自考.工程经济学.（F/P,8%,5）=1.469(P/F,8%,5)=0.6806(F/A,8%,5)=5.867(A/F,8%,5)=0.17046（A/P,8%,5）=0.25046(P/F,8%,5)=3.993(F/P,8%,10)=2.159(P/F,8%,10)=0.4632（F/A,8%,10）=14.487(A/F,8%,10)=0.06903(A/P,8%,10)=0.14903(P/A,8%,10)=

Question

自考.工程经济学.
（F/P,8%,5）=1.469 (P/F,8%,5)=0.6806 (F/A,8%,5)=5.867 (A/F,8%,5)=0.17046
（A/P,8%,5）=0.25046 (P/F,8%,5)=3.993 (F/P,8%,10)=2.159 (P/F,8%,10)=0.4632
（F/A,8%,10）=14.487 (A/F,8%,10)=0.06903 (A/P,8%,10)=0.14903 (P/A,8%,10)=6.710
（F/P,8%,1）=1.080 (P/F,8%,1)=0.9259 (F/A,8%,1)=1.000 (A/F,8%,1)=1.000
36．(本题6分)某企业计划10年后进行生产设备的技术改造,需要经费40万元,若年
利率为8％,如果每年存入相同数量的金额,则在每年末存款时,应存入资金多少万
当改为每年初存款时,又应存入资金多少万元?(计算结果保留小数点后两位)

Answer 1

每年末存款A= 40x(A/F,8%,10)=2.76
每年初存款A= 40x(A/F,8%,10)x (P/F,8%,10)=1.28