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

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LeI>|©3=1)HAPTER6JLomyw|CHAPTER1SECHOT6.4voeoeeerereressssesssrsivssseeneesessessessissssssssssmssensinensesses120|CHAPTER7;LOSv{||SECTION1.1SECHOD7.2ooevrroererseeesseresesosssssensersensesssssssssssssssssmmsssesssssneees133ee:1.1.1Balancesare10,000(1.04)=10,400afteroneyear,CHAPTER8]:.14110,000(1.04)*=10,816after2years,andSECHOM8.1coovvireeerereevcusessseareessesreerrasansss.Section8.2AA8.3wevrvreerresrssssressmsesmssesessssereresssssessssesssseninnes14210,000(1.04)°=11,248.64after3years.SECHONBibrrrInterestamountsare400attheendofthe1%year,416attheend:ofthe2"year,and432.64attheendofthe3year.CHAPTER9SECHOR9.1ooeeeooerrsreeeeseresesesssnsesenesisessnsseneesssssesssssnicsinnianss145|Section9.2ANd9.3everest1!1.12(a)2500[1+(.04)(10)]=3500HON91oveeeeresessseesessemenssesrassrarereterssansa:Section(5)2500(1.04)'°=3700.61’|(cy250001.02)*=3714.87:(d)25000.00)%=3722.16.1.13Balanceafter12monthsis10,000(1.01)1.0075)"=11,019.70.:Averagemonthlyinterestrateisj,where:10,0000+7)=11,019.70..Solvingforjresultsin0081244..A,

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Ei2»MATHEMATICSOFINVESTMENTANDCREDITSOLUTIONSTOTEXTBOOKEX31.1.4Therearetwo(equivalent)waystoapproachthisproblem.Wecan1.15(a)Over5yearstheunitvaluehasgrownbyaofupdatethebalanceintheaccountatthetimeofeachtransaction(1.10)(1.16)(1—.07)(1.04)(1.32)=1.629074.Thegeunlwereachtheodof10years,andsetthebalanceequaltoannual(compound)growthis(1.629074)°=1.1025,orXd))averageannualgrowthof10.25%for5years.Balanceat¢=4(afterinterestandwithdrawal)is."(b)Five-yearaverageannualreturnfromJanuary1,1996to10,000(1.04)"—(1.05)K;Lo’balanceat15isDecember31,2005isj,where(1+(1L17)°=(1.13)"°,so.that7=.0914.Annualreturnfor2004isk,where[10,00001.04)~(LOS)K}(1.04)~(1.05)K;;4A(1+k)(1.22)=(1.15),sothat&=.084.balanceatt=6is[[10,000(1.04)*-(L05)KJ(1.04)~105)K0.04)K;(¢)Overnyearsthegrowthis(I+)(L+iy)~(I+i,)=1+)",at¢=7iswheretheaverageannual(compound)interestrateis7,so-§at{+1isthegeometricf1+ii[[[10,0000.08)"~(1.05)K|eeFilthThus,1+i<(ryeene(by)yobritort,<=.(1.04)~(1L05)K|(1.04)—K(1.04)-K;at¢=10thereis3yearsofcompoundingfromtime7,sothat:yep2|1.1.6WeequatetheaccumulatedvalueofJoe'sdepositswiththatof[[[[r0.000004)"—(L.05)K1(1.04)-(1.05)K]:Tina’s.Notethatitisassumedthatforsimpleinterest,cachnewdepositisconsideredseparatelyandbeginscarningsimpleBa.05-K](1.04)K0.047=10,000.interestfromthetimeofthenewdeposit..10[1+10(.D]+30{1+5(.D]=SolvingforKfromthisequationresultsinK'=979.93.[0][1+5(0]=67.5Alternatively,wecanaccumulatetotime10theinitialdeposit=101.0915)"+30(1.0915)"%".andthewithdrawalsseparately.Thebalanceattime10isThiscanbesolvedbysubstitutinginthepossiblevaluesofn10,000(1.0:14)—K(1.05)(1.04)°_K(1.05)(1.04)°untiltheequationissatisfied.Alternatively,theequationcanberewrittenas_K(1.04)"—K(1.04)>=10,000.67.5(1.0195)*"—10(1.0915)'°1.0915)"-30(1.0915)°=0Thisisthesameequationasinthefirstapproach(andmust.whichisaquadraticequationin(1.0915)".ThesolutionisresultinthesamevalueofK).Ingeneral,whenusingcompound.no.244141.5:interest,foraseriesofdepositsandwithdrawalsthatoccurat.1.0915)"=gisvariouspointsintime,thebalanceinanaccountatanygiven2Weignorethenegatin_timepointistheaccumulatedvaluesofalldepositsminusthe:ennegativerootandget1.0915)"=1.226>n=23.accumulatedvaluesofallwithdrawalstothattimepoint.Thisisbalsotheideabehindthe“dollar-weightedrateofreturn,”which.willbediscussedlater.0el

