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1、第一章 利息的基本概念1.)()0()(taAtA2. ,11)0(ba180)5(100a508)8()5(300aa35.用公式(1-4b) 79.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 (321iiik14.nnnniiii)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
2、论 24. 用公式(1-32) 25. 4 4 216%1(1)(1 10%)11 8%45%12i 26.对于 c)及 d),c)中,nena)(1 111) 1 (vdieavlnd)中,ed128.tdxxeta0)()(29.;4411jihej 131.(1)902 天39. ,两边同时求导,类似tetAdr10)1ln(0tdrt AttA11)()(tB46.,10009200.081000d920)2108. 01 (288)08. 01 (x第二章 年金4.解:12010.087110.0870.08712160001000110.0871212 12A5.解:2221111
3、1111(*)n nnnniaxixiiixiayii 将代入(*)1did7.解:1000 100010000117185 12 1810001 0.0839169.84s8.解:10 0.110 0.15000sRa9.解:10 0.110 0.155000sRa14.解:永续年金每年支付R11 2nnRaRai17.解: 解得 即正常还款次数为 95 次0.0081500100000ma95.6m 解得95 95 0.0081500(1 0.008)100000af965.74f 19.解:(2)(2)(2) 105522210510002000170001117 1150iiisssi
4、ii令105( )1715f tttt0(1.03)(1.035)(1.03) 1.031.035 1.03fff i(1.032)0.003186f 23.解:,46 6 0.04111.04iai40.04114i 24.解:R1.1025R1.205R01423得4321.051.1025 1.051.1025 1.051.205 1.0511000RRRR2212.147R 25.解:1211111nnnn nainiiiaiii 其中通过公式(2-76)得到0.1020.1 16.8670.10.002nnn niaaain29.解: 7 7 77111vaviaiKi 类似地,11
5、18 11181111viaiLviaiM ,从而71118(1)(1)1v vviKiLiM LKMiKL31.解: (2)(12) (2)(12)(12)11 12nnnnnvviiaaaidi,32.解: 500lim110000tninai 半半,12 2111111iiidd 半半1 211id半1120tii半半36.解: 2020 201195.36n nanvainiIaii37.解:1101231该永续年金现值为1 i110123 3 6541该永续年金现值为:24111(2)iii i所求年金现值为:113 (2)(2)i ii ii i39.解: 01ntkt v dtf
6、gh11limlimnnnnvfa1(1)ngknv40.解:01 1( )1tdrra tet 1001( )ln(1)1nnnaat dtdtnt42.解:后五年等比551051111000105011k issiiik43.解:120567 3 10983414684468111vvvvaaaiiiiiiivd45.解:230 0.015251.0215KsKa46.解:1010 120180180 3003001.031.03iii iiaaaaa 月月新月新月月11x110000047.解: 01 1( )1tdrra tet 231414212111(0)(1)( )(1)84.5
7、1vtat dttdtt48.解:11tntnvvaa, 001111144010%tnnntnvva dtdtnna49.解:1) 11tnn t t ttatvIai第三章收益率2.解:234000 1.12000 0.933823.解:237000100040005500(0)vvvvv110.090.11.091.1iviv时,;时,令(0)0vvi 及7.解:81.5 16.510 (1)11.995%xxii 8.解:11100.250.751 (1)1 (1)1 (1)100000150002000011000kkkdtdtdtt kt kt keee解得:0.14117k 10
8、.解:01234567810911111i2i3i4i5i5i5i5i5i5i本金利息56 0.04 5 0.04610001.04550.04siis13.解:50000068000060000500055000ABI,29.78%IiABI14.解:1 114 43200001 12%5000180001112%196104Bi 15.解: 书后答案是,不知我对它对。12121231ktdttek1k 16.解:802 85%1.05 12dtje17.解:10654310000 1.041.05 1.041.05 1.041.041.0410000kkkk19.解:(1) 解得:2100
9、0 11000 12200ii6.52%i (2) 解得:21200220011000 1200 1000i9.54%i 20.解:30 30 0.0420 0.0420 0.04210000 1.04k ssksk1010 0.04108881000014.4%ksii 22.解: 选择第一种付款方式2 0.124000 150005003350000a24.解:2 54 34 81000 111990.76241000 12203.764jkk 25.解: 11273000 6%1.06118000810007.92181273000 1.061.06372000ttCCtC个月128t2
10、6.解:(1)231.0522 11.06216.71%ff(2)2331.0621 11.06667.339%ff第四章债务偿还1.解:5 510000 1.1220004917.72s2.解:1010 0.081.081468.0510xxxa3.解:设共需还款次n415001200na最初贷款额1500na4.解:10 0.0810000 10000 1.510000 2XiP Ps XiP 5.解:过去法:7251051510524000300020001400013000aaaaaisis未来法:3 3530002000aa v7.解:11 4812111.5ii月80 120100
11、000i iaa月月8.解:由于不知利率上调后偿还期的变化,因此用过去法比较简便 121212 12 0.0312 0.03523115 1 0.031 0.03510001.0351000ss9.解:2012011kkvv 10.解:61261000 11366.87 110001iii31366.87 1 i11.解:20 16 120 17 120 18 120 19 120 20 1vvvvv;5 543211 vvvvvvi;20 3 1100PvP 51 vPi12.解: (为剩余还款数)20 8 17 720(1)ksvaia k解得:=12k原利息:; 现利息:2020a207
12、xka节省利息 13131313121xkvv 13.解: 35 8 1135 22 114144113511080.25PPvPvv 第 29 次 35 29 1172Pv14.解:每次还款额为,L030B aN 每次还款本金为,第 次还款额为0 30Bt000 030(1)3030BBBBtia15.解:30 121 121 121.0021 1.05125000 1.051 0.002P 16.解:3 10 8 112iv 17.解:3 13000 12000nvn 6 13000nv 18.解:41 8%(1 i季)40500 40ia 季19.解:每月还款额360100000iPa月120 120325.40)(1)100000120ik iPaPaik 月月月(利息支出为:120(325.40) 120 10000066261.2PP20.解:104 44104 410410 1.054 1.05200101.05200101.050.050.05aaBaa 第 5 次还款中的利息为4iB21.解:10 10 1010avPaPiPi22.解:105 55105 10510510( 1)(1)10( 1)(1)avavaiaiiii 23.解:(1) 331.0411.120004001.11287.760.06 (2)第二年末贷款余额为