第1章 利息理论基础08
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第一章利息的基本概念1.)()0()(t a A t A =2.,11)0(=∴=b a 180)5(100=a 508)8()5(300=a a 3~5.用公式(1-4b)7~9.用公式(1-5)、(1-6)11.第三个月单利利息1%,复利利息23%)11(%)11(+−+12.1000)1)(1)(1(321=+++i i i k 14.nn nni i i i −−+⋅+>+++)1()1(2)1()1(16.用p.6公式17.用P.7最后两个公式19.用公式(1-26)20.(1)用公式(1-20);(2)用公式(1-23)22.用公式(1-29)23.(1)用公式(1-32);(2)用公式(1-34)及题6(2)结论24.用公式(1-32)25.44216%1(1)(110%)118%45%12i ⎛⎞+=++⎜⎟−⎝⎠⎛⎞−⎜⎟⎝⎠26.对于c)及d),,,c)中,,δn e n a =)(1111)1(−=−=+==∴v di e a δ∴v ln −=δd)中,δ−−=ed 128.∫=tdxx e t a 0)()(δ29.;4411⎟⎠⎞⎜⎝⎛+=+j i h e j =+131.(1)902天39.,两边同时求导,,类似t e tA dr +=∫10δ)1ln(0t dr tA +=∫∴δtt A +=11)(δ)(t B δ46.,10009200.081000d −==9202108.01(288)08.01(=×−+−x 第二章年金4.解:12010.087110.0870.08712160001000110.087121212A −−⎛⎞−+⎜⎟⎛⎞⎛⎞⎝⎠=+⋅++⎜⎟⎜⎟⎝⎠⎝⎠5.解:()()()()22211111111(*)nnn nn i a x i xiii xi a y i i −−−−+==⇒+=−−+−−===将代入(*)1d i d=−7.解:100010001000011718…()51218100010.0839169.84s −+=&&8.解:100.1100.15000s Ra =&&&&9.解:100.1100.155000s Ra =&&&&14.解:永续年金每年支付R112n n Ra R a i ⎛⎞=−⎜⎟⎝⎠17.解:解得即正常还款次数为95次0.0081500100000m a =95.6m ≈解得95950.0081500(10.008)100000a f −++=965.74f =19.解:()()()(2)(2)(2)1055222105100020001700011171150i i i s s s i i i ⎛⎞−+=⎜⎟⎜⎟⎝⎠∴+++−++=令105()1715f t t t t =+−+0(1.03)(1.035)(1.03)1.03 1.035 1.03f f f i −−=−−(1.032)0.003186f =−23.解:,()4660.0411 1.04i a i −−−++40.04114i ⎛⎞+=+⎜⎟⎝⎠24.解:R 1.1025R 1.205R 01423得4321.05 1.1025 1.05 1.1025 1.05 1.205 1.0511000R R R R ×+++=2212.147R =25.解:()()()1211111nn nn n a i n i i i a iii −−−−∂−++−++=∴=∂其中通过公式(2-76)得到0.1020.116.8670.10.002n n n n i a a a i==∂−∴==∂L n29.解:7777111v a v i a iKi−=∴=−=−类似地,111811181111v ia iL v ia iM=−=−=−=−,从而71118(1)(1)1v v v iK iL iM =∴−−=−Q L K M i KL+−=31.解:(2)(12)(2)(12)(12)1112nn nnnv v i i aaa id i−−⎛⎞===+⎜⎟⎝⎠&&,32.解:()500lim 110000tn i n a i −→∞+=&&半半,()()122111111i i i d d−+==+⇒+=−−半半()1211i d −=−−半()1120ti i −+∴=半半36.解:()()()2020201195.36n n anv a i n i Ia ii−−+−+=∴=&&37.解:110123……1该永续年金现值为1i11123……6541该永续年金现值为:()()24111(2)i i i i−−++++=+L ∴所求年金现值为:113(2)(2)i i i i i i++=++39.解:()01ntkt v dt f g h−=−−∫11lim lim n n n n v f a δδ→∞→∞−===1(1)ng kn v δ=−⋅40.解:011()1tdrr a t e t+∫==+1001()ln(1)1nnn a a t dt dt n t−===++∫∫42.解:后五年等比()()()551051111000105011k i s s i i i k+⎛⎞−⎜⎟+⎝⎠−+×++−&&&&43.解:120567……10983…414684468111v v v v a a a i i i i i i i vd−+−+−+=+++=−L L 45.解:2300.015251.0215KsKa−=+&&&&46.解:1010120180180300300 1.03 1.03i i i iia a a a a −−++=月月新月新月月11x110000047.解:011()1tdrr a t e t+∫==+231414212111(0)(1)()(1)84.51v t a t dt t dt t−=−=−=+∫∫48.解:11tn t n v v a a δδ−−==,()001111144010%t n nnt n v v a dt dt n n a δδδδ⎛⎞−−==−=−=×=⎜⎟⎝⎠∫∫49.解:1)()11t n nt tt t atv Ia i==−=∑∑&&第三章收益率2.解:234000 1.120000.93382×−×=3.解:237000100040005500(0)v v v v v −−++=110.090.11.09 1.1i v i v ====时,;时,令(0)0v v i=⇒及7.解:81.516.510(1)11.995%x x i i ⋅⋅=+⇒=8.解:11100.250.751(1)1(1)1(1)100000150002000011000kkkdtdtdtt k t k t k e ee+−+−+−∫∫∫+−=解得:0.14117k =10.解:1234567810911111i 2i 3i 4i 5i5i5i5i5i5i本金利息560.0450.0461000 1.04550.04s i is −⎛⎞++⎜⎟⎝⎠13.解:50000068000060000500055000A B I ===−=,,29.78%Ii A B I=≈+−14.解:()11144320000112%5000180001112%196104B i −⎛⎞⎡⎤⎛⎞=×++×+−×+−×=⎜⎟⎜⎟⎢⎥⎝⎠⎣⎦⎝⎠15.解:书后答案是,不知我对它对。