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TT4»MATHEMATICSOFINVESTMENTANDCREDITSOLUTIONSTOTEXTBOOKEXER54>MATHEMATICSOFINVESTMENTANDCRED117@)1000=850[1+i($L)]—i=10735107.35%)11.10(@)10000112)=3000—t=—2)9694(9yea)365In(1.12)(6)1000=900[1+i(£%)]—i=.6759(67.59%)approximately253days).b)Attheendof9yearstheaccumulatedvalueis9001+(09(LL)|=913.32(©[((4%)1000(1.12)°=2773.08.Attimesduringthe10"year,the@900]1+(09)(5%5)=1000—&=451;accumulatedvaluebasedonsimpleinterestwithinthe10%yearis2773.08(1+.125).Settingthisequalto3000and*dayi(25%)!118ItistoSmith’sadvantagetotaketheloanof975onthe7"dayifsolvingforsresultsins=-2""32=6819yearstheamountpayableonthe30thdayislessthantheamountdue12tothesupplier:(approximately249days)aftertheendof9years.._3)o75[1+:<1000—i£4069.(€)10000103)’=3000>¢=AEs=110.41months(about9yearsand2monthsand13days).1.19(a)Maturityvalueof180-daycertificateis(d)10000+i)'®=3000—i=3"°~1=.1161(11.61%per100,000(1-+.075(182)=103,698.63.year).daysi(e)1000(1+/)"=3000—i=3"1=009197_Interimbookvalueafter120aysis-(:9197%permonth).100,000(1-+.075(322))=102,463.75.BankwillpayXafter120dayssothatLIL(2)(00757=1.0299<1.0360.1)=82.;x(1+-09(£))=103,698.63—X=10218682(but(1.007917=(1.0075)*=1.0303)Thepenaltychargedis102,465.75—102,186.82=278.93.:|®(1.01577=1.0604>1.06(b)1.08=(+2)(1+5)>i=.0819::1.1.12(a)SmithbuysA=227.5unitsafterthefront-endloadispaid.-Sixmonthslatershereceives(227.5)(5)(.985)=1120.4375..Smith’s6-monthrateofreturnis12.04%onherinitial1000..(b)Ifunitvaluehaddroppedto3.50,shereceives.(227.5)(3.5)(.985)=784.30625,whichisa6-montheffectiveirateof—21.57%.

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6»MATHEMATICSOFINVESTMENTANDCREDITSOLUTIONSTOTEXTBOOKEX71.1.13Weusethefollowingresultfromcalculus:iffandgareGraphof¢In(l.1)differentiablefunctionssuchthatf(a)=g(a)andf(x)<g'(®)fora<x<b,thenf(b)<g(b).In(8)(i)Suppose0<7<l.Letf(i)=(1+)andg(i)=I+i-z.ThenFO)=g(0)=1.Tfwecanshowthatf'(i)<g'())foranyIn(4)i>0,thenwecanusethecalculusresultabovetoconcludethatfi)<g()foranyi>0.Firstnotethatf'())=(i)In(2)andg'(i)=t.Sincei>0,itfollowsthat1+i>1,andsincei(<1,itfollowsthat¢—1<0.Then(M+i)7<l,so[.fa<g@.:[3Thiscompletestheproofofpart(i).:i"CLAelei(iy=(141)..(ii)Supposethat¢>1.Letf(i)=1+i-tand8(1+)SECTIONS1.2AND1.3Againf(0)=g(0).Ifwecanshowthatf'(i)<g'(§)forany:—_—i>0,thenwecanusethecalculusresultabovetoconclude:21PpAthatf(i)<g()foranyi>0.Since¢>1andi>0it.“-resentvalueisfollowsthat£~1>0and1+>1.Thus(144)>1,andit5000etoto732553-ted_i106(1.06)(106°(1.06)-followsthatf'(7)=¢<¢-(1+)""=g'(i).Thiscompletes:theproofofpart(ii).i::1.2.2Amountnowrequiredis1.1.14Originalgraphisy=(1+).Newgraphs107=(1+),or,25,0000”+v'%+v'21+100,000[%+v'®+v*]=75,686equivalently,y=tSe,sothatyisnowalinearfunctionof#o:311.2328=15+16.50v—v=T8779—i=.26921241000-Vvivy=494.62Graphof(LI)Vos“Vor“Voy=494.641.2.5EquationofvalueonJuly1,2013is2:2001.04)+300=100(1.04)*+X—X=379.48ib51015-t:

