Ngày đăng: 13/09/2018, 10:28
Chapter 14: Inflation and Price Change 14-1 During times of inflation, the purchasing power of a monetary unit is reduced In this way the currency itself is less valuable on a per unit basis In the USA, what this means is that during inflationary times our dollars have less purchasing power, and thus we can purchase less products, goods and services with the same $1, $10, or $100 dollar bill as we did in the past 14-2 Actual dollars are the cash dollars that we use to make transactions in our economy These are the dollars that we carry around in our wallets and purses, and have in our savings accounts Real dollars represent dollars that not carry with them the effects of inflation, these are sometimes called “inflation free” dollars Real dollars are expressed as of purchasing power base, such as Year-2000-based-dollars The inflation rate captures the loss in purchasing power of money in a percentage rate form The real interest rate captures the growth of purchasing power, it does not include the effects of inflation is sometimes called the “inflation free” interest rate The market interest rate, also called the combined rate, combines the inflation and real rates into a single rate 14-3 There are a number of mechanisms that cause prices to rise In the chapter the authors talk about how money supply, exchange rates, cost-push, and demand pull effects can contribute to inflation 14-4 Yes Dollars, and interest rates, are used in engineering economic analyses to evaluate projects As such, the purchasing power of dollars, and the effects of inflation on interest rates, are important The important principle in considering effects of inflation is not to mix-and-match dollars and interest rates that include, or not include, the effect of inflation A constant dollar analysis uses real dollars and a real interest rate, a then-current (or actual) dollar analysis uses actual dollars and a market interest rate In much of this book actual dollars (cash flows) are used along with a market interest rate to evaluate projects this is an example of the later type of analysis 14-5 The Consumer Price Index (CPI) is a composite price index that is managed by the US Department of Labor Statistics It measures the historical cost of a bundle of “consumer goods” over time The goods included in this index are those commonly purchased by consumers in the US economy (e.g food, clothing, entertainment, housing, etc.) Composite indexes measure a collection of items that are related The CPI and Producers Price Index (PPI) are examples of composite indexes The PPI measures the cost to produce goods and services by companies in our economy (items in the PPI include materials, wages, overhead, etc.) Commodity specific indexes track the costs of specific and individual items, such as a labor cost index, a material cost index, a “football ticket” index, etc Both commodity specific and composite indexes can be used in engineering economic analyses Their use depends on how the index is being used to measure (or predict) cash flows If, in the analysis, we are interested in estimating the labor costs of a new production process, we would use a specific labor cost commodity index to develop the estimate Much along the same lines, if we wanted to know the cost of treated lumber years from