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Solution Manual For Mathematics Of Investment And Credit, 5th Edition

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    Le I > | © 3 = 1) HAPTER 6 JL om yw | CHAPTER 1 SECHOT 6.4 voeoeeerereressssesssrsivssseeneesessessessissssssssssmssensinensesses 120 | CHAPTER 7 ; LO Sv {| | SECTION 1.1 SECHOD 7.2 ooevrroererseeesseresesosssssensersensesssssssssssssssssmmsssesssssneees 133 ee : 1.1.1 Balances are 10,000(1.04) =10,400 after one year, CHAPTER 8 ] : . 141 10,000(1.04)* =10,816 after 2 years, and SECHOM 8.1 coovvireeerereevcusessseare esses reer rasan sss . Section 8.2 AA 8.3 wevrvreerresrssssressmsesmssesessssereresssssessssesssseninnes 142 10,000(1.04)° =11,248.64 after 3 years. SECHON Bib rrr Interest amounts are 400 at the end of the 1% year, 416 at the end : of the 2" year, and 432.64 at the end of the 3 year. CHAPTER 9 SECHOR 9.1 ooeeeooerrsreeeeseresesesssnsesenesisessnsseneesssssesssssnicsinnianss 145 | Section 9.2 ANd 9.3 everest 1 ! 1.12 (a) 2500[1+(.04)(10)]=3500 HON 9 1oveeeeresessseesessemens se srass rarer eters sansa : Section (5) 2500(1.04)'° =3700.61 ’ | (cy 250001.02)* =3714.87 : (d) 25000.00)% =3722.16 . 1.13 Balance after 12 months is 10,000(1.01) 1.0075)" =11,019.70. : Average monthly interest rate is j, where : 10,0000+ 7) =11,019.70. . Solving for j results in 0081244. . A ,

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    Ei 2 » MATHEMATICS OF INVESTMENT AND CREDIT SOLUTIONS TO TEXTBOOK EX 3 1.1.4 There are two (equivalent) ways to approach this problem. We can 1.15 (a) Over 5 years the unit value has grown by a of update the balance in the account at the time of each transaction (1.10)(1.16)(1 — .07)(1.04)(1.32) = 1.629074. The ge unl we reach the od of 10 years, and set the balance equal to annual (compound) growth is (1.629074)° =1.1025, or X d ) ) average annual growth of 10.25% for 5 years. Balance at ¢ =4 (after interest and withdrawal) is . " (b) Five-year average annual return from January 1, 1996 to 10,000(1.04)" — (1.05)K; Lo ’ balance at 15 is December 31, 2005 is j, where (1+ (1L17)° =(1.13)"°, so . that 7=.0914. Annual return for 2004 is k, where [10,00001.04) ~(LOS)K }(1.04) ~ (1.05)K; ; 4 A (1+k)(1.22) = (1.15), so that & =.084. balance at t=6 is [[10,000(1.04)* -(L05)K J(1.04)~ 1 05)K 0.04) K; (¢) Over n years the growth is (I+ )(L+iy)~(I+i,) = 1+)", at ¢=7 is where the average annual (compound) interest rate is 7, so - § at {+1 is the geometric f 1+ i i [[[10,0000.08)" ~(1.05)K | ee Filth Thus, 1+i < (rye ene (by) yo bri tort, < = . (1.04) ~ (1L05)K |(1.04)—K (1.04) - K; at ¢ = 10 there is 3 years of compounding from time 7, so that : ye p 2 | 1.1.6 We equate the accumulated value of Joe's deposits with that of [[[[r0.0000 04)" —(L.05)K 1 (1.04)-(1.05)K ] : Tina’s. Note that it is assumed that for simple interest, cach new deposit is considered separately and begins carning simple B a.05-K] (1.04) K 0.047 = 10,000. interest from the time of the new deposit. . 10[1+10(.D]+30{1+5(.D] = Solving for K from this equation results in K' = 979.93. [ 0] [1+5(0]= 67.5 Alternatively, we can accumulate to time 10 the initial deposit = 101.0915)" +30(1.0915)"%". and the withdrawals separately. The balance at time 10 is This can be solved by substituting in the possible values of n 10,000(1.0: 14) — K(1.05)(1.0 4)° _ K(1.05)(1.04)° until the equation is satisfied. Alternatively, the equation can be rewritten as _ K(1.04)" — K(1.04)> =10,000. 67.5(1.0195)*" —10(1.0915)'° 1.0915)" -30(1.0915)° = 0 This is the same equation as in the first approach (and must . which is a quadratic equation in (1.0915)". The solution is result in the same value of K). In general, when using compound . no . 244141.5 : interest, for a series of deposits and withdrawals that occur at . 1.0915)" = gis various points in time, the balance in an account at any given 2 We ignore the negati n_ time point is the accumulated values of all deposits minus the : en negative root and get 1.0915)" =1.226 > n =23. accumulated values of all withdrawals to that time point. This is b also the idea behind the “dollar-weighted rate of return,” which . will be discussed later. 0 el

