1. 1/10/2016rd1 Engineering Economic Analysis Chapter 14  Inflation and Price Change

2. Inflation Rate Factors Cost of energy Interest rates Availability and cost of skilled people Scarcity of materials Political stability Inflation is the increase in the amount of money necessary to buy the same amount of a product or service over time. Government prints more dollars (weakens value) while the supply of goods does not increase. Same dollar amount buys less. 1/10/2016rd2

3. Inflation Problems Inflation causes a redistribution of wealth. Some profit while other suffer. (Oil companies, truck drivers) Price Effects – Oil companies make obscene profits Wealth effects – Hurts lenders but benefits borrowers. 1/10/2016rd3

4. Combating Inflation Fix costs and allow income to increase. Fix costs by entering into long-term contracts for material and labor, by buying materials before they are needed by stockpiling products for sale later. Borrowing becomes more attractive in inflationary times since debt is repaid in cheaper dollars. 1/10/2016rd4

5. 1/10/2016rd5 Inflation A$ Actual Dollars, nominal, inflated, future, then-current R$ Real Dollars, constant-value, today, relative to base year f General price inflation rate i m Market interest rate i r Real or effective interest rate Base time period

6. 1/10/2016rd6 Inflation Inflation Rate ~ f Market Rate ~ i m Real Rate ~ i r R$ = A$(1 + f) -n A$ brought back to the base year A$ = R$(1 + f) n R$ inflated to the current year Bank interest rates reflect a real rate plus an expected inflation rate. Real rate is typically near the rate of a US Treasury bond (safest investment).

7. 1/10/2016rd7 Inflation and Price Indexes Construction Cost Index Average % YearIndex valuerate/year 1950510--- 19556605.3 660 = 510(1 + i) 5 => i = 5.29 19608244.5 824 = 660(1 + i) 5 => i = 4.54 …. 199554716.11 5471 = 510(1 + i) 40 => i = 6.11 1955-1995 (40 years)6.11 General inflation rate ~ average rate CPI of items in market basket Specific inflation rate ~ pertains to a segment of the economy CPI k = [(CPI) k - (CPI) k-1 ](100)/ (CPI) k-1

8. Prices over Time Consumer Purchase2003 2004 2005 f (%) Big Mac$2.22 $2.29 $2.39 3.76 Birth$6696 $7187 $7907 8.67 Movie ticket$8.08 $8.39 $8.52 2.69 Hospital stay$3889 $4416 $4848 11.65 Pair of jeans$39.50 $39.50 $39.50 0 Unleaded gasoline$1.61 $1.90 $2.32 20.04 Single family home$170K $184.1K $215.9K 12.69 Year in college$15,441 $16,862 $17,799 7.36 Funeral$6366 $6530 $6725 2.781 Toyota Camry$19560 $19835 $20125 1.433 Clear clogged sink$189 $201 $212 5.9 1/10/2016rd8

9. 1/10/2016rd9 Price Relatives Year Cost ($) Gallon Price Relatives (Base 1976 =100) 19760.58 100 19770.63 (0.63/0.58)(100) = 109 19780.68 (0.68/0.58)(100) = 117 19790.93 (0.93/0.58)(100) = 160 19801.20 (1.20/0.58)(100) = 207 Price Relative in Period t = Price in Period t (100) Base Period Price CPI 1980 = 1.20 * 100 / 0.58 = 206.9 = {[(1.2 – 0.58) / 0.58 ] + 1} * 100 CPI measures changes in food, shelter, medical care, transportation, apparel, and other selected goods and services used by us. No VCRs in 1970 is an example that the CPI Basket has to change.

10. 1/10/2016rd10 CPI Overstates Inflation The CPI does not allow for a substitution effect. When the price of soda rises, consumers may switch to water, Snapple, or milk. The CPI market basket is fixed, so it misses this substitution effect. The CPI does not measure quality changes. Cars are more expensive today than in 1980, but they are also safer, more mechanically reliable, pollute less and use less gasoline. The CPI attempts to adjust for quality changes but does so imperfectly, and with a time lag. Some 2008 vehicles are cheaper than 2007. COLAs are then higher than inflation.