today, we might use a commodity index that tracks costs of treated lumber In the absence of commodity indexes, or in cases where we are more interested in capturing aggregate effects of inflation (such as with the CPI or PPI) one would use a composite index to incorporate/estimate how purchasing power is affected 14-6 The stable price assumption is really the same as analyzing a problem in Year dollars, where all the costs and benefits change at the same rate Allowable depreciation charges are based on the original equipment cost and not increase Thus the stable price assumption may be suitable in some before-tax computations, but is not satisfactory where depreciation affects the income tax computations 14-7 F = P (F/P, f%, 10 yrs) = $10 (F/P, 7%, 10) = $10 (1.967) = $19.67 14-8 iequivalent = i’inflation corrected + f% + (i'inflation corrected) (f%) In this problem: iequivalent = 5% f% = +2% iinflation corrected = unknown 0.05 = i’inflation corrected + 0.02 + (i'inflation corrected) (0.02) i'inflation corrected = (0.05 – 0.02)/(1 + 0.02) = 0.02941 = 2.941% That this is correct may be proved by the year-by-year computations Year Cash Flow -$1,000 +$50 +$50 (1 + f)-n (P/F, f%, n) 0.9804 0.9612 Cash Flow in Year dollars -$1,000.00 +$49.02 +$48.06 PW at 2.941% -$1,000.00 +$47.62 +$45.35 10 11 12 13 14 15 16 17 18 19 20 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$50 +$1,000 0.9423 0.9238 0.9057 0.8880 0.8706 0.8535 0.8368 0.8203 0.8043 0.7885 0.7730 0.7579 0.7430 0.7284 0.7142 0.7002 0.6864 0.6730 +$47.12 +$46.19 +$45.29 +$44.40 +$43.53 +$42.68 +$41.84 +$41.02 +$40.22 +$39.43 +$38.65 +$37.90 +$37.15 +$36.42 +$35.71 +$35.01 +$34.32 +$706.65 +$43.20 +$41.13 +$39.18 +$37.31 +$35.54 +$33.85 +$32.23 +$30.70 +$29.24 +$27.85 +$26.52 +$25.26 +$24.05 +$22.90 +$21.82 +$20.78 +$19.79 +$395.76 +$0.08 Therefore, iinflation corrected = 2.94% 14-9 $20,000 in Year dollars n = 14 yrs P = Lump Sum deposit Actual Dollars 14 years hence = $20,000 (1 + f%)n = $20,000 (1 + 0.08)14 = $58,744 At 5% interest: P = F (1 + i)-n = $58,744 (1 + 0.05)-14 = $29,670 Since the inflation rate (8%) exceeds the interest rate (5%), the money is annual losing purchasing power Deposit $29,670 14-10 To buy $1 worth of goods today will require: F = P (F/P, f%, n) n years hence F = $1 (1 + 0.05) = $1.47 years hence For the subsequent years the amount required will increase to: $1.47 (F/P, f%, n) = $1.47 (1 + 0.06)5 = $1.97 Thus for the ten year period $1 must be increased to $1.97 The average price change per year is: ($1.97 - $1.00)/10 yrs = 9.7% per year 14-11 (1 + f)5 (1 + f) f = 1.50 = 1.501/5 = 0.845 = 1.0845 = 8.45% 14-12 Number of dollars required five years hence to have the buying power of one dollar today = $1 (F/P, 7%, 5) = $1.403 Number of cruzados required five years hence to have the buying power of 15 cruzados today = 15 (F/P, 25%, 5) = 45.78 cruzados Combining: $1.403 = 45.78 cruzados $1 = 32.6 cruzados (Brazil uses cruzados.) 14-13 Price increase = (1 + 0.12)8 Therefore, required fuel rating = 2.476 x present price = 10 x 2.476 = 24.76 km/liter 14-14 P= 1.00 F = 1.80 n = 10 1.80 = 1.00 (F/P, f%, 10) (F/P, f%, 10) = 1.80 From tables, f is slightly greater than 6% (f = 6.05% exactly) 14-15 i = i' + f + (i') (f) 0.15 = i' + 0.12 + 0.12 (i') 1.12 i' = 0.03 i' = 0.03/1.12 = 0.027 = 2.7% f = ? 