The future and present cash flow series are shown for a project. How long is the simple payback period?

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1 ENGM 401 & 620 X1 Fundamentals of Engineering Finance Fall 2010 Lecture 27: Effects of Inflation on Present Worth; Introduction to Sensitivity Analysis Analysis A weak currency is the sign of a weak economy, and a weak economy leads to a weak nation. - Ross Perot M.G. Lipsett University of Alberta Other Analysis Techniques (Review) The future and present cash flow series are shown for a project. How long is the simple payback period? a) Less than 3 years. b) Just over 7 years. c) Just over 9 years. d) 20 years. e) 22 years. IRR = 13.26% Year FV Sum FV PV Sum PV MG Lipsett,

2 Other Analysis Techniques (Review) (2) True or False: Discounted payback is calculated using the present value series. A new bridge is proposed, which will cost $2 million and last 20 years. Operating and maintenance costs will be $180,000 per year, and overall benefit to the public is estimated at $650,000 per year. At a 10% return, the Benefit-Cost ratio is: a) 650,000 / 2,000,000(A P,10%,20) + 180,000 b) 2,000,000(A P,10%,20) + 180,000 / 650,000 c) 650,000 / [ 2,000,000(A P,10%,20) + 180,000 ] d) 650,000 (P A,10%,20) / 2,000, ,000(P A,10%,20) e) 650,000 (A P,10%,20) / [ 2,000, ,000(A P,10%,20) ] MG Lipsett, Revisiting Inflation Inflation is the concept where as time moves forward, the purchasing power of money decreases A dollar tomorrow will buy less than it does today In Canada, inflation is often measured by the Consumer Price Index (CPI) The real interest rate realized by a lender is often given as the difference between the interest rate paid by the borrower (nominal or market interest rate) and inflation (f): i r i n - f But this is only an approximation MG Lipsett,

3 Real Interest A more accurate calculation of real interest rate is: nominal inflation real (approx.) real (accurate) error 10.00% 0.00% 10.00% 10.00% 0.00% 10.00% 1.00% 9.00% 8.91% 0.09% 10.00% 2.00% 8.00% 7.84% 0.16% 10.00% 3.00% 7.00% 6.80% 0.20% 10.00% 4.00% 6.00% 5.77% 0.23% 10.00% 5.00% 5.00% 4.76% 0.24% 10.00% 6.00% 4.00% 3.77% 0.23% 10.00% 7.00% 3.00% 2.80% 0.20% 10.00% 8.00% 2.00% 1.85% 0.15% 10.00% 9.00% 1.00% 0.92% 0.08% 10.00% 10.00% 0.00% 0.00% 0.00% i r 1 + in = f nominal inflation real (approx.) real (accurate) error 20.00% 0.00% 20.00% 20.00% 0.00% 20.00% 2.00% 18.00% 17.65% 0.35% 20.00% 4.00% 16.00% 15.38% 0.62% 20.00% 6.00% 14.00% 13.21% 0.79% 20.00% 8.00% 12.00% 11.11% 0.89% 20.00% 10.00% 10.00% 9.09% 0.91% 20.00% 12.00% 8.00% 7.14% 0.86% 20.00% 14.00% 6.00% 5.26% 0.74% 20.00% 16.00% 4.00% 3.45% 0.55% 20.00% 18.00% 2.00% 1.69% 0.31% 20.00% 20.00% 0.00% 0.00% 0.00% MG Lipsett, Inflation Example #1 You put $1000 into an account with an interest rate of 12%. How much money will you have 5 years from now and what will its purchasing power be if inflation is 5% per year? Actual money: F actual = P (1+i n ) n = 1000 x (1.12) 5 = $ (the actual dollars that you d get after five years) Purchasing power: i r = 1.12/1.05-1= F real = P (1+i r ) n = 1000 x ( ) 5 = $ (the real purchasing power you ll have in five years, in terms of today s dollars) MG Lipsett,

