MIE 353 Engineering Economics Todays Goals

1
MIE 353 Engineering EconomicsTodays Goals

• Review for Final
• Breakeven points
• Uncertainty Analysis
• Sample Final Exam
• Presentation Monday Dec. 10
• Hitchcock Center Presentation Thursday Dec. 13,
  530
• Final Tuesday Dec. 18 4pm Elab 305
• Report due Fri. Dec. 14

2
Homework EVPI
0
Without information
858
CD
p.5
.5
3
stock
real estate
522
.3
292
.5
1030
.2
3
Homework EVPI
0
Without information
858
CD
p.5
430
.5
3
stock
real estate
522
.3
292
.5
509
1030
.2
4
Homework EVPI
0
Without information
858
CD
p.5
Choose Real Estate for an expected value of 509
430
.5
3
stock
real estate
3444
.3
292
.5
509
1030
.2
5
Homework EVPI
With information learn then act
CD
0
stock
858
p.5
509
.5
CD
0
stock
3
509
6
Homework EVPI
With information learn then act
If the stock goes up, choose stock. If the stock
goes down, choose real estate.
CD
0
stock
858
p.5
509
.5
CD
0
stock
3
509
7
Homework EVPI
With information learn then act
If the stock goes up, choose stock. If the stock
goes down, choose real estate. EV with
information is 683
CD
0
stock
858
p.5
509
.5
CD
0
683
stock
3
509
8
Homework EVPI
With information learn then act
If the stock goes up, choose stock. If the stock
goes down, choose real estate. EV with
information is 683
CD
0
stock
858
p.5
509
.5
CD
0
683
EVPI 683 509 174
stock
3
509
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ENGINEERING ECONOMIC ANALYSIS PROCEDURE

• Recognize Opportunity (Problem definition).
• Selection of a criterion ( or criteria).
• 3. Development of alternatives.
• 4. Development of the prospective outcomes
  (typically cash flows).
• 5. Analysis and comparison of the alternatives.
• 6. Selection of the preferred alternative.
• 7. Performance monitoring and post-evaluation
  results.

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Decision Making Steps

• Define Values
• What you will use to evaluate the alternatives
• Generate Alternatives
• Gather information and develop cash flows
• Evaluate and compare cash flows (and other
  non-monetized items)
• Perform sensitivity analysis

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Calculate PW of an arbitrary cash flow

• Find the interest rate per cash flow.
• Discount each cash flow to the present and add
  them up.
• ak (P/F,i,k) ak/(1i)k
• ak cash flow at period k
• i interest rate per cash flow period
• k period
• PW

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PW arbitrary

• You can always just find the PW of each cash flow
  and add them.
• As weve seen, once you consider taxes, you often
  lose nice annuity properties.
• Example A firm has MARR 10 and the following
  annual cash flows, starting at time 0
• (-100 5 10 12 13 13 40)
• PW -100 5/1.1 10/1.12 12/1.1313/1.14
  13/1.15 40/1.16

13
Formulas

• We use the formulas for P/A and A/P when we
  happen to have a nice annuity. They are short
  cuts. You can always find the PW as shown on the
  previous page.
• We also use the formulas when we want to find an
  annuity.

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Example

• A firm would like to know the annual worth of the
  following cash flow over its useful life. Their
  MARR is 10.
• (-100 10,10,50,50,50)
• Find PW -100 10/1.1 10/1.12
  50/1.1350/1.14 50/1.15 20.12
• AW (A/P,.1,5) 20.12.2638 5.31

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Groupwork example

• A person needs 18,000 immediately. They get a
  loan, and are required to repay the loan in equal
  payments every six months over the next 12 years.
  The interest is 10 nominal, compounded monthly.
  What is amount of each payment?
• If she wants to pay it back over 10 years, what
  would her semi annual payment be?
• If she wants to pay 6000 every six months, how
  long will it take her to pay it off?
• If she wants to pay 1000 a month, how long will
  it take to pay off?

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Solutions

• Compounding monthly payment period 6-months.
  Calc interest i (1.1/12)6 -1 .051 N 24 6
  month periods
• Semi-annual payment is A 18(A/P, i ,24)
• Over 10 years A 18(A/P, i ,20)
• Solve for N, plugging in i above
• 6(P/A, i, N) 18

N 3.33 half years, about 20 months.
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Solutions

• Solve for N,
• use i .1/12
• 1(P/A, i, N) 18

N 19.6 months
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Comparing Projects

• Use an equivalent worth method.
• You cannot compare projects by simply comparing
  the IRR. If you do this, you are implicitly
  assuming that any excess cash is being
  re-invested at the IRR, rather than the MARR. In
  order to use IRR, you must do an incremental
  comparison.
• Choose project whose PW is highest.
• equivalently, choose project whose AW or FW is
  highest.

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Comparing Projects with different useful lives

• If projects have different useful lives, you must
  make an assumption. Make one assumption or the
  other they are different and will give you
  different answers.
• Assumption 1 Repeatability
• Assumption 2 Co-termination

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Assumptions

• Repeatability assumes that the alternatives can
  be repeated an indefinite number of times.
• This is a convenient assumption
• This is often a very reasonable assumption.
• Equipment that is used for the basic line of
  business
• Projects that are very typical for the firm

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Assumptions

• Repeatability 2 equivalent methods.
• Method 1 Find the AW over the respective useful
  lives. This is often the easiest method.
• Method 2 Repeat each project until they end at
  the same time and calculate PW.

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Example

• Don is deciding between buying a new car and
  selling it after 10 years, or buying a used car
  and selling it after 7 years. The new car cost
  20,000, can be sold for 2000 after 10 years and
  is estimated to have annual running costs of
  5000. The used car will cost 7000, can be sold
  for 2500 after 7 years, and is estimated to have
  annual running costs of 7000. Dons Marr is 5.
  Which car should he buy?