14-16 Compute equivalent interest/3 mo =x = (1 + x)n – = (1 + x)4 – = 1.19250.25 = 1.045 = 0.045 = 4.5%/3 mo ieff 0.1925 (1 + x) x $3.00 n=? i = 4.5% $2.50 $2.50 = $3.00 (P/F, 4.5%, n) (P/F, 4.5%, n) = $2.50/$3.00 = 0.833 n is slightly greater than So purchase pads of paper- one for immediate use plus extra pads 14-17 f i' i = 0.06 = 0.10 = 0.10 + 0.06 + (0.10) (0.06) = 16.6% 14-18 (a) $109.6 1981 1986 n=5 $90.9 $109.6 (F/P, f%, 5) f% = $90.9 (F/P, f%, 5) = $109.6/$90.9 = 1.2057 = 3.81% (b) CPI 1996 n=9 f = 3.81% $113.6 CPI1996 = $113.6 (F/P, 3.81%, 9) = $113.6 (1 + 0.0381)9 = $159.0 14-19 F = $20,000 (F/P, 4%, 10) = $29,600 14-20 Compute an equivalent i: iequivalent = i' + f + (i') (f) = 0.05 + 0.06 + (0.05) (0.06) = 0.113 = 11.3% Compute the PW of Benefits of the annuity: PW of Benefits = $2,500 (P/A, 11.3%, 10) = $2,500 [((1.113)10 – 1)/(0.113 (1.113)10)] = $14,540 Since the cost is $15,000, the benefits are less than the cost computed at a 5% real rate of return Thus the actual real rate of return is less than 5% and the annuity should not be purchased 14-21 log (1/0.20) = 0.20 (1.06)n = n log (1.06) n = 27.62 years 14-22 Use $97,000 (1 + f%)n, where f%=7% and n=15 $97,000 (1 + 0.07)15 = $97,000 (F/P, 7%, 15) = $97,000 (2.759) = $268,000 If there is 7% inflation per year, a $97,000 house today is equivalent to $268,000 15 years hence But will one have “profited” from the inflation? Whether one will profit from owning the house depends somewhat on an examination of the alternate use of the money Only the differences between alternatives are relevant If the alterate is a 5% savings account, neglecting income taxes, the profit from owning the house, rather than the savings account, would be: $268,000 - $97,000 (F/P, 5%, 15) = $66,300 On the other hand, compared to an alternative investment at 7%, the profit is $0 And if the alternative investment is at 9% there is a loss If “profit” means an enrichment, or being better off, then multiplying the price of everything does no enrich one in real terms 14-23 Let x = selling price Then long-term capital gain Tax After-Tax cash flow in year 10 Year 10 = x- $18,000 = 0.15 (x - $18,000) = x – 0.15 (x - $18,000) = 0.85x + $2,700 ATCF -$18,000 +0.85x + $2,700 Multiply by 1.06-10 Year $ ATCF -$18,000 0.4743x + $1,508 For a 10% rate of return: $18,000 = (0.4746x + $1,508) (P/F, 10%, 10) = 0.1830 x + $581 x = $95,186 Alternate Solution using an equivalent interest rate iequiv = i' + f + (i') (f) = 0.10 + 0.06 + (0.10) (0.06) = 0.166 So $18,000 (1 + 0.166)10 = 0.85x + $2,700 $83,610 = 0.85x + $2,700 Selling price of the lot =x = ($83,610 - $2,700)/0.85 14-24 Cash Flow: Year $500 Kit $900 Kit -$500 -$900 -$500 $0 (a) PW$500 kit PW$900 kit = $500 + $500 (P/F, 10%, 5) = $810 = $900 To minimize PW of Cost, choose $500 kit = $95,188 (b) Replacement cost of $500 kit, five years hence = $500 (F/P, 7%, 5) = $701.5 PW$500 kit PW$900 kit = $500 + $701.5 (P/F, 10%, 5) = $900 = $935.60 To minimize PW of Cost, choose $900 kit 14-25 If one assumes the 5-year hence cost of the Filterco unit is: $7,000 (F/P, 8%, 5) = $10,283 in Actual Dollars and $7,000 in Yr dollars, the year $ cash flows are: Year Filterco Duro Duro – Filterco -$7,000 -$10,000 -$3,000 -$7,000 $0 +$7,000 ∆ROR = 18.5% Therefore, buy Filterco 14-26 Year Cost to City (Year $) -$50,000 1- 10 -$5,000/yr Benefits