ENGM 401 & 620 –X1 Fundamentals of Engineering Finance Fall 2010 Lecture 2 6 :

ENGM 401 & 620 –X1ENGM 401 & 620 –X1Fundamentals of Engineering FinanceFundamentals of Engineering Finance

Fall 2010Fall 2010

Lecture 2Lecture 266::

Other Analysis TechniquesOther Analysis TechniquesIf you work just for money, you'll never make it, but if you love what you're doing If you work just for money, you'll never make it, but if you love what you're doing

and you always put the customer first, success will be yours.and you always put the customer first, success will be yours.

- Ray - Ray KrocKroc

M.G. Lipsett

Department of Mechanical Engineering

University of Albertahttp://www.ualberta.ca/~mlipsett/ENGM401_620/ENGM401_620.htm

© MG Lipsett, 2010 2


Engineering Management Group

ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

An Interesting News Item

• On Nov. 7/09, The Edmonton Journal reported that Peter Pocklington owes $884 daily on a $13,000,000 provincial loan. What interest rate is the government charging?

(this works out to 0.00680% as the daily interest rate)

a) 2.48%

b) 2.50%

c) 2.51%

d) 2.68%

e) 6.80%





Choosing a Rate of Return i (Review)Choosing a Rate of Return i (Review)

• Many broad factors affect choice of rate of return i • For many projects or investments, companies identify a

“hurdle rate” such as– minimum acceptable rate of returnminimum acceptable rate of return (MARR), which is the lowest

return the company is willing to earn on an investment– weighted average cost of capitalweighted average cost of capital (WACC), which is the cost of the

company’s mix of financing (so an investment that doesn’t give at least this rate of return for no effective risk is not worthwhile for the company)

• Internal Rate of ReturnInternal Rate of Return (IRR) is the effective interest rate at which the NPV of an investment’s cash flows (including both benefits and costs) is zero

• Spreadsheets have software tools such as “goal seek” or “Solver” to solve for the IRR (interest rate at which NPV = 0)





Incremental Investment Analysis (Review)Incremental Investment Analysis (Review)

• We have used NPV to evaluate an investment opportunity compared against an hurdle rate for interest (MARR, WACC, etc.)

• IRR tells us when the investment benefits match the costs• IIRR (also referred to as ΔIRR) tells us whether one

alternative has a better incremental return than another





Other Investment Evaluation Techniques:

• Benefit-Cost Ratio Analysis • Incremental Benefit-Cost Ratio Analysis• Payback Analysis• Sensitivity Analysis• Breakeven Analysis





cos

1benefits

ts

PWBCR

PW

Benefit-Cost Ratio AnalysisBenefit-Cost Ratio Analysis

• Benefit-Cost Ratio AnalysisBenefit-Cost Ratio Analysis compares the value gained from a project to the cost of the project– We expect the benefits to outweigh the costs (i.e., the total value of the

benefits should be greater than the total value of the costs)

• The idea is that a project with a large benefit-cost ratiobenefit-cost ratio (BCR) will be a good project (but bigger isn’t always better)

• Often, BCR is given as ratio of equivalent uniform annual benefits (EUAB) versus equivalent uniform annual cost (EUAC).

• Benefit-cost ratio analysis is frequently used for analyzing public sector projects– and can include costs or benefits that may not necessarily be considered if it

was a private investment• Can consider lots of other factors, including qualitative benefits/costs,

etc.• For example, could have environmental impacts, far-reaching economic

effects, etc.

cosbenefits tsPW PW





Incremental Benefit-Cost Ratio ExampleIncremental Benefit-Cost Ratio Example

• Incremental BCR is used to select between options.• You can select between the Baseline model or the Gold

Standard model for a new piece of equipment your company needs. The equivalent uniform annual costs and benefits of each are:– Baseline model: EUAC = $50k, EUAB = $75k– Gold Standard model: EUAC = $125k, EUAB = $140k

• Which model should you choose?








BCRbase = 75 / 50 = 1.25 ;BCRgold_standard = 140 / 125 = 1.12

• Which model should you choose?• Since the BCR of each is greater than one, we do an

incremental BCR








BCRbase = 75 / 50 = 1.25 ;BCRgold_standard = 140 / 125 = 1.12

• Which model should you choose?– Since the BCR of each is greater than one, then each is individually

acceptable, so calculate the incremental BCR (using cheaper cost option as the basis for comparison:

ΔBCR = 140 – 75 < 1 125 - 50

Since the ∆BCR is less than one, it means that the incremental BCR of choosing the more expensive option over the less expensive option is below the threshold to make it a good option.





