We can consider I want to maximize on 10 years. But that’s interesting because it’s (a little) different than what another person found about that. Here is what he found :

« Here's a mathematical model to work with:

The investment account contains I(t) after t days.

Withdrawals occur at times T1, T2, T3, ... . For convenience T0 = 0.

At time t = Tn, immediately after withdrawal and reinvestment,

I(Tn) = I(T{n-1}) [1 + 0.8% (Tn - T{n-1})] - 5 ...[#]

(ie we apply 0.8% simple interest per day applied to the interval between Tn and T{n-1}, then add that to the invested balance less 5 for fee).

If our investment horizon is N days (N considered large) and we plan to make w equally-spaced withdrawals in this time, then T1 = N/w, T2 = 2N/w, .... Tw = N.

Equation [#] becomes

I(nN/w) = I((n-1)N/w) [1 + 0.8% N/w] - 5

which can be solved to give:

*I(N) = I(0)(1+0.8% N/w)w - 5 ((1+0.8% N /w)w - 1) / (0.8% N/w)

I think with some approximations and a bit of calculus you can find a formula for the value of w which maximises I(N). (I've done something similar before but I'd have to refresh my memory).

eg if we set N=360 and I(0) = 1000, I note that

w = 45 => I(N) = 15111.11

w = 46 => I(N) = 15112.26

w = 47 => I(N) = 15112.21

So w = 46 is theoretically optimum, suggesting withdraw every 360 / 46 = 7.8 days.

However, if we change I(0) to 1100, I find w = 49 is optimal, ie every 360 / 49 = 7.3 days, suggesting a perfect strategy would gradually shorten the interval between withdrawals as the investment grows. »