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8»MATHEMATICSOFINVESTMENTANDCREDIT.SOLUTIONSTOTEXTBOOKEX)<91264830=50+100(v+v?++v")+X07,wherev=rk,1.2.121000(+)%+1092=2000(1+4)sothatX=67.57.Ifinterestis.01permonth,thenv=odSolvingthequadraticequationfor1+resultsinnoreSs.andX=67.98.]12.43(@)£0+)"=a+)"©Za+)=(+0)In)on1012710042000"+300v>"=600v'"—(b)Ly=wm(d)Ly=~v"In(l+i)6000=100+200(.75941)+300(.75941)o>V0=708155~»i=(708155)~1=0351.:..1.2.14Withanannualyieldratequotedtothenearest.01%,theannual:yield7isintheinterval.11065<i<11075.128(a)(20)2000)[v+v*+9+--+]=1,607,391(at75%)Sincethequotedannualyieldrateis365-109=Ptceitfoliows48_irice(b)1,607,391200,000vN1,747,114that11065<365.100=Fdee«11075,or,equivalently,©X=1,607,391+.156™—X=1,795,55194.767<Price<94.771.129750=367.85[1+(1+/)]=>j=0389isthe2-monthrate.-[L+@ep}=71215(a)P=2000000_95550.52|L1051.2.10WithXinitiallystocked,thenumberafter4yearsis|_100,000dP100,0001824155_Lo0)P===mwgesX(14)-5000[(14)+(1.4)°]=X>X=4997.-EZ{voiL=-45,239.03ifi=.10-100,1000._400,1000frispor..12111000=290+4008,and1000={95+30%.TtisnoZL28D=-45239.03—AP=—45239.03-Al.possiblethatj=k,sincethetwopresentvaluescouldnotbothaIfAF=00Lthbeequalto1000(unlessj=k=0,whichisnottrue).Ifj>k,.¢=001thenAP=—45.24.then(1+)?>1+kand(14+)>(+k),inwhichcasethefirst:100.000presentvaluewouldhavetobelessthanthesecondpresent©P=T75rdbo1000092373334ifi=.10.value.Sincebothpresentvaluesare1000,itmustbethecaseLo765(14-2)thatj<k(j=.0333andk=.0345).:AstheT-billapproachesitsduedatethe&goesto0.