11. 1/10/2016rd11 How Inflation Works Pay for work = $10/hour Price of bread = $1/loaf One year later Pay for work = $11/hour ~ a 10% increase Price of bread = $1.10/loaf~ a 10% increase Conclude: Still can only buy 10 loaves of bread. Are you better off? Keeping pace with inflation? Demand-pull inflation ~ Too many dollars chasing too few goods. Cost-push inflation ~ price hikes like oil.

12. 1/10/2016rd12 Inflation Calculations Year f = 4%; i r = 10% A$ i m = 14.4% R$ PW(R$) R$(P/F, I r, n) 0 5000 5000 3415 15000*0.04=200 52005000 4545 25200*0.04=208 54085000 4132 35408*0.04=216 56245000 3757 45624*0.04=225 58495000 3415

13. 1/10/2016rd13 Gain in Purchasing Power Suppose you want a real return (i r ) of 4% on $1000 investment for one year with inflation f at 10%. Compute Fat one year. F = 1000(1 + f) = 1K(1.1) = $1100 to keep pace with inflation. F = 1000(1 + f)(1 + 0.04) = $1144. (1 + f)(1 + r) = [1 + f + r + fr] = 1 + m Thus (1 + market rate) = (1 + inflation)(1 + real) (1 + i m ) = (1 + f)(1 + i r ) => i m = (1 + f)(1 + i r ) – 1 = 1.1 * 1.04 – 1 = 14.4% i m = f + i r + f * i r = 0.1 + 0.04 + 0.1* 0.04 = 14.4%

14. 1/10/2016rd14 Real Interest Rate i m = (1 + f)(1 + i r ) – 1 i r =

15. 1/10/2016rd15 Example 1 Given P = $3000; F = $5000, N = 5 years, f = 5%: Find the market RoR and the real rate of return. Method I: 5/3 = (1 + i m ) 5 => i m = 10.76% i r = (1 + i m )/(1 + f) – 1 = 1.1076/1.05 - 1 = 5.49% real interest gain Method II F 5-real = 5K(P/F, 5%, 5) = $3917.63 (IGPFN 3000 3917.63 5)  5.49%

16. 1/10/2016rd16 Example 2 Given P = -$3000, A = $400, F 4 = $2000, N = 4, f = 5% Find i m and i r. PW(i m ) = -3000 + 400(P/A, i m, 4) + 2000(P/F, i m, 4) = 0 (IRR ‘(-3000 400 400 400 2400))  5.677% => i m = 5.68% (UIRR 3000 400 4 2000)  5.68% Adjusting the cash flow each year for inflation, (IRR ‘(-3000 380.95 362.81 345.54 1974.49))  0.65% i r = 0.65% at base year 0.

17. Jack vs. Phil Nicklaus won the Masters for $20K in 1963; Mickelson in 2004 for $1.17M. Use CPI of 91.7 in 1963 and 561.23 in 2004. Average inflation is 4.517% = (IGPFN 91.7 561.23 41) 20K(1.04517) 41 = $122,405.67 << $1.17M, but Jack had 41 years to invest his money. 1/10/2016rd17

18. 1/10/2016rd18 Example 3 A B N = 20 years First Cost$600K$1000K AOC 30K 10K a)No inflation B – A: 400 = 20(P/A, i m, 20) => i m = 0% Select A for any positive MARR. b)Assume inflation f at 5% per year. Select A.

19. 1/10/2016rd19 Example 4 You borrow $100K today to repay in 3 years at i m = 11%. Find A$ after 3 years, i r and R$ equivalent to purchasing power with f = 5%. (A$) 3 = 100K(F/P, 11%, 3) = $136,763.10 i r = 5.714% = (1 + i m )/(1 + f) - 1 (R$) 3 = 100K(1.05714) 3 = $118,140.15 = 100K(F/P, 11%, 3)(P/F, 5%, 3) = $118,140.15

20. 1/10/2016rd20 Fixed vs. Responsive Annuity Fixed f = 6% Responsive n A$ R$ A$R$ 1 2000 188721202000 22000 178022472000 32000 16792382 2000 42000 15842525 2000 …… … … …