to City Description of Benefits 10 +$50,000 i +A Fixed annual sum in thencurrent dollars In then-current dollars = i' + f + i'f = 0.03 + 0.07 + 0.03 (0.07) = 0.1021 = 10.21% PW of Cost $50,000 + $5,000 (P/A, 3%, 10) $50,000 + $5,000 (8.530) $92,650 A = ($92,650 - $18,915)/6.0895 = $12,109 * Computed on hand calculator = PW of Benefits = A(P/A, 10.21%, 10) +$50,000 (P/F,10.21%,10) = A (6.0895*) + $50,000 (0.3783*) = 6.0895A + $18,915 14-27 Month 1- 36 36 BTCF $0 -$1,000 +$40,365 $1,000 (F/A, i%, 36 mo) (F/A, i%, 36) = $40,365 = 40.365 Performing linear interpolation: (F/A, i%, 36)) i 41.153 ¾ % 39.336 ½ % i = 0.50% + 0.25% [(40.365 – 39.336)/(41.153 – 39.336)] = 0.6416% per month Equivalent annual interest rate i per year = (1 + 0.006416)12 – = 0.080 = 8% So, we know that i = 8% and f = 8% Find i' i = i' + f + (i') (f) 0.08 = i' + 0.08 + (i') (0.08) i' = 0% Thus, before-Tax Rate of Return = 0% 14-28 Actual Dollars: F= $10,000 (F/P, 10%, 15) = $41,770 Real Dollars: Year 1- 6- 10 11- 15 R$ in today’s base Inflation 3% 5% 8% = $41,770 (P/F, 8%, 5) (P/F, 5%, 5) (P/F, 3%, 5) = $18,968 Thus, the real growth in purchasing power has been: $18,968 = $10,000 (1 + i*)15 i* = 4.36% 14-29 (a) F = $2,500 (1.10)50 = $293,477 in A$ today (b) R$ today in (-50) purchasing power = $293,477 (P/F, 4%, 50) = $41,296 14-30 (a) PW = $2,000 (P/A, ic, 8) icombined = ireal + f + (ireal) (f) = 0.0815 PW (b) PW = 0.03 + 0.05 + (0.03) (0.05) = $2,000 (P/A, 8.15%, 9) = $11,428 = $2,000 (P/A, 3%, 8) = $14,040 14-31 Find PW of each plan over the next 5-year period ir = (ic – f)/(1 + f) = (0.08 – 0.06)/1.06 PW(A) PW(B) PW(C) = 1.19% = $50,000 (P/A, 11.5%, 5) = $236,359 = $45,000 (P/A, 8%, 5) + $2,500 (P/G, 8%, 5) = $65,000 (P/A, 1.19, 5) (P/F, 6%, 5) = $229,612 = $198,115 Here we choose Company A’s salary to maximize PW 14-32 (a) R today $ in year 15 = $10,000 (P/F, ir%, 15) ir = (0.15 – 0.08)/1.08 = 6.5% R today $ in year 15 = $10,000 (1.065)15 = $25,718 (b) ic = 15%f = 8% F = $10,000 (1.15)15 = $81,371 14-33 No Inflation Situation Alternative A: Alternative B: PW of Cost PW of Cost Alternative C: PW of Cost = $6,000 = $4,500 + $2,500 (P/F, 8%, 8) = $4,500 + $2,500 (0.5403) = $5,851 = $2,500 + $2,500 (P/F, 8%, 4) + $2,500 (P/F, 8%, 8) = $2,500 ( + 0.7350 + 0.5403) = $5,688 To minimize PW of Cost, choose Alternative C For f = +5% (Inflation) Alternative A: Alternative B: PW of Cost PW of Cost Alternative C: PW of Cost = $6,000 = $4,500 + $2,500 (F/P, 5%, 8) (P/F, 8%, 8) = $4,500 + $2,500 (1 + f%)8 (P/F, 8%, 8) = $4,500 + $2,500 (1.477) (0.5403) = $6,495 = $2,500 + $2,500 (F/P, 5%, 4) (P/F, 8%, 4) + $2,500 (F/P, 5%, 8) (P/F, 8%, 8) = $2,500 + $2,500 (1.216) (0.7350) + $2,500 (1.477) (0.5403) = $6,729 To minimize PW of Cost in year dollars, choose Alternative A This problem illustrates the fact that the prospect of future inflation encourages current expenditures to be able to avoid higher future expenditures 14-34 Alternative I: Continue to Rent the Duplex Home Compute the Present Worth of renting and utility costs in Year dollars Assuming end-of-year payments, the Year payment is: = ($750 + $139) (12) = $7,068 The equivalent Year payment in Year dollars is: $7,068 (1 + 0.05)-1 = $6,713.40 Compute an equivalent i iequivalent = i' + f + (i') (f) Where i' = interest rate without inflation = 15.5% f = inflation rate = 5% iequivalent = 0.155 + 0.05 + (0.155) (0.05) = 0.21275 = 21.275% PW of 10 years of rent plus utilities: = $6,731.40 (P/A, 21.275%, 10) = $6,731.40 [(1 + 0.21275)(10-1))/(0.21275 (1 + 0.21275)10)] = $6,731.40 (4.9246) = $33,149 An Alternative computation, but a lot more work: Compute the PW of the 10 years of inflation adjusted rent plus utilities using 15.5% interest PWyear = 12[$589 (1 + 0.155)-1 + $619 (1 + 0.155)-2 + … + $914 (1 + 0.155)-10] = 12 ($2,762.44) = $33,149 Alternative II: Buying a House $3,750 down payment plus about $750 in closing costs for a cash requirement of $4,500 Mortgage interest rate per month = (1+I)6 = 1.04 I = 0.656% n = 30 years x 12 = 360 payments Monthly Payment: A = ($75,000 - $3,750) (A/P, 0.656%, 360) = $71,250 [(0.00656 (1.00656)360)/((1.00656)360 – 1)] = -$516.36 Mortgage Balance After the 10-year Comparison Period: A’ = $523 (P/A, 0.656%, 240) = $523 [((1.00656)240 – 1)/(0.00656 (1.00656)240)] = $62,335 Thus: $523 x 12 x 10 $71,250 - $62,504 = $62,760 = $8,746 = $54,014 total payments principal repayments (12.28% of loan) interest payments Sale of the property at 6% appreciation per year in year 10: F = $75,000 (1.06)10 = $134,314 Less 5% commission = -$6,716 Less mortgage balance = -$62,335 Net Income from the sale = $65,263 Assuming no capital gain tax is imposed, the Present Worth of Cost is: PW= $4,500 [Down payment + closing costs in constant dollars] + $516.36 x 12 (P/A, 15.5%, 10) [actual dollar mortgage] + $160 x 12 (P/A, 10%, 10) [constant dollar utilities] + $50 x 12 (P/A, 10%, 10) [constant dollar insurance & maint.] - $65,094 (P/F, 15.5%, 10) [actual dollar net income from sale] PW= $4,500 + $516.36 x 12 (4.9246) + $160 x 12 (6.145) + $50 x 12 (6.145)- $65,263 (0.2367) = $35,051 The PW of Cost of owning the house for 10 years = $35,051 in Year dollars Thus $33,149 < $35,329 and so buying a house is the more attractive alternative 14-35 Year Cost- Cost- Cost- Cost- TOTAL $4,500 $4,613 $4,728 $4,846 $4,967 $7,000 $7,700 $8,470 $9,317 $10,249 $10,000 $10,650 $11,342 $12,079 $12,865 $8,500 $8,288 $8,080 $7,878 $7,681 $30,000 $31,250 $32,620 $34,121 $35,762 PWTOTAL $24,000 $20,000 $16,702 $13,976 $11,718 10 $5,091 $5,219 $5,349 $5,483 $5,620 $11,274 $12,401 $13,641 $15,005 $16,506 $13,701 $14,591 $15,540 $16,550 $17,626 $7,489 $7,302 $7,120 $6,942 $6,768 $37,555 $39,513 $41,649 $43,979 $46,519 $9,845 $8,286 $6,988 $5,903 $4,995 PW = -$60,000 – ($24,000 + $20,000 + $16,702 + … +$4,995) + $15,000 (P/F, 25%, 10) = $180,802 14-36 (a) Unknown Quantities are calculated as follows: % change = [($100 - $89)/$89] x 100% PSI = 100 (1.04) % change = ($107 - $104)/$104 % change = ($116 - $107)/$107 PSI = 116 (1.0517) (b) = 12.36% = 104 = 2.88% = 8.41% = 122 The base year is 1993 This is the year of which the index has a value of 100 (c) (i) (ii) PSI (1991) PSI (1995) h i* i* = 82 = 107 = years =? = (107/82)0.25 – = 6.88% PSI (1992) PSI (1998) n i* i* = 89 = 132 = years =? = (132/89)(1/6) – = 6.79% 14-37 (a) LCI(-1970) LCI(-1979) n i* i* (b) LCI(1980) LCI(1989) n i* i* = 100 = 250 =9 =? = (250/100)(1/9) – = 250 = 417 =9 =? = (417/250)(1/9) – (c) LCI(1990) LCI(1998) = 417 = 550 = 10.7% = 5.85% n i* i* =8 =? = (550/417)(1/8) – = 3.12% 14-38 (a) Overall LCI change (b) Overall LCI change (c) Overall LCI change = [(250 – 100)/100] x 100% = [(415 – 