,000 220,000
  • Net Annual receipts 15,200 31,900
    35,900 41,500
  • Salvage value (SV) 10,000
    0 15,000
    20,000
  • Useful life 10 10 10 10
  • If the MARR is 12, use the PW method to select
    the best alternative
  • PW(12) -I A(PA, 12, 10) SV(PF, 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.

10
Cost Alternatives - Example
  • Cost Alternatives
  • Study Period Useful Life
  • A B C
  • Initial cost (I) -85,600 -63,200 -71,800
  • Annual expenses
  • years 1-7 -7,400 -12,100 -10,050
  • 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.

11
Rate-of-Return Analysis Multiple Alternatives
  • Assume we have two or more mutually exclusive
    Alt.
  • Objective Which, if any of the alternatives is
    preferred?
  • 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.
  • iA 0.20 20, iB 0.15 15, PWA(10)
    9.09, PWB(10) 24.89
  • Using ROR Ranking _ is superior to _
  • Using a PW(10) approach _ is superior to _
  • The two methods do not rank the same?

12
Using the IRR Method Another Example
  • Why not select the investment opportunity that
    maximizes IRR?
  • Consider 2 alternatives A B
  • Investment -100 -10,000
  • Lump-Sum Receipt 1,000 15,000
  • IRR 900 50
  • If MARR 20, would you rather have A or B if
    comparable risk is involved?
  • If MARR 20, PWA 733 and PWB 2,500

13
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
    IRR
  • 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(PF, i',
    1)
  • i' A-B 41.4 gt MARR

14
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
    larger investment at time t 0

15
Example
  • Given three MEAs and MARR 15 per year
  • 1 2 3
  • Investment (FC) -28,000 -16,000
    -23,500
  • Net Cash Flow/year 5,500 3,300
    4,800
  • Salvage Value 1,500 0
    500
  • Useful Life 10 yrs 10 yrs 10 yrs
  • Use the Incremental IRR procedure to choose the
    best alternative

16
Incremental Investment Analysis Procedure
  • 1. Order the feasible alternatives
  • 2. Establish a base alternative
  • a. Cost alternatives -- The first (LCI)
    alternative is the base
  • b. Investment alternatives - If the first (LCI)
    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 exist
  • 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

17
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18
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 acceptable.
  • 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.

19
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 ¹ useful life
  • Repeatability Assumption (page 244)
  • 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.

20
Example
  • Cost Alternatives Study Period gt Useful Life
    MARR 15. A B
  • Investment cost 14,000 65,000
  • Annual costs 14,000 9,000
  • Useful life 5 20
  • SV (MKT value) 8,000 13,000
  • If the study period 20 years, which alternative
    is preferred?

21
Different Lives
  • Comparison must be made over equal time periods
  • 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
    alternatives

22
AW for Unequal Lives
  • Consider the AW over the useful life of
    Alternative A
  • AWA -14,000(AP, 15, 5)- 14,000 8,000(AF,
    15, 5) -16,990
  • Life 1 AW 1-5 -16,990
  • Life 2 AW 6-10 -14,000(AP, 15, 5)- 14,000
    8,000(AF, 15, 5) -16,990
  • Life 3 AW 11-15 -16,990
  • Life 4 AW 16-20 -16,990
  • AWB -65,000(AP,15,20) 9000
    13,000(AF,15,20) -19,259.6
  • 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 valid). In this
    case, Alt. A is preferred.

23
PW Approach
  • LCM 20 years MARR 15
  • Because Repeatability Assumption applies then
  • Identical replacement of Alternative A at EOY of
    5,10, 15
  • No replacement for Alternative B
  • Draw CFD
  • PW(15)A -14,000 -6,000(P/F,15,5) -
    6,000(P/F,10,10) - 6,000 (P/F,10,15) -
    14,000(P/A,15,20) -106,834
  • PW(15)B -65,000 13,000(P/F,15,20) -
    9,000(P/A,15,20) -120,539
  • Therefore, alternative A is preferred

24
Example 6-8
25
Example 6-8
  • LCM 12 years MARR 10 Study Period 12 years
  • If Repeatability assumption applies then
  • Identical replacement of Alternative A at EOY of
    4 and 8. Last 2 years reinvestment at MARR
  • Identical replacement of Alternative B at EOY of
    6.
  • Draw CFD
  • PW(10)A -3,500 -3,500(P/F,10,4)
    -3,500(P/F,10,8) 1,255(P/A,10,12) 1,028
  • PW(10)B -5,000 -5,000(P/F,10,6)
    1,480(P/A,10,12) 2,262
  • Therefore, alternative B is preferred

26
Problem?
  • 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 alternatives
  • 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

27
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 period
  • 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).

28
Study Period lt 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
29
Example of Cotermination
  • Suppose the study period had been stated to be 20
    years. Which boiler would you recommend? Boil
    er A Boiler B
  • Investment cost 50,000 120,000
  • Useful life 20 yrs. 40 yrs.
  • SV_at_end of useful life 10,000 20,000
  • Annual costs 9,000 6,000
  • Useful life of A 20 years study period
  • Useful life of B 40 years gt study period
  • Assume MV B _at_ EOY 20 50,000
  • The MARR is 10 per year.

30
Solution
  • AWA (10) -14,700, AWB (10) -19,225
  • Select _
  • What would the market value of Boiler B _at_ EOY 20
    have to be in order to select Boiler B instead of
    A?
  • Set AWA AWB
  • -14,700 -6,000 - 120,000(AP, 10, 20) -
    X(AF,10, 20)
  • X 308,571 therefore, MVB gt 308,571 to favor B
  • Such a value is very unlikely because X is more
    than the initial cost of Boiler B.

31
Problem Revisited with Cotermination
  • If the study period 10 years and the estimated
    market value for alternate B 25,000 _at_ EOY 10,
    which alternative is preferred?

32
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

33
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
  • Next, let N approach infinity
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