21. Example A company borrows $100,000 today to repay at the end of 3 years at a market rate i m = 11%. Compute the A$ owed at the end of 3 years, the actual and real rate of return to the lender, and the R$ at the end of the third year with inflation f at 5% per year. (A$) = 100,000(F/P, 11%, 3) = $136,763.10. (R$) = 100,000(F/P, 5.714%, 3) = $118,141.11 IRR (actual) = 11% vs. IRR(real) = (0.11 - 0.05) / 1.05 = 5.714% = i r. Note that $136,763(P/F, 5%, 3) = $118,140.15 1/10/2016rd21

22. Bond Rate You want to buy a $1000 par-value bond at 9.5% annual coupon rate with inflation at 4%. Compute your real rate of return if you sell the bond for $1080 after 2 years. n012 cf -1000 95 1175 (IRR '(-1000 95 1175))  13.25% is the market rate RoR. (UIRR 1000 95 2 1080)  13.251436% Find the real RoR with f = 4%. (IRR '(-1000 91.36 1086.35))  8.8961679% = i r or use f = 4 and i m = 13.25%. 1/10/2016rd22

23. Problem 14-11 In last 5 years prices have increased a total of 50%. Find f. F = P(1 + i) n 1.5 = 1(1 + f) 5 => f = 8.4472% 1/10/2016rd23

24. Problem 14-15 Sally loaned a friend $10K at 15%/yr to be repaid in 5 uniform payments. Inflation f = 12%. Compute Sally's real RoR with this inflation rate. A = 10K(A/P, 15%, 5) = $2983.16 (UIRR 10e3 2983.16 5)  15% = i m Use f and i m to get i r = 2.6786% (UIRR 10e3 2163.55 5)  2.6786% where 2163.55 = 10K(A/P, 2.6786%, 5) 1/10/2016rd24

25. Problem 14-21 With inflation at 6%/yr, how long in years for the purchasing power to be 1/5 of the present value? 1 = (1/5)(1 + 0.06) n (NGPFI 1/5 1 6)  27.62 years 1/10/2016rd25

26. Problem 14-23 Sal invests in a lot for $18K for 10 years seeking an i r = 10% After Tax RoR. f = 6% and the capital gains tax rate is 15%. Find Sal’s selling price. Let x = selling price. Sal pay 15% of (X – 18K) = 0.15X - 2700 n BTCF TI 15%Tax RateATCF 0 -18K -18K 10 X (X – 18K) -0.15X + 2700 0.85X + 2700 i m = (i r + f + i r * f) in percent = 16.6% 18K = (0.85X + 2700)(P/F, 16.6%, 10) => X = $95,188.19; or 18K(F/P, 16.6%, 10) = 0.85X + 2700 => X = $95,188.19 1/10/2016rd26

27. Problem 14-36 Year PSIChange in PSI 1991 82 3.22% 1992 89 8.50 1993 100 aa = (100-89)/89 = 12.36% 1994 b 4.00(b-100)100 = 0.04; b = 104 1995 107 c(107-104)/104 = c = 2.88% 1996 116 d9/107 = d = 8.41% 1997 e 5.17 (e-116)/116 = 5.17 => e = 122 1998 132 7.58 b) Base year is 1993 where PSI = 100 c) 1991-1995 107/82 = (1 + f) 4 => f = 6.88% 1992-1998 132/89 = (1 + f) 6 => f = 6.79% 1/10/2016rd27

28. Problem 14-37 nCost 1 st MailLCI 1970 $0.06100 ……… 1979 0.15250 250 = 100 (1 + f) 9 => i = 10.72% (IGPFN 100 250 9)  10.72% 1/10/2016rd28

29. Problem 14-41 Ten years ago the EAT index was 330 and averaged an increase of 12%/year, calculate current value of index. CI= 330(1.12) 10 = 1024.93 1/10/2016rd29

30. Problem 14-42 ItemCost nowYear 1Year 2Year 3 Structural$120K4.3%3.2%6.6% Roofing 140K2.02.53.0 Heating 35K1.62.13.6 Insulation 9K5.86.07.5 Labor 85K5.04.5 1/10/2016rd30 I m = 25% Find labor costs for years 1-3. 85K(1.05) = $89,250, 93,266.25; 97,463.23 Average % increase for labor (IGPFN 85 97.46323 3)  4.67% Find PW(insulation) = $9K Find future worth of insulation and labor Insulation = 9K(1.058)(1.06)(1.075) = $10,850.32 vs. $17,578.13 at 25% Labor = 97,463.23 => Future cost of (Labor + Insulation) = $108,313.55 Future worth 94K(F/P, 25%, 3) = $183,593.75