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    TT 4 » MATHEMATICS OF INVESTMENT AND CREDIT SOLUTIONS TO TEXTBOOK EXER 5 4 > MATHEMATICS OF INVESTMENT AND CRED 117 @) 1000=850[1+i($L)] — i=10735107.35%) 11.10 (@) 10000112) =3000 — t=—2) 9694 (9 yea ) 365 In(1.12) (6) 1000 =900[1+i(£%)] — i= .6759(67.59%) approximately 253 days). b) At the end of 9 years the accumulated value is 900 1+(09(LL) | = 913.32 ( © [ ( (4%) 1000(1.12)° =2773.08. At time s during the 10" year, the @ 900] 1+ (09)(5%5) = 1000 — & = 451 ; accumulated value based on simple interest within the 10% year is 2773.08(1+.125). Setting this equal to 3000 and * day i (25%)! 118 Itis to Smith’s advantage to take the loan of 975 on the 7" day if solving for s results in s=-2""32 = 6819 years the amount payable on the 30th day is less than the amount due 12 to the supplier: (approximately 249 days) after the end of 9 years. . _ 3) o75[ 1+: < 1000 — i £ 4069. (€) 10000103)’ =3000 > ¢ = AEs =110.41 months (about 9 years and 2 months and 13 days). 1.19 (a) Maturity value of 180-day certificate is (d) 10000+i)'® = 3000 — i = 3"°~1 = .1161 (11.61% per 100,000(1-+.075( 182) = 103,698.63. year). days i (e) 1000(1+/)" = 3000 — i = 3" 1 = 009197 _ Interim book value after 120 ays is - (:9197% per month). 100,000(1-+.075(322)) = 102,463.75. Bank will pay X after 120 days so that LIL (2) (00757 = 1.0299 < 1.03 60.1) = 82. ; x(1+-09(£)) = 103,698.63 — X = 10218682 (but (1.007917 = (1.0075)* = 1.0303) The penalty charged is 102,465.75— 102,186.82 = 278.93. : | ® (1.01577 = 1.0604 > 1.06 (b) 1.08 = (+2) (1+5) > i=.0819 : : 1.1.12 (a) Smith buys A =227.5 units after the front-end load is paid. - Six months later she receives (227.5)(5)(.985)=1120.4375. . Smith’s 6-month rate of return is 12.04% on her initial 1000. . (b) If unit value had dropped to 3.50, she receives . (227.5)(3.5)(.985) = 784.30625, which is a 6-month effective i rate of —21.57%.

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    6 » MATHEMATICS OF INVESTMENT AND CREDIT SOLUTIONS TO TEXTBOOK EX 7 1.1.13 We use the following result from calculus: if f and g are Graph of ¢In(l.1) differentiable functions such that f(a) = g(a) and f(x) <g'(®) for a<x<b, then f(b)<g(b). In(8) (i) Suppose 0<7<l. Let f(i)=(1+) and g(i) = I+i-z. Then FO) =g(0)=1. Tf we can show that f'(i)<g'()) for any In(4) i>0, then we can use the calculus result above to conclude that fi) <g() for any i>0. First note that f'()) = (i) In(2) and g'(i)=t. Since i>0, it follows that 1+i>1, and since i (<1, it follows that ¢—1<0. Then (M+i)7<l, so [ . fa < g@. : [3 This completes the proof of part (i). : i" CL Ae lei (iy = (141). . (ii) Suppose that ¢>1. Let f(i) = 1+i-t and 8 (1+) SECTIONS 1.2 AND 1.3 Again f(0) = g(0). If we can show that f'(i)< g'(§) for any : —_— i>0, then we can use the calculus result above to conclude : 21 Pp A that f(i)<g() for any i>0. Since ¢>1 and i>0 it . “- resent value is follows that £~1 > 0 and 1+ >1. Thus (144) >1, and it 5000 eto to 732553 - ted _ i 106 (1.06) (106° (1.06) - follows that f'(7) = ¢ < ¢-(1+)"" = g'(i). This completes : the proof of part (ii). i : : 1.2.2 Amount now required is 1.1.14 Original graph is y = (1+). New graphs 107 = (1+), or, 25,0000” +v'% +v'21+100,000[% + v'® + v*] = 75,686 equivalently, y=t Se, so that y is now a linear function of # o : 3 1 1.23 28 = 15+16.50v — v = T8779 — i = .2692 124 1000-V vi vy = 494.62 Graph of (LI) Vos “Vor “Voy = 494.6 4 1.2.5 Equation of value on July 1, 2013 is 2 : 2001.04) +300 = 100(1.04)* + X — X = 379.48 ib 510 15 - t :

