ࡱ> lnk]` 5bjbj ,)TTTTH H H \ @@@88@@|\ w2lBdE"EEEMP;OtO<wwwwwww$xhF{9w-H PoM|MPP9wTTEEfwVVVPTELH EwVPwVVn bH Kr E`B 젔&@QVp8Ws|w0wq4{Sn{pWr{H WrO" PV%P9POOO9w9w\V^OOOwPPPP\ \ \ 4@\ \ \ @\ \ \ TTTTTT CHAPTER 3 How to Calculate Present Values Answers to Practice Questions 1. a.PV = $100/1.0110 = $90.53b.PV = $100/1.1310 = $29.46c.PV = $100/1.2515 = $ 3.52d.PV = $100/1.12 + $100/1.122 + $100/1.123 = $240.18 2. a.  EMBED Equation.3 r1 = 0.1050 = 10.50% b.  EMBED Equation.3  PV of an annuity = C ( [Annuity factor at r% for t years] Here: $24.65 = $10 ( [AF3] AF3 = 2.465 AF3 = DF1 + DF2 + DF3 2.465 = 0.905 + 0.819 + DF3 DF3 = 0.741 The present value of the 10-year stream of cash inflows is:  EMBED Equation.3  Thus: NPV = $800,000 + $886,739.66 = +$86,739.66 At the end of five years, the factorys value will be the present value of the five remaining $170,000 cash flows:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  or 1.57 billion. Since the NPV is positive, Kuro-biro should accept the project. a. Let St = salary in year t  EMBED Equation.3  EMBED Equation.3   EMBED Equation.3  PV(salary) x 0.05 = $38,018.96 Future value = $38,018.96 x (1.08)30 = $382,571.75  EMBED Equation.3  a. PV = $100,000 PV = $180,000/1.125 = $102,137 PV = $11,400/0.12 = $95,000  EMBED Equation.3  PV = $6,500/(0.12 0.05) = $92,857 Prize (d) is the most valuable because it has the highest present value. We can break this down into several different cash flows, such that the sum of these separate cash flows is the total cash flow. Then, the sum of the present values of the separate cash flows is the net present value of the entire project. (All dollar figures are in millions.) Cost of the ship is $8 million PV = ($8 million Revenue is $5 million per year, operating expenses are $4 million. Thus, operating cash flow is $1 million per year for 15 years.  EMBED Equation.3  Major refits cost $2 million each, and will occur at times t = 5 and t = 10. PV = (($2 million)/1.085 + (($2 million)/1.0810 = ($2.288 million Sale for scrap brings in revenue of $1.5 million at t = 15. PV = $1.5 million/1.0815 = $0.473 million Adding these present values gives the net present value of the entire project: NPV = ($8 million + $8.559 million ( $2.288 million + $0.473 million NPV = ($1.256 million Assume the Zhangs will put aside the same amount each year. One approach to solving this problem is to find the present value of the cost of the boat and then equate that to the present value of the money saved. From this equation, we can solve for the amount to be put aside each year. PV(boat) = $20,000/(1.10)5 = $12,418 PV(savings) = Annual savings EMBED Equation.3  Because PV(savings) must equal PV(boat): Annual savings EMBED Equation.3  Annual savings EMBED Equation.3  Another approach is to find the value of the savings at the time the boat is purchased. Because the amount in the savings account at the end of five years must be the price of the boat, or $20,000, we can solve for the amount to be put aside each year. If x is the amount to be put aside each year, then: x(1.10)4 + x(1.10)3 + x(1.10)2 + x(1.10)1 + x = $20,000x(1.464 + 1.331 + 1.210 + 1.10 + 1) = $20,000x(6.105) = $20,000x = $ 