Benefit-Cost Ratio In-Class Problem #1Benefit-Cost Ratio In-Class Problem #1

• You can consider two types of equipment for your company, each with different costs, net annual benefits, salvage values, and useful lives:

#1 #2

Purchase cost $ 200k $ 700k

Annual benefit $ 95k $ 120k

Salvage value $ 50k $ 150k

Useful life 6 years 12 years

• Assuming a 10% interest rate, what are their benefit-cost ratios?

• Which would you choose? Why?





Benefit-Cost Ratio In-Class Problem #1 (2)Benefit-Cost Ratio In-Class Problem #1 (2)

• Use equivalent annual costs and benefits to calculate BCRs

Cash Flow Diagram (Benefits and Costs) Option 1

0 1 2 3 4 5 6 7 8 9 10 11 12

Cash Flow Diagram (Benefits and Costs)Option 2

0 1 2 3 4 5 6 7 8 9 10 11 12

-200 -700

95

-55=-200+50+95

145=50+95120

270=120+150








0 1 2 3 4 5 6 7 8 9 10 11 12


0 1 2 3 4 5 6 7 8 9 10 11 12

#1 #2

-200 -700

95

-55=-200+50+95

145=50+95120

270=120+150

EUAC2 = 700(A|P,10%,12)-150(A|F,10%,12) = 96EUAB2 = 120 (given)EUAB2 = 120 = 1.25EUAC2 96









0 1 2 3 4 5 6 7 8 9 10 11 12


0 1 2 3 4 5 6 7 8 9 10 11 12

#1 #2

-200 -700

95

-55=-200+50+95

145=50+95120

270=120+150



ΔEUAB = EUAB2 - EUAB1 = 25 = 0.45 < 1ΔEUAC EUAC2 - EUAC1 56

Therefore choose lower cost option #1





Benefit-Cost Ratio Analysis, Multiple OptionsBenefit-Cost Ratio Analysis, Multiple Options

• When comparing more than one alternative project, we don’t necessarily want to choose the one with the largest BCR– For reasons similar to those for not automatically selecting the

project with the largest IRR

• The method is to calculate incremental costs (∆C) and incremental benefits (∆B) for each progressively more expensive pair, which will allow us to calculate incremental benefit-cost ratios (∆BCR) for those pairs– If ∆BCR ≥ 1, then the more expensive project is worthwhile– This is because the incremental benefit is worth more than the

incremental cost above the less expensive option





Benefit-Cost Ratio In-Class Problem #2Benefit-Cost Ratio In-Class Problem #2

• Consider that you can choose one of six alternatives, each with the same useful life and with no salvage value:

A B C D E F

PWcost $4000 $2000 $6000 $1000 $9000 $10000

PWbenefit $7330 $4700 $8730 $1340 $9000 $9500

BCR 1.83 2.35 1.46 1.34 1.00 0.95

• Question:Question: Which alternative should you select?

– Note: We don’t need to know i in this case because we’re given the PW of all of our cash flows. But if all we had were the actual cash flows, we’d need to calculate the PWcost and PWbenefit ourselves, so we’d need to know i





Benefit-Cost Ratio In-Class Problem #2 (2)Benefit-Cost Ratio In-Class Problem #2 (2)Given:

A B C D E F PWcost $4,000 $2,000 $6,000 $1,000 $9,000 $10,000PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500BCR 1.83 2.35 1.46 1.34 1 0.95

First arrange in ascending order of costs, after discarding Option F (which has BCR < 1):

D B A C EPWcost $1,000 $2,000 $4,000 $6,000 $9,000PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000BCR 1.34 2.35 1.83 1.46 1

Then calculate incremental BCR, dicarding any undesireable BCRs as analysis progresses:B - D A - B C - A E - A

delta cost $1,000 $2,000 $2,000 $5,000delta benefit $3,360 $2,630 $1,400 $1,670delta BCR 3.36 1.315 0.7 0.334

reject C so do E - Areject E

choose option with highest of the acceptable incremental BCRs (in this case Option A)

Note Option A didn't have the best BCR !





Payback PeriodPayback Period• Payback period is the amount of time that elapses before

the net benefits of an investment equal its cost– For projects with high uncertainty, payback period becomes very

important (why?) faster recovery of initial costs reduces risk

• Two main types of payback period analysis– simple: using FV series (no discounting)– discounted, using PV series (with MARR, etc.),

often called the breakeven point• There are other versions of payback:

– can consider depreciation, inflation, taxes, etc.

