Hey all,

I’m trying to figure out how to run a compounded yearly interest rate on a principal (P) where each year, a dollar amount (X) is removed from P.

For instance year 1
P is 100,000 at 5%. End of year we have 105,000.
At end of year, 10,000 is removed.
Next year is run with 95k. We end with 99750
At end of year 10k is removed.
New starting principal is 89750.

And so on for 10 years.

I’m looking for the total account value after 10 years of this process.

https://en.wikipedia.org/wiki/Annuity

Your problem is this;

A(n)=A(n-1)*f-b (E1) where f=1+r and r is the interest rate per year compounded only once a year and b is the amount deduced.

This is not a geometric sequence but it can be converted into one and solved that way.

Define A(n)=B(n)+K (E2)

Substitute E2 to E1 B(n)+K=(B(n-1)+K)*f-b or B(n)=f*B(n-1)+f*K-K-b or

B(n)=f*B(n-1)+(f*K-K-b)

Now choose K such that the last term above is zero to convert the problem to a geometric one for which you have the solution ie B(n)=B(1)*f^(n-1)

this means f*K-K-b=0 or K= b/(f-1)=b/r (E3)

So A(n)=B(n)+K= B(1)*f^(n-1)+b/r so

A(n)=(A(1)-b/r)*(1+r)^(n-1)+b/r (E4)


where A(1) is initial position and b is what you take out and r the interest rate.

Eg A(n)=(100000-10000/0.05)*(1.05)^(n-1)+10000/0.05

A(10)=44867.2

Keep in mind this means 10 years later in the sense that at the beginning of year 1 you have 100000 and in the beginning of year 10 you have 44867.2. That of course is only 9 years later. If you mean 10 years entirely later you need (ie after you have deducted 10k 10 times) A(11)=37110.5. On the 15th year after 14 10k removals you will run out of money.

You can force A(N)=0 in E4 and find what r or what b is needed for this to happen after N-1 withdrawals.


Eg imagine someone wants to live at a respectable 50k cost of living per year for at least 40 years after getting retired at 60 and they use a historical stock market index or some other combination of real estate investments plus stock market or other more stable objects and had a 7% rate then they need

0=(A(1)-50000/0.07)*(1.07)^(41-1)+50000/0.07 or A(1)=666700


So start investing when you first get a job 1k per month and use for b =-1000 now in above equation and 40 years later at 60 you will have at the same stock market equivalent management system; n=40*12 compounded monthly at r=0.07/12


A(40*12+1)=(1000+1000/(0.07/12))*(1+0.07/12)^(40*12+1-1)-1000/(0.07/12)=2.64 mil more than enough to be super well prepared.

Now how many actually do that? 1k per month is easily doable given most decent jobs and wise trading investing can yield over 10-15% per year even. You will have deposited only 480k of that 2.6 mil. And you will never run out of money.

To be fair even at some stupid sub history 5% returns with 2k per month from your first good job you will be very comfortable eventually in a low inflation world. Go get a drip investing account or a no commissions account and start buying 1-10 shares per week (eg of the 20 most productive or promising decent companies you can find and keep researching for new ones). Easiest job in the world to always invest in the biggest names of every decade the moment a first hint of success triggers your attention. 5% crypto every week, 30% index or a real estate equivalent fund, 30% big names, 35% new well researched names.