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Example

• Assume repeatability.
• AW1 20,000(A/P,.05,10) 5000
  2000(A/F,.05,10)
• 20,000.1295 5000 2000.0795 7431
• AW2 7000(A/P, .05, 7) 7000 2500(A/F,.05,7)
  
• 7000.1728 7000 -2500.1228 7902
• Choose Car A, lower AW over useful life.
• What is the breakeven running cost for car 2?

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Example

• Assume repeatability.
• AW1 20,000(A/P,.05,10) 5000
  2000(A/F,.05,10)
• 20,000.1295 5000 2000.0795 7431
• AW2 7000(A/P, .05, 7) 7000 2500(A/F,.05,7)
  
• 7000.1728 7000 -2500.1228 7902
• Choose Car A, lower AW over useful life.
• What is the breakeven running cost for car 2?
• 7000-(7902-7431) 7000-4716529

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Assumptions

• Co-termination the projects cannot be repeated.
  We need to make appropriate assumptions in order
  that both projects end at the same time.
• Specialized equipment that can only be used for a
  specific project.
• Specialized projects that are atypical.
• A known end date, i.e. graduation from college a
  tax benefit that is ending etc.

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Cotermination AssumptionUseful life lt Study
Period

• Cost alternatives -- each cost alternative must
  provide same level of service as study period
• contract for service or lease equipment for
  remaining time
• repeat part of useful life of original
  alternative until study period ends
• investment alternatives -- assume all cash flows
  reinvested in other opportunities at MARR to end
  of study period. This is just equivalent to
  assuming the project ends at its useful life.

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Cotermination AssumptionUseful life gt Study
Period

• Truncate the alternative at the end of the study
  period using an estimated market value. This
  method assumes disposable assets will be sold at
  the end of the study period at that value.

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Sensitivity Analysis Breakeven points

• If you are considering one project, the breakeven
  point is where the PW0.
• If you are comparing projects, the breakeven
  point is where the PWs of the two projects are
  equal.

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Sensitivity Analysis Breakeven points

• If you are considering one project, the breakeven
  point is where the PW0.
• If you are comparing projects, the breakeven
  point is where the PWs of the two projects are
  equal.
• Simply find an expression for the PW(s), set them
  equal and solve.

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Example

• A firm is choosing a generator. They want to
  minimize the total cost over 10 years. They use
  about 100 KwH per year. MARR 10.
• Find which generator is best
• What is breakeven for KwH
• What is breakeven for initial cost of Generator
  1?
• What is breakeven for salvage cost of generator 1?

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Calc PW

• PW I c KwH (P/A,i,10) SV(P/F,i,10)
• PW110002100(6.1446) 120(.3855)
• 2275
• PW220001100(6.1446) 150(.3855)
• 2672

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Calc PW

• PW I c KwH (P/A,i,10) SV(P/F,i,10)
• PW110002100(6.1446) 120(.3855)
• 2275
• PW220001100(6.1446) 150(.3855)
• 2672

Choose Generator 1 lower cost.
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Breakeven KwH

• PW I c KwH (P/A,i,10) SV(P/F,i,10)
• PW110002X(6.1446) 120(.3855)
• 104612.28X
• PW220001X(6.1446) 150(.3855)
• 20586.14X
• 104612X 20586X
• 6X1012
• X169
• If KwH are greater than 169, which alt is better?

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Breakeven for initial cost of Gen 1

• PW I c KwH (P/A,i,10) SV(P/F,i,10)
• PW1X2100(6.1446) 120(.3855)
• 1275X
• PW220001100(6.1446) 150(.3855)
• 2672
• 1275X 2672
• X1397
• What does this tell us?

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Uncertainty Analysis

• If uncertainty is important, then choose the
  alternative with the highest expected PW
• Example say a parameter can be high with
  probability .22 or low with probability .78.
  Assume we want to maximize PW.

P.22
Value if high 27,000
1-P.78
Value if low 12,000
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Uncertainty Analysis

• If uncertainty is important, then choose the
  alternative with the highest expected PW
• Example say a parameter can be high with
  probability .22 or low with probability .78

P.22
Expected value of alternative 15,300
Value if high 27,000
1-P.78
Value if low 12,000
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Uncertainty Analysis

• Example say a parameter can be high with
  probability .22 or low with probability .78
• Alternative 2 has a certain payoff of 20,000
• Choose Alternative 2, since the expected payoff
  is higher

P.22
Expected value of alternative 15,300
Value if high 27,000
1-P.78
Value if low 12,000
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Uncertainty Analysis EVPI

• Consider the problem if you learn then act
• Alternative 2 has a certain payoff of 20,000
• Choose Alternative 2, since the expected payoff
  is higher
• The value of the decision problem is 20,000

P.22
Expected value of alternative 15,300
Value if high 27,000
1-P.78
Value if low 12,000
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EVPI Learn then act

• EVPI 21,540 20,000 540

Alt 1
27,000
P.22
20,000
Alt 2
Alt 1
12,000
1-P.78
20,000
Alt 2
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EVPI

• EVPI 21,540 20,000 540

Alt 1
27,000
27,000
P.22
20,000
21,540
Alt 2
Alt 1
12,000
1-P.78
20,000
20,000
Alt 2
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Final

• Decision analysis and EVPI (ch18.4-5)
• Taxes (ch16-17)
• Interest calculations (ch 4)
• Cost estimations (ch15)
• Choosing between alternatives with different
  useful lives (ch 5.3 6)
• Breakeven analysis (ch13)
• IRR (ch 7-8)
• Decision Making Steps