ENGM 401 620 X 1 Fundamentals of Engineering

ENGM 401 & 620 –X 1 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/ENGM 401_620. htm

Department of Mechanical Engineering Management Group An Interesting News Item • On Nov. 7/09, The Edmonton Journal reported that Peter Pocklington owes $884 daily on a $13, 000 provincial loan. What interest rate is the government charging? (this works out to 0. 00680% as the daily interest rate) a) b) c) d) e) © MG Lipsett, 2010 2. 48% 2. 50% 2. 51% 2. 68% 6. 80% 2 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

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

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

Department of Mechanical Engineering Management Group Benefit-Cost Ratio Analysis • Benefit-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 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. © MG Lipsett, 2010 6 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group 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 = $50 k, EUAB = $75 k – Gold Standard model: EUAC = $125 k, EUAB = $140 k • Which model should you choose? © MG Lipsett, 2010 7 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group 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 = $50 k, EUAB = $75 k – Gold Standard model: EUAC = $125 k, EUAB = $140 k 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 © MG Lipsett, 2010 8 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group 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 = $50 k, EUAB = $75 k – Gold Standard model: EUAC = $125 k, EUAB = $140 k 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. ENGM 401 & 620 – Fundamentals of Engineering Finance, © MG Lipsett, 2010 9 Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group 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: Purchase cost Annual benefit Salvage value Useful life #1 $ 200 k $ 95 k $ 50 k 6 years #2 $ 700 k $ 120 k $ 150 k 12 years • Assuming a 10% interest rate, what are their benefit-cost ratios? • Which would you choose? Why? © MG Lipsett, 2010 10 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

Department of Mechanical Engineering Management Group Benefit-Cost Ratio In-Class Problem #1 (2) 95 145=50+95 270=120+150 120 -55=-200+50+95 -700 -200 • Use equivalent annual costs and benefits to calculate BCRs #1 #2 EUAC 1 = = EUAB 1 = EUAC 1 © MG Lipsett, 2010 EUAC 2 = = EUAB 2 = EUAC 2 200(A|P, 10%, 6)-50(A|F, 10%, 6) 40 95 (given) 95 = 2. 41 40 12 700(A|P, 10%, 12)-150(A|F, 10%, 12) 96 120 (given) 120 = 1. 25 96 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group Benefit-Cost Ratio In-Class Problem #1 (2) 95 145=50+95 270=120+150 120 -55=-200+50+95 -700 -200 • Use equivalent annual costs and benefits to calculate BCRs #1 #2 EUAC 1 = = EUAB 1 = EUAC 1 EUAC 2 = = EUAB 2 = EUAC 2 200(A|P, 10%, 6)-50(A|F, 10%, 6) 40 95 (given) 95 = 2. 41 40 ΔEUAB = EUAB 2 - EUAB 1 = 25 ΔEUAC 2 - EUAC 1 56 700(A|P, 10%, 12)-150(A|F, 10%, 12) 96 120 (given) 120 = 1. 25 96 = 0. 45 < 1 Therefore choose lower cost option #1 © MG Lipsett, 2010 13 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

Department of Mechanical Engineering Management Group 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: PWcost PWbenefit BCR A B C D E F $4000 $2000 $6000 $1000 $9000 $10000 $7330 $4700 $8730 $1340 $9000 $9500 1. 83 2. 35 1. 46 1. 34 1. 00 0. 95 • 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 © MG Lipsett, 2010 15 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

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

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

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

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

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

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

Department of Mechanical Engineering Management Group Payback Period Example In simple payback, we use future values directly (not discounted values) In this case, there is value created in so we start the clock in year 1. © MG Lipsett, 2010 23 Payback of 6 years the first 5. 24 year, (actually years) ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group Payback Period Example Payback of 4 years (actually 3. 71 years) © MG Lipsett, 2010 24 Payback of 6 years (actually 5. 24 years) ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group Payback Period Example Payback of 4 years (actually 3. 71 years) © MG Lipsett, 2010 25 In discounted payback, we use present values ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group Payback Period Example Payback of 4 years (actually 3. 71 years) © MG Lipsett, 2010 26 Payback of 6 years (actually 5. 24 years) ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

Department of Mechanical Engineering Management Group 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 $140 k. – Option B is to build in two phases: • Phase 1 provides the capacity you need for the first few years at a cost of $100 k. • Phase 2 provides the remaining additional capacity at a cost of $120 k. – 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 © MG Lipsett, 2010 28 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group Sensitivity and Break-Even Analysis Example (2) WACC = 8% Option B both options are equivalent here (between 14 & 15 years out) 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. Option A This plot shows the effect on net present cost of option B phase 2 ($120 k using discounted dollars), thus cheaper as phase 2 gets delayed © MG Lipsett, 2010 29 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group Sensitivity and Break-Even Analysis Example (3) Option B with i = 10%, options are equivalent here (11. 4 years) The decision will also depend on our WACC, or the interest rate we use to calculate our discounted cash flows. Option A If money is more costly, then Option B becomes preferable at an earlier point (discounted faster). 30 © MG Lipsett, 2010 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

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

Department of Mechanical Engineering Management Group 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 NPVA = PWb – PWc NPV B = PWb –PWc Solve for the cost of B with the same NPV as for Option A © MG Lipsett, 2010 33 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

Department of Mechanical Engineering Management Group 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 NPVA = 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 © MG Lipsett, 2010 34 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques

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

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

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

Department of Mechanical Engineering Management Group 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: A/P, 10%, 10 A/P, 10%, n $80. 36 = $500 × (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. : © MG Lipsett, 2010 A/P, 10%, 10 38 A/P, 10%, 11 ENGM 401 & 620 – Fundamentals of Engineering Finance, Lecture 26: Other Analysis Techniques