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10>MATHOMATICSOTINVESTMENTANDCREDIT:SouuTions19TexTs0okExecsfill_Nocr1.2.18(a)MinimummonthlybalanceforJanuary2005is25i1216)B=By(i+)+2a,[140-4]February2005is6000,andforMarch2005is9500.tuiearnedis(10)(f5)r2s00+6000+9500]=150.=By+Xa+5+2%a0]k=lr=BalanceonMarch31is2500(4)--1000(3)+150=13,150.(b)ThebalanceisB,for1years,By+a,for¢,—1years,(b)Minimumdailybalanceis2500forJanuary1-15,3500forBy+a;+a,fort;~t,years,By+a+a,+--+a,for1-4,£January16-31,6000forFebruary1-15,7000forFebruaryyears.Theaveragebalanceis16-28,9500forMarch1-15,and10,500forMarch16-31.Interestcarnedis=_Bot+(BoraY=)+(Borar+ayta—ty)+++(Boratay+--+,Y1-1,)Lyw=cerEressrerBai(10){355)((2500(15)+3500(16)+6000(15)+7000(13)+9500(15)+10,500(16))=160.27.=oroi)raln)bta,(=)BalanceonMarch31is©)=By+2a,(1-1).2500(4)+1000(3)+160.27=13,160.27.(c)MinimummonthlybalanceforJanuary2005is2500,sointerest(c)Followsdirectlyfrom(a)and(b).:onJanuary31is422500)=20.83,sobalanceonJanuary31(afterdepositandinterest)is6020.83.1.2.17Thedifferencebetweenthetwopaymentplansisthatthefirst2;MinimummonthlybalanceforFebruary2005is6020.83,so’paymentsaredeferredfor2months,sothesavingis[:io210;interestonFebruary28is43(6020.83)=50.17,sobalance30[0)—(H+)~12.68.;onFebruary28is9571.00.N3MinimummonthlybalanceforMarch2005is9571,sointerestAlternatively,thepresentvalueunderthecurrenttplanis§2v,thepntpaymentplan1s[onMarch31is409571)=79.76,sobalanceonMarch31is300+v+v?+P]=643.67.13,150.76.ThepresentvalueunderSmith’sproposedpaymentplanis:(d)Minimumdailybalanceis2500forJanuary1-15and3500|forJanuary16-31,sointerestonJanuary31is25.62.300°+--+v5]=643.677=630.99.:|Minimumdailybalanceis6025.62forFebruary1-15andSavingis12.68.i7025.62forFebruary16-28,sointerestonFebruary28is49.79.Minimumdailybalanceis9575.41forMarch1-15and|10,575.41forMarch16-31,sointerestonMarch31is85.71.iBalanceonMarch31is13,161.12.

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12»MATHEMATICSOFINVESTMENTANDCREDIT.SOLUTIONSTOTEXTBOOKEXERCISSECTION1.41.43Equivalenteffectiveannualratesare}]]MountainBank:(1.075)%—1=.1556251.4.1m=1impliesinterestconvertibleammually(m=1timeperyear),RiverBank:whichimpliestheoffectiveannualinterestrate¢M=i=.12.Weuse365Equation(1.5)tosolvefor7fortheothervaluesof,asshownbelow.6+52)—-12.155625je)361/365-TearcffectiveiN(1+)>(1.155625)=1.000396356-.fm7Effective.EE|Cayinterestrate£7’[i-"I—i21446701(1year)LedZego12)-1=122(6months)|Sr=+12=.06(1.06)%~1=.123614.4Thelast6mouthsofthe8%yearisthetimefromtheendofthezzs|15"totheendofthe16%half-year.3(4months)|Sr=:12=.04(1.04)°~1=.124864|43months)02031.03%1=.12550001H2H=1Y3H4H=2Y...14H=T7Y15H16H=8YEYOSEEHt6@months)|fg=2=02(1.02)°1=.126162I8(1.5months)===0151.015)—1=.126593TheamountofinterestearnedinEric’saccountinthe16”half-on1yearisthechangeinbalancefromtime15Htotime16H.12:1months)|45=+2=01(10D-1=126825:.Y125iThebalancesatthosepointsareX(1+4)andX(1+4).s2(tweek)|5p=r=0023|(1442)<1=127341;Co365iTheamountofinterestearnedbyEricintheperiodis2-12=00033|(14427%=..INTO365(1day)|565=365(1+42)1=127475(14)x(e4)_X14).d2Vqop12iIEim(1+y)=eol=127497iThebalanceinMike'saccountattheendofthe15%half-yeari(7.5years)is2X(1+7.5/),andthebalanceattheendofthe16”.half-year(8years)is2X(1481).142(a)1000vhys=414.64.:TheinterestearnedbyMikeinthatperiodis0!.(b)1000v55=409.30|2X18)2X(147.50)=2X(50)=2X(4).(©)1000v5%s=407.94WearetoldthatEricandMikeearnthesameamountofinterest.|INTO.\I5Therefore,X(144){£)=2(£),sothat(1+4)"=2..Theni=2[2""°—1]=.0946.5

