ENGG 401 X2ENGG 401 X2Fundamentals of Engineering ManagementFundamentals of Engineering Management
Spring 2008Spring 2008
Chapter 7:Chapter 7:
Other Analysis TechniquesOther Analysis Techniques
Dave LudwickDave Ludwick
Dept. of Mechanical EngineeringDept. of Mechanical Engineering
University of AlbertaUniversity of Albertahttp://members.shaw.ca/dave_ludwick/
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20082
ENGG 401 X2 – Fundamentals of Engineering Management
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).
• Benefit-cost ratio analysis is frequently used for analyzing pubic sector projects.– and can include costs or benefits that may not necessarily be
considered if it was a private investment
cosbenefits tsPW PW cos
1benefits
ts
PWBCR
PW
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20083
ENGG 401 X2 – Fundamentals of Engineering Management
Benefit-Cost Ratio ExampleBenefit-Cost Ratio Example
• You can select between the Baseline model or Gold Standard model for a new piece of equipment your company needs. The equivalent uniform annual costs and benefits of each are as follows:– Baseline model: EUAC = $50k, EUAB = $75k– Gold Standard model: EUAC = $125k, EUAB = $140k
• Which model should you choose?– Since the BCR of each is greater than one, then each is individually
acceptable, so calculate the incremental BCR:
– Since ∆BCR < 1, the extra benefits we get from the Gold Standard model isn’t worth the extra costs. Choose the Baseline model.Choose the Baseline model.
$140 $75 $651
$125 $50 $75
BBCR
C
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20084
ENGG 401 X2 – Fundamentals of Engineering Management
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?
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20085
ENGG 401 X2 – Fundamentals of Engineering Management
Benefit-Cost Ratio Analysis with Multiple OptionsBenefit-Cost Ratio Analysis with 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, since
the incremental benefit is worth more than the incremental cost.
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20086
ENGG 401 X2 – Fundamentals of Engineering Management
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 $9500
BCR 1.83 2.35 1.46 1.34 1.00 0.95
• Question:Question: Which alternative should you select?
– Note: We don’t need to know i in this case because we’re given the PW of all of our cash flows. But if all we had were the actual cash flows, we’d need to calculate the PWcost and PWbenefit ourselves, so we’d need to know i.
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20087
ENGG 401 X2 – Fundamentals of Engineering Management
Payback PeriodPayback Period
• Payback period is the amount of time that elapses before the net benefits of an investment equal its cost.– For projects with high uncertainty, payback period becomes very
important.
• Two main types of payback period analysis:– Discounted payback
• Uses the specified discount rate, MARR, interest rate, etc.
– Simple payback• Doesn’t use any discount rate at all (i.e., i = 0, or FV = PV)
• There are other versions payback:– can consider depreciation, inflation, taxes, etc.
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20088
ENGG 401 X2 – Fundamentals of Engineering Management
Payback Period ExamplePayback Period Example
Year FV PV NPV PV NPV
0 -10,000$ -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 4 years(actually 3.71 years)
Payback of 6 years(actually 5.24 years)
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 20089
ENGG 401 X2 – Fundamentals of Engineering Management
Payback Period – In-Class ProblemPayback Period – In-Class Problem
• What is the payback period for the following two investment options?
Year Cash Flow Year Cash Flow
1 -212,000$ 12 108,000$ 2 -461,000$ 13 110,000$ 3 90,000$ 14 112,000$ 4 92,000$ 15 114,000$ 5 94,000$ 16 116,000$ 6 96,000$ 17 119,000$ 7 97,000$ 18 121,000$ 8 99,000$ 19 124,000$ 9 101,000$ 20 126,000$
10 103,000$ 21 129,000$ 11 105,000$ 22 131,000$
Early OptionYear Cash Flow Year Cash Flow
1 -212,000$ 12 228,000$ 2 -461,000$ 13 232,000$ 3 12,000$ 14 237,000$ 4 15,000$ 15 242,000$ 5 20,000$ 16 247,000$ 6 30,000$ 17 252,000$ 7 50,000$ 18 257,000$ 8 70,000$ 19 262,000$ 9 100,000$ 20 267,000$
10 120,000$ 21 272,000$ 11 150,000$ 22 278,000$
Late Option
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200810
ENGG 401 X2 – Fundamentals of Engineering Management
Sensitivity and Break-Even AnalysisSensitivity and Break-Even Analysis
• Sensitivity and break-even analysisSensitivity and break-even analysis is used to determine which value of a particular parameter will result in a break-even scenario (and which parameters a decision is sensitive to).– At break-even, costs will equal revenues, NPV equals zero, two
options are equivalent, etc.
• 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 projects?– How will the useful life of a piece of equipment impact a decision?
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200811
ENGG 401 X2 – Fundamentals of Engineering Management
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?
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200812
ENGG 401 X2 – Fundamentals of Engineering Management
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 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.
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200813
ENGG 401 X2 – Fundamentals of Engineering Management
Sensitivity and Break-Even Analysis Example (3)Sensitivity and Break-Even Analysis Example (3)
$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
with i = 10%, options are
equivalent here (13 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.
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200814
ENGG 401 X2 – Fundamentals of Engineering Management
Sensitivity and Break-Even Analysis Example (4)Sensitivity and Break-Even Analysis Example (4)
$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
with i = 6%, options are
equivalent here (19 years)
Option BOption B
Option AOption A
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200815
ENGG 401 X2 – Fundamentals of Engineering Management
Break-Even – In-Class Problem #1Break-Even – In-Class Problem #1
• Suppose you can choose between two options. – Option A has a cost that is known with some precision to be $5000,
and will provide an net annual benefit of $700.– Option B has an unknown cost, but we know 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?
Dave Ludwick, Dept. of Mech. Eng.Ch 7 – Other Analysis Techniques
Summer 200816
ENGG 401 X2 – Fundamentals of Engineering Management
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 highly conducive 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 a WACC of 10%.
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