Renter - page 26
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avtomat:
a percentage of what?
We have defined k as the percentage of funds withdrawn from the current deposit amount.
We have defined k as the percentage of the current deposit amount that can be withdrawn.
percentage of what?
There was a deposit of 100, q=0.3 part of the deposit was accrued, i.e. +30%. It became 130. Withdrawn k=6.1% of the full amount (by the way, Sergey, let's correct the solution, because we withdraw the full amount, right?). So, it means that we withdrew 0.061*130=7.93. The share to the accrued amount is 7.93/30 = 0.264333.
Yes, the answer formula has to be corrected. And it should be like this:
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. We accrued q, so we have (1+q)D(1+q)(1-k). We withdrew k(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 there are no geometrical progressions.
It's a typical postnumerando annuity, how come we forgot about it, idiots...
The maximum is obtained when there is a minimum of (1-k)^t, i.e. at k=1. Well, since k is bounded from above - and not even by the value of q, but a little bit smaller:
(1+q)(1-k_boundary) >=1, i.e.
1-k_boundary = 1/(1+q)
The removed error is
D(1+q)^t*{1-(1-k_boundary)^t} = D(1+q)^t*{1-(1+q)^(-t)} = D(1+q)^t - D
In other words, only the initial - D will remain on the deposit, since if nothing was withdrawn, there would be D(1+q)^t.
is that the right thing to do?
or is it?
.......... .......... ....
In this case it will be removed
D(1+q)^t*{1-(1-k_boundary)^t} = D (1+q)^t*{1-(1+q)^(-t)} = D(1+q)^t - D
In other words, only the initial - D will remain on the deposit , because if nothing was withdrawn, there would beD(1+q)^t.
I don't see how this is possible at all. If the withdrawal is always less than the accrual (by convention).
Or did I misunderstand something in your text?
All right, less. By how much? That's what the top starter said:
Мне разрешается каждый месяц снимать некоторый процент k со счёта которая не превышает величину q.
We won't even be able to withdraw q, because we'll take more than we've accrued. At most we may withdraw q/(1+q), i.e. still less than q. At the end of each month the deposit will be equal to the initial one: we will withdraw all the profit.
It looks like Yura was right after all. And I should have double-checked the calculations...
Find the error in my reasoning, MD. If you find it, I'll bounce you.
OK, let it be less. By how much? That's what the topic-starter said:
We won't be able to withdraw q either, as we'll withdraw more than what's accrued. At most we may withdraw q/(1+q). In this case, at the end of each month the deposit will be equal to the initial one: we withdraw all the profit.
It seems that Yura was right after all. And I should have double-checked the calculations...
This is the only way to leave an even D at the end.
But it seems that in the process of disassembly it was found out that at high q it is optimal to take off less. // By the way, there should be a threshold, when this extremum appears at smaller removal.