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

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Engineering Economic AnalysisEngineering Economic Analysis

Chapter 14 Inflation and Price Change

Inflation Rate FactorsInflation 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.

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Inflation ProblemsInflation 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.

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Combating InflationCombating 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.

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InflationInflation

A$ Actual Dollars, nominal, inflated, future, then-current

R$ Real Dollars, constant-value, today, relative to base year

f General price inflation rate

im Market interest rate

ir Real or effective interest rate

Base time period

(1 )(1 )

(1 ) 1m m

r r

i i fi i

f f

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InflationInflation

Inflation Rate ~ f

Market Rate ~ im

Real Rate ~ ir

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).

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Inflation and Price IndexesInflation and Price Indexes

Construction Cost Index Average % Year Index value rate/year1950 510 ---1955 660 5.3 660 = 510(1 + i)5 => i = 5.291960 824 4.5 824 = 660(1 + i)5 => i = 4.54….1995 5471 6.11 5471 = 510(1 + i)40 => i = 6.111955-1995 (40 years) 6.11

General inflation rate ~ average rate CPI of items in market basketSpecific inflation rate ~ pertains to a segment of the economy

CPIk = [(CPI)k - (CPI)k-1](100)/ (CPI)k-1

Prices over TimePrices over Time

Consumer Purchase 2003 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

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Price RelativesPrice Relatives

Year Cost ($) Gallon Price Relatives (Base 1976 =100)1976 0.58 1001977 0.63 (0.63/0.58)(100) = 1091978 0.68 (0.68/0.58)(100) = 1171979 0.93 (0.93/0.58)(100) = 1601980 1.20 (1.20/0.58)(100) = 207

Price Relative in Period t = Price in Period t (100) Base Period Price

CPI1980 = 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.

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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.

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How Inflation WorksHow Inflation Works

Pay for work = $10/hourPrice of bread = $1/loaf

One year later

Pay for work = $11/hour ~ a 10% increasePrice 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.

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Inflation CalculationsInflation Calculations

Year f = 4%; ir = 10% A$

im = 14.4%

R$ PW(R$) R$(P/F, Ir, n)

0 5000 5000 5000 3415

1 5000*0.04=200 5200 5000 4545

2 5200*0.04=208 5408 5000 4132

3 5408*0.04=216 5624 5000 3757

4 5624*0.04=225 5849 5000 3415

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Gain in Purchasing PowerGain in Purchasing Power

Suppose you want a real return (ir) 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 + im) = (1 + f)(1 + ir) => im = (1 + f)(1 + ir) – 1= 1.1 * 1.04 – 1 = 14.4%

im = f + ir + f * ir = 0.1 + 0.04 + 0.1* 0.04 = 14.4%

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Real Interest RateReal Interest Rate

im = (1 + f)(1 + ir) – 1

ir = m m f

f f

1+i i -i-1 =

1+i 1+i

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Example 1Example 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 + im)5 => im = 10.76%

ir = (1 + im)/(1 + f) – 1

= 1.1076/1.05 - 1 = 5.49% real interest gain

Method II F5-real = 5K(P/F, 5%, 5) = $3917.63

(IGPFN 3000 3917.63 5) 5.49%

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Example 2Example 2

Given P = -$3000, A = $400, F4 = $2000, N = 4, f = 5%Find im and ir.

PW(im) = -3000 + 400(P/A, im, 4) + 2000(P/F, im, 4) = 0

(IRR ‘(-3000 400 400 400 2400)) 5.677% => im = 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%

ir = 0.65% at base year 0.

Jack vs. PhilJack 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.

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Example 3Example 3

A B N = 20 yearsFirst Cost $600K $1000KAOC 30K 10K

a) No inflation B – A: 400 = 20(P/A, im, 20) => im = 0%Select A for any positive MARR.

b) Assume inflation f at 5% per year. Select A.

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Example 4Example 4

You borrow $100K today to repay in 3 years at im = 11%. Find A$ after 3 years, ir and R$ equivalent to purchasing power with f = 5%.

(A$)3 = 100K(F/P, 11%, 3) = $136,763.10

ir = 5.714% = (1 + im)/(1 + f) - 1

(R$)3 = 100K(1.05714)3 = $118,140.15

= 100K(F/P, 11%, 3)(P/F, 5%, 3)

= $118,140.15

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Fixed vs. Responsive AnnuityFixed vs. Responsive Annuity

Fixed f = 6% Responsive

n A$ R$ A$ R$

1 2000 1887 2120 20002 2000 1780 2247 20003 2000 1679 2382 20004 2000 1584 2525 2000

… … … … …

ExampleExampleA company borrows $100,000 today to repay at the end of 3 years at a market rate im = 11%. Compute the A$ 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% = ir.

Note that $136,763(P/F, 5%, 3) = $118,140.15

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Bond RateBond 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.

n 0 1 2

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% = ir or

use f = 4 and im = 13.25%.