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14»MATHEMATICSOFINVESTMENTANDCREDIT;.SOLUTIONST0TEXTROOKEXERCIS:14.5Quariedyeffectiverateis323008125.InitialamountL1.49(a)Wewishtoshowthatinvestedaftercommissionis.99.Attheendof3months,the!Fim)=FANEiaccumulatedvalueis.99(1.008125).Thisisthensubjecttothe:on)=finf1+7)reg|>0First,fm)01%commissionfortherolloverandthenthe3-monthinterest-sincej>0.Also,ifrateof.008125.Attheendoftheyear,theaccumulatedvalueis450,ifx>0andh(x)=In(l+x)-73.then.(x)=5en.[99(1.008125)]°=992198=1-078.Theeffectiveaflr-(x)=qi>0,andsince'h@©)=0,itfollowsthatcommissionreturnis—.78%.:h(x)>0forallx>0.Lettingx=1>0,weseethat|Ndin{1+Z)-—L>0,whichimpliy1.46FOIsla-1]~105:(2)[ra,whichimpliesthatf(m)>0.Trg-NY1+)iJId1]=116025b)g'm)=+H"1Gene)LhAY_a.#0=10[10)1]=159374-=ap[tne]-1.0=oi](10)-1]=137.796.Butx{1~Inx]hasamaximumof1atx=1,sothatwith.(+)=x,weseethatg'(m)<0form>1.1.47FromNovember9toJanuaty1(53days)Smithearns(twofull.-(©)ConsiderTn[f(m)]=m-In(1+FA)_In(i+4)hmonths)interestof£(.1125)(1000)=18.75.Thus,Smithearnsa.[aEAen53-dayeffectiverate.ofinterestof.01875.Theequivalenteffec-de.in(1+4)oRaelim1=fima)LfOTLme?.Liveannualrateofinterestisi=(1.01875)=1=1365.ooJimInfGn]=Jim=p=im7a)=J.Thuslimf(m)=e’.14.8Leftondepositforayearat0%=09,Xaccumulatesto-©zm)=GafniCo[i2)X(1.0075)2.Ifthemonthlyinterestisreinvestedatmonthly.IJimg(m)=limpyrate75%,theaccumulatedvalueattheendoftheyearisE=In(l+),sincelim(1+/)""=1.X+X(0075)[(1.0075)+(1.0075)+--+(1.0075)+1)...1.4.10Wewanttofindthesmallestinteger7sothat.2kop._mSince147472++rt=,itfollowsthatthetotalattheoofm)=[14:2]2118,f(2)=1.1772,f(3)=1.1798,endoftheyearwithreinvestmentofinterestis.Caf)=11811=m=4.Ey9,:167"#1coors)SIRS!=xa.0079)"aa::=netierbowmanytimesperyearcompoundingtakesplace,a=.nalrateofinterestof16%cannotaccumulati|.rateofmorethan17.35%.atetoaneffective