250)/250] x 100% = [(650 – 417)/417] x 100% = 150% = 66.8% = 31.9% 14-39 (a) CPI (1978) CPI (1982) n i* i* = 43.6 = 65.3 =4 =? = (65.3/43.6)(1/4) – = 9.8% (b) CPI (1980) CPI (1989) n i* i* = 52.4 = 89.0 =9 =? = (89.0/52.4)(1/9) – = 6.1% (c) CPI (1985) CPI (1997) n i* i* = 75.0 = 107.6 = 12 =? = (107.6/75.0)(1/12) – = 3.1% 14-40 (a) Year Brick Cost CBI 1970 2.10 442 1998 X 618 x/2.10 = 618/442 x = $2.94 Total Material Cost = 800 x $2.94 = $2,350 (b) Here we need f% of brick cost CBI(1970) = 442 CBI(1998) = 618 n = 18 i* =? i* = (618/442)(1/18) – = 1.9% We assume the past average inflation rate continues for 10 more years Brick Unit Cost in 2008 = 2.94 (F/P, 1.9%, 10) = $3.54 Total Material Cost = 800 x $3.54 = $2.833 14-41 EAT(today) = $330 (F/P, 12$, 10) = $1,025 14-42 Item Structural Roofing Heat etc Insulating Labor Total Year $125,160 $14,280 $35,560 $9,522 $89,250 $273,772 Year $129,165 $14,637 $36,306 $10,093 $93,266 $283,467 Year $137,690 $15,076 $37,614 $10,850 $97,463 $298,693 (a) $89,250; $93,266; $97,463 (b) PW = $9,522 (P/F, 25%, 1) + $10,093 (P/F, 25%, 2) + $10,850 (P/F, 25%, 3) = $19,632 (c) FW = ($9,522 + $89,250) (F/P, 25%, 2) + ($10,093 + $93,266) (F/P, 25%, 1) + ($10,850 + $97,463) = $391,843 (d) PW = $273,772 (P/F, 25%, 1) + $283,467 (P/F, 25%, 2) + $298,693 (P/F, 25%, 3) = $553,367 14-43 The total cost of the bike 10 years from today would be $2,770 Item Frame Wheels Gearing Braking Saddle Finishes Sum= Current Cost 800 350 200 150 70 125 1695 Inflation 2.0% 10.0% 5.0% 3.0% 2.5% 8.0% Sum= Future Cost 975.2 907.8 325.8 201.6 89.6 269.9 2769.8 14-44 To minimize purchase price Mary Clare should select the vehicle from company X Car X Y Z Current Price 27500 30000 25000 Inflation 4.0% 1.5% 8.0% Min= Future Price 30933.8 31370.4 31492.8 30933.8 14-45 FYEAR = $100 (F/A, 12/4=3%, x 4=20) = $2,687 FYEAR 10 = $2,687 (F/P, 4%, 20) + $100 (F/A, 4%, 20) = $8,865 FYEAR 15 (TODAY) = $8,865 (F/P, 2%, 20) + $100 (F/A, 2%, 20) = $15,603 14-46 To pay off the loan Andrew will need to write a check for $ 18,116 Year Amt due Begin yr 15000 15750.0 16773.8 Inflation 5.0% 6.5% 8.0% Due= Amt due End yr 15750.0 16773.8 18115.7 18115.7 14-47 See the table below for (a) through (e) Year years ago years ago years ago years ago last year This year Ave Price 165000.0 167000.0 172000.0 180000.0 183000.0 190000.0 Inflation for year (a) = 1.2% (b) = 3.0% (c) = 4.7% (d) = 1.7% (e) = 3.8% (f) see below One could predict the inflation (appreciation) in the home prices this year using a number of approaches One simple rule might involve using the average of the last years inflation rates This rate would be (1.2+3+4.7+1.7+3.8)/5 = 2.9% 14-48 Depreciation charges that a firm makes in its accounting records allow a profitable firm to have that amount of money available for replacement equipment without any deduction for income taxes If the money available from depreciation charges is inadequate to purchase needed replacement equipment, then the firm may need also to use after-tax profit for this purpose Depreciation charges produce a tax-free source of money; profit has been subjected to income taxes Thus substantial inflation forces a firm to increasingly finance replacement equipment out of (costly) after-tax profit 14-49 (a) Year BTCF -$85,000 $8,000 $8,000 $8,000 $8,000 $8,000 $77,500 Sum SL Deprec