An Interesting News Item

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1 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 University of Alberta 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 % as the daily interest rate) a) 2.48% b) 2.50% c) 2.51% d) 2.68% e) 6.80% MG Lipsett,

2 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, 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 IRR) tells us whether one alternative has a better incremental return than another MG Lipsett,

3 Other Investment Evaluation Techniques: Benefit-Cost Ratio Analysis Incremental Benefit-Cost Ratio Analysis Payback Analysis Sensitivity Analysis Breakeven Analysis MG Lipsett, 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) PW benefits PW costs PWbenefits BCR = 1 PW 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,

4 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? MG Lipsett, 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 MG Lipsett,

5 Incremental Benefit-Cost Ratio Example MG Lipsett, 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: BCR = < 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 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? MG Lipsett,

6 Benefit-Cost Ratio In Cost Ratio In-Class Problem #1 (2) 95 Cash Flow Diagram (Benefits and Costs) Option 1 145= Cash Flow Diagram (Benefits and Costs) Option 2 270= = Use equivalent annual costs and benefits to calculate BCRs MG Lipsett, Benefit-Cost Ratio In Cost Ratio In-Class Problem #1 (2) 95 Cash Flow Diagram (Benefits and Costs) Option 1 145= Cash Flow Diagram (Benefits and Costs) Option 2 270= = Use equivalent annual costs and benefits to calculate BCRs #1 #2 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 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 MG Lipsett,

7 Benefit-Cost Ratio In Cost Ratio In-Class Problem #1 (2) 95 Cash Flow Diagram (Benefits and Costs) Option 1 145= Cash Flow Diagram (Benefits and Costs) Option 2 270= = Use equivalent annual costs and benefits to calculate BCRs #1 #2 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 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 EUAB = EUAB 2 - EUAB 1 = 25 = 0.45 < 1 EUAC EUAC 2 - EUAC 1 56 Therefore choose lower cost option #1 MG Lipsett, 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,

8 Benefit-Cost Ratio In 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 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 MG Lipsett, Benefit-Cost Ratio In 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 $10,000 PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500 BCR First arrange in ascending order of costs, after discarding Option F (which has BCR < 1): D B A C E PWcost $1,000 $2,000 $4,000 $6,000 $9,000 PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000 BCR 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 $5,000 delta benefit $3,360 $2,630 $1,400 $1,670 delta BCR reject C so do E - A reject E choose option with highest of the acceptable incremental BCRs (in this case Option A) Note Option A didn't have the best BCR! MG Lipsett,

9 Benefit-Cost Ratio In 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 $10,000 PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500 BCR First arrange in ascending order of costs, after discarding Option F (which has BCR < 1): D B A C E PWcost $1,000 $2,000 $4,000 $6,000 $9,000 PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000 BCR 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 $5,000 delta benefit $3,360 $2,630 $1,400 $1,670 delta BCR reject C so do E - A reject E choose option with highest of the acceptable incremental BCRs (in this case Option A) Note Option A didn't have the best BCR! MG Lipsett, Benefit-Cost Ratio In 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 $10,000 PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500 BCR First arrange in ascending order of costs, after discarding Option F (which has BCR < 1): D B A C E PWcost $1,000 $2,000 $4,000 $6,000 $9,000 PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000 BCR 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 $5,000 delta benefit $3,360 $2,630 $1,400 $1,670 delta BCR reject C so do E - A reject E choose option with highest of the acceptable incremental BCRs (in this case Option A) Note Option A didn't have the best BCR! MG Lipsett,

