Renter - page 25
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It is necessary to specify the period t as well.
Alexey defined it above: t=50
It must be said that this expression for
gives a good approximation up to t>30:
Who has a better approximation?
Recall that the red line in the figure shows the original function whose maximum is to be found. In blue is its derivative (zero of the derivative coincides with the first-order maximum). Black is the approximation for the derivative represented as a quadratic polynomial and its zero, which gives an expression for kOpt in analytic form (approximately).
I don't think you're interpreting the picture quite right...
.
the upper horizontal line (red), corresponds to the maximum calculated according to my method.
the bottom horizontal line (blue), corresponds to the maximum calculated by your method.
Oleg, I understood your algorithm. Judging by it, x in the Σ function is the fraction of a month's total accrual, which is withdrawn by the trader. Proceeding from the sense of the problem, it is exactly the alpha=k/q.
How you managed to put my k (withdrawable percentage) in there - I don't understand. That's a completely different value - economically different.
According to the sense of the problem, k should be divided by 0.3 and the result should be substituted into your function of x:
k/q= 0.0280638338/0.3 = 0.093546.
So substitute this, 0.093546, into your function(q=0.3, t=50). What's the output? I get 17256.1236, which is more than you... Your algorithm is a bit inaccurate.
Oleg, where to look in your figure. Where is the analytical expression for the optimum percentage of removal?
Let's be clear: you need an "analytical" expression, even at the expense of accuracy?
For t=30, q=0.15 the share of removal is ~0.338,
value k=0.061, appearing in your calculations, can not be called optimal
Oleg, I understood your algorithm. Judging by it, x in it is a fraction of the accrued for the month, which is withdrawn by the trader. Based on the meaning of the problem, this is exactly alpha=k/q.
How you've managed to put my k in there (the percentage to be taken off) - I don't understand. It is an entirely different value - economically different.
In order to solve the problem, divide k by 0.3 and apply the result to the function on x:
k/q= 0.0280638338/0.3 = 0.093546.
So substitute this, 0.093546, into your function(q=0.3, t=50). What's the output? I get 17256.1236, which is more than you...
The problem is that k is a fraction of q... that's the way I see it... maybe I'm wrong...
but why k/q -- I don't get it!
Once again, I suggest that you define the values!
Oleg, you're confusing k and your x again.
k is the percentage of removal, and the removal fraction is k/q=0.061/0.15=0.4067. You have to admit, as a first approximation, it's not bad at all...
Once again, Oleg:
k is the percentage of withdrawals in values relative to 1, i.e. say if it is 6.1%, then 0.061.
k/q= x in your problem is the fraction of the rent charged for the month.
Вот это, 0.093546, и подставь в свою функцию (q=0.3, t=50). Сколько выйдет? У меня выходит 17256.1236, т.е. поболее твоего...
Oleg, you are again confusing k and your x.
k is the percentage of removal, and the percentage of removal would be k/q=0.061/0.15=0.4067. Admittedly, as a first approximation it's not bad at all...
Let's be clear: you need an "analytical" expression, even at the expense of accuracy?
Well, there is no problem with the numerical solution, but getting an analytical approximation is a yes!