Engineering Economic Analysis Canadian Edition

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Engineering Economic AnalysisCanadian Edition

• Chapter 6 Annual Cash Flow Analysis

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Chapter 6. . .

• defines equivalent uniform annual costs (EUAC)
  and equivalent uniform annual benefits (EUAB).
• conducts an annual worth analysis for a single
  project.
• converts engineering economic problems to EUAC
  and EUAB and compares competing alternatives.
• develops and uses spreadsheets to analyze loans.

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Assumptions in Solving Economic Analysis Problems

• End-of-year convention
• simplifies calculations
• Viewpoint
• generally the firm
• Sunk costs
• past has no bearing
• Owner provided capital
• no debt capital

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• Stable prices
• No depreciation, income taxes
• Annual and within-year interest compounding
• Frequency mismatches
• Cash flows and compounding

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Annual Cash Flow CalculationsFrom Present Cost
to Annual Cost

• Simplest case is to convert the NPW to a series
  of EUAW (equivalent uniform annual worth)
• EUAW EUAB EUAC
• A P(A/P,i,n)
• A is -PMT in Excel
• Where there is salvage value?
• A will be reduced
• A F(A/F,i,n)

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Cash Flow Equivalency
Present Worth (PW)
Annual Worth (EUAW)
Future Worth (FW)
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Annual Cash Flow Four Essential Points

• 1. EUAW (EUAB-EUAC) NPW(A/P,i,n).
• 2. EUAW is decreased by a cost or increased by a
  benefit.
• 3. In Excel use -PMT to calculate EUAW
  (remember the minus sign).
• 4. For an irregular cash flow over the analysis
  period, first determine the PW then convert to
  EUAW.

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EUAW Formulas

• 1. EUAW P(A/P,i,N) - SV(A/F,i,N) (most
  common), given that (A/P,i,N) (A/F,i,N) i
• 2. EUAW (P - SV)(A/P,i,N) SV.i
• EUAW (P - SV)(A/F,i,N) P.i, where SV
  Salvage Value

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Annual Cash Flow Analysis
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Independent Projects

• Select all projects with non-negative annual
  equivalent worth, up to the budgetary constraint,
  if any.
• If all projects have EUAW lt 0, select the status
  quo (that is, invest available money at the going
  interest rate)
• If EUAW gt 0, accept project
• If EUAW lt 0, reject project
• If EUAW 0, indifferent (accept or reject
  project)

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• Decision Select all projects with non-negative
  EUAWs (if no budget constraint)

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Mutually Exclusive Projects

• Select the project (only one project is to be
  selected) with the largest non-negative EUAW
• If no project has a non-negative EUAW, select the
  status quo option
• invest available funds at the MARR

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Analysis Period Considerations

• 1. Analysis period is equal to alternative lives.
• 2. Analysis period is a common multiple of
  alternative lives.
• 3. Analysis period is for a continuing
  requirement.
• 4. Some other period such as project life may
  need to be considered.

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Analysis Period Equal to Alternative Lives

• Base the comparison on the life of the
  alternatives
• This is the case we have most often considered in
  our examples.
• This is rarely the case in real-life
  organizations.

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Finite Projects of Equal Duration
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• EUAW (A) -2500(A/P,10,5) - 900 1800
  200(A/F,10,5) 273
• EUAW (B) -3500(A/P,10,5) - 700 1900
  350(A/F,10,5) 335
• EUAW (C) -5000(A/P,10,5) - 1000 -
  100(A/G,10,5) 2100(P/C,i,k,5) (A/P,10,5)
  255

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Decision

• If projects A, B, and C are independent (no
  capital rationing), then select projects A, B, C
• If projects A, B, and C are mutually exclusive,
  then select project B (largest non-negative EUAW)

The same independent projects or mutually
exclusive project would be selected using the PW
and EUAW criteria
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Analysis Period is a Common Multiple of
Alternative Lives

• A common period of analysis may not be required
  to determine the validity of independent
  projects.
• The initial parameters of the project must remain
  intact throughout the period of analysis.
• If a project has a 6-year useful life, its EUAW
  for the 6-year period of analysis would be equal
  to its EUAW for a 12-year period if project
  parameters remained intact during the 12-year
  period.

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• When the lives of two alternatives vary, one can
  use a common multiple of the two lives to
  determine the better project

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Alternatives with Unequal Lives
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• EUAW (A) - 30,000(A/P,10,6) 12,000 25000
  2,000(A/F,10,6) 6,371
• EUAW (B) -40,000(A/P,10,10) - 10,000 21,000
  3,000(A/F,10,10) 6,600

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Decision

• Projects A and B are acceptable (valid) since
  EUAW 0
• If projects A and B are mutually exclusive,
  select project B since EUAWB6,600 gt EUAWA
  6,371

The PW method would lead to the same answer.
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Alternatives with Unequal Lives
30-year common period of analysis
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• From the repeatability method (period of analysis
  30 years),
• EUAW (A 30 yrs) 6,371
• EUAW (B 30 yrs) 6,600
• Above results are based on constant project
  parameters throughout the 30-year period of
  analysis

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• Its possible that one or more parameters of
  alternatives A and B would change during the
  period of analysis.
• Use repeatability method but account for specific
  changes in project parameters.

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Analysis Period for a Continuing Requirement

• Where the project will last forever (nothing
  does) use an infinite time period.
• In most analyses, organizations often use a
  representatively long time period to get a
  reasonable estimate.

