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ENGM 401 & 620 –X1 Fundamentals of Engineering Finance Fall 2010 Lecture 26: Other Analysis Techniques 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. - Ray Kroc M.G. Lipsett Department of Mechanical Engineering University of Alberta http://www.ualberta.ca/~mlipsett/ENGM401_620/ENGM401_620.htm

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© MG Lipsett, 2010 2 Department of Mechanical Engineering 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%

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© MG Lipsett, 2010 3 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques 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 return –minimum acceptable rate of return (MARR), which is the lowest return the company is willing to earn on an investment –weighted average cost of capital –weighted 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)

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© MG Lipsett, 2010 4 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques 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

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© MG Lipsett, 2010 5 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Other Investment Evaluation Techniques: Benefit-Cost Ratio Analysis Incremental Benefit-Cost Ratio Analysis Payback Analysis Sensitivity Analysis Breakeven Analysis

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© MG Lipsett, 2010 6 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-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) benefit-cost ratioThe idea is that a project with a large benefit-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.

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© MG Lipsett, 2010 7 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Incremental 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?

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© MG Lipsett, 2010 8 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Incremental 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 BCR base = 75 / 50 = 1.25 ; BCR gold_standard = 140 / 125 = 1.12 Which model should you choose? Since the BCR of each is greater than one, we do an incremental BCR

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© MG Lipsett, 2010 9 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Incremental 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 BCR base = 75 / 50 = 1.25 ; BCR gold_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.

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© MG Lipsett, 2010 10 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-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 years12 years Assuming a 10% interest rate, what are their benefit-cost ratios? Which would you choose? Why?

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© MG Lipsett, 2010 11 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #1 (2) Use equivalent annual costs and benefits to calculate BCRs -200 -700 95 -55=-200+50+95 145=50+95 120 270=120+150

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© MG Lipsett, 2010 12 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #1 (2) Use equivalent annual costs and benefits to calculate BCRs #1#2 -200 -700 95 -55=-200+50+95 145=50+95 120 270=120+150 EUAC 2 = 700(A|P,10%,12)-150(A|F,10%,12) = 96 EUAB 2 = 120 (given) EUAB 2 = 120 = 1.25 EUAC 2 96 EUAC 1 = 200(A|P,10%,6)-50(A|F,10%,6) = 40 EUAB 1 = 95 (given) EUAB 1 = 95 = 2.41 EUAC 1 40

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© MG Lipsett, 2010 13 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #1 (2) Use equivalent annual costs and benefits to calculate BCRs #1#2 -200 -700 95 -55=-200+50+95 145=50+95 120 270=120+150 EUAC 2 = 700(A|P,10%,12)-150(A|F,10%,12) = 96 EUAB 2 = 120 (given) EUAB 2 = 120 = 1.25 EUAC 2 96 EUAC 1 = 200(A|P,10%,6)-50(A|F,10%,6) = 40 EUAB 1 = 95 (given) EUAB 1 = 95 = 2.41 EUAC 1 40 ΔEUAB = EUAB 2 - EUAB 1 = 25 = 0.45 < 1 ΔEUAC EUAC 2 - EUAC 1 56 Therefore choose lower cost option #1

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© MG Lipsett, 2010 14 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-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

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© MG Lipsett, 2010 15 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-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 PW cost $4000$2000$6000$1000$9000$10000 PW benefit $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 PW cost and PW benefit ourselves, so we’d need to know i

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© MG Lipsett, 2010 16 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #2 (2)

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© MG Lipsett, 2010 17 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #2 (2)

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© MG Lipsett, 2010 18 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #2 (2)

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© MG Lipsett, 2010 19 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #2 (2)

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© MG Lipsett, 2010 20 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #2 (2)

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© MG Lipsett, 2010 21 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Benefit-Cost Ratio In-Class Problem #2 (2)

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© MG Lipsett, 2010 22 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Payback 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

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© MG Lipsett, 2010 23 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Payback Period Example 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)

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© MG Lipsett, 2010 24 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Payback Period Example Payback of 4 years (actually 3.71 years) Payback of 6 years (actually 5.24 years)

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© MG Lipsett, 2010 25 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Payback Period Example Payback of 4 years (actually 3.71 years) In discounted payback, we use present values

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© MG Lipsett, 2010 26 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Payback Period Example Payback of 4 years (actually 3.71 years) Payback of 6 years (actually 5.24 years)

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© MG Lipsett, 2010 27 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Sensitivity 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

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© MG Lipsett, 2010 28 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Sensitivity 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

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© MG Lipsett, 2010 29 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Sensitivity and Break-Even Analysis Example (2) both options are equivalent here (between 14 & 15 years out) Option B Option 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

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© MG Lipsett, 2010 30 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Sensitivity and Break-Even Analysis Example (3) with i = 10%, options are equivalent here (11.4 years) Option B Option 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).

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© MG Lipsett, 2010 31 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Sensitivity and Break-Even Analysis Example (4) with i = 6%, options are equivalent here (19 years) Option B Option A If money is les costly, then Option B becomes preferable at an later point (not discounting as much).

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© MG Lipsett, 2010 32 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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?

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© MG Lipsett, 2010 33 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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 options And then solve for the unknown (unknown cost of B): Option A Option B NPV A = PWb – PWc NPV B = PWb –PWc Solve for the cost of B with the same NPV as for Option A

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© MG Lipsett, 2010 34 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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 options And then solve for the unknown (unknown cost of B): Option A Option B NPV A = PWb – PWc NPV B = PWb –PWc = 700(P|A,6%,20) – 5000 = 639(P|A,6%,20) – PWc = 700(11.470) – 5000 = 7329 – PWc = $3029 Solve 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

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© MG Lipsett, 2010 35 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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%

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© MG Lipsett, 2010 36 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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

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© MG Lipsett, 2010 37 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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 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

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© MG Lipsett, 2010 38 Department of Mechanical Engineering Engineering Management Group ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques Break-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 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.: A/P, 10%, 10 A/P, 10%, 11 A/P, 10%, n

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