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    8 » MATHEMATICS OF INVESTMENT AND CREDIT . SOLUTIONS TO TEXTBOOK EX) <9 126 4830 = 50+100(v+v? + +v") + X07, where v=rk, 1.2.12 1000(+)% +1092 = 2000(1+4) so that X = 67.57. If interest is .01 per month, then v = od Solving the quadratic equation for 1+ results in no re Ss. and X =67.98. ] 12.43 (@) £0+)" = a+)" © Za+) = (+0) In) on 10 127 10042000" +300v>" = 600v'" — (b) Ly = wm (d) Ly = ~v" In(l+i) 6000 = 100+ 200(.75941) + 300(.75941) o> V0 = 708155 ~» i = (708155) ~1 = 0351. : . . 1.2.14 With an annual yield rate quoted to the nearest .01%, the annual : yield 7 is in the interval .11065 < i < 11075. 128 (a) (20)2000)[v+v* +9 +--+] = 1,607,391 (at 75%) Since the quoted annual yield rate is 365-109=Ptce it foliows 48 _ i rice (b) 1,607,391 200,000v N 1,747,114 that 11065 < 365. 100=Fdee «11075, or, equivalently, © X = 1,607,391+.156™ — X = 1,795,551 94.767 < Price < 94.771. 129 750 = 367.85[1+ (1+ /)] => j = 0389 isthe 2-month rate. - [L+@ep} = 7 1215 (a) P = 2000000 _ 95 550.52 | L105 1.2.10 With X initially stocked, the number after 4 years is | _ 100,000 dP 100,000 182 4 15 5 _ Lo 0) P= = = mw ges X(14) -5000[ (14) +(1.4)°] = X > X = 4997. - EZ {voi L = -45,239.03 if i = .10 - 100, 1000. _ 400, 1000 fr is por . . 1211 1000 = 290+ 4008, and 1000 = {95+ 30%. Tt is no ZL 2 8D = -45239.03 — AP = —45239.03- Al. possible that j =k, since the two present values could not both a If AF = 00L th be equal to 1000 (unless j =k =0, which is not true). If j>k, . ¢ = 001 then AP = —45.24. then (1+)? >1+k and (14+) > (+k), in which case the first : 100.000 present value would have to be less than the second present © P=T75r dbo 10000 92373334 if i=.10. value. Since both present values are 1000, it must be the case Lo 765 (14-2) that j <k (j=.0333 and k=.0345). : As the T-bill approaches its due date the & goes to 0.

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    10 > MATHOMATICS OT INVESTMENT ANDCREDIT : SouuTions 19 TexTs0ok Execs fill _ No cr 1.2.18 (a) Minimum monthly balance for January 2005 is 25 i 1216) B = By(i+)+ 2 a, [140-4] February 2005 is 6000, and for March 2005 is 9500. t u i earned is (10)(f5)r2s00+ 6000 +9500] = 150. = By+ Xa +5 + 2% a0] k=l r= Balance on March 31 is 2500(4) --1000(3) +150 = 13,150. (b) The balance is B, for 1 years, By +a, for ¢, —1 years, (b) Minimum daily balance is 2500 for January 1-15, 3500 for By +a; +a, for t;~t, years, By +a +a, +--+a, for 1-4, £ January 16-31, 6000 for February 1-15, 7000 for February years. The average balance is 16-28, 9500 for March 1-15, and 10,500 for March 16-31. Interest carned is = _ Bot + (Bora Y=) + (Borar+ay ta—ty) +++ (Bora tay +--+, Y1-1,) Ly w= cer Eres sr er Ba i (10){355)((2500(15)+ 3500(16) + 6000(15) + 7000(13) +9500(15) +10,500(16)) = 160.27. =o roi) ra ln) bt a, (=) Balance on March 31 is © ) =By+ 2 a, (1-1). 2500(4) + 1000(3) + 160.27 = 13,160.27. (c) Minimum monthly balance for January 2005 is 2500, so interest (c) Follows directly from (a) and (b). : on January 31 is 422500) = 20.83, so balance on January 31 (after deposit and interest) is 6020.83. 1.2.17 The difference between the two payment plans is that the first 2 ; Minimum monthly balance for February 2005 is 6020.83, so’ payments are deferred for 2 months, so the saving is [ : io 210 ; interest on February 28 is 43 (6020.83) = 50.17, so balance 30[ 0) — (H+) ~ 12.68. ; on February 28 is 9571.00. N 3 Minimum monthly balance for March 2005 is 9571, so interest Alternatively, the present value under the current t plan is § 2 v, thep nt payment plan 1s [ on March 31 is 409571) = 79.76, so balance on March 31 is 300+ v+v? +P] = 643.67. 13,150.76. The present value under Smith’s proposed payment plan is : (d) Minimum daily balance is 2500 for January 1-15 and 3500 | for January 16-31, so interest on January 31 is 25.62. 300° +--+ v5] = 643.677 = 630.99. : | Minimum daily balance is 6025.62 for February 1-15 and Saving is 12.68. i 7025.62 for February 16-28, so interest on February 28 is 49.79. Minimum daily balance is 9575.41 for March 1-15 and | 10,575.41 for March 16-31, so interest on March 31 is 85.71. i Balance on March 31 is 13,161.12.

