\fEngineering Economy Chapter 11: Breakeven and Sensitivity Analysis The objective of Chapter 11 is to illustrate breakeven and sensitivity methods for investigating variability in outcomes of engineering projects. To this point we have assumed a high degree of confidence in estimated values. The degree of confidence is sometimes called assumed certainty, and decisions made on the basis of this kind of analysis are called decisions under certainty. In virtually all situations, ultimate economic results are unknown. Breakeven and sensitivity analysis are used to help understand how our decision might be affected if our original estimates are incorrect. The breakeven point is the value of a key factor at which we are indifferent between two alternatives (one may be \"do nothing\"). The breakeven point is the value of y where Common factors to consider for breakeven analysis. annual revenue and expenses rate of return market (or salvage) value equipment life capacity utilization Should Jim sell his gas-guzzler? Jim's1998 minivan is quite functional, but it only averages 20 miles per gallon (mpg). He has found a somewhat newer vehicle (roughly the same functionality) that averages 26 mpg. He can sell hiscurrent minivan for $2800 and purchase the newer vehicle for $4,000. Assume a cost of gasoline $4.00 per gallon How many miles per year must Jim drive if he wants to recover his investment in three years? Assume an interest rate of 6%, zero salvage value for either vehicle after three years, and identical maintenance cost. Gas-guzzler solution Current minivan New vehicle Equating these, and solving for x, we find Pause and solve The \"one time good deal\" Cash-For-Clunkers program offered by the federal government proved a temporary boon for car dealers. In addition to this program, dealers were eager to add their own incentives. Bill Mitselfik was considering two different deals he could make for his new car. He can finance the purchase price, $25,000, entirely through the dealer at a 1.9% APR (compounded monthly) for 5 years, with payments monthly. Alternatively, the dealer will give Bill a cash rebate and provide financing at 9% APR (compounded monthly) for 5 years, with monthly payments. What is the value of the rebate for which Bill would be indifferent between the two financing options? We use sensitivity analysis to see what happens to project profitability when the estimated value of study factors are changed. What if expenses are 10% higher than expected is the project profitable? What if sales revenue is 15% lower than expected? What change in either expenses or revenues will cause the project to be unprofitable (decision reversal)? Reconsidering Jim's gas-guzzler. Considering that Jim drives about 10,000 miles per year, our previous analysis would indicate that he should purchase the vehicle that gets better mileage. However, what if gas prices drop by 10%? Should Jim still sell his gas-guzzling minivan? So, if gas prices drop by 10%, Jim should keep his minivan. Pause and solve Acme Delivery is considering a proposal for new package tracking technology. The system has an estimated initial cost of $1.9 million and will require upgrades and maintenance of $140,000 each year. Acme estimates that improved tracking will save approximately $680,000 per year, after system operating expenses. Acme has a MARR of 15% per year, and the study period for this technology is 6 years, after which time Acme expects the entire system will need to be replaced. The PW of this proposal is PW(15%) = -$1,900,000+($680,000 - $140,000)(P/A,15%,6) = $143,630 Determine how sensitive the decision to invest in the system is to the estimates of initial investment cost and annual savings. Spreadsheets are very useful in performing sensitivity analysis. Formulas easily reflect changes in parameter values. Tables and plots can provide quick answers and visual cues to the effect of changes. A spider plot can be especially useful in sensitivity studies. It can be useful to examine more than one alternative on a plot, or to examine sensitivity of incremental cash flows. [Note that the steeper the curve, the more sensitive is the PW to the factor.] Changing the value of more than one factor at a time. To this point we have only looked at changes in one factor at a time. In reality, each factor considered can change, so it is useful to look at the effect of simultaneous changes in factors of interest. One way to accomplish this is to use the Optimistic-Most Likely-Pessimistic (O-ML-P) technique. Optimistic-Most Likely-Pessimistic Establish optimistic (the most favorable), most likely, and pessimistic (the least favorable) estimates for each factor. The optimistic condition, which should occur about 1 time out of twenty, is when all factors are at their optimistic levels. Similarly for pessimistic condition. The most likely condition should occur roughly 18 times out of 20. Perform EW calculations under each condition for insight into the sensitivity of the solution. The results can be seen on a spider plot for further insight. Consider investment in a new crane. Assume a MARR of 8%. Estimation Condition Optimistic (O) Most Likely (M) Pessimistic (P) Investment, I $240,000 $270,000 $340,000 Useful life, N 10 yr 8 yr 5 yr $20,000 $15,000 $8,000 Annual revenues, R $100,000 $80,000 $50,000 Annual expenses, E $10,000 $15,000 $20,000 Market value, MV Considering O-ML-P for I and R (fix E, MV, and life at their ML levels). Value in each cell is the PW for the project. Investment, I Revenues, R Optimistic (O) Most Likely (M) Pessimistic (P) Optimistic, (O) $256,568 $226,568 $156,568 Most Likely, (M) $141,636 $111,636 $41,636 Pessimistic, (P) -$30,764 -$60,764 -$130,764 This suggests that perhaps some additional effort should be place on getting refined estimates of revenues. Of course, the complete study needs to consider the other factors. \fEngineering Economy Chapter 9: Replacement Analysis The objective of Chapter 9 is to address the question of whether a currently owned asset should be kept in service or immediately replaced. What to do with an existing asset? Keep it Abandon it (do not replace) Replace it, but keep it for backup purposes Augment the capacity of the asset Dispose of it, and replace it with another Three reasons to consider a change. Physical impairment (deterioration) Altered requirements New and improved technology is now available. The second and third reasons are sometimes referred to as different categories of obsolescence. Some important terms for replacement analysis Economic life: the period of time (years) that yields the minimum equivalent uniform annual cost (EUAC) of owning and operating as asset. Ownership life: the period between acquisition and disposal by a specific owner. Physical life: period between original acquisition and final disposal over the entire life of an asset. Useful life: the time period an asset is kept in productive service (primary or backup). Replacement: past estimation errors Any study today is about the futurepast estimation \"errors\" related to the defender are irrelevant. The only exception to the above is if there are income tax implications forthcoming that were not foreseen. Replacement: watch out for the sunk-cost trap Only present and future cash flows are considered in replacement studies. Past decisions are relevant only to the extent that they resulted in the current situation. Sunk costsused here as the difference between an asset's BV and MV at a particular point in timehave no relevance except to the extent they affect income taxes. Replacement: the outsider viewpoint The outsider viewpoint is the perspective taken by an impartial third party to establish the fair MV of the defender. Also called the opportunity cost approach. The opportunity cost is the opportunity foregone by deciding to keep an asset. If an upgrade of the defender is required to have a competitive service level with the challenger, this should be added to the present realizable MV. Replacement: economic lives of the challenger and defender The economic life of the challenger minimizes the EUAC. The economic life of the defender is often one year, so a proper analysis may be between different-lived alternatives. The defender may be kept longer than it's apparent economic life as long as it's marginal cost is less than the minimum EUAC of the challenger over it's economic life. Replacement: income taxes Replacement often results in gains or losses from the sale of depreciable property. Studies must be made on an after-tax basis for an accurate economic analysis since this can have a considerable effect on the resulting decision. Before-tax PW example Acme owns a CNC machine that it is considering replacing. Its current market value is $25,000, but it can be productively used for four more years at which time its market value will be zero. Operating and maintenance expenses are $50,000 per year Acme can purchase a new CNC machine, with the same functionality as the current machine, for $90,000. In four years the market value of the new machine is estimated to be $45,000. Annual operating and maintenance costs will be $35,000 per year. Should the old CNC machine be replaced using a beforetax MARR of 15% and a study period of four years? Example solution Defender Challenger PW of the challenger is greater than PW of the defender (but it is close). Proper analysis requires knowing the economic life (minimum EUAC) of the alternatives. The EUAC of a new asset can be computed if the capital investment, annual expenses, and year-by-year market values are known or can be estimated. The difficulties in estimating these values are encountered in most engineering economy studies, and can be overcome in most cases. Finding the EUAC of the challenger requires finding the total marginal cost of the challenger, for each year. The minimum such value identifies the economic life. This equation represents the present worth, through year k, of total costs. (Although the sign is positive, it is a cost. Eq. 9-1.) Total marginal cost formula The total marginal cost is the equivalent worth, at the end of year k, of the increase in PW of total cost from year k-1 to year k. This can be simplified to (eq. 9-2) Finding the economic life of the new CNC machine. O&M costs Market value Year 1 Year 2 Year 3 Year 4 $35,000 $35,000 $35,000 $35,000 $75,000 $60,000 $50,000 $45,000 Marginal costs: O&M Depreciation Int. on capital Year 1 Year 2 Year 3 Year 4 $35,000 $35,000 $35,000 $35,000 $15,000 $15,000 $10,000 $5,000 $13,500 $11,250 $9,000 $7,500 TC $63,500 $61,250 $54,000 $47,500 Pause and solve In a replacement analysis for an industrial saw, the following data are