TI 34% Income Taxes $1,500 $1,500 $1,500 $1,500 $1,500 $6,500 $6,500 $6,500 $6,500 $6,500 $0 -$2,210 -$2,210 -$2,210 -$2,210 -$2,210 ATCF -$85,000 $5,790 $5,790 $5,790 $5,790 $83,290 $7,500 SL Depreciation = ($67,500 - $0)/45 = $1,500 Book Value at end of years = $85,000 – ($1,500) = $77,500 After-Tax Rate of Return = 5.2% (b) Year BTCF -$85,000 $8,560 $9,159 $9,800 $10,486 $11,220 $136,935* Sum SL Deprec TI 34% Income Taxes $1,500 $1,500 $1,500 $1,500 $1,500 $7,060 $7,659 $8,300 $8,986 $9,720 -$2,400 -$2,604 -$2,822 -$3,055 -$3,305 -$16,242** Actual Dollars ATCF -$85,000 $6,160 $6,555 $6,978 $7,431 $131,913 $7,500 *Selling Price = $85,000 (F/P, 10%, 5) = $85,000 (1.611) = $136,935 ** On disposal, there are capital gains and depreciation recapture Capital Gain = $136,935 - $85,00 = $51,935 Tax on Cap Gain = (20%) ($51,935) = $10,387 Recaptured Depr = $85,000 - $77,500 = $7,500 Tax on Recap Depr = (34%)($7,500) = $2,550 Total Tax on Disposal = $10,387 + $2,550 = $12,937 After Tax IRR = 14.9% After-Tax Rate of Return in Year Dollars Year Sum Actual Dollars ATCF -$85,000 $6,160 $6,555 $6,978 $7,431 $131,913 Multiply by 1.07-1 1.07-2 1.07-3 1.07-4 1.07-5 In year dollars, After-Tax Rate of Return Year $ ATCF -$85,000 $5,757 $5,725 $5,696 $5,669 $94,052 = 7.4% 14-50 Year BTCF TI 42% Income Taxes -$10,000 $1,200 $1,200 $1,200 $1,200 $1,200 $10,000 $1,200 $1,200 $1,200 $1,200 $1,200 -$504 -$504 -$504 -$504 -$504 ATCF Multiply by Year $ ATCF -$10,000 $696 $696 $696 $696 $10,696 1.07-1 1.07-2 1.07-3 1.07-4 1.07-5 -$10,000 $650 $608 $568 $531 $7,626 Sum -$17 (a) Before-Tax Rate of Return ignoring inflation Since the $10,000 principal is returned unchanged, i = A/P = $1,200/$10,000 = 12% If this is not observed, then the rate of return may be computed by conventional means $10,000 = $1,200 (P/A, i%, 5) + $10,000 (P/F, i%, 5) Rate of Return = 12% (b) After-Tax Rate of Return ignoring inflation Solved in the same manner as Part (a): i = A/P = $696/$10,000 = 6.96% (c) After-Tax Rate of Return after accounting for inflation An examination of the Year dollars after-tax cash flow shows the algebraic sum of the cash flow is -$17 Stated in Year dollars, the total receipts are less than the cost, hence there is no positive rate of return 14-51 Now: Taxable Income Income Taxes After-Tax Income = $60,000 = $35,000x0.16+25,000x0.22 = $11,100 = $60,000 - $11,100 = $48,900 Twenty Years Hence: To have some buying power, need: After-Tax Income = $48,900(1.07)20 = $189,227.60 = Taxable Income – Income Taxes Income Taxes = $24,689.04+0.29(Taxable Income - $113,804) Taxable Income = After-Tax Income + Income Taxes = $189,227.60+$24,689.04+0.22(TI - $113,804) = $242,153.50 14-52 P= CCA= t= S= $10,000 30% 50% f= 7% Year Actual $s Received -$10,000 $2,000 $3,000 $4,000 $5,000 $6,000 $7,000 $8,000 Actual CCA Actual $s Tax Net Salvage -$1,500 -$2,550 -$1,785 -$1,250 -$875 -$612 -$429 $250 $225 $1,108 $1,875 $2,563 $3,194 $3,786 $500 Actual $s ATCF -$10,000 $1,750 $2,775 $2,893 $3,125 $3,437 $3,806 $4,714 IRR= Real $s ATCF -$10,000 $1,636 $2,424 $2,361 $2,384 $2,451 $2,536 $2,936 13.71% Net Salvage Calculation from Equation 12-7, 12-8, & 11-8 UCC7 = $1,000 =$B$1*(1-$B$2/2)*(1-$B$2)^(7-1) Net Salvage end of yr = $500 =$B$4+$B$3*IF($B$4>$B$1,$C$18-$B$1,$C$18-$B$4)-1/2*$B$3*MAX(($B$4-$B$1),0) 14-53 P= CCA= t= S= $ $103,500 10% 35% 103,500 f= 0% Actual $s Received -$103,500 $15,750 $15,750 $15,750 $15,750 $15,750 Year Actual CCA -$5,175 -$9,833 -$8,849 -$7,964 -$7,168 Actual $s Tax $3,701 $2,071 $2,415 $2,725 $3,004 Net Salvage -$46,500 $136,354 Actual $s ATCF -$150,000 $12,049 $13,679 $13,335 $13,025 $149,100 Real $s ATCF -$150,000 $12,049 $13,679 $13,335 $13,025 $149,100 7.06% =IRR Net Salvage Calculation from Equation 12-7, 12-8, & 11-8 UCC5 =$64,511 =$B$1*(1-$B$2/2)*(1-$B$2)^(5-1) Net Salvage end of yr = $89,854 =$B$4+$B$3*IF($B$4>$B$1,$C$18-$B$1,$C$18-$B$4)-1/2*$B$3*MAX(($B$4-$B$1),0) 14-54 P= CCA= t= S= f= Year $ $103,500 10% 35% 103,500 10% Actual $s Received -$103,500 $12,000 $13,440 $15,053 $16,859 $18,882 Actual CCA Actual $s Tax Net Salvage -$46,500 -$5,175 -$9,833 -$8,849 -$7,964 -$7,168 $2,389 $1,263 $2,171 $3,113 $4,100 $230,694 Actual $s ATCF -$150,000 $9,611 $12,177 $12,882 $13,746 $245,476 16.01% Real $s ATCF -$150,000 $8,738 $10,064 $9,678 $9,389 $152,421 5.46% =IRR Net Salvage Calculation from Equation 12-7, 12-8, & 11-8 UCC5 = $64,511 =$B$1*(1-$B$2/2)*(1-$B$2)^(5-1) Net Salvage end of yr = $89,854 =$B$4+$B$3*IF($B$4>$B$1,$C$18-$B$1,$C$18-$B$4)-1/2*$B$3*MAX(($B$4-$B$1),0) Market Value in years = 150000 x (1+12%)^5 house value land sale capital gain's tax net land salvage $264,351.25 $103,500.00 $160,851.25 $ 20,011.47 $140,839.78 14-55 Alternative A Year Cash Flow in Year $ Cash Flow in Actual $ SL Deprec TI 25% Income Tax -$420 -$420 $200 $200 $200 $210 $220.5 $231.5 $140 $140 $140 $70 $80.5 $91.5 -$17.5 -$20.1 -$22.9 SL Deprec TI 25% Income Tax $100 $100 $100 $57.5 $65.4 $73.6 -$14.4 -$16.4 -$18.4 ATCF in Actual $ ATCF in Year $ -$420 -$420 $192.5 $200.4 $208.6 $183.3 $181.8 $180.2 ATCF in Actual $ ATCF in Year $ -$300 -$300 $143.1 $149.0 $155.2 $136.3 $135.1 $134.1 Alternative B Year Cash Flow in Year $ Cash Flow in Actual $ -$300 -$300 $150 $150 $150 $157.5 $165.4 $173.6 Quick Approximation of Rates of Return: Alternative A: $420 = $182 (P/A, i%, 3) (P/A, i%, 3) = $420/$182 = 2.31 12% < ROR < 15% (Actual ROR = 14.3%) Alternative B: $300 = $135 (P/A, i%, 3) (P/A, i%, 3) = $300/$135 = 2.22 15% < ROR < 18% (Actual ROR = 16.8%) Incremental ROR Analysis for A- B Year A -$420 $183.3 $181.8 $180.2 B -$300 $136.3 $135.1 $134.1 A- B -$120 $47 $46.7 $46.1 Try i = 7% NPW = -$120 + $47 (P/F, 7%, 1) + $46.7 (P/F, 7%, 2) + $46.1 (P/F, 7%, 3) = +$2.3 So the rate of return for the increment A- B is greater than 7% (actually 8.1%) Choose the higher cost alternative: choose Alternative A ... iequivalent = i inflation corrected + f% + (i 'inflation corrected) (f%) In this problem: iequivalent = 5% f% = +2% iinflation corrected = unknown 0.05 = i inflation corrected + 0.02 + (i 'inflation. .. dollars, where all the costs and benefits change at the same rate Allowable depreciation charges are based on the original equipment cost and not increase Thus the stable price assumption may be suitable... specific and individual items, such as a labor cost index, a material cost index, a “football ticket” index, etc Both commodity specific and composite indexes can be used in engineering economic
- Xem thêm -
Xem thêm: Solution manual engineering economic analysis 9th edition ch14 INflation and price change , Solution manual engineering economic analysis 9th edition ch14 INflation and price change