3,276 Mr. Erikkson is buying a security worth SK500,000 now. That is its present value. The unknown is the annual payment. Using the present value of an annuity formula, we have:  EMBED Equation.3  The fact that Kangaroo Autos is offering free credit tells us what the cash payments are; it does not change the fact that money has time value. A 10 percent annual rate of interest is equivalent to a monthly rate of 0.83 percent: rmonthly = rannual /12 = 0.10/12 = 0.0083 = 0.83% The present value of the payments to Kangaroo Autos is:  EMBED Equation.3  A car from Turtle Motors costs $9,000 cash. Therefore, Kangaroo Autos offers the better deal, i.e., the lower present value of cost. The NPVs are: at 5 percent  EMBED Equation.3  at 10 percent  EMBED Equation.3  at 15 percent  EMBED Equation.3  The figure below shows that the project has zero NPV at about 11 percent.  As a check, NPV at 11 percent is:  EMBED Equation.3  a. PV = $100,000/0.08 = $1,250,000 PV = $100,000/(0.08 0.04) = $2,500,000 c.  EMBED Equation.3  The continuously compounded equivalent to an 8 percent annually compounded rate is approximately 7.7 percent , because: e0.0770 = 1.0800 Thus:  EMBED Equation.3  This result is greater than the answer in Part (c) because the endowment is now earning interest during the entire year. a. This is the usual annuity, and hence:  EMBED Equation.3  This is worth the PV of a three-year annuity plus the immediate payment of 10,000:  EMBED Equation.3  The continuously compounded equivalent to a 7 percent annually compounded rate is approximately 6.77 percent, because: e0.0677 = 1.0700 Thus:  EMBED Equation.3  Note that the pattern of payments in part (c) is more valuable than the pattern of payments in part (a) and less valuable than the pattern of payments in part (b). It is preferable to receive cash flows at the start of every year than to spread the receipt of cash evenly over the year; with the former pattern of payment, you receive the cash more quickly. One way to approach this problem is to solve for the present value of: (1) $100 per year for 10 years, and (2) $100 per year in perpetuity, with the first cash flow at year 11. If this is a fair deal, these present values must be equal, and thus we can solve for the interest rate (r). The present value of $100 per year for 10 years is:  EMBED Equation.3  The present value, as of year 10, of $100 per year forever, with the first payment in year 11, is: PV10 = $100/r At t = 0, the present value of PV10 is:  EMBED Equation.3  Equating these two expressions for present value, we have:  EMBED Equation.3  Using trial and error or algebraic solution, we find that r = 7.18%. FV = PV ( (1 + r)t 360,033.33 = 100 ( (1 + r)180  EMBED Equation.3 r = 4.654% Let A represent the investment at 12percent, compounded annually. Let B represent the investment at 11.7percent, compounded semiannually. Let C represent the investment at 11.5 percent, compounded continuously. After one year: FVA = $1,000 ( (1 + 0.12)1 = $1,120.00 FVB = $1,000 ( (1 + 0.0585)2 = $1,120.42 FVC = $1,000 ( e(0.115 ( 1) = $1,121.87 After five years: FVA = $1,000 ( (1 + 0.12)5 = $1,762.34 FVB = $1,000 ( (1 + 0.0585)10 = $1,765.67 FVC = $1,000 ( e(0.115 ( 5) = $1,777.13 After twenty years: FVA = $1,000 ( (1 + 0.12)20 = $9,646.29 FVB = $1,000 ( (1 + 0.0585)40 = $9,719.29 FVC = $1,000 ( e(0.115 ( 20) = $9,974.18 The preferred investment is C. 1 + rnominal = (1 + rreal) ( (1 + inflation rate) Nominal Interest RateInflation RateReal Interest Rate6.00%1.00%4.95%23.20%10.00%12.00%9.00%5.83%3.00% Because the cash flows occur every six months, we use a six-month discount rate, here 8%/2, or 4%. Thus:  EMBED Equation.3  The total elapsed time is 113 years. At 5%: FV = $100 ( (1 + 0.05)113 = $24,797 At 10%: FV = $100 ( (1 + 0.10)113 = $4,757,441 a. Each installment is: $9,420,713/19 = $495,827  EMBED Equation.3  b. If ERC is willing to pay $4.2 million, then:  EMBED Equation.3  Using Excel or a financial calculator, we find that r = 9.81%. a.  EMBED Equation.3  b. YearBeginning-of-Year BalanceYear-end Interest on BalanceTotal Year-end PaymentAmortization of LoanEnd-of-Year Balance1402,264.7332,181.1870,000.0037,818.82364,445.912364,445.9129,155.6770,000.0040,844.33323,601.583323,601.5825,888.1370,000.0044,111.87279,489.714279,489.7122,359.1870,000.0047,640.82231,848.885231,848.8818,547.9170,000.0051,452.09180,396.796180,396.7914,431.7470,000.0055,568.26124,828.547124,828.549,986.2870,000.0060,013.7264,814.82864,814.825,185.1970,000.0064,814.810.01 First, with nominal cash flows: The nominal cash flows form a growing perpetuity at the rate of inflation, 4%. Thus, the cash flow in one year will be $416,000 and: PV = $416,000/(0.10 0.04) = $6,933,333 The nominal cash flows form a growing annuity for 20 years, with an additional payment of $5 million at year 20:  EMBED Equation.3  Second, with real cash flows: Here, the real cash flows are $400,000 per year in perpetuity, and we can find the real rate (r) by solving the following equation: (1 + 0.10) = (1 + r) ( (1 + 0.04) ( r = 0.05769 = 5.769% PV = $400,000/(0.05769) = $6,933,611 b. Now, the real cash flows are $400,000 per year for 20 years and $5 million (nominal) in 20 years. In real terms, the $5 million dollar payment is: $5,000,000/(1.04)20 = $2,281,935 Thus, the present value of the project is:  EMBED Equation.3  [As noted in the statement of the problem, the answers agree, to within rounding errors.] This is an annuity problem with the present value of the annuity equal to $2million (as of your retirement date), and the interest rate equal to 8percent, with 15 time periods. Thus, your annual level of expenditure (C) is determined as follows:  EMBED Equation.3  With an inflation rate of 4 percent per year, we will still accumulate $2 million as of our retirement date. However, because we want to spend a constant amount per year in real terms (R, constant for all t), the nominal amount (C t ) must increase each year. For each year t: R = C t /(1 + inflation rate)t Therefore: PV [all C t ] = PV [all R ( (1 + inflation rate)t] = $2,000,000  EMBED Equation.3  R ( [0.9630 + 0.9273 + . . . + 0.5677] = $2,000,000 R ( 11.2390 = $2,000,000 R = $177,952 Thus C1 = ($177,952 ( 1.04) = $185,070, C2 = $192,473, etc. Let x be the fraction of Ms. Pools salary to be set aside each year. At any point in the future, t, her real income will