known about the challenger. Initial investment is $18,000. Annual maintenance costs begin at the end of year three, with a cost at that time of $1,000, with $1,000 at the end of year four, increasing by $8,600 each year thereafter. The salvage value is $0 at all times. Using a MARR of 6% per year, what is the economic life of the challenger? Solution From the economic life table below, we see the EUAC reaches a minimum in year 4, which is the economic life of this challenger. EOY 0 MV@ EOY Loss in MV Cost of Capital Annual Expense TC for year EUAC 18,000 1 0 2 0 3 18,000 1,080 0 19,080 19,080 0 0 0 0 9,818 0 0 0 1,000 1,000 7,048 4 0 0 0 1,000 1,000 5,666 5 0 0 0 9,600 9,600 6,364 6 0 0 0 18,200 18,200 8,060 The economic life of the defender If a major overhaul is needed, the life yielding the minimum EUAC is likely the time to the next major overhaul. If the MV is zero (and will be so later), and operating expenses are expected to increase, the economic life will the one year. The defender should be kept as long as its marginal cost is less than the minimum EUAC of the best challenger. Finding the economic life of the defender CNC machine. O&M costs Market value Year 1 Year 2 Year 3 Year 4 $50,000 $50,000 $50,000 $50,000 $15,000 $10,000 $5,000 $0 O&M Depreciation Int. on capital Year 1 Year 2 Year 3 Year 4 $50,000 $50,000 $50,000 $50,000 $10,000 $5,000 $5,000 $5,000 $3,750 $2,250 $1,500 $750 TC $63,750 $57,250 $56,500 $55,750 Replacement cautions. In general, if a defender is kept beyond where the TC exceeds the minimum EUAC for the challenger, the replacement becomes more urgent. Rapidly changing technology, bringing about significant improvement in performance, can lead to postponing replacement decisions. When the defender and challenger have different useful lives, often the analysis is really to determine if now is the time to replace the defender. Repeatability or cotermination can be used where appropriate. Abandonment is retirement without replacement. For projects having positive net cash flows (following an initial investment) and a finite period of required service. Should the project be undertaken? If so, and given market (abandonment) values for each year, what is the best year to abandon the project? What is its economic life? These are similar to determining the economic life of an asset, but where benefits instead of costs dominate. Abandon the year PW is a maximum. Abandonment example A machine lathe has a current market value of $60,000 and can be kept in service for 4 more years. With an MARR of 12%/year, when should it be abandoned? The following data are projected for future years. Year 1 Year 2 Year 3 Year 4 Net receipts $50,000 $40,000 $15,000 $10,000 Market value $35,000 $20,000 $15,000 $5,000 Abandonment solution Keep for one year Keep for two years Keep for three years (BEST!) Keep for four years Taxes can affect replacement decisions. Most replacement analyses should consider taxes. Taxes must be considered not only for each year of operation of an asset, but also in relation to the sale of an asset. Since depreciation amounts generally change each year, spreadsheets are an especially important tool to use. The effect of taxes. The economic life of an asset becomes, after taxes, (eq. 93) which reflects not only annual taxes but also tax effects of the sale of the asset. The total marginal cost, for each year, is (eq. 9-4) We must also consider the possible tax effects of the sale of the defender. The MV of the asset must be compared to the BV to assess the possible tax implications, and this should be reflected in the opportunity cost of keeping the defender. The net ATCF, if the defender is kept, after taxes, is Pause and solve Acme Cycles purchased a bending machine two years ago for $45,000. Depreciation deductions have followed the MACRS (GDS, 3-year recovery period) method. Acme can sell the bending machine now for $12,000. Assuming an effective income tax rate of 45% compute the after-tax investment value of the bending machine if it is kept. Solution First, find the current book value. BV = $45,000 (1 - 0.3333 - 0.4445) = $10,000 Next, find the ATCF if the machine is kept. BTCF -$12,000 Depreciation Taxable income $0 -($12,000$10,000)=$2,00 0 Cash flow for tax -0.45*$2,000 = -$900 ATCF -$12,900 \"AUM American IlnMnllyOl'l'hoIIddoEas'l Problem 4 - Open questions (25 points) Answer each question clearly with your own words. Question 4.1 - If a certain project has a sensitivity of x=-0.13 for Initial Investment. What was the original decision of the project? Explain your reasoning in your own words. (5 points) Question 4.2 - If a certain project has a negative AW. The sensitivity (x=?) for annual expenses will be positive or negative? Explain your reasoning in your own words. (5 points) Question 4.3 - 3 projects have the following IRR: Project 1 IRR = 12.4% Project 2 IRR = 6.8% Project 3 IRR = 10.2% Which project should be selected? Explain your reasoning in your own words. (5 points} Question 4.4 - 3 projects have the following Benefits, when evaluated using PW: Project 1 PW of Benefits: 6,700 Project 2 PW of Benefits: 8,000 Project 3 PW of Benefits: 16,500 Which project should be selected? Explain your reasoning in your own words. (5 points) Question 4.5 Explain the difference in between the MARR and IRR of a given project. (5 points}