刘占国利息理论答案全套1 5章

1、第一章利息的基本概念1. A(t) = A(O)a(t)300a(8508a(5)2. a(0) =1二 b =1 100a(5) =180,35.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(1+1%)3 -(1 +1%)212. k(1 +(1 +i2)(1 5) =100014.(1 +i)n +(1 +i)2j(1 +i)n (1 +i)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）;用公式（1-23）22. 用公式（1-29）23. （1）用公式（1-32）;（2）用公式（1-3

2、4）及题6（ 2）结论24. 用公式（1-32）41 6% V25. (1+i)4 =(1+10%)1+2丿 1-8% I 4 丿 5%Y-1=VC)中，6 = - In V，I 2丿26.对于 c)及 d)，a(n) =en&，二 a(1)=e5 = 1+i =11 - dd)中，d =1 -ej&x) dx28.a(t) =e029.1 +i； 1 + j =e31. (1) 902 天39.e=1+t 二 0仇4=1 n（1+t），两边同时求导,S(t)=1丙3B（t）类似46.d =1000 920 =0.08，x(1 -0.08) +288(1 0.08xJ) =9201000 2第

3、二章年金an - X -i二(1+i) =1-Xi_2n21(1 +i)1 一(1 -xi )a2n - y -iipl将i =代入（*）_n5.解:(*)1-d_n1-(1 +i 4.解:+0.087 l16000 =A +1000 ”0.08712+0087舊+占12人 12丿1000 100010001718_51000鵝(1+0.08)P =39169.848.解：5000齬0.1 =9.解：5000s.0.1514.解：永续年金每年支付 R17.解：15OOam0.008 =100000 解得m止95.6 即正常还款次数为 95次1500a埶.008 + f (1 +0.008)5

4、=100000 解得 f =965.7419.解:+2000S i(2)LS肚-s5ec1.054 +1.1025R1.053 +1.1025R1.052 +1.205R1.05 =11000 得 R = 2212.1471 -(1 +i f25.解：an =ann(1+i f i-1 + (1 + icici_ Hi _ 弔0.102 an0.1甞6宀=0.002其中n通过公式（2-76）得到29.解：a7 =1-v7/. v7 =1-i a7 =1 -iK类似地，V11:v7v11= v18/. (1-iK)(1-iL) =1-iM从而 i = L + K-MKL

5、31.解:= 1-vnC_ 1 - V12丿anV18 =1 -ia诃=1 -iMi32.解:5001也轴半(1 + i半)=100001 +i =1 -d1=(1 +i半 2 = 1+i半=(1 d P，1 -L3_ = 2036.解 ： g卄屮-20.95.36mm37.解: 0123该永续年金现值为1 111该永续年金现值为:4+(1+i)所求年金现值为:1+(2 + i)i (2+i)in t39.解：ghf =lim a-i =limnC nin_ , n .1 -v 1(2+i)ig =(1-kn)- vn040.解：a(t)=e0dr=1 +tJ0a(t)dt = J。1+tdj

6、(1+n)42.解：后五年等比1000(漏) + 1050咒n+k Y Ji丿122330411 1+-a6 +-i 6 i45.解：251.02 2 4146.解：aa月a300i 月47.解：a(t) =eZv(0)=We48.解：-1 -vt 卄6，43.解:468计川十吟钏10v4i -vd- 5 + Ka00.0151+ a300b10陆月新03十100000陆月新1.03=1JO1421dr(t yy84.5n1 -vanrnn1 v1 f 1-v0 dt = 0 dt =- (n 第三章收益率1 _ 1=(n an =X 4 ：= 405丿盯 可丿10%49.解:虻tvt2 34

7、000 咒 1.1 -2000x0.9 =33823.解:J000v 1000v2 +4000v +5500v3= v(0)7.解:8.解:解得:k =0.14117i09时，v=而；y。本金 时，v=1.1令 v(0) =0= V 及 iX 1.5 16.5 =10x(1+i)8= i =11.995%1dtfdtdt1OOOO(fe01*4k +15000e.251E)k-2OOO0e0.751 卡=1100010.解:利息i 2i3i4i5i 5i 5i 5i 5i 5i012345678910s60.0414.解:B =20000x(1+12%)+5000X 11+ jI F丿f. 3