be: ($40,000)(1 + 0.02) t The real amount saved each year will be: (x)($40,000)(1 + 0.02) t The present value of this amount is:  Ms. Pool wants to have $500,000, in real terms, 30 years from now. The present value of this amount (at a real rate of 5 percent) is: $500,000/(1 + 0.05)30 Thus:   $115,688.72 = (x)($790,012.82) x = 0.146 Challenge Questions 1. a. Using the Rule of 72, the time for money to double at 12 percent is 72/12, or 6 years. More precisely, if x is the number of years for money to double, then: (1.12)x = 2 Using logarithms, we find: x (ln 1.12) = ln 2 x = 6.12 years With continuous compounding for interest rate r and time period x: e r x = 2 Taking the natural logarithm of each side: r x = ln(2) = 0.693 Thus, if r is expressed as a percent, then x (the time for money to double) is: x = 69.3/(interest rate, in percent). Spreadsheet exercise. a. 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+~$~8~5~,~0~0~0~1~.~1~2 ~5 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ +~$~9~2~,~0~0~0~1~.~1~2 ~6 +~$~9~2~,~0~0~0~1~.~1~2 ~7 +~$~8~0~,~0~0~0~1~.~1~2 ~8 +~$~6~8~,~0~0~0~1~.~1~2 ~9 +~$~5~0~,~0~0~0~1~.~1~2 ~1~0 =~$~2~3~,~6~9~6~.~1~5 FMicrosoft Equation 3.0 DS Equation Equation.39q H\ Arial~N~P~V=~C ~t ~(~1~.~0~6~) ~t~t=~0~1~0 " =ObjInfo!<Equation Native ='_1171368450'$F@Ġ&hƠ&Ole F"~1~2+~2~.~3~1~.~0~6+~2~.~6~1~.~0~6 ~2 +~2~.~8~1~.~0~6 ~3 +~3~.~0~1~.~0~6 ~4 ~ +~3~.~1~1~.~0~6 ~5 +~2~.~9~1~.~0~6 ~6 =~1~.~5~7 FMicrosoft Equation 3.0 DS EqCompObj#%GfObjInfo&IEquation Native Jk_1171368466)FhƠ&hƠ&uation Equation.39q:O8N Arial~P~V=~S ~t ~(~1~.~0~8~) ~t~t=~1~3~0 " =~4~0~,~0~0~0~(~1~.~0~5~) ~t"~1 ~(~1~.~0~8~) ~t~t=~1~3~0 "Ole PCompObj(*QfObjInfo+SEquation Native T FMicrosoft Equation 3.0 DS Equation Equation.39q:cJ =Arial~(~4~0~,~0~0~0~/~1~.~0~5~)~(~1~.~0~8~/~1~.~0~5~) ~t~t=~1~3~0 " =~3~8~,~0~9~5~.~2~4~(~1~.~0~2~8~6~) ~t~t=~1~3~0 " FMicrosoft Equation 3.0 DS Equation Equation.39q98' =Arial~3~8~,~0~9~5~.~2~4~ ~ ~1~0~._11693137461.FhƠ&Ǡ&Ole ZCompObj-/[fObjInfo0]Equation Native ^U_11693162173FǠ& vɠ&Ole dCompObj24ef~0~2~8~6"~1~0~.~0~2~8~6~(~1~.~0~2~8~6~) ~3~0 []=~$~7~6~0~,~3~7~9~.~2~0 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo5gEquation Native h_1151936703;@8F vɠ& vɠ&Ole t1D Arial e~P~V =~C~ ~ ~1~r"~1~r~(~1+~r~) ~t [] e~$~3~8~2~,~5~7~1~.~7~5 =~C~ ~ ~1~0~.~0~8"~1~0~.~0~8~(~1~.~0~8~) ~2~0 [] e~C=)~$~3~8~2~,~5~7~1~.~7~5~1~0~.~0~8"~1~0~.~0~8~(~1~.~0~8~) e e e~2~0 e[] ~ ~ =~$~3~8~,~9~6~5~.~7~8 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj79ufObjInfo:wEquation Native x._1151936289=F vɠ&ʠ&c("  ePV =$Arial~19,000 ~1~0~.~1~2"~1~0~.~1~2~(~1~.~1~2~) ~1~0 []=~$~1~0~7~,~3~5~4 FMicrosoft Equation 3.0 DS Equation Equation.39qOle }CompObj<>~fObjInfo?Equation Native Zc>"  ePV =$Arial~1 million~ ~ ~1~0~.~0~8"~1~0~.~0~8~(~1~.~0~8~) ~1~5 []=~$~8~.~5~5~9 million_1152353752BFʠ&ʠ&Ole CompObjACfObjInfoD FMicrosoft Equation 3.0 DS Equation Equation.39q6 |   Arial~1~0~.~1~0"~1~0~.~1~0~(~1~.~1~0~) ~5 []Equation Native _11523538356YGFʠ&`̠&Ole CompObjFHf FMicrosoft Equation 3.0 DS Equation Equation.39q6|   Arial~1~0~.~1~0"~1~0~.~1~0~(~1~.~1~0~) ~5 []=$12,418ObjInfoIEquation Native _1160655218LF`̠&`̠&Ole CompObjKMfObjInfoNEquation Native  _1171193920mQF`̠& Π& FMicrosoft Equation 3.0 DS Equation Equation.39qX  =)Arial~$~1~2~,~4~1~8~1~0~.