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16>MATHEMATICSOFINVESTMENTANDCREDIT.SOLUTIONSTOTEXTBOOKEXERCISESSECTION1.5:1.5.6Thepresentvalueof1dueinnyearsis(I-d)",sotheacc:latedvalueafterryearsofaninitialinvestmentof1is—-_X=-LSI(0)4992=gf—X=5187.84==0-0)=—X=5191.68(b)4992Tecan].(©)4992=X(-08)"—>X=$204.521.5.7Theinitialdepositof10growsto10(1~4)""attheendof10_|B|years(40quarters),andthencontinuestogrowat3%perhalf(d)4992=X[--co)(3)]-X=5200|yearafterthat.Theaccumulatedvalueoftheinitialdepositof10:.40;attheendof30yearsis10(1-4)x(1.03)*(20moreyears,1.52Withaquoteddiscountrateof.940,thepriceofa91-dayT-Bill[40morehalf-yearsat3%perhalf-year).91I.;shouldbe10001-575>-0113)=-99.714asquoted.|Theseconddepositis20madeattime15.Theaccumulated:valueoftheseconddepositattime30(15yearsafterthesecondTheinvestmentrateisfoundas(55%-1)«368=0115,as|deposit)is20(1.03)*®(15yearsis30half-years).quoted.:Thetotalaccumulatedvalueattheendof30yearsis:0(1-4)x(1.03)*+20(1.03)°=100.A{r+T)—AfAl|1554=ADHDADagSolvingfordresultsind=.0453.AGT)4AQT|J=delA)=2x-:ThisquestionisfromtheMay2003Course2examthatwasAr)conductedjointlybytheSocietyofActuariesandtheCasualty=1-d;=FTrynd:ActuarialSociety.Ttshouldbenotedthatthenominalinterest.a:ratenotationi")andnominaldiscountratenotationisnot@d;=eaand(b)J==|alwaysspecificallyusedontheprofessionalactuarialexams.InJ’Lthisexample,thenotationdwasanominalannualrateof=discountcompoundedquarterly.154115=(1-d)1.3)—»d=.1154|.1.5.8(a)Bankpays:dogs=mgt=We]=1.5.5Bruce'sinterestinyear11:100(1—-d)™°{a-ay*-1]=X.’3+363ndsIdeRobbie'sinterestinyear17::As7increases,iincreases.&_1icl500-d)7[aay-1]X=1001-2)[1=ay1]|(b)From(a)1-dt=friedd=TaIfi=.11then: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»MATHEMATICSOFINVESTMENTANDCREDIT:coSOLUTIONSTOTEXTBOOKEXERCISES18»MATHEMATICSOFINVESTMENTANDCRED1.5.9SupposethattheT-Bill’sfaceamountis$100.ThenSmith;1.62(122)?=dnlasde_004045day0339.purchasesthebillfor100]1-482.(10)]=94.94(nearest01).!791dayslater,thevalueoftheTBillis:I2i.100[1-£(10)]=97.47,:163exp(f34dn)=(1-047>P=150414i125_=Smithsreturnforthe91daysis3237—1=0266(2.66%).!—422=In(1.50414)—k=102..»mL,:1.6.4EffectiveannualrateforTawnyisi=(1.05)-1=.1025.53hitoeEg,and£2=J,50|1.5.10FromExercise1.3.3,webaveT==4":Tawny:&=In(1.1625)=09758.am_god&rd=Fand:Fabio:Simpleinterestratej—5,=1L.@SedmnTeThE"[7|Attime5,wi~_JPe=He©—Lo09758=pd=j=.1906->Z=10001+51=1953.Ed:::15.111000=1200(g;)=i)—©=0909165oofora-ehota],xlhota~1)-x.>1007=e)=X(2-¢%)—>X=7846.15.121000(+/)=1000+40(1+j)?=(1+))°~25(+/)+25=0—1+=1.043561or23.9564.SinceJ<.10,itfollowsthati1.6.6Bruce’s6-monthrateofinterestisiand7.25yearsis14.5J=.0436.I6-monthperiods.Bruce'saccumulatedvalueafter7.25yearsis.NTE]|100(1+4)=200.Solvingfori,wegetiSECTION1.6|(14)=25i=0979.1.6.1Accumulatedvalueattime1is:Peter’saccountgrowsto100¢”%%%=200,sothat'os;10,0005el™"=10,000x¢%=10,512.71,|6=-LzIn2=.0956.Theni~6=.0023=23%.Accumulatedvalueattime2is|10,000xJo05H105+026-0141.10,000x9%=11,162.78167(79)=1.36086~>¢’0=1.36086—1+i=¢=1.045.=Bat

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20»MATHEMATICSOFINVESTMENTANDCREDIT:coSOLUTIONSTOTEXTBOOKEXERCISI.s.:JsSW_os168(a)(1+)=oxp|[;(08-9230)dr]=1:616407>=1008!(©For0<t=4,6=gy=1x09Tofind8,44,let7=td,orf=r-L.(®)1+i=exp(108+828)9=1.091629—>1;=.091629Thenthi,=oxp[I](08-0231)dt|=1.0995091,=.099509Ll55oS)iertid,=Op=greyfy=102751,iy=.104532,is=.105659S(r+3)‘Eo1000(1.02)(.08)_08(©)1000-expJ:(08+)ar|=821.00SO)Tovacoofiecosr)]|TTCO%[Thesameoccursfortkand+31.69Ke?=960,Ke®=1200—¢™=1-d=.80—K=1500)andd=.20.Ifdchangesto.10,thenthepresentvaluebecomesA-neAh)1500(1-.10)°=1215.;L612@=m)|1)1)a1610[=e~1,6=26>+)Arey).i_1:myyiA(eHR)=A(t)A:melmPel=(I)=2i>2,©Letb=.Thenlima=fimAGRZ50=465.d==l-e¥=1-(1-@)*=2d~d*<2d.a)L613(@)6=40=adm"fim5=0Taragrrat1.6.11(a)1000(1.02)%[1+co9)(3%)]=1044.73(applyI’Hospital’srule)(b)For0<¢<&,A()=100001+(.08)]|(b)4)=ox[0,a]=exp[k-2-£"%].|pry11=1oA.fordsrl,40)=100001.02){1+(.08){z1]lim£9=lin|opfor&<1<3,A@)=100001.02)2]1+(08)(tL:CANEs24[(3)]my=Immafor3<1<1,Ar)=1000(1.02)°|1+(.08){r—2:-Ei1_4[(3]|ergpryBl|