8、-8000m2%196101000 i 1.045 +5s引0 04 +50.0450.0413. 解：A =500000, B =680000, I =60000 5000 =55000 i*；I8%15.解:16.解:I 5%dte0e t +=2= k =1 书后答案是k =1，不知我对它对。2810654317.解:解得：i=6.52%10000X1.04-1.05k1.04 -1.05k1.04 -k1.04 -k1.04 = 1000020.解:kS-100.0422.解:24.解:1200 1000k (s300.042200 . 21200 +1000 二1 中)解得：i =9

9、.54%30-s叭.04 )+2ks.04 =10000沢 1.04 = k=m10+10888 =10000x(1 +i)= i =4.4%24000+1500080=50033 50000 选择第一种付款方式251000 M+丄 1 M +k f =1990.76I 2丿I 4丿4:810001+ k=2203.764丿25.解:1273000x6%+C(1.061t1 )=18000fC =81000; 彳I 273000x1.06 +C1.061 丄=372000L7.92181P18个月2 326. 解：(1) 1.0522(1 +f ) =1.06213= f=6.71%3 23(

10、2) 1.06213 (1 + f ) =1.06663= f =7.339%第四章债务偿还3.解:设共需还款n次1.解:10000x1.125 -2000s5 =4917.72aT00.08x(1.O810 -1 ) = 468.05+ 10-x1500a口 =1200最初贷款额=1500弔4.解:X =10000i +PjP si00.08 100001.5X =10000X2i + P5.解：过去法:B000a5 +3000 臨a5 )+2000 (a15 ai0 )(1 +i ) 4000S5 (1 + i )2 -3000s2未来法:3000a3 +2000a5v31 17.解：1+

11、i月=(1 +i y2 =1.548 100000 a騙月20!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12 12 1223115(1 +0.03 ) (1 + 0.035 ) -1000甩0.031.035 -1000衍。隔9.解：V20X=1-V20S6 12 610.解：1000(1 +i )+1366.87(1 +i )一 =1000=(1+冷一=川1366.87(1 +i 厂=11111.解：V20 斗 +v20T +V2m Jv20 斗丰=V5 +v4 +v3 +v2 +V1 ：=Pv2 =100= P ;P7i12.解：3 +v20+ =a羽(1 +i)7

12、 ak( k 为剩余还款数)解得：k=i2原利息：20 a刃；现利息:7 + x+k-a2013-V13” .节省禾利息 13Xk=13V 12=113.解:IP (1 v35 )=135 P =144= 丄P (1v352十)=108v =0.25祈第 29 次 P (1-V35卜 7214.解：L每次还款额为 鱼a301BoN每次还款本金为30，第t次还款额为帚心釣+30弋1.0021 -15.解：P11.05。=1250001.0512 -1 -0.00216.解T叮17.解：3000（1 V心甲）= 2000 =n3000v518.解：1+8%=（1+i 季）500（40 -a40季）

13、19.解：每月还款额P = 100000a360i月Pa闻月（P +325.40）a砸（1+i月尸20 =100000= k =120利息支出为：P X120 + （P + 325.40） X120 -100000 = 66261.220.解:B4 =”003 诃 +10怖TO05】0.051.054 -200a4 +10 a4 41.050.05X1.054第5次还款中的利息为iB421.解:Pa+P22.解:陆+（异10（1可一血+（一1）ILI La5V23.解:（1）2000 400弋41.13 =1287.760.06(2)第二年末贷款余额为:i12000400-=1564第三次还款

14、中本金为1564-1287.76=276.2424.此题较麻烦25.解:105106Btdt = L In 1.05a25d2.865827. 解:Bt=1-0.lt(1)B5=1 -0.1x 5 = 0.5前5年还款本金为：1 - B5 = 0.555(2) L 6Btdt = L0.1(1-0.1t)dt =0.37528.解：100005% -10000s20.04O.O4 = 355.69%50.04第9年偿债基金增长额为I0000 +l0000s80.04O.O4 = 328.6150.04%50.0429.解：10000-10000S40.031.O3+100005%si0l0.0

15、330.解:(1) Li -10000 X10% = 1000(2)1500-1000=500 Li -5000j =10005000x0.08 = 600(4) 5000(1 +j)+5005000 =900(5) 5000(1 +j)+500 =590032.解：XS10|007 - 1000033.解：DSi00.04 Exa1010.0 - 2rx*100034.解:12000 - (280 S眄0.04 1.04 10 + 400 s何)35.解:400000+400000 =36000S310.0336.解:-_20Xa200.03 +50 g；031-03第五章.0320 =10