4 Actual versus Real Dollars Actual dollars are the money we ordinarily think of. They are the amount of money in an account, in your wallet, etc. The first part of the last example is asking about actual dollars Actual dollars (also called inflated dollars) have their value reduced due to inflation, so you need more to have equivalent value to today s dollars Real dollars, on the other hand, are a more abstract concept. They are defined relative to an equivalent sum at an earlier point in time Real dollars (inflation-free dollars) keep their value The second part of the last example is asking about real dollars MG Lipsett, Actual versus Real Dollars (2) Figures from Zimbabwe's Central Statistical Office showed inflation reached 7,634.8 per cent in July, the highest in the world. (Source Telegraph 2007/08/03) (Other sources suggest it s much higher) If it takes ZMD$1000 to buy a 1 litre bottle of water today, how many ZMD will it take this time next week? Annual Inflation rate = > daily inflation rate of 21 % F = P(1.2i)^7 = 1000 * = $3,797. in actual ZBD or you can think of it as next week you could get 1000/3797 = 263 ml of water for $1000 ZMD So in one week the ZMD loses almost three quarters of its purchasing power. MG Lipsett,

5 Present Worth Analysis with Inflation NPV is typically done using a nominal interest rate. If inflation f is a factor, then calculate NPV using inflationcorrected future values, as follows: 1. Adjust the Future Values in each period for inflation f The future values will go up, reflecting the equivalence associated with the declining value of money The real money needed to have equivalent purchasing value has to be inflated into actual future dollars 1. Calculate the PVs using these inflation-corrected future values with respect to the nominal interest rate (or discount rate, or return rate) In the case of a series of taxable incomes, First do the inflation calculation to get the inflation-corrected FV of revenue for the particular year Then do the tax calculation MG Lipsett, Method 1 Example (3) Now calculate the Present Value for inflation-corrected Future Values F f using standard procedure with the nominal discount rate i nom : "quoted" Inflation corrected Year F (FV) F f (FV) P (PV) P = F f x (1 + i nom ) -m 0 $1, $1, $1, $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ MG Lipsett,

6 Alternative Present Worth Analysis with Inflation An alternative approach is to use the nominal future values and an interest rate that is corrected for inflation: First, list the future values as quoted, without any inflation correction Then, use the effective real interest rate i r to calculate the Present Values: i r = in 1 + f Caution: This alternative short-cut method only works if there is no intermediate calculation to do (such as taxes) MG Lipsett, Method 2 Example Calculate the Present Value for inflation-corrected Future Values using standard procedure using i r, the real discount rate that is corrected for inflation: P = F x (1 + i r ) -m "quoted" Year F (FV) P (PV) 0 $1, $1, $ $ $ $ $ $ $ $ $ $ P 1 = F 1 x (1 + i r ) -1 Here i r = MG Lipsett,

7 Inflation In-Class Problem #1 Your city spent $2.5 million to build a public facility 50 years ago, and put aside an equal sum at the same time for its eventual replacement. The money was invested in an account earning 8% interest, and inflation has been 6% per year. Questions: What are the actual and real dollars of the investment? What value facility will the city be able to purchase? What value facility will the city be able to purchase relative to dollars from 50 years ago? Will the city be able to purchase an equivalent facility today and still put aside enough money to replace it in the future? MG Lipsett, Problem #1 (2) $2.5M 50 years ago, 8% interest, 6 % inflation Actual: A$=P(1+i n ) n =2.5M(1.08) 50 = $117,254,031 (actual means inflated, so reduced purchasing power compared to real dollars with respect to past point in time) Real: real interest rate: i r = % R$ = P(1+i r ) n = 2.5M( ) 50 = $6,365,529 (real value with respect to dollars of 50 years ago) Equivalent inflated amount needed for new facility today: F = P(1+f) n = 2.5M(1.06) 50 = $46,050,386 So there is not only enough to pay for new facility, but there is also $71M to invest in a replacement facility in future MG Lipsett,

8 Inflation In-Class Problem #2 On the birth of your first child, you decide to put aside money for her first year s tuition to the UofA. Tuition is currently $5,730 per year, and the provincial government just passed legislation mandating that tuition increases are tied to inflation (projected to be 3.5%). What lump sum do you need to invest today in a GIC paying out 5% interest to cover her first year of tuition 18 years from now? (this example is not solved in the lecture: see the solution posted in the Examples directory) MG Lipsett, 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,

9 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, 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,

10 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, 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,

11 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, 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 Now we 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 (NPV B > NPV A ) otherwise, choose Option A (NPV A > NPV B ) MG Lipsett,

12 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,