1. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 1 Lecture slides to accompany Basics of Engineering Economy by Leland Blank and Anthony Tarquin Chapter 10 Effects of Inflation

2. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 2 Chapter 10 – Effects of Inflation PURPOSE Consider the effects of inflation in PW, AW and FW equivalence calculations TOPICS Definition and impact PW adjusted for inflation FW with inflation; real interest rate; inflation adjusted MARR AW with inflation Spreadsheet usage

3. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 3 Sec 10.1 – Understanding Inflation Inflation definition – A decrease in the value of a currency Inflation impact – An increase in the amount of money required to purchase the same amount of goods or services over time Directly associated with inflation – an increase in the money supply, i. e., more (government) money is printed to counteract inflationary impacts To perform economic comparisons - conversion to a common basis of money at different times must be included What to do in engineering economy?

4. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 4 Sec 10.1 – Understanding Inflation Constant-value (CV) – Money represents same purchasing power regardless of when in time –Another term used for CV: Today’s dollars Future dollars – amount of money in the future –Other terms used for future dollars: Then-current dollars Inflated dollars

5. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 5 Sec 10.1 – Understanding Inflation  f - inflation rate per period (year), e.g., 5% per year  n – number of periods (years) considered Example: A $100 bill five years from now at 4% inflation will purchase only $82.20 worth of goods

6. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 6 Sec 10.1 – Understanding Inflation Viewed from a future perspective Future amount = CV amount (1+f) n Example: If it costs $82.20 today, in five years it will cost $100 if inflation is 4% per year 82.20(1+0.04) 5 = 82.20(1.2167)= $100 Two (equivalent) approaches to adjust for inflation: 1.Convert all amounts (estimates) to have same value (purchasing power) by equivalency relations (this uses CV relation shown above prior to time-value-of-money calculations) 2.Change interest rates to account for inflation, as well as time value of money (this approach follows)

7. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 7 Sec 10.1 – Understanding Inflation Consider the erosive impact of even moderate inflation Save $25,000/year for 40 years of employment (ages 22 to 62) Assume 7% per year return on your investments Future total amount of retirement portfolio No return considered: 25,000(40) = $1 million 7% return considered FW = 25,000(F/A,7%,40) = 25,000(199.6351) = $4.99 million A return of 7% per year means a lot over time

8. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 8 Sec 10.1 – Understanding Inflation Now, assume 4% per year inflation What is the effective purchasing power of your retirement portfolio 40 years from now? FW = $1.835 million This means that purchasing power is $1.835 million, not $4.99 million  Inflation is a real killer of investment returns Conclusions:  The real return on savings is significantly lower when inflation is accounted for  If inflation rate exceeds return rate ( f > i), money loses purchasing power over time Computations discussed later

9. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 9 Sec 10.1 – Inflation-related Terminology  i Real (inflation-free) rate – Rate at which interest is earned or paid when effects of inflation are removed  Usually considered the ‘safe investment’ rate. Approximately 3.5% per year, historically, but varies depending upon health of economy  i f Inflation-adjusted rate – Also called market rate. Interest rate with effect of inflation accounted for  This rate is the one quoted daily,.e.g., 7% per year or effective 7.08%. Combination of real rate i and inflation rate f  f Inflation rate – Rate of change in value of currency  Formulas for and use of these terms follow

10. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 10 Sec 10.1 - Deflation Deflation is opposite of inflation Deflation rate is -f in % per year Purchasing power of deflated currency is greater in future than at present Sounds good after inflationary period, however … There are fewer jobs, less credit and fewer loans available; overall ‘tighter’ money situation Can disrupt national economies faster and more severely than an equivalent inflation rate

11. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 11 Sec 10.1 - Deflation Equivalency computations use -f, instead of f Example: Asset costs $10,000 today. If 2% per year deflation is assumed, future estimated cost of repurchase in 5 years is FW = 10,000(1-0.02) 5 = 10,000(0.9039) = $9,039 International Dumping ■Import materials (e.g., cars, cement, steel) into a country from international sources at very low prices ■Causes temporary deflation in the targeted sector ■Domestic producers must reduce costs; some go broke, if financially weak ■Once competition is beaten, prices return to normal or above previous levels

12. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 12 Sec 10.2 – PW Adjusted for Inflation As described earlier, there are two (equivalent) approaches to adjust for inflation. These are correct for all PW computations 1. Convert everything to current value amounts by PW equivalency relations before applying the real interest rate i in PW computations 2. Determine the inflation-adjusted interest rate i f and use it in all PW calculations Example: Assume a first cost of $5,000 now (CV amount). Will purchase item during next 4 years. Let Inflation = 4% per yearf = 4% Real rate of return = 10% per yeari = 10%

13. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 13 Sec 10.2 – PW Adjusted for Inflation Approach 1--Use CV dollars and real rate i CV = PW = $5,000 now Future costs at f = 4% are in column (3) FW = 5,000(1+ f) n = 5,000(1.04) n

14. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 14 Sec 10.2 – PW Adjusted for Inflation Approach 1 – Now find equivalent PW of $5,000 CV over 4 years at i = 10% Year, t Future cost in CV $ PW at real i = 10% 5,000(P/F,10%,t) 05,000 1 4,545 25,0004,132 35,0003,757 45,0003,415

15. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 15 Sec 10.2 – PW Adjusted for Inflation Effects of 4% inflation and 10% real interest rate over 4 years

16. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 16 Sec 10.2 – PW Adjusted for Inflation Approach 2 – Use inflation-adjusted interest rate i f in all PW calculations  To derive a formula for i f, calculate PW from FW amounts using real i and determine CV amounts using f  Let i f = i + f + if be the inflation-adjusted (market) interest rate This approach is more commonly used

17. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 17 Sec 10.2 – PW Adjusted for Inflation Resulting PW formula Interest rate formula Requires no conversion to CV amounts to obtain PW values Example: Real i = 8%Inflation = 6% Inflation-adjusted (market) rate = 0.08 + 0.06 + (0.08)(0.06) i f = 0.1448 or 14.48%

18. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 18 Sec 10.2 – PW Adjusted for Inflation From previous example P = $5,000 is cost now; cost increases to $5,849 in 4 years  Inflation rate = 4%/year  Real interest rate = 10%/year Use i f to determine equivalent PW now of the cost 4 years in future: i f = 0.10 + 0.04 + (0.10)(0.04) = 14.4% P = 5,849(P/F,14.4%,4) = 5,849(0.5838) = $3,415 Conclusion: At market rate (inflation considered) of 14.4%, paying $5,849 in four years for something that costs $5,000 now is equivalent to paying $3,415 now in CV terms

19. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 19 Sec 10.2 – PW Inflation-Adjusted Example #1 Lottery winnings can be taken in 1 of 3 ways. Which is better financially if f = 4% and i = 6%? 1: $100,000 immediately 2: 8 payments of $15,000 each starting next year 3: 3 payments of $45,000 each in years 0, 4 and 8 Use approach 2 with i f : i f = 0.06 + 0.04 + (0.06)(0.04) = 10.24% Plan 1 PWPlan 2 PWPlan 3 PW $100,000 15,000(P/A,10.24%,8) = $79,329 45,000[(1 + (P/F,10.24%,4) + (P/F,10.24%,8)] = $96,099 Winner

20. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 20 Sec 10.2 – PW Inflation-Adjusted Example #2 Find PW with and without 11% inflation considered cont →

21. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 21 Use (P/A,g,i,n) = (P/A,12%,interest rate,9) factor for geometric series to determine PW, where ‘interest rate’ is 1. i = 15% (without inflation) 2. i f = 0.15+0.11+(0.15)(0.11) = 27.65% (with inflation) Without inflation considered PW = -35,000 - 7,000(P/A,15%,4) - 7,000(P/A,12%,15%,9)(P/F,15%,4) = $-83,232 Sec 10.2 – PW Inflation-Adjusted Example #2 cont →

22. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 22 Sec 10.2 – PW Inflation-Adjusted Example #2 2. Inflation accounted for using i f = 27.65% PW = -35,000 - 7,000(P/A,27.65%,4) - 7,000(P/A,12%,27.65%,9)(P/F,27.65%,4) = $-62,436 Conclusions:  As f and i f increase, the P/A and P/F factors decrease, thus making the PW values smaller  Moving debt to the future and paying with future (inflated) money seems financially ‘smart’, BUT …  If cash is not available in the future, debt-laden people, companies and countries will suffer and may go bankrupt  ‘Buy now, pay later’ philosophy can be a dangerous strategy

23. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 23 Sec 10.3 – FW Adjusted for Inflation Like PW computations, possible to consider or neglect inflation rate f on top of real interest rate i  Let P = $1,000; n = 7; find F under different conditions  Market return i f = 10%  Inflation rate f = 4% Cases 1 and 4: Actual future amount accumulated at market rate i f (both purchasing power and return included) F = P(F/P,i f,n) = 1,000(F/P,10%,7) = $1,948 Interpretation: This case correctly projects amount of capital needed in future to overcome ongoing inflation plus make a stated return on the investment

24. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 24 Sec 10.3 – FW Adjusted for Inflation Case 2: Purchasing power maintained, but no inflation considered requires use of real interest rate i  Usually the market (inflation-adjusted) rate i f and inflation rate f are estimated  Determine real interest rate i by dividing market rate by inflation factor (1+f)

25. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 25 Sec 10.3 – FW Adjusted for Inflation Case 2 (cont):  Let P = $1,000; n = 7; find F  Market return i f = 10%  Inflation rate f = 4%  Therefore, real return i = (0.10 – 0.04) / (1.04) = 5.77% F = 1,000(F/P,5.77%,7) = $1,481 Interpretation: This case correctly estimates future purchasing power with inflation effects removed. An inflation rate of 4% reduces the future 10% return amount that can be purchased from $1,948 to $1,481 -- a 24% reduction

26. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 26 Sec 10.3 – FW Adjusted for Inflation Case 3: Future amount with no return uses only the inflation rate f  Let P = $1,000; n = 7; find F  Inflation rate f = 4% F = P(F/P,f%,n) = 1,000(F/P,4%,7) = $1,316 Interpretation: This case correctly estimates the future amount needed to keep up with inflation only. A price of $1,000 grows by 31+% in 7 years at 4% per year inflation

27. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 27 Sec 10.3 – Inflation-Adjusted MARR Most corporations determine MARR using both inflation and a return to cover capital increases and expected return This is i f used in cases 1 and 4 Inflation-adjusted MARR is: MARR f = i + f + if The real rate of return i is the corporate return requirement at or above the following:  ‘safe’ investment, which is usually ~3.5%, or  cost of capital

28. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 28 Sec 10.3 – Inflation-Adjusted MARR Example: Cost of capital = 10% Required return = 3% Inflation rate projected = 4%  Perform PW, FW and AW computations at MARR f = 0.13 + 0.04 + (0.13)(0.04) = 17.52% Real return i = 13%

29. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 29 Sec 10.3 – FW Inflation-Adjusted Example Alternatives: purchase now for $200,000, or in 3 years at an estimated $340,000 MARR is 12% and inflation averages 6.75% per year WITHOUT INFLATION MARR = 12% Purchase now FW = -200,000(F/P,12%,3) = $-280,986 Purchase later FW = $-340,000 Buy now is more economic INFLATION CONSIDERED MARR f = 12% + 6.75% + 12%(6.75%) = 19.56% Purchase now FW = -200,000(F/P,19.56%,3) = $-341,812 Purchase later FW = $-340,000 Buy later is slightly more economic

30. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 30 Sec 10.3 - Hyperinflation  Inflation usually averages 2% to 8% per year  Political and/or financial instability, overspending, serious trade imbalance can increase inflation dramatically  Inflation rates above 50%, 75% and 100% per year are considered hyperinflation  FW estimates skyrocket as inflation increases become larger  But, future estimates are so uncertain that a dependable economic analysis can not be performed

31. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 31 Sec 10.3 - Hyperinflation Previous example: P = $200,000 MARR = 12% per year If f = 10%/month = 120%/ year (without considering compounding of inflation) MARR f = 12% + 120% + 12%(120%) = 146.4% Future equivalent cost estimate with inflation is huge FW = -200,000(F/P,146.4%,3) = $-2.99 million Yet, estimated cost in 3 years is unpredictable What is to be done from the economic perspective????

32. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 32 Sec 10.4 - AW Adjusted for Inflation  Include inflation in AW computations because:  Capital must be recovered with future, inflated dollars  Less buying power in future means more money needed to recover present investments plus a return To find future inflated amount needed per year – Use market rate i f to determine A, given P If future amount is known, fewer annual dollars are needed, since their current buying power is greater – Again, use i f to determine A, given F

33. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 33 Sec 10.4 - AW Adjusted for Inflation Example: Spend $1,000,000 now for 3-year IT service contract Expect 4% return over cost of capital of 8.5% Inflation averages 6% per year What annual amount is needed to recover cost? Expected real return = 4% + 8.5% = 12.5% MARR f = 12.5% + 6% + 12.5%(6%) = 19.25% Required annual recovery amount: AW = 1 million(A/P,19.25%,3) = $469,159 / year

34. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 34 Sec 10.4 - AW Adjusted for Inflation For comparison purposes only, assume contract provider agreed to one payment of $1 million after 3 years. What is AW now?  Again, MARR f = 19.25%  Use A/F factor to find AW AW = 1 million(A/F,19.25%,3) = $276,659 There is a large reduction of ~$192,000 per year to recover the (fixed) $1 million expenditure after 3 years

35. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 35 Sec 10.4 - AW Inflation-Adjusted Example For retirement, desire is to invest equal amounts annually for 5 years to maintain purchasing power of $10,000 today Expected (market) return is 10% Assume inflation is relatively high at 8% Step 1: Required total after 5 years when f = 8% (case 3): F = 10,000(F/P,8%,5) = $14,693.28 Step 2: Annual amount for 5 years at i f = 10% (cases 1, 4): A = 14,693.28(A/F,10%,5) = $2406.76 Note: real return is only i = (0.10-0.08)/(1.08) = 1.85%/year

36. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 36 Sec 10.4 – Inflationary Period Reactions THINGS THAT OCCUR WHEN INFLATION INCREASES More needed annually to recover capital investments + required return Lenders tend to increase interest rates to individuals (credit cards, mortgages) and corporations (loans) People make lower payments on cards and loans (money is used to buy other, necessary items to live ) Lenders need more money to cover their higher costs of making loans Bankruptcies increase Spiraling effects of inflation can lead to national recession

37. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 37 Sec 10.5 – Spreadsheet Usage Lottery winnings can be taken in 1 of 3 ways 1: $100,000 immediately 2: 8 payments of $15,000 each starting next year 3: 3 payments of $45,000 each in years 0, 4 and 8 Assume f = 4% and i = 6% per year Questions about the 3 plans: A.Best plan based on PW values? B.FW in 8 years with inflation considered? C.FW in 8 years in terms of today’s purchasing power? cont →

38. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 38 Sec 10.5 – Spreadsheet Usage A.Calculate i f = 10.24%, enter cash flows and use NPV function (row 13); select Plan 1 cont →

39. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008, McGraw-Hill All rights reserved 10 - 39 Sec 10.5 – Spreadsheet Usage B. Determine FW with inflation considered using FV function at 10.24% (row 16 – note minus sign) C. Determine FW without inflation considered using FV function at 6% (row 18- note minus sign) As expected, Plan 1 has best PW and FW values Note: The impact of inflation is very clear. For example, Plan 3 generates equivalent value in 8 years of ~$209,600. However, this will purchase only about ~ $153,100 of goods in terms of today’s dollars