10 Benefit-Cost Ratio In 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 $10,000 PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500 BCR First arrange in ascending order of costs, after discarding Option F (which has BCR < 1): D B A C E PWcost $1,000 $2,000 $4,000 $6,000 $9,000 PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000 BCR 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 $5,000 delta benefit $3,360 $2,630 $1,400 $1,670 delta BCR reject C so do E - A reject E choose option with highest of the acceptable incremental BCRs (in this case Option A) Note Option A didn't have the best BCR! MG Lipsett, Benefit-Cost Ratio In 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 $10,000 PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500 BCR First arrange in ascending order of costs, after discarding Option F (which has BCR < 1): D B A C E PWcost $1,000 $2,000 $4,000 $6,000 $9,000 PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000 BCR 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 $5,000 delta benefit $3,360 $2,630 $1,400 $1,670 delta BCR reject C so do E - A reject E choose option with highest of the acceptable incremental BCRs (in this case Option A) Note Option A didn't have the best BCR! MG Lipsett,

11 Benefit-Cost Ratio In 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 $10,000 PWbenefit $7,330 $4,700 $8,730 $1,340 $9,000 $9,500 BCR First arrange in ascending order of costs, after discarding Option F (which has BCR < 1): D B A C E PWcost $1,000 $2,000 $4,000 $6,000 $9,000 PWbenefit $1,340 $4,700 $7,330 $8,730 $9,000 BCR 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 $5,000 delta benefit $3,360 $2,630 $1,400 $1,670 delta BCR reject C so do E - A reject E choose option with highest of the acceptable incremental BCRs (in this case Option A) Note Option A didn't have the best BCR! MG Lipsett, 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,

12 Payback Period Example Simple Discounted (15%) 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 $ In -5,500 simple $ payback, 1,890 $ we -6,371 use 3 $ 3,000 $ 3,000 $ future -2,500 values $ 1,973 directly $ -4,398 4 $ 3,500 $ 3,500 $ 1,000 $ 2,001 $ -2,397 (not discounted values) 5 $ 4,000 $ 4,000 $ 5,000 $ 1,989 $ $ 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, $ 4,000 $ 4,000 $ 25,000 $ 989 $ 6,258 Payback of 6 years In this case, there is value created in the (actually first 5.24 year, years) so we start the clock in year 1. MG Lipsett, Payback Period Example Simple Discounted (15%) 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 $ $ 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, $ 4,000 $ 4,000 $ 25,000 $ 989 $ 6,258 Payback of 4 years (actually 3.71 years) Payback of 6 years (actually 5.24 years) MG Lipsett,

13 Payback Period Example Simple Discounted (15%) 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 $ $ 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, $ 4,000 $ 4,000 $ 25,000 $ 989 $ 6,258 Payback of 4 years (actually 3.71 years) In discounted payback, we use present values MG Lipsett, Payback Period Example Simple Discounted (15%) 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 $ $ 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, $ 4,000 $ 4,000 $ 25,000 $ 989 $ 6,258 Payback of 4 years (actually 3.71 years) Payback of 6 years (actually 5.24 years) MG Lipsett,

14 Sensitivity and Break-Even Analysis Sensitivity and break-even 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, 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 MG Lipsett,

15 Sensitivity and Break-Even Analysis Example (2) NPV of Costs $220,000 $200,000 $180,000 $160,000 $140,000 Option B Option A WACC = 8% 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. $120,000 $100, Year when Option B's Phase 2 constructed 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 MG Lipsett, Sensitivity and Break-Even Analysis Example (3) $220,000 NPV of Costs $200,000 $180,000 $160,000 $140,000 Option B Option A 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. $120,000 $100, Year when Option B's Phase 2 constructed If money is more costly, then Option B becomes preferable at an earlier point (discounted faster). MG Lipsett,

16 Sensitivity and Break-Even Analysis Example (4) NPV of Costs $220,000 $200,000 $180,000 $160,000 $140,000 $120,000 Option B Option A with i = 6%, options are equivalent here (19 years) $100, Year when Option B's Phase 2 constructed If money is les costly, then Option B becomes preferable at an later point (not discounting as much). MG Lipsett, 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,

17 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 MG Lipsett, 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 MG Lipsett,

18 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, 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,

19 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 = 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, 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 = n = 10 + = 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 years A/P, 10%, 11 A/P, 10%, 10 A/P, 10%, n MG Lipsett,