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    12 » MATHEMATICS OF INVESTMENT AND CREDIT . SOLUTIONS TO TEXTBOOK EXERCIS SECTION 1.4 1.43 Equivalent effective annual rates are } ] ] Mountain Bank: (1.075)% —1 = .155625 1.4.1 m=1 implies interest convertible ammually (m=1 time per year), River Bank: which implies the offective annual interest rate ¢M=i=.12. We use 365 Equation (1.5) to solve for 7 for the other values of , as shown below. 6 +52) —-1 2.155625 je) 36 1/365 - Tear cffective iN (1+) > (1.155625) = 1.000396356 - . fm 7 Effective. EE | Cay interest rate £7 ’ [i- " I — i 2 144670 1(1 year) LedZe go 12) -1 = 12 2 (6 months) | Sr =+12=.06 (1.06)% ~1 = .1236 14.4 The last 6 mouths of the 8% year is the time from the end of the z z s | 15" to the end of the 16% half-year. 3 (4 months) | Sr =:12=.04 (1.04)° ~1 = .124864 | 43 months) 0203 1.03% 1 = .125500 0 1H 2H=1Y 3H 4H=2Y ... 14H=T7Y 15H 16H=8Y EY OS EE Ht 6 @months) | fg =2=02 (1.02)° 1 = .126162 I 8 (1.5 months) = == 015 1.015) —1 = .126593 The amount of interest earned in Eric’s account in the 16” half- on 1 year is the change in balance from time 15H to time 16H. 12:1 months) | 45 =+2 = 01 (10D -1 = 126825 : . Y 12 5 i The balances at those points are X (1+4) and X (1+4) . s2(tweek) | 5p =r =0023 | (1442) <1 = 127341 ; Co 365 i The amount of interest earned by Eric in the period is 2-12 = 00033 | (14427% = . . INTO 365 (1 day) | 565 =365 (1+42) 1 = 127475 (14) x(e4) _ X14) . d2V qo p12 i IE im (1+ y ) =e ol = 127497 i The balance in Mike's account at the end of the 15% half-year i (7.5 years) is 2X(1+7.5/), and the balance at the end of the 16” . half-year (8 years) is 2X (1481). 142 (a) 1000vhys = 414.64 . : The interest earned by Mike in that period is 0 ! . (b) 1000v55 = 409.30 | 2X18) 2X (147.50) = 2X(50) = 2X (4). (©) 1000v5%s =407.94 We are told that Eric and Mike earn the same amount of interest. | INTO . \I5 Therefore, X (144) {£) = 2 (£), so that (1+4)" =2. . Then i = 2[2""° —1] = .0946. 5