• In projects, payback period is usually calculated from the time of start-up, not the time of the beginning of the project doesn’t consider timing of cash flows before or after the payback period





Payback Period ExamplePayback Period Example

Year FV PV NPV PV NPV

0 -10,000$ -10,000$ -10,000$ -10,000$ -10,000$ 1 2,000$ 2,000$ -8,000$ 1,739$ -8,261$ 2 2,500$ 2,500$ -5,500$ 1,890$ -6,371$ 3 3,000$ 3,000$ -2,500$ 1,973$ -4,398$ 4 3,500$ 3,500$ 1,000$ 2,001$ -2,397$ 5 4,000$ 4,000$ 5,000$ 1,989$ -408$ 6 4,000$ 4,000$ 9,000$ 1,729$ 1,321$ 7 4,000$ 4,000$ 13,000$ 1,504$ 2,825$ 8 4,000$ 4,000$ 17,000$ 1,308$ 4,133$ 9 4,000$ 4,000$ 21,000$ 1,137$ 5,270$

10 4,000$ 4,000$ 25,000$ 989$ 6,258$

Simple Discounted (15%)

Payback of 6 years(actually 5.24 years)In this case, there is value created in the first year,

so we start the clock in year 1.

In simple payback, we use future values directly (not discounted values)







0 -10,000$ -10,000$ -10,000$ -10,000$ -10,000$ 1 2,000$ 2,000$ -8,000$ 1,739$ -8,261$ 2 2,500$ 2,500$ -5,500$ 1,890$ -6,371$ 3 3,000$ 3,000$ -2,500$ 1,973$ -4,398$ 4 3,500$ 3,500$ 1,000$ 2,001$ -2,397$ 5 4,000$ 4,000$ 5,000$ 1,989$ -408$ 6 4,000$ 4,000$ 9,000$ 1,729$ 1,321$ 7 4,000$ 4,000$ 13,000$ 1,504$ 2,825$ 8 4,000$ 4,000$ 17,000$ 1,308$ 4,133$ 9 4,000$ 4,000$ 21,000$ 1,137$ 5,270$

10 4,000$ 4,000$ 25,000$ 989$ 6,258$


Payback of 4 years(actually 3.71 years)








0 -10,000$ -10,000$ -10,000$ -10,000$ -10,000$ 1 2,000$ 2,000$ -8,000$ 1,739$ -8,261$ 2 2,500$ 2,500$ -5,500$ 1,890$ -6,371$ 3 3,000$ 3,000$ -2,500$ 1,973$ -4,398$ 4 3,500$ 3,500$ 1,000$ 2,001$ -2,397$ 5 4,000$ 4,000$ 5,000$ 1,989$ -408$ 6 4,000$ 4,000$ 9,000$ 1,729$ 1,321$ 7 4,000$ 4,000$ 13,000$ 1,504$ 2,825$ 8 4,000$ 4,000$ 17,000$ 1,308$ 4,133$ 9 4,000$ 4,000$ 21,000$ 1,137$ 5,270$

10 4,000$ 4,000$ 25,000$ 989$ 6,258$



In discounted payback, we use present values







0 -10,000$ -10,000$ -10,000$ -10,000$ -10,000$ 1 2,000$ 2,000$ -8,000$ 1,739$ -8,261$ 2 2,500$ 2,500$ -5,500$ 1,890$ -6,371$ 3 3,000$ 3,000$ -2,500$ 1,973$ -4,398$ 4 3,500$ 3,500$ 1,000$ 2,001$ -2,397$ 5 4,000$ 4,000$ 5,000$ 1,989$ -408$ 6 4,000$ 4,000$ 9,000$ 1,729$ 1,321$ 7 4,000$ 4,000$ 13,000$ 1,504$ 2,825$ 8 4,000$ 4,000$ 17,000$ 1,308$ 4,133$ 9 4,000$ 4,000$ 21,000$ 1,137$ 5,270$

10 4,000$ 4,000$ 25,000$ 989$ 6,258$








Sensitivity and Break-Even AnalysisSensitivity and Break-Even Analysis

• Sensitivity and break-even analysisSensitivity and break-even analysis determines which value of a particular parameter will result in a break-even scenario (and to which parameters a decision is sensitive)

• At break-even, costs equal revenues, NPV equals zero, two options are equivalent, etc.

At this point the decision can go either way• Example uses:

– What cost do we set for a particular project so that it is equivalent to another project?

– What timing should be used to build a multi-phase project?– How will the useful life of a piece of equipment impact a decision?

• Sensitivity concerns how much a parameter can change before it would affect a decision





Sensitivity and Break-Even Analysis ExampleSensitivity and Break-Even Analysis Example

• Your company needs to build a new plant.– Option A is to build all at once:

• The plant has the capacity you will need years from now, at a cost of $140k.

– Option B is to build in two phases:• Phase 1 provides the capacity you need for the first few years at a cost

of $100k.

