Renter - page 27
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We were relying, imho, on an incorrect formula as we went along. I simply suggested, in my opinion, a more logical method of calculation - not by the initial deposit of the month, but by the final one, after the accrual of q.
It's interesting, it seems that Oleg derived his formulas independently. And he also found a certain optimum. I don't get it...
As we went along, we were relying, imho, on an incorrect formula. I simply suggested, in my opinion, a more logical method of calculation - not by the initial deposit of the month, but by the final one, after the accrual of q.
It's interesting, it seems that Oleg derived his formulas independently. And he also found some kind of optimum. I don't get it...
Examination on my scooter (Excel) revealed a simple fact - the extremum becomes acceptable for taking into account at muchhigher q, at 50% p.a. it is barely pronounced (k~ 45% p.a.).
// I.e. at 50% per annum it is easier to withdraw the same 50% and not bother, if q is even less - definitely withdraw the entire increment.
The graphs at the beginning of the thread show a monthly growth of 50%. //That's when it's YES.
--zy. Ah yes Alexey, you're wrong somewhere, the vapcheta extremum has a place. At high yields you have to keep in mind and count.
--
But don't expect any analytical formula from me. You shouldn't intimidate me with diphurcs and MatCad. :)))
What difference does it make what yields, Volodya. The main formula.
And the total removed would be D(1+q)^t - D((1+q)(1-k))^t = D(1+q)^t*{1-(1-k)^t}.
I deduced without any restrictions at all. The maximum on k in this formula is obvious. And then, given Sergei's constraints, I simply calculated the maximum possible k_max = q/(1+q) < q.
Look for a mistake "somewhere", I don't see it myself yet. The reasoning is elementary, but it is more detailed than Sergei did.
Well, we are not solving diphurcs or integrals here; everything is simpler, at 7th grade school level...
There was a deposit of 100, q=0.3 part of the deposit was accrued, i.e. +30%. It was 130. It was withdrawn k=6.1% of the full amount (by the way, Sergey, let's correct solution, because we withdraw the full amount, right?) So, 0.061*130=7.93. The share to the accrued amount equals 7.93/30 = 0.264333.
Yes, the answer formula has to be corrected. And it should be:
Let the deposit at the beginning of month 1 be D. Accruing interest q gives us deposit D(1+q). Then we withdraw interest k, i.e. kD(1+q). That leaves D(1+q)(1-k).
Second month. Accrued q, left (1+q)D(1+q)(1-k). D(1+q)D(1+q)(1-k), D((1+q)(1-k))^2 is left.
At the end of the t-th month, the account (by induction) will have D((1+q)(1-k))^t.
And the total withdrawal will be D(1+q)^t - D((1+q)(1-k))^t = D(1+q)^t*{1-(1-k)^t}.
That's how it works. And no geometric progressions here.
And where did you get the idea that "And the total removed would be... " ??? Exactly the first term is not clear. // D(1+q)^t is kinda like a deposit grown without withdrawal?
It's not obvious to me in any way. Double-check. You missed something.
// Excel is a bastard, of course, but it stubbornly shows the extremum
Well yes, this is the deposit that would have grown from D if we had not withdrawn anything. But since we did, we have withdrawn exactly the difference between what would have been if we hadn't withdrawn, minus what's really left. Where else is the money going to go?
But there is one serious problem.
Well, the maximum is obtained when the minimum is (1-k)^t, i.e. at k=1.
And this maximum, according to my stupid formula, equals D(1+q)^t. It cannot be so, because we withdraw the whole deposit in 1st month, and it is only D(1+q). There is nothing to grow further.
Oh, one more inconsistency: at boundary k = q/ (1+q) we withdraw not D(1+q)^t - D, as I've calculated here, but only k_boundary*D(1+q)t = Dqt: the deposit will simply increase by q% every month, we withdraw the whole amount and the new month starts again with D
OK, let's calculate the removed directly, by summation. Removed:
kD(1+q)^1 + kD(1+q)^2*(1-k)^1 + kD(1+q)^3*(1-k)^2 + ... + kD(1+q)^t*(1-k)^(t-1) =
= kD(1+q) + kD(1+q)*Sum( i=1..t-1; ((1+q)(1-k))^i ) =
= kD(1+q){1 + r + rr + ... + r^(t-1)}
Here r=(1+q)(1-k)
Now let's be more careful. If k=1, then r=0, and the whole bracket equals 1, since there is only 1 non-zero term. The answer here is D(1+q) - everything converges. Not our case, we want to work longer.
If r=1 (boundary k=q/(1+q)), then the bracket equals t, and the whole removed equals k_boundary*D(1+q)*t = Dqt. Everything converges again.
If r<1 (k is smaller than the boundary), then everything sums normally: we get kD(1+q)*(1-r^t)/(1-r). By the way, this formula can also be used in the previous case, going to the limit at r->1 and calculating it by Lopital's rule. One more thing: this formula works even for the very first case!
Still it is not clear why "sincewe withdraw, we withdraw exactly the difference between what would have been if we had not withdrawn, minus what is actually left. Where else is the money going to go?"Wrong? I think it's time for a material balance equation...
So, withdrawn equals kD(1+q) * (1-(1+q-k-qk)^t) / (qk+k-q)
OK, let's calculate what was taken directly, by summing it up.
Mathemat:
It is still unclear why "since it was withdrawn, they withdrew exactly the difference between what would have been if they had not withdrawn, minus what was actually left. Where else is the money going to go?"Wrong? I think it's time to draw up a material balance equation...
So, the withdrawn equals kD(1+q) * (1-(1+q-k-qk)^t) / (qk+k-q)
Well of course it's wrong. To illustrate.
Suppose we have a 10% increase per month, i.e. q=0.1;
Then in 12 months, the deposit without withdrawal would become D*(1.1)^12 = D*3.13843
If one withdrew per month k=q=0.1, then in total D*0.1*12=D*1.2, while the deposit remained = D, i.e. in total D*1.2+D=D*2.2
I'm sure 3.13843 > 2.2.
Your material balance equation doesn't add up, oh it doesn't add up....
;)
mmm.... I honestly, er... don't understand why such an "analytical" solution is more beautiful than the formula I gave you...
(which, by the way, looks quite analytical)
.
.
for comparison:
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for given values:
.
there's something to reduce-simplify, but multiplying by t...
Last time I made a little mistake with the substitution... now it's right:
Oleg, explain your formulas. Write in human language (in general form, not with substituted numbers) the withdrawal formula you used. If you can't write - then I'm not at all sure that you made the program correctly :)
Just don't do it in ASAP language, please. The simpler the better. Let me remind you my formula (initial deposit equals 1, k is withdrawal percentage, q is accrual percentage, t is time in months):
So, the withdrawal equals k(1+q) * (1-(1+q-k-qk)^t) / (qk+k-q)
MD: Уверен, что 3.13843 > 2.2
Your material balance equation doesn't add up, oh it doesn't add up....
I don't get it either, where did the rest go, MD?