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    14 » MATHEMATICS OF INVESTMENT AND CREDIT ; . SOLUTIONS T0 TEXTROOK EXERCIS : 14.5 Quariedy effective rate is 323 008125. Initial amount L 1.49 (a) We wish to show that invested after commission is .99. At the end of 3 months, the ! Fim) = FANE i accumulated value is .99(1.008125). This is then subject to the : on) = fi nf1+7) reg | > 0 First, fm) 0 1% commission for the rollover and then the 3-month interest - since j>0. Also, if rate of .008125. At the end of the year, the accumulated value is 4 50, if x>0 and h(x) = In(l+x)-73. then . (x) = 5 en . [99(1.008125)]° = 992198 = 1-078. The effective aflr- (x) = qi > 0, and since 'h@©)=0, it follows that commission return is —.78%. : h(x)>0 for all x>0. Letting x=1>0, we see that | Nd in{1+Z)-—L > 0, which impli y 1.46 FOI sla -1] ~ 105 : ( 2) [ra , which implies that f(m) > 0. Tr g - NY 1+) i J Id 1] = 116025 b) g'm) = +H" 1 Gene) Lh AY _ a . #0 = 10[ 10) 1] = 159374 - = ap [tne ]-1. 0 = oi] (10) -1] = 137.796 . But x{1~Inx] has a maximum of 1 at x=1, so that with . (+) = x, we see that g'(m) <0 for m>1. 1.47 From November 9 to Januaty 1 (53 days) Smith earns (two full . - (©) Consider Tn[f(m)] = m-In (1+ FA ) _ In(i+4) h months) interest of £(.1125)(1000) =18.75. Thus, Smith earns a . [a EA en 53-day effective rate.of interest of .01875. The equivalent effec- de . in(1+4) oR ae lim 1 = fim a) Lf OTL me? . Live annual rate of interest is i= (1.01875) =1 = 1365. oo Jim Inf Gn] = Jim =p = im 7 a) = J . Thus lim f(m)=e’. 14.8 Left on deposit for a year at 0% = 09, X accumulates to - © zm) = Gafni Co [i 2) X(1.0075)2. If the monthly interest is reinvested at monthly . I Jim g(m) = lim py rate 75%, the accumulated value at the end of the year is E = In(l+), since lim (1+ /)"" = 1. X + X(0075)[ (1.0075) + (1.0075) +--+ (1.0075) +1). . . 1.4.10 We want to find the smallest integer 7 so that . 2 kop . _ m Since 147472 ++ rt =, it follows that the total at the oo fm) = [14:2] 2 118, f(2) = 1.1772, f(3) = 1.1798, end of the year with reinvestment of interest is . Caf) = 11811 = m= 4. Ey 9, : 167" #1 coors) SIRS! = xa.0079)" a a : : = ne tier bow many times per year compounding takes place, a =. nal rate of interest of 16% cannot accumulat i | . rate of more than 17.35%. ate to an effective

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    16 > MATHEMATICS OF INVESTMENT AND CREDIT . SOLUTIONS TO TEXTBOOK EXERCISES SECTION 1.5 : 1.5.6 The present value of 1 due in n years is (I-d)", so the acc : lated value after r years of an initial investment of 1 is —- _X = - LSI (0) 4992 = gf — X = 5187.84 = = 0-0) = — X = 5191.68 (b) 4992 Tecan] . (©) 4992 = X(-08)" —> X = $204.52 1.5.7 The initial deposit of 10 grows to 10(1~4)"" at the end of 10 _ | B | years (40 quarters), and then continues to grow at 3% per half (d) 4992 = X [--co)(3)] - X= 5200 | year after that. The accumulated value of the initial deposit of 10 : . 40 ; at the end of 30 years is 10(1- 4) x(1.03)* (20 more years, 1.52 With a quoted discount rate of .940, the price of a 91-day T-Bill [ 40 more half-years at 3% per half-year). 91 I. ; should be 10001 -575>-0113) =-99.714 as quoted. | The second deposit is 20 made at time 15. The accumulated : value of the second deposit at time 30 (15 years after the second The investment rate is found as (55%- 1) «368 = 0115, as | deposit) is 20(1.03)*® (15 years is 30 half-years). quoted. : The total accumulated value at the end of 30 years is : 0 (1-4) x(1.03)* +20(1.03)° = 100. A{r+T)— Af Al | 155 4 = ADHD AD ag Solving for d results in d =.0453. AGT) 4 AQT | J = del A) = 2x - : This question is from the May 2003 Course 2 exam that was Ar) conducted jointly by the Society of Actuaries and the Casualty = 1-d; = FT ry nd : Actuarial Society. Tt should be noted that the nominal interest . a : rate notation i") and nominal discount rate notation is not @d; = ea and (b) J = = | always specifically used on the professional actuarial exams. In J ’ L this example, the notation d was a nominal annual rate of = discount compounded quarterly. 154 115 = (1-d)1.3) —» d = .1154 | . 1.5.8 (a) Bank pays : dogs = mg t= We] = 1.5.5 Bruce's interest in year 11: 100(1—-d)™° {a-ay* -1] =X. ’ 3 + 363 nds Ide Robbie's interest in year 17: : As 7 increases, i increases. & _ 1 i cl 500-d)7 [aay -1] X= 1001-2) [1=ay 1] | (b) From (a) 1-dt = fried d= Ta If i=.11 then : S00) = 1000)" > (=d)f = 5 => d = 109] : t=1->d =.099099, tr = 50 > d =.104265, . 1 = -» X = 389 | t=13 ~>d =.109001. Bo