~1~0"~1~0~.~1~0~(~1~.~1~0~) ~5 []=~$~3~,~2~7~6Ole CompObjPRfObjInfoSEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39q`dd Arial e~P~V =~C~ ~ ~1~r"~1~r~(~1+~r~) ~t [] e~S~K~5~0~0~,~0~0~0 =~C~ ~ ~1~0~.~0~6"~1~0~.~0~6~(~1~.~0~6~) ~1~5 [] e~C=)~S~K~5~0~0~,~0~0~0~1~0~.~0~6"~1~0~.~0~6~(~1~.~0~6~) e e e~1~5 e[] ~ ~ =~S~K~5~1~,~4~8~1~.~3~8_1152354305VF Π&Ϡ&Ole CompObjUWfObjInfoX FMicrosoft Equation 3.0 DS Equation Equation.39q6!|  Arial~$~1~,~0~0~0+~$~3~0~0~1~0~.~0~0~8~3"~1~0~.~0~0~8~3~(~1~.~0~0~8~3~) ~3~0Equation Native =_1152354375T^[FϠ&Ϡ&Ole CompObjZ\f []=~$~8~,~9~38 FMicrosoft Equation 3.0 DS Equation Equation.39q6 @̖ !Arial~N~P~V="~$~1~7~0~,~0~0~0"~$~1~0~0~,~0~0~0~1~.~0~5+~$~ObjInfo]Equation Native &_1152354402c`FϠ&@Ѡ&Ole 3~2~0~,~0~0~0~(~1~.~0~5~) ~2 =~$~2~5~,~0~1~1 FMicrosoft Equation 3.0 DS Equation Equation.39q6Pdw !Arial~N~P~V="~$~1~7~0~,~0~0~0"~$~1~0~0~,CompObj_afObjInfobEquation Native _1152354428eF@Ѡ&@Ѡ&~0~0~0~1~.~1~0+~3~2~0~,~0~0~0~(~1~.~1~0~) ~2 =~$~3~,~5~5~4 FMicrosoft Equation 3.0 DS Equation Equation.39q6 ؒ1 !Arial~N~P~V=Ole CompObjdffObjInfogEquation Native &"~$~1~7~0~,~0~0~0"~$~1~0~0~,~0~0~0~1~.~1~5+~3~2~0~,~0~0~0~(~1~.~1~5~) ~2 ="~$~1~4~,~9~9~1 FMicrosoft Equation 3.0 DS Equation Equation.39q_1160655518jF@Ѡ&Ҡ&Ole CompObjikfObjInfolEquation Native _1171194325oFҠ&Ҡ&Ole CompObjnpf8 Arial~N~P~V="~$~1~7~0~,~0~0~0"~$~1~0~0~,~0~0~0~1~.~1~1+~3~2~0~,~0~0~0~(~1~.~1~1~) ~2 ="~$~3~7~1 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoqEquation Native -_1171194343OtFҠ&$Ԡ&Ole Zdd Arial~P~V=~$~1~0~0~,~0~0~0~1~0~.~0~8"~1~0~.~0~8~(~1~.~0~8~) ~2~0 []=~$~9~8~1~,~8~1~5 FMicrosoft Equation 3.0 DS EqCompObjsufObjInfovEquation Native M_1171197683yF$Ԡ&$Ԡ&uation Equation.39q1hX Arial~P~V=~$~1~0~0~,~0~0~0~1~0~.~0~7~7"~1~0~.~0~7~7~e ~(~0~.~0~7~7~)~(~2~0~) []=~$~1~,~0~2~0~,~2~8~4Ole CompObjxzfObjInfo{Equation Native   FMicrosoft Equation 3.0 DS Equation Equation.39qS Arial~P~V=~1~0~,~0~0~0~1~0~.~0~7"~1~0~.~0~7~(~1~.~0~7~) ~4 []=~3~3~,~8~7~2 FMicrosoft Equation 3.0 DS Equation Equation.39qS Arial~P~V=~1~0~,~0~0~0+~1~0~,~0~0~0~1~0~.~0~7"~1~0~._1171197890w~F$Ԡ& ՠ&Ole CompObj}fObjInfoEquation Native 5_1171198637F ՠ&1נ&Ole CompObjf    !"#&+./012589:;>ABCDGJKLMPSTUVWXYZ[\]^adefghilopqrstuvwxy|~0~7~(~1~.~0~7~) ~3 []=~3~6~,~2~4~3 FMicrosoft Equation 3.0 DS Equation Equation.39q a Arial~P~V=~1~0~,~0~0~0~1~0~.~0~6~7ObjInfoEquation Native 9_1085995783F1נ&1נ&Ole  ~7"~1~0~.~0~6~7~7~e ~(~0~.~0~6~7~7~)~(~2~0~) []=~3~5~,~0~4~2 FMicrosoft Equation 3.0 DS Equation Equation.39qP\! Arial~P~V=~$~1CompObj fObjInfoEquation Native _1085996192F1נ&`ؠ&~0~0~1~r"~1~(~r~)~(~1+~r~) ~1~0 [] FMicrosoft Equation 3.0 DS Equation Equation.39qP\! Arial~P~V=Ole CompObjfObjInfoEquation Native ~1~(~1+~r~) ~1~0 []~$~1~0~0~r[] FMicrosoft Equation 3.0 DS Equation Equation.39qP7  Arial~$~1~0~0~1~r"~1~(~r_1085996201F`ؠ&`ؠ&Ole CompObjfObjInfoEquation Native S_1171199107|F`ؠ&?ڠ&Ole $CompObj%f~)~(~1+~r~) ~1~0 []=~1~(~1+~r~) ~1~0 []~$~1~0~0~r[] FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo'Equation Native ()_1171200174F?ڠ&۠&Ole ) a ! 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