Theory of Interest Solution Manual

Theory of Interest Solution Manual

ID:40476251

大小:783.15 KB

页数:159页

时间:2019-08-03

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1、TheTheoryofInterest-SolutionsManualChapter11.(a)Applyingformula(1.1)2Att()=++2t3andA(0)=3sothatAt()At()1()2at()===+t23.t+kA()03(b)Thethreepropertiesarelistedonp.2.1(1)a()03==()1.31(2)at′()=+()2t2>0fort≥0,3sothatat()isanincreasingfunction.(3)at()isapolynomialandthusiscon

2、tinuous.(c)Applyingformula(1.2)22IA=−−()nA(n12)=+[nn+31]−−+−⎡⎣()()n2n1+3⎤⎦n22=++−+−−+−nnnnn2321223=+21n.2.(a)Applingformula(1.2)II+++=…"IAA[()1021−()]+[A()−A()]++[AnAn()()−−1]12n=−AnA()()0.(b)TheLHSistheincrementinthefundoverthenperiods,whichisentirelyattributabletothei

3、nterestearned.TheRHSisthesumoftheinterestearnedduringeachofthenperiods.3.Usingratioandproportion5000()12,153.9611,575.20−=$260.11,13024.Wehaveatatb()=+,sothatab()01==aa()3=+=91.72.b1TheTheoryofInterest-SolutionsManualChapter1Solvingtwoequationsintwounknownsab=.08and=1.T

4、hus,()2()a55.=0813+=()2()a10=10.08+=19.a(10)9andtheansweris100==100300.a()535.(a)Fromformula(1.4b)andA(tt)=1005+AA()5−−(4)12512051i====.5A()412012024AA(10)−−(9)15014551(b)i====.10A()914514529t6.(a)At()=1001.1and()54AA()5−−()41001.1⎡⎤⎣⎦()()1.1i===1.11.1.−=54A()41001.1()1

5、09AA()()10−−91001.1⎡⎤⎣⎦()()1.1(b)i===1.11.1.−=109A()91001.1()7.Fromformula(1.4b)AnAn()−(−1)i=nAn()−1sothatAnAn()−(−=11)iAn(−)nandAn()=(11+−iAn)().n8.Wehaveiii===.05,.06,.07,andusingtheresultderivedinExercise7567AAiii()74=+()(111)(+)(+)567==10001.051.061.07()()()$1190.91

6、.9.(a)Applyingformula(1.5)61550012.5=+=+(ii)5001250sothat2TheTheoryofInterest-SolutionsManualChapter11250ii===115and115/1250.092,or9.2%.(b)Similarly,6305001.078=+=+(tt)50039sothat39tt==130and130/3910/33==13years.10.Wehave111010001=+(it)=+10001000it1000it=110andit=.11sot

7、hat⎡⎤⎛⎞35001⎢⎥+=⎜⎟()()it250011.5[]+it⎣⎦⎝⎠4=+5001[]()1.5.11()=$582.50.11.Applyingformula(1.6)i.04i==and.025n11+−in()1+−.04()n1sothat.025.001+−()nnn1=.04,.001=.016,and=16.12.Wehaveiiiii=====.01.02.03.04.0512345andadaptingformula(1.5)10001⎡⎤+++++=(iiiii)10001.15()=$1150.⎣⎦

8、1234513.Applyingformula(1.8)26001()+=+=i600264864whichgives2()1+=ii864/6001.44,1=+=1.2,andi=.2sothat3320001()+

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