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    18 » MATHEMATICS OF INVESTMENT AND CREDIT : co SOLUTIONS TO TEXTBOOK EXERCISES 18 » MATHEMATICS OF INVESTMENT AND CRED 1.5.9 Suppose that the T-Bill’s face amount is $100. Then Smith ; 1.62 (122)? = dnl asde _ 004045 day 0339. purchases the bill for 100] 1-482.(10)] = 94.94 (nearest 01). ! 7 91 days later, the value of the T Bill is : I 2 i. 100[1-£(10)] = 97.47, : 163 exp(f34 dn) = (1-047 > P= 150414 i 125 _ = Smiths return for the 91 days is 3237 —1 = 0266 (2.66%). ! — 422 = In(1.50414) — k =102. . » mL, : 1.6.4 Effective annual rate for Tawny is i=(1.05) -1=.1025. 53 hit oe Eg, and £2 = J, 50 | 1.5.10 From Exercise 1.3.3, webave T= =4 " : Tawny: & = In(1.1625) = 09758. am _ god & rd =F and : Fabio: Simple interest rate j — 5, =1L. @ Sed mn Te ThE" [ 7 | At time 5, w i ~_J Pe = He © — Lo 09758 = pd = j=.1906 -> Z=10001+51=1953. Ed : : : 15.11 1000 = 1200(g;) =i) — © = 0909 165 oof ora -ehota], xl hota ~1)-x . > 1007 =e) = X(2-¢%) —> X = 7846. 15.12 1000(+/) = 1000+40(1+j)? = (1+))° ~25(+/)+25 = 0 — 1+ = 1.043561 or 23.9564. Since J <.10, it follows that i 1.6.6 Bruce’s 6-month rate of interest is i and 7.25 years is 14.5 J=.0436. I 6-month periods. Bruce's accumulated value after 7.25 years is . NTE] | 100(1+4) = 200. Solving for i, we get i SECTION 1.6 | (14) = 25 i = 0979. 1.6.1 Accumulated value at time 1 is : Peter’s account grows to 100¢”%%% =200, so that 'os ; 10,0005el™" = 10,000x¢% = 10,512.71, | 6=-LzIn2=.0956. Then i~6 = .0023 = 23%. Accumulated value at time 2 is | 10,000 x Jo 05H 105+ 026-0141. 10,000x 9% = 11,162.78 167 (79) = 1.36086 ~> ¢’0 = 1.36086 — 1+i = ¢ = 1.045 . = Bat

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    20 » MATHEMATICS OF INVESTMENT AND CREDIT : co SOLUTIONS TO TEXTBOOK EXERCISI . s . : Js SW _ os 168 (a) (1+) =oxp|[; (08-9230) dr] =1:616407 > = 1008 ! (© For 0 <t = 4, 6 = gy = 1x09 To find 8,44, let 7 = td, or f= r-L. (®) 1+i = exp (108+828) 9 =1.091629 —> 1; =.091629 Then thi, = oxp[ I] (08-0231) dt |=1.099509 1, =.099509 Ll 5 5 oS) i ertid, = Op = grey fy =102751, iy =.104532, is =.105659 S(r+3) ‘ Eo 1000(1.02)(.08) _ 08 (©) 1000- exp J: (08+) ar | = 821.00 SO) Tovacoofiecosr)] | TTCO% [ The same occurs for tk and +3 1.69 Ke? =960, Ke® =1200 — ¢™ =1-d =.80 — K =1500 ) and d =.20. If d changes to .10, then the present value becomes A - ne Ah) 1500(1-.10)° =1215. ; L612 @ =m ) | 1) 1) a 1610 [=e ~1,6=26 > +) Arey) . i _1 : my yi A(eHR) = A(t) A : me lm Pel = (I) = 2i > 2, © Let b=. Then lim a = fim AGRZ50 = 465. d= =l-e¥ =1-(1-@)* = 2d ~d* <2d . a ) L613 (@) 6 = 40 = adm "fim 5 =0 Taragrrat 1.6.11 (a) 1000(1.02)% [1+co9)(3%)] = 1044.73 (apply I’Hospital’s rule) (b) For 0 < ¢ < &, A() = 100001 +(.08)] | (b) 4) = ox [0, a] = exp[k-2-£"%]. | pry 1 1 = 1 oA . ford srl, 40)= 100001.02){1+(.08){z 1] lim £9 = lin | op for & <1 <3, A@)=100001.02)2] 1+ (08) (tL : CAN Es 2 4 [ ( 3)] my = Imma for 3 <1 <1, Ar) = 1000(1.02)°| 1+ (.08){r—2 : -E i 1 _ 4 [ ( 3] | erg pry Bl |