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22»MATHEMATICSOFINVESTMENTANDCREDITSOLUTIONSTOTEXTBOOKEXERCNSECTION1.71.7.7Smithneeds{38=917.4312USnowifheinvestsinNair_0205usars;account,Thisisequivalentto2812=12567551Cdn,whichleteal=157=I+.05|growsto1382.4306inoneyearinaCanadiandollaraccounteaming:L10%.Theimplicationisthatoneyearfromnow,1000US=1382.4306Cdn.,or,equivalently,723364US=1Cdn.172After-taxretumisS2639710=0309.178(+r)0"=(zyJESBEN173(a)SmithsATIthisyearwillbeI+[ECbE21,000(.75)+21,000(.50)=26,250|—i+rI+randtaxespaidwillbe15,750.Therealgrowthfromlastyear|tothisyearinSmith'sATVis20250230001.00,andthe|hb:1.7.9(a)Realafter-taxrateofreturnonstandardtermditiBL1s.25,250/15,000|epositiswihintaxsmi=1.00.:i(1=,)~.realgrowthintaxespaidis“227771.00i|Mor”,andontheindexedtermdepositis|asitus(L084=1-1)._(b)Continuingtheoldtaxationscheme,Smith’staxespaidthisiyearwillbe(25)(20,000)-+(.50)(22,000)=16,000,andhisi.CoATIwillbeSoe,Therealgrowthi"taxespaidwillbev(b)Settingthetwoexpressionsinpart(a)equalandsolvingfor16,000715,000i|i,wehave=i'"(1+7)+5-L.EE=1.015873(1.59%)andtherealgrowthinI=.26,000/25,000.,ATTis28000/55.000990476=1-.009524(~95%)..If#'=.02andr=.12,theni=(i).1424,(i).1824,(iii)2224,(iv).3224.17.4Smithsellstheitemsfor100,000x1.15=115,000attheendof.theyearandmustpayback100,000x1.10=110,000.Netgain.is5,000(inyear-enddollars).-175dm=lei=Ho=GoSF|18-14_i=l.|176“yr=52»i=1070175(107%)-

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-x=}=2i;wv-CHAPTER2:jiSECTION2.1:2.1.1OptionIaccumulatedvalueis50,000(1+1)*".|iOption2annuitypaymentisX,whereK=50,000=5564.99J[aoo!Then556499553=50,0000+1)sothat =6.9%...2.1.2Supposetheannualeffectiverateisi.Then900s;=1000a,..|10_.|sothatsuo(7-1)=1000/1].andthen(14°=12.|..|WethenhaveKsg,x(1+)=1000a;,sothat.:i'K=—J000151942.|(+2)(141):203sp=[AH+Uh)bn(0)+(1)[+05QFen(4)+1]=df[ayi)er(4)+1],=ifs,+5;I31000s,|b21410005,=Vag,—Y=——sbv=1000(1.0D*°=19,788.47

