Info

此问题已关闭。 请重新打开它进行编辑或回答。

ı want to solve this question by excel formulas?

1 次查看（过去 30 天）
matlab coder
matlab coder 2022-5-9
关闭： Rena Berman 2022-5-24
此 个问题 已被 Steven Lord 标记
显示标记
You initially put $5,000 into the bank at 4% annual interest rate. Each year you put in an additional $2000. After 20 years of doing this, you start taking out bank disbursements (basically payments *from* the bank) to pay for college. If you will take out 4 equally spaced disbursements to pay for 4 years of college and the final value of the college loan will be 0 after taking those 4 disbursements, what will be the value of each of those 4 disbursements?
  7 个评论
显示 5更早的评论
Walter Roberson
Walter Roberson 2022-5-10
User explicitly wants excel... "But actually ı dont want its formula for matlab. I couldnt find its result and ı want to see if anyone could help for excel solution. Because there is no platform that ı know to discuss about excel problems, ı am new on excel and ı want to improve myself about it."
Torsten
Torsten 2022-5-11
Sorry - I only read the title " I want to solve this question by Excel formulas ".

此问题已关闭。

回答（1 个）

Torsten
Torsten 2022-5-10
编辑：Torsten 2022-5-10
Without guarantee:
disbursement = (5000*1.04^23 + 2000*1.04^4*(1.04^19-1)/0.04)/((1.04^4-1)/0.04)
thus approximately 18150 $.
  3 个评论
显示 1更早的评论
Torsten
Torsten 2022-5-10
编辑：Torsten 2022-5-10
Start year 1: 5000
After 1 year: 5000*1.04 + 2000
After 2 years: 5000*1.04^2 + 2000*1.04 + 2000
...
After 20 years: 5000*1.04^20 + 2000*(1.04+1.04^2+...+1.04^19)
(Don't know if after 20 years, 2000 $ are still put in - I assumed no).
Beginning of 21st year:
5000*1.04^20 + 2000*(1.04+1.04^2+...+1.04^19) - dbm
Beginning of 22nd year:
5000*1.04^21 + 2000*(1.04^2+1.04^2+...+1.04^20) - dbm*1.04 - dbm
Beginning of 23rd year:
5000*1.04^22 + 2000*(1.04^3+1.04^3+...+1.04^21) - dbm*1.04^2 - dbm*1.04 - dbm
Beginning of 24th year:
5000*1.04^23 + 2000*(1.04^4+1.04^4+...+1.04^22) - dbm*1.04^3 - dbm*1.04^2 - dbm*1.04 - dbm =
5000*1.04^23 + 2000*1.04^4*(1.04^19-1)/0.04 - dbm*((1.04^4-1)/0.04) == 0
Solve for dbm.
Sam Chak
Sam Chak 2022-5-10
编辑：Sam Chak 2022-5-10
I think @Torsten is correct. (+1)
The compound interest after 20 years is $65,876.70. Then
Year 21: $65876.70*1.04 - x
Year 22: ($65876.70*1.04 - x)*1.04 - x
Year 23: (($65876.70*1.04 - x)*1.04 - x)*1.04 - x
Year 24: ((($65876.70*1.04 - x)*1.04 - x)*1.04 - x)*1.04 - x ... the 4th disbursement empties the account
format longg
f = @(x) (((65876.70*1.04 - x)*1.04 - x)*1.04 - x)*1.04 - x;
dbm = fsolve(f, 20000)
dbm = 18148.37506946083

此问题已关闭。

标签

产品


版本

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by