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    22 » MATHEMATICS OF INVESTMENT AND CREDIT SOLUTIONS TO TEXTBOOK EXERC N SECTION 1.7 1.7.7 Smith needs {38 =917.4312 US now if he invests in N a ir _ 0205 usars ; account, This is equivalent to 2812 = 12567551 Cdn, which le teal = 157 = I+.05 | grows to 1382.4306 in one year in a Canadian dollar account eaming : L 10%. The implication is that one year from now, 1000 US =1382.4306 Cdn., or, equivalently, 723364 US=1 Cdn. 172 After-tax retum is S2639710 = 0309. 178 (+r) 0" = (zy JES BEN 173 (a) Smiths ATI this year will be I+ [EC bE 21,000(.75) + 21,000(.50) =26,250 | — i +r I+r and taxes paid will be 15,750. The real growth from last year | to this year in Smith's ATV is 2025023000 1.00, and the | hb : 1.7.9 (a) Real after-tax rate of return on standard term d it i B L 1s. 25,250/15,000 | eposit is wih in tax smi = 1.00. : i(1=,)~ . real growth in taxes paid is “22777 1.00 i | Mor ”, and on the indexed term deposit is | asitus (L084 = 1-1). _ (b) Continuing the old taxation scheme, Smith’s taxes paid this i year will be (25)(20,000)-+ (.50)(22,000) =16,000, and his i . Co ATI will be Soe, The real growth i" taxes paid will be v (b) Setting the two expressions in part (a) equal and solving for 16,000715,000 i | i, we have =i'"(1+7) + 5-L. EE = 1.015873 (1.59%) and the real growth in I= . 26,000/25,000 . , ATTis 28000/55.000 990476 = 1-.009524(~95%). . If #'=.02 and r=.12, then i= (i) .1424, (i) .1824, (iii) 2224, (iv) .3224. 17.4 Smith sells the items for 100,000x1.15 =1 15,000 at the end of . the year and must pay back 100,000x 1.10 = 110,000. Net gain . is 5,000 (in year-end dollars). - 175 dm = lei = Ho= Go SF | 18-14 _ i=l . | 176 “yr = 52» i = 1070175(107%) -

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    - x =} = 2 i; wv - CHAPTER 2 : ji SECTION 2.1 : 2.1.1 Option I accumulated value is 50,000(1+1)*". | i Option 2 annuity payment is X, where K = 50,000 =5564.99 J [ a oo ! Then 556499553 = 50,0000 +1) so that =6.9%. . . 2.1.2 Suppose the annual effective rate is i. Then 900s; =1000a, . . | 10 _ . | so that suo (7-1) = 1000/1]. and then (14° =12. | . . | We then have Ksg, x (1+) =1000a;, so that . : i ' K=—J000 151942. | (+2) (141) : 203 sp = [AH + Uh) bn (0) + (1) [+05 QF en (4) +1] = df [ay i) er (4) +1], = ifs, +5; I 3 1000s, | b 214 10005, = Vag, — Y= ——s bv = 1000(1.0D*° = 19,788.47