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26»MATHEMATICSOFINVESTMENTANDCREDIT:SOLUTIONSTOTEXTBOOKEXERCIS|215(i)10087,=715.952.1.10AfternyearsSmith'sAVis[80s+200](1.06)"(i)10085314095=2033.87:Brown'sAVis4085-55146+P.Thus,ifn=13,thenP=1%.93,0,n=20»P=17.19,n=25—20.75.(iif)100]5153075(1.00875)(LOY+sg)7(LOY)+5201].=3665.125teint[ml(HPLny2210(iv)3665.12(.01)=36.652H@gh=yoy=Hl==;on(yd:=>+)"=2,70=sz=2169855+)"+19655=8000,(L+)"=2o>0ok=014286,55=5m(14)55,=490==[S;T+"1==:Ili~2P—196]2s+557]=8000—255755=40.82®)J_Sr=Pr)1=L20wr2GHPen—»quadraticequationinz=(1+i)":22+z5K=0¢:i~1%ITERBNi=>(+H)=z=—y—t—(discardnegativeroot)>243=40.82>i=.1225.|I2ICTIpa2251);“2B:217(@)100105)sg;+20005si50,©sg=I)"sym+(HD+1+3001.03)"+575+40875,=2328.82IL=3634(1+i)+(1+)-47.99=0®10sg555+2s)+Hsggr=syg)+dsp|—147=1.1355,or—1.1630(discardnegativeroot)=0$55105*555105+550105*Si7l0s|’2102AV=sy+(L1Dsg;=128+(111)(34)©ofFsincesz,=ELE©128,itfollowsthar(1.11)"=15.08218Zsgy=Zqg=10]dg~10]=11S-100[>AV=640.72:_PEn=2.1.13Weaccumulatethepaymentstothebeginningofthe6”year219I=i=(iy1pra(time5)andthenaccumulatethemforanother5years.=Ela1]=spn.2055,(1.1)°+Xz(1.1=2000.04—X=8.92Totalinterest=totalaccumulatedvalue—totaldepositl

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28»MATHEMATICSOFINVESTMENTANDCREDITiSOLUTIONSTOTEXTBOOKEXERCISE:2.1.14Aninvestmentofamount1isequaltothepresentvalueofthe12.1.18Option1correspondstoasingledepositearningorreturnofprincipalin»yearsplusthepresentvalueoftheinterestcompoundinterest(compoundedannually),andtheaccumygeneratedoverthenyears.valueattheendof24yearsis10,000(1+1)%..UnderOption2,the10,000purchasesanannuity-immediateat2.1152825.49£10%payingKperyear,sothat10,000=Kaz;|(thepurchasepriceof10,000isthepresentvalueoftheannuity-immediate2.1.16Theequivalenteffectiveannualrateofinterestis|beingpurchased).SolvingforKresultsinP=2-1=i|=10000_10000CH_1000i=(1.04)*—~1=.0816.ThebalanceonJanuary1,2010is:K=Cn1,113.20_t)100,000(1.04)™+5000775,12,0005,(1.04)=109,926.:UnderOption2,thepaymentsof1,113willbereceivedattheIendofeachyearfor24years(itisimplicitlyunderstoodthat|withanannuity-immediatethepaymentsbeginoneperiodafter2.1.17Annuity(a)haspresentvalue55a55,..theannuityispurchased—thisisreferredtoasthe“end”oftheCa:year).If,asthepaymentsarereceived,theyaredepositedintoanThepresentvalueofannuity(b)canbeformulatedas.accountearninginterestateffectiveannualinterestrate5%,then30a,+600g,+90varg.Notethatannuity(2)canalsobe.theaccumulatedvalueoftheaccountattheendof24yearsis.”24writtenas55azy=SSagy,+55v'%ary,.Bothannuitieshavethe3s,=(113)G01=(1113)(44.502)=49,531.-samepresentvalueX,sothatiSinceOption1resultsinthesameaccumulatedvalue,wehave10,000(1+4)*=49,531,fromwhichitfollowsthati=.0689.SSacg,+55%,=agg,+60vag,+90v7ay.000(t+1)331,fromwhichitfollowsthat{=0689,Aftercancelingthefactorag,theequationbecomes-2.1.19Thephrase“attheendofeachyear”indicatesanannuityimmediate.101020:Xam=X(150)=493,3Xam=3(1202)=274855+550'%=30+60v'7+90v*..a=X35)=493,3a=3X|1=7—)=2748.Cei-Usingthefactorization1—v¥"=(1—v/")(1+"),wehaveWithv"=y,thisbecomesthequadraticequation-3Xag,30_Hn")=Z95=5574>v7=858295=beXa;vyT4937EE90y%+5y-25=0,Lwo(=v)orequivalently18)+y—5=0.Therootsarey=.50,356.1ThisideahasariseninexamquestionsanumberoftimesoverWeignorethenegativerootforv'®=.Therefore,V0=50so.theyears.Asimilarfactorizationcouldbeappliedif53;andthatv=(.50)",andtheni=0718.Finally,X=55455,=575.4S51weregiven.Amoreinvolvedsituationarisesifazand!aa5;aregiven.Inthatcase,weusethefactorization.1-07=(1)(+0+27).BefF

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