• Phase 2 provides the remaining additional capacity at a cost of $120k.

– Both options have the same total useful lifetime, the same operation and maintenance costs, and no salvage value.

• With a WACC of 8%, at what time will the cost of both options be equivalent?– What does this mean?– At the time of equivalence the decision could be made for

either option





Sensitivity and Break-Even Analysis Example (2)Sensitivity and Break-Even Analysis Example (2)

$100,000

$120,000

$140,000

$160,000

$180,000

$200,000

$220,000

0 5 10 15 20 25 30

Year when Option B's Phase 2 constructed

NP

V o

f C

os

ts

both options are equivalent here (between 14 & 15 years out)

Option BOption B

Option AOption A

The decision of which option to use is only sensitive to the timing if the range of estimates is in the area of 15 years.

WACC = 8%

This plot shows the effect on net present cost of option B phase 2 ($120k using discounted dollars), thus cheaper as phase 2 gets delayed






$100,000

$120,000

$140,000

$160,000

$180,000

$200,000

$220,000

0 5 10 15 20 25 30


NP

V o

f C

os

ts

with i = 10%, options are

equivalent here (11.4 years)

Option BOption B

Option AOption A

The decision will also depend on our WACC, or the interest rate we use to calculate our discounted cash flows.

If money is more costly, then Option B becomes preferable at an earlier point (discounted faster).






$100,000

$120,000

$140,000

$160,000

$180,000

$200,000

$220,000

0 5 10 15 20 25 30


NP

V o

f C

os

ts

with i = 6%, options are

equivalent here (19 years)

Option BOption B

Option AOption A

If money is les costly, then Option B becomes preferable at an later point (not discounting as much).





Break-Even – In-Class Problem #1Break-Even – In-Class Problem #1• Choosing between two options:

– Option A has a cost known to be $5000, with an net annual benefit of $700

– Option B has an unknown cost, but it will provide an net annual benefit of $639

– Both options have a 20-year useful life with no salvage value

• With a WACC of 6%, which option should we choose?










Solve by writing expressions that are equivalent for the two optionsAnd then solve for the unknown (unknown cost of B):Option A Option BNPVA = PWb – PWc NPVB = PWb –PWc

Solve for the cost of B with the same NPV as for Option A










Solve by writing expressions that are equivalent for the two optionsAnd then solve for the unknown (unknown cost of B):Option A Option BNPVA = PWb – PWc NPVB = PWb –PWc = 700(P|A,6%,20) – 5000 = 639(P|A,6%,20) – PWc = 700(11.470) – 5000 = 7329 – PWc = $3029Solve for the cost of B with the same NPV as for Option A

NPVA = NPVB 3029 = 7329 – PWc => PWc = $4300 Therefore, choose Option B if present cost < $4300





Break-Even – In-Class Problem #2Break-Even – In-Class Problem #2• You need to replace a component in a piece of equipment used in an

environment that is highly susceptible to corrosion. An ordinary part has a cost of $350, a useful life of only 6 years, and no salvage value. How long a useful life must a more expensive ($500) corrosion resistant part have if it is preferred over the ordinary part? Assume WACC= 10%







Let’s find the breakeven life for the corrosion resistant part, using equivalent uniform annual costs.

The annual cost of the untreated part:

$350 × (A/P, 10%, 6) = $350 × (0.2296) = $80.36

The annual cost of the treated part must be at least this low, for breakeven









$350 × (A/P, 10%, 6) = $350 × (0.2296) = $80.36

The annual cost of the treated part must be at least this low so, for breakeven, we use the annual cost of the other option to solve for the time period:

$80.36 = $500 × (A/P, 10%, n)

(A/P, 10%, n) = $80.36/$500 = 0.1607

If we look this value up in the Capital Recovery Factor table, we don’t see this exact value, but we can interpolate within the tables

and solve for n









$350 × (A/P, 10%, 6) = $350 × (0.2296) = $80.36

The annual cost of the treated part must be at least this low so, for breakeven, we use the annual cost of the other option to solve for the time period:

$80.36 = $500 × (A/P, 10%, n)

(A/P, 10%, n) = $80.36/$500 = 0.1607

If we look this value up in the Capital Recovery Factor table, we don’t see this exact value, but we can interpolate within the tables

and solve for n to be just over 10 years.:

0.1627 0.1607

10 10.23 0.1627 0.1540

n years

A/P, 10%, 10

A/P, 10%, 10 A/P, 10%, 11

A/P, 10%, n

ENGM 401 & 620 –X1 Fundamentals of Engineering Finance Fall 2010 Lecture 2 6 :

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