31. Problem 14-50 Tom bought a $10K 5-year bond with coupon rate of 12%/year. Tax rate is 42% with inflation f = 7% per year. Find his Before Tax RoR without inflation. 12% Find AT RoR without inflation and with inflation. n BTCF TI Tax ATCF f = 7% 0 -10K -10K-10K 1 1200 1200 -504 696 650.47 6.96% w/o inflation 21200 " " " 607.91 (1– 0.42)*0.12 = 6.96% 31200 " " " 568.14 41200 " " " 530.98 51200 " " " 496.24 5 10K10K$7129.86+ 496.24 = 74843.67 Before tax RoR is just the 12%. After Tax RoR is 696/10K = 6.96% With inflation, i r is not positive since the sum is less than $10K. (IRR '(-10000 650.47 607.91 568.14 530.98 7483.67))  -0.3645% 1/10/2016rd31

32. 1/10/2016rd32 Problem 14-55 F = 5% using straight line depreciation, MARR = 7%. nBTCFBTCF(A$ 5%)DepTITax-25% ATCF(A$) R$ 0-420-420 -$420 -$420 1 200 21014070 -17.5 192.5 183.33 2 200 220.5014080.5-20.125 200.375 181.75 3 200231.5314091.53-22.88 208.65 180.24 nBTCFBTCF(A$ 5%)DepTITax-25% ATCF(A$) R$ 0-300-300 -$300 -$300 1 150 157.510057.5 -14.4 143.1 136.3 2 150 165.410065.4-16.4 149.0 135.1 3 150 173.610073.6-18.41 155.2 134.1 A-B: (IRR '(-120 47 46.7 46.1))  8.07% > 7% => Choose A.

33. 1/10/2016rd33 Deflation or Negative Inflation The deflation rate is 2% per year for the next 5 years. A par value bond of $10K with an annual coupon rate of 5% is bought and had a real rate of return of 4%. How much was paid for the bond? i m = (1 – 0.02)(1 + 0.04) - 1 = 1.92% PW = 500(P/A, 1.92%, 5) + 10K(P/F, 1.92%, 5) = $11,455.12 (IRR '(-11455.12 500 500 500 500 10500))  1.92% (UIRR 11455.12 500 5 10000)  1.92%

34. Deflation In 1993 the semiconductor producer price index was 141.9 and today is 69.5. Compute the average rate of deflation. 69.5 = 141.9(1 + f) 14 => f = -5.231% => deflation. 1/10/2016rd34

35. Example An engineer earns $50K now with a 4.3% annual increase. Inflation is at 2.3%. Compute the engineer's actual dollars over the next 5 years. Engineer's A$ A$1 = 50K(1.043) 0 (P/F,1.023%,1) = $48,875.86 A$2 = 50K(1.043) 1 (P/F,1.023%,2) = $49,831.40 A$3 = 50K(1.043) 2 (P/F,1.023%,3) = $50,805.62 A$4 = 50K(1.043) 3 (P/F,1.023%,4) = $51,798.88 A$5 = 50K(1.043) 4 (P/F,1.023%,5) = $52,811.57 Real rate of annual increase is not 4.3%, but 1.95%. 1/10/2016rd35

36. Exchange Rate The exchange rate for a European firm in 2007 was 0.6868 euro to purchase $1 US. Suppose a contract showed that a US company was to receive 11M Euros being earned equal payments quarterly over 1 and ½ years with an increase of 0.125% per quarter in the exchange rate. Show payments to US firm. 11M(A/P, 0.125%, 6) = 1,841,362.52  euro A 1 = 1.841M(P/F, 0.125%, 1) / 0.6868 = $2,684,427 A 2 = 1.841M(P/F, 0.125%, 2) / 0.6868 = $2,673,859 A 3 = 1.841M(P/F, 0.125%, 3) / 0.6868 = $2,670,520 shrinking US A 4 = 1.841M(P/F, 0.125%, 4) / 0.6868 = $2,667,187 dollar A 5 = 1.841M(P/F, 0.125%, 5) / 0.6868 = $2,663,857 A 6 = 1.841M(P/F, 0.125%, 6) / 0.6868 = $2,660,531 Notice shrinking US dollars with time due to increasing exchange rate 1/10/2016rd36