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    26 » MATHEMATICS OF INVESTMENT AND CREDIT : SOLUTIONS TO TEXTBOOK EXERCIS| 215 (i) 10087, =715.95 2.1.10 After n years Smith's AV is [80s +200](1.06)" (i) 10085314095 =2033.87 : Brown's AV is 4085-551 46 +P. Thus, if n=13, then P=1%.93, 0 , n=20 » P=17.19, n=25 — 20.75. (iif) 100] 5153075 (1.00875) (LOY + sg) 7 (LOY) +5201] . = 3665.12 5 teint [ ml (HPL ny 2 210 (iv) 3665.12(.01) = 36.65 2H @ gh = yoy = Hl = = ; on (yd : => +)" = 2,70 = sz = 216 9855+)" +19655 = 8000, (L+)" = 2 o> 0 ok = 014286, 55 = 5m (14) 55, = 490 = = [ S; T+" 1 = = : Ili ~2 P — 196] 2s +557] = 8000 — 255755 = 40.82 ®) J _ Sr = Pr) 1 = L20wr 2 GHP en —» quadratic equation in z= (1+i)" : 22 +z 5K =0 ¢ : i ~1% ITER BN i => (+H) =z = —y—t— (discard negative root) > 243 = 40.82 > i=.1225. | I 2 ICT Ip a 2251 ) ; “2 B : 217 (@) 100105) sg; + 20005 si50, © sg = I)" sym + (HD +1 +3001.03)" +575 + 40875, = 2328.82 IL = 3634(1+i) + (1+) - 47.99 = 0 ® 10 sg 555+ 2s) + Hsggr=syg) + dsp | — 147 = 1.1355, or —1.1630 (discard negative root) = 0 $55105 * 555105 + 550105 * Si7l0s | ’ 2102 AV = sy +(L1D sg; = 128+ (111) (34) © of F since sz, = ELE © 128, it follows thar (1.11)" = 15.08 218 Zsgy = Zqg = 10] dg ~10] = 11S -100 [ > AV =640.72 : _ PE n = 2.1.13 We accumulate the payments to the beginning of the 6” year 219 I =i = (iy 1 pra (time 5) and then accumulate them for another 5 years. = Ela 1] = spn . 2055, (1.1)° + Xz (1.1 = 2000.04 — X = 8.92 Total interest = total accumulated value — total deposit l

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    28 » MATHEMATICS OF INVESTMENT AND CREDIT i SOLUTIONS TO TEXTBOOK EXERCISE: 2.1.14 An investment of amount 1 is equal to the present value of the 1 2.1.18 Option 1 corresponds to a single deposit earning or return of principal in » years plus the present value of the interest compound interest (compounded annually), and the accumy generated over the n years. value at the end of 24 years is 10,000(1+1)%. . Under Option 2, the 10,000 purchases an annuity-immediate at 2.115 2825.49 £ 10% paying K per year, so that 10,000 = Kaz; | (the purchase price of 10,000 is the present value of the annuity-immediate 2.1.16 The equivalent effective annual rate of interest is | being purchased). Solving for K results in P= 2-1 = i | = 10000 _ 10000CH _ 1000 i = (1.04)* —~1 = .0816. The balance on January 1, 2010 is : K = Cn 1,113. 20 _ t ) 100,000(1.04)™ + 5000775, 12,0005, (1.04) = 109,926. : Under Option 2, the payments of 1,113 will be received at the I end of each year for 24 years (it is implicitly understood that | with an annuity-immediate the payments begin one period after 2.1.17 Annuity (a) has present value 55a55,. . the annuity is purchased — this is referred to as the “end” of the Ca : year). If, as the payments are received, they are deposited into an The present value of annuity (b) can be formulated as . account earning interest at effective annual interest rate 5%, then 30a, + 600g, + 90v arg. Note that annuity (2) can also be . the accumulated value of the account at the end of 24 years is . ” 24 written as 55azy = SSagy, + 55v'%ary,. Both annuities have the 3s, = (113) G01 = (1113)(44.502) = 49,531. - same present value X, so that i Since Option 1 results in the same accumulated value, we have 10,000(1+4)* = 49,531, from which it follows that i =.0689. SSacg, + 55%, = agg, +60v ag, +90v7 ay . 000(t+1) 331, from which it follows that {= 0689, After canceling the factor ag, the equation becomes - 2.1.19 The phrase “at the end of each year” indicates an annuity immediate. 10 10 20 : Xam = X (150) = 493, 3 Xam = 3 (1202) = 2748 55+550'% = 30+60v'7 +90v*. . a = X35) = 493, 3a =3X | 1=7—) = 2748. Ce i - Using the factorization 1 —v¥" = (1—v/")(1+"), we have With v" = y, this becomes the quadratic equation - 3Xag, 30 _ Hn") = Z95 = 5574 > v7 = 858 2 95 = be Xa; vy T4937 EE 90y% +5y-25 = 0, L wo (=v) or equivalently 18) + y—5 = 0. The roots are y =.50, 356. 1 This idea has arisen in exam questions a number of times over We ignore the negative root for v'® =. Therefore, V0=50 so . the years. A similar factorization could be applied if 53; and that v=(.50)", and then i = 0718. Finally, X = 55455, = 575. 4 S51 were given. A more involved situation arises if az and ! a a5; are given. In that case, we use the factorization . 1-07 = (1) (+0 +27). Be fF

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