供45年
每年回報率 6%
淨本金 54萬
最後可拿幾多錢??
show step
更新1:
唔岩播,都唔係275萬(answer) http://www.mpfa.org.hk/tc_chi/mpf_edu/mpf_edu_yc/mpf_edu_yc_em/mpf_edu_yc_em_2.html
[40分]利息計算問題
回答 (2)
2011-06-13 1:57 am
✔ 最佳答案
sorry , E 度只能post 尾10期(字數所限), 請check e mailEvent Deposit Interest Balance
Deposit 530 $1,000.00 $12,990.93 $2,612,177.42
Deposit 531 $1,000.00 $13,060.89 $2,626,238.31
Deposit 532 $1,000.00 $13,131.19 $2,640,369.50
Deposit 533 $1,000.00 $13,201.85 $2,654,571.35
Deposit 534 $1,000.00 $13,272.86 $2,668,844.21
Deposit 535 $1,000.00 $13,344.22 $2,683,188.43
Deposit 536 $1,000.00 $13,415.94 $2,697,604.37
Deposit 537 $1,000.00 $13,488.02 $2,712,092.39
Deposit 538 $1,000.00 $13,560.46 $2,726,652.85
Deposit 539 $1,000.00 $13,633.26 $2,741,286.11
Deposit 540 $1,000.00 $13,706.43 $2,755,992.54
Grand Total $540,000.00 $2,215,992.54 $2,755,992.54
參考: excel
2011-05-31 1:50 am
8,746,997.04, simplified to annual contribution of $12000.
We'll write c for the annual contribution; and to keep things cleaner we'll write "z" for (1 + r). Now we start writing down the account balance for the first few years: YearBalanceNowP1(P + c)z2((P + c)z + c)zIn other words, to go from one year's balance to the next, you add on the contribution c, then multiply by z to get the interest. If you multiply these terms out, you'll start to see the pattern emerge: YearBalanceNowP1Pz + cz2Pz2 + c(z + z2)YPzY + c(z + z2 + . . . + zY)The second part of that last line is just c times the sum of a geometric series. So you can boil the whole thing down to: 1.Balance(n) = PzY + c[(zY + 1 - z)/(z - 1)] Finally, write z out in terms of r, to get the formula we're looking for: 2.Balance(Y) = P(1 + r)Y + c[ ((1 + r)Y + 1 - (1 + r)) / r ] Timing IssuesThe formula above assumes contributions occur at the start of each year. If you want them to happen at the end instead, then you get one less interest period per contribution: YearBalanceNowP1Pz + c2Pz2 + c(1 + z)YPzY + c(1 + z + z2 + . . . + zY-1)and the formula becomes: 2a.Balance(Y) = P(1 + r)Y + c[ ((1 + r)Y - 1) / r ]
2011-05-30 17:52:23 補充：
And also assuemed that interest is calculated at every year end only.
2011-06-11 16:14:43 補充：
樓主你講嗰個淨本金嘅意思原來係咁...一般財務學上"本金"係指投資期開始時投入嘅首筆款項。強迫金網嗰個所謂"淨本金"我哋通常叫"總供款"。
以上答案係指喺開始嘅時候你有$54萬本金，跟住月供$1000。
如果本金係$0，套用公式，得出期終值$2,706,097.50。
We'll write c for the annual contribution; and to keep things cleaner we'll write "z" for (1 + r). Now we start writing down the account balance for the first few years: YearBalanceNowP1(P + c)z2((P + c)z + c)zIn other words, to go from one year's balance to the next, you add on the contribution c, then multiply by z to get the interest. If you multiply these terms out, you'll start to see the pattern emerge: YearBalanceNowP1Pz + cz2Pz2 + c(z + z2)YPzY + c(z + z2 + . . . + zY)The second part of that last line is just c times the sum of a geometric series. So you can boil the whole thing down to: 1.Balance(n) = PzY + c[(zY + 1 - z)/(z - 1)] Finally, write z out in terms of r, to get the formula we're looking for: 2.Balance(Y) = P(1 + r)Y + c[ ((1 + r)Y + 1 - (1 + r)) / r ] Timing IssuesThe formula above assumes contributions occur at the start of each year. If you want them to happen at the end instead, then you get one less interest period per contribution: YearBalanceNowP1Pz + c2Pz2 + c(1 + z)YPzY + c(1 + z + z2 + . . . + zY-1)and the formula becomes: 2a.Balance(Y) = P(1 + r)Y + c[ ((1 + r)Y - 1) / r ]
2011-05-30 17:52:23 補充：
And also assuemed that interest is calculated at every year end only.
2011-06-11 16:14:43 補充：
樓主你講嗰個淨本金嘅意思原來係咁...一般財務學上"本金"係指投資期開始時投入嘅首筆款項。強迫金網嗰個所謂"淨本金"我哋通常叫"總供款"。
以上答案係指喺開始嘅時候你有$54萬本金，跟住月供$1000。
如果本金係$0，套用公式，得出期終值$2,706,097.50。
收錄日期: 2021-04-23 23:29:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110530000051KK00584