37. Monthly Payment You bought an auto for $25K to be repaid for 4 years at 12% interest compounded monthly. If general inflation is 6% compounded monthly, find actual and real value of 20 th payment. A$ 20 = $25K(A/P, 1%, 48) = $658.35 R$ 20 = 658.35(P/F, ½%, 20) = $595.84 1/10/2016rd37

38. Geometric Gradient and Inflation Four payments beginning at $7,000 at the end of year 1 and growing at the rate of 8% per year with market rate at 13%/year and general inflation at 7%/year. Find the present worth of the series. (PGGG 7000 8 5.60747 4)  $27,428.25, where i m = 13% and f = 7% yielding i r = 5.60747%. Note real cash flow is 7000 7560 8165 8818 actual cash flow is 7490 8655 10,002 11,559 PW = 7490(1.13) -1 + … + 11599(1.13) -4 = $27,428.25 1/10/2016rd38

39. Monthly Payments under Inflation You borrow $20K at 12% compounded monthly for 5 years. Average monthly general inflation rate is 0.5%. Determine equivalent equal monthly payment in real dollars. A$ = 20K(A/P, 1, 60) = $444.89 R$ = 20K(A/P, 0.4975%, 60) = $386.38 where i r = 0.4975%. 1/10/2016rd39

40. 1/10/2016rd40 Inflation 1.A machine costing $2550 4 years ago now costs $3930 with general inflation at 7% per year. Calculate the true percentage increase in the cost of the machine. a) 14.95% b) 54.12% c) 35.11% d) 7% e) 17.58% 2.If you want a 7% inflation-free return on your investment with f = 9% per year, your actual interest rate must be closest to a)16% b) 20% c) 12% d) 15% e) 14% f) 17% 3.I want $25K per year forever in R$ when I die for my family. Insurer offers 7% per year while inflation is expected to be 4% per year. First payment is 1 year after my death. How much insurance do I need? a) $867K b) $357K c) $625K d) $841K e) $750K 1/10/2016rd40

41. 1/10/2016rd41 4.If the market rate is x% when the inflation rate is y% and the real rate of return is z%, what is the market rate if f = z% and i r = y%? a) > x%b) < x%c) same as x%d) cannot determine 5. Your brother needs a $5,000 loan to go to college. Because of his poverty, he will pay nothing for the next four years. Five years from today he will begin paying you $2500 a year for the next 4 years. The first payment occurs 5 years from today and the total of the four payments will be $10,000. a. First consider the investment without inflation. If your minimum rate of return is 8%, is this an acceptable investment? F 4 = 5K(1.08) 4 = $6802.44 vs. 2500(P/A, 8%, 4) = $8280.32 YES b. For the same payment schedule but with a 5% rate of inflation, is this an acceptable investment? Note that your brother pays you $2500 a year regardless of the inflation rate. Your real MARR is 8%. i M = 13.4% => F 4 = 5K(1.134) 4 = $8268.41 vs. 2500(P/A,13.4%,4) = $7374.80. NO

42. 1/10/2016rd42 6.The cost of utilities for a fixed amount of power is shown below. Find the inflation rate for each year and then f-bar for the 3 years. n0123 cost$350K374.5K397.7421.6K 7% 6.2%6% F-bar = (IGPFN 350 421.6 3)  6.4% 7.Find the present worth of the cash flow in #6 with i r = 5%. PW = 350 + 350(P/A, 5%, 3) = $1303.137K

43. 1/10/2016rd43 8. First cost $125123 Annual Expenses 404244.146.31 Annual revenue 100105110.25115.76 Salvage value 5052.555.12557.88 Depreciation 3-yr MACRS 41.662504 55.5625 18.5125 9.2625 Inflation 5% NPW ? Net Income63 66.15 69.45 Depreciation 41.66 55.56 9.26 Taxable Income22.66 10.59 60.19 Tax Rate (40%) 9.06 4.24 24.08 Net Income 13.6 6.35 36.11 Depreciation 41.66 55.56 9.26 Salvage at 5% 57.88 Gains Tax Net cash flow (A$)-12555.26 61.91 47.37 Depreciation costs are understated and taxable income is overstated (higher taxes resulting)