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Problem 14-11Problem 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%

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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% = im

Use f and im to get ir = 2.6786%

(UIRR 10e3 2163.55 5) 2.6786%

where 2163.55 = 10K(A/P, 2.6786%, 5)

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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

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Sal invests in a lot for $18K for 10 years seeking an ir = 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 Rate ATCF

0 -18K -18K

10 X (X – 18K) -0.15X + 2700 0.85X + 2700

im = (ir + f + ir * 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

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Year PSI Change in PSI

1991 82 3.22%

1992 89 8.50

1993 100 a a = (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 d 9/107 = 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%

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n Cost 1st Mail LCI

1970 $0.06 100

… … …

1979 0.15 250

250 = 100 (1 + f)9 => i = 10.72%

(IGPFN 100 250 9) 10.72%

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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

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Item Cost now Year 1 Year 2 Year 3

Structural $120K 4.3% 3.2% 6.6%

Roofing 140K 2.0 2.5 3.0

Heating 35K 1.6 2.1 3.6

Insulation 9K 5.8 6.0 7.5

Labor 85K 5.0 4.5 4.5

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Im = 25% Find labor costs for years 1-3. 85K(1.05) = $89,250, 93,266.25; 97,463.23Average % increase for labor (IGPFN 85 97.46323 3) 4.67%Find PW(insulation) = $9KFind 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) = $108,313.55Future worth 94K(F/P, 25%, 3) = $183,593.75

Problem 14-50Problem 14-50Tom 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 -10K1 1200 1200 -504 696 650.47 6.96% w/o inflation2 1200 " " " 607.91 (1– 0.42)*0.12 = 6.96%3 1200 " " " 568.144 1200 " " " 530.985 1200 " " " 496.245 10K 10K $7129.86+ 496.24 = 74843.67

Before tax RoR is just the 12%. After Tax RoR is 696/10K = 6.96%

With inflation, ir is not positive since the sum is less than $10K.

(IRR '(-10000 650.47 607.91 568.14 530.98 7483.67)) -0.3645%

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Problem 14-55F = 5% using straight line depreciation, MARR = 7%.

n BTCF BTCF(A$ 5%) Dep TI Tax-25% ATCF(A$) R$

0 -420 -420 -$420 -$4201 200 210 140 70 -17.5 192.5 183.332 200 220.50 140 80.5 -20.125 200.375 181.753 200 231.53 140 91.53 -22.88 208.65 180.24

n BTCF BTCF(A$ 5%) Dep TI Tax-25% ATCF(A$) R$

0 -300 -300 -$300 -$3001 150 157.5 100 57.5 -14.4 143.1 136.32 150 165.4 100 65.4 -16.4 149.0 135.13 150 173.6 100 73.6 -18.41 155.2 134.1

A-B: (IRR '(-120 47 46.7 46.1)) 8.07% > 7% => Choose A.

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Deflation or Negative InflationDeflation 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?

im = (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%

DeflationDeflation

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.

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ExampleExample

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%.

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Exchange RateExchange 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

A1 = 1.841M(P/F, 0.125%, 1) / 0.6868 = $2,684,427

A2 = 1.841M(P/F, 0.125%, 2) / 0.6868 = $2,673,859

A3 = 1.841M(P/F, 0.125%, 3) / 0.6868 = $2,670,520 shrinking US

A4 = 1.841M(P/F, 0.125%, 4) / 0.6868 = $2,667,187 dollar

A5 = 1.841M(P/F, 0.125%, 5) / 0.6868 = $2,663,857

A6 = 1.841M(P/F, 0.125%, 6) / 0.6868 = $2,660,531

Notice shrinking US dollars with time due to increasing exchange rate

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Monthly PaymentMonthly 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 20th payment.

A$20 = $25K(A/P, 1%, 48) = $658.35

R$20 = 658.35(P/F, ½%, 20) = $595.84

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Geometric Gradient and InflationGeometric 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 im = 13% and f = 7% yielding ir = 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

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Monthly Payments under InflationMonthly 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 ir = 0.4975%.

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InflationInflation1. 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

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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 ir = 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?

F4 = 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%.

iM = 13.4% => F4 = 5K(1.134)4 = $8268.41 vs. 2500(P/A,13.4%,4) = $7374.80. NO

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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.n 0 1 2 3

cost $350K 374.5K 397.7 421.6K7% 6.2% 6%

F-bar = (IGPFN 350 421.6 3) 6.4%

7. Find the present worth of the cash flow in #6 with ir = 5%.

PW = 350 + 350(P/A, 5%, 3) = $1303.137K

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8. First cost $125 1 2 3 Annual Expenses 40 42 44.1 46.31 Annual revenue 100 105 110.25 115.76 Salvage value 50 52.5 55.125 57.88 Depreciation 3-yr MACRS 41.662504 55.5625 18.5125 9.2625 Inflation 5% NPW ? Net Income 63 66.15 69.45 Depreciation 41.66 55.56 9.26 Taxable Income 22.66 10.59 60.19 Tax Rate (40%) 9.06 4.24 24.08 Net Income 13.6 6.35 36.11 Depreciation 41.66 55.56 9.26 Salvage at 5% 57.88 Gains Tax Net cash flow (A$) -125 55.26 61.91 47.37

Depreciation costs are understated and taxable income is overstated (higher taxes resulting)

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

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