16、0000J债券与其他证券1解：(1)N =100, n =10, C =105,=0.05, i ：= 0.04, h = 0.3,上2 = 0.2Nr6100x0.051- _21105K =Cvn =105x1.040二 p =K + g(1-t1)(C -K) =95.487 i(1-t2)K +g(1t1)(C-K)(2) P =i=94.5311-t2K/C2.解：Pa =100 1+(0.04-0.05 100.05P 7.72%100P3 =100+(0.040.05 隔0.05P 12.46%1003.解：C =1000, i =0.02, P = 1136.78 g 0.02

17、5g2 =0.0125P =C 1 +(g1 -i )ani ，解得 a, =27.356P2 =C0 +(g2-i)3帀=794.834.解：N =1000, r =0.045, C =1125, Cvn =225,i =0.05P =C1 +(g-i总卜112512251+曾-7= 9455.解：K =Cvn2=450 =0.45C 10000.5566P =Nra帀 +K”1110 =1001+450 削。1 -0.45 +450，解得 i ii100=- =12000.556.解:。叫“伽卜卄0.1卩一100=总=vn =6-0.005 P1-v2np+g。00% 和 i000;01

18、益= 10001 +0.2(1-(6-0.005P )2 )解得：p =11007.解：C =1, gi =1.5, g2 =0.75iP =C(gi -i)% = i% =2PP2 =Cp +(g2-i)anJ = 1-0.5 P8. 解：本金调整=票息利息5= 票息0.75票息票息=209. 解：总票息收入=总票息-总本金调整=ncg -c(g -用1 _1仃応亠0=仞1000905-1000(0.05-0.06是= 573.6012.解：P =c1+(gi)ani-24I1 -(1 +i )即：110 =100 1+ (0.05 -i)Ii所求为2i13.解：P (1+i)24 =Nrs

19、勿j+C1 035241EW24,00.0500解得：i021%年收益率为8.042%14. 解：P G = 2(C2 P ) = -2( P2 -C2 )二 G (1 -i )a401 = -2C2 - i )a40i500(0.09 i 戸羽= 2x1000(0.03 i 戸码15. 解：C =100, n= 15, P= 92, i=g+0.01 92 =100 卩+(-0.01 )a15i16.解：(1) i g , n应最大1 _1 0530 fP = 1000( +(0.04-0.05)；.0517.解：不管是否可赎回，债券价格P是相同的N =1000, 季=0.02, i季=0.

20、0151 1015401P +(g-i)a轲=10001+(0.02-0.015) 一。帖 jJ20P = Nr季a刃季+ x(1 +i季) 解得x =108618.解：N =1000, i =0.025, g =0.02, ig, n 应大n = 4,5,6, C =1050,取 n = 6P =1050+ (20-1050天 0.025)3臥.025 =985.89n =7, 8, 9, C =1025,取 n =9R =1025 +(20-1025x0.025)a诃0.025 =944.26n =10 , C =1000F3 =1000 + (20-1000 x0.025)a顽.025

21、=922.0519.解：票息率为每年计息两次的年名义利率6%P= 1000 1 +(0.03 -0.035)a币0.035P =1000x0.03a珂0.035 +(1000 + X )1.035-解得 X =58.6520.解：00003词0 04 + (-1000)陥.04 -10“04冋0.040.04江 O.06 + 1000a10 0.04 = P21.解:P =10000咒6%a50.1 +扁0003羽0.1 -500 吩0.1 一鳥1x6%x1.宀 500a四胪1.1“= 7057.2522.解：10000 F 0.2iae+10000 -0.2ia冋+10000 呂-0.2 i

22、a羽+20000 卩0.2ia冋+ 20000 1 -0.2ia刑+30000 P -0.2ia4010 川182812、V (1 -v )V (1 V )40= 800000 +2000-_+4000-_+ 6000v1 -V1 -V46402810= 800000 +2000。一vv 厂。一v)1-v6=800000 + 20003a46 a40 窃 知a623.解:12 11 10987654321567891011121314151678000% 4%a5 +12000咒 1.05 + (78000 12000)% 4%咒 1.05+(78000 12000 11000) X 4% 咒 1.05 + 11000X 1.05+(78000 -12000 -11000 -10000) X 4% X 1.05 +10000