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(A/P,i,N?8)

• (A/P,i,N) where N is finite
• i(1i)N / (1i)N - 1
  ... 1)
• Divide 1) by (1i)N
• i(1i)N / (1i)N
• -----------------------------------
  2)
• (1i)N/(1i)N 1/ (1i)N
• i (when n ? 8)
  3)

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Finite and Infinite Life Projects, and Infinite
Periods of Analysis
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Infinite Life Projects

• The wood structure has finite life (N 50 years)
  and will be repeated every fifty years
  (repeatability assumption).
• Since the projects have identical annual
  revenues, the decision maker wants to minimize
  cost.

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• EUAW (Wood Structure) 3,000,000(A/P,10,50)
  40,000 302,700 80,00 382,700
• EUAW (Cement Structure) 6,000,000(A/P,10,infin
  ity) 15,000 6,000,000(10) 15,000
  615,000
• Decision Select the wood structure

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Some Other Period Such AsProject Life

• Physical equipment usually has a useful life that
  is different from the project life.
• In this case, use the project life as the
  analysis period.
• This is the most common case in real-life
  organizations.

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Some Other Period Such AsProject Life

• EUACA 20,000 (25,000 - 2,000) (P/F,10,7) -
  500(P/F,10,10)(A/P,10,10)
• EUACB 30,000-5,000)(P/F,10,10) (A/P,10,10)

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Amortization Schedule

• An engineer borrows 8,000 at 10 repayable
  monthly over 10 months. Calculate
• a) the monthly repayment, and
• b) the composition of each repayment.

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Loan Amortization

• The engineer in the previous slide was the lucky
  winner of a 100,000 jackpot exactly five months
  after getting the 8,000 loan and decides to pay
  off the loan at that point (after making the 5th
  repayment). What is the value of the outstanding
  loan repayments?

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• PW A(P/A,i,N) 837.12(P/A,0.5,5)
  837.12(4.9259) 4,123.57
• The interest portion of the 4,123.57 repayment
  is 5(837.12) - 4123.57 52.03

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EUAW Summary Problem Valid Projects? Best
Project?
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EUAW

• EUAW(A N10) -P(A/P,i,N) A SV(A/F,i,N)
• -10,000(A/P,10,10) 2,000 0(A/F,10,10)
• 373
• EUAW(B N10) -P(A/P,i,N) A SV(A/F,i,N)
• -18,000(A/P,10,10) 3,850 - 1,000
  (A/F,10,10)
• 858

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• EUAW(C N 20) -P(A/P,i,N) A G(A/G,i,N)
  SV(A/F,i,N)
• -30,000(A/P,10,20) 2,800 400(A/G,10,20)
  3,000(A/F,10,20)
• 1,932

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EUAW and Mutually Exclusive Projects

• From the previous slide, EUAW (A) 373 and EUAW
  (B) 858
• Conclusion project B is better than project A

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• Projects B and C compare over 20 years
• EUAW (B N20) -P1 P(P/F,i,10)
  (A/P,10,20) A SV(P/F,i,10)
  (P/F,i,20)(A/F,10,20) -18,000
  1(P/F,10,10) 3,850 - 1,000
  (P/F,10,10)(P/F,10,20)(A/F,10,20) 858

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• EUAW(C N20) -P(A/P,i,20) A1 G(A/G,i,N)
  SV(A/F,i,N) -30,000 2,800
  400(A/G,10,20) 3,000 (A/F,10,20) 1,932
• Conclusion C is best project

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Alternative Projects

• If A, B, and C are mutually exclusive projects,
  you must select the BEST project (as long as its
  PW 0)
• The selection of the best alternative project
  must be based on a common period of analysis
  when a single sum method is used.
• In this example, (NA NB) ? NC

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Alternative Projects

• Since projects A, B, and C have different lives,
  a common period of analysis must be used to
  determine the best project when a single sum
  method (PW and FW) is applied.

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• Using the repeatability assumption, the least
  common denominator is 20 years
• Projects A and B must be repeated a second time
  to be compatible with the 20-year life of Project
  C.

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Alternative Projects
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• Since projects A and B have a common duration
  (life), we can determine the better project based
  on a 10-year period of analysis.
• PW(A) -P A(P/A,i,N) SV(P/F,i,N)
• -10,000 2,000(P/A,10,10) 0(P/F,10,10)
• 2,289

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• PW(B) -P A(P/A,i,N) SV(P/F,i,N)
• -18,000 3,850(P/A,10,10) -
  1,000(P/F,10,10)
• 5,271
• Conclusion Project B is better than A.

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Alternative Projects

• From a previous slide, NPWA 2,289 and NPWB
  5,271
• Conclusion project B is better than project B.

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• Projects B and C compared over 20 years
• PW(B) -P1 P(P/F,i,10) A(P/A,i,20)
  SV(P/F,i,10) (P/F,i,20) -18,000 1
  (P/F,10,10) 3,850(P/A,10,20) - 1,000
  (P/F,10,10) (P/F,10,20) 7,303

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• PW(C) -P A(P/A,i,N) G(P/G,i,N)
  SV(P/F,i,N) -30,000 2,800 (P/A,10,20)
  400(F/G,10,20) 3,000 (P/F,10,20) 16,447
• Conclusion Project C is best project.

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Conclusion NPW and EUAW

• The NPW and EUAW measures of merit provide the
  same decision as to the
• acceptability of projects
• better project (mutually exclusive)
• best project (competing alternatives)