Renter - page 4

 
Reshetov:

Tried again to get a solution and it's the only one:

balance(t) = balance(0) * (1 + q - x)^t

where:

0 < x < q

There are no extrema here.


Yura, we are not solving the problem for balance! We're solving it for the sum of all the money withdrawn in period t.

Do you understand the difference, or are you just going to wave your gun around?

For our case (and in your notation), karman(t)=x*balance(0)*(1-(1 + q - x)^t)/(x-q).

Try to understand what's being discussed first, and then give your opinion.

Lord_Shadows:


See answer for Reshetov.
 
Neutron:

See answer for Reshetov.

Seen it.

 
Neutron:
Lord_Shadows:


See answer for Reshetov.
Sothe problem is to maximize the amount of money withdrawn over a period of t months - that's your problem condition... If it is, then making up something else is silly. The answer with formulas is in my post. Reshetov is also right... Or ask the question correctly.
 
Lord_Shadows:
So the problem is to maximise the amount of money withdrawn over a period of t months - that's your problem condition... If it is, it's silly to come up with something else. The answer with formulas is in my post. Reshetov is also right... Or ask the question correctly.

Then it is a paradox!

Reshetov and I cannot be right at the same time.
 

We look at the condition and see only what is written. But if there were another condition, such as the amount Y I need to keep my trousers on in each month of period t. Then yes, we would have to look for an optimum of funds withdrawn ( k*100/X ) and left ( (q-k)*100/X ). But this condition may break the problem, because nobody knows all conditions. Initial deposit, interest, and most importantly how much we need for these very trousers... Otherwise, under certain conditions Y > k > q and consequently the problem has no solutions. In the same case, if you need maximum money, the formula is simple. There is nothing else to reason about.

P.S. Under condition of minimal withdrawal each month of sum Y. the problem is solved simply Max = X0*(1+(q-min_k)*t/100)^t, where min_k = Y*100/X0.

P.P.S. Everything else is bogus.

 
Lord_Shadows:

2. With compound interest (initial deposit (X0) + interest (q) = (X) ) the maximum will be reached at the end of period t. Max = X0*(1+(q-k)*t/100), I think it is easy to see that at k=0 the maximal value is reached.

Once again.

At k=0 you will have zero in your pocket, not maximum! Is it clear?

We maximize the amount of money withdrawn and do not consider (do not touch) the value of the deposit. That is how the condition was set.

 

From an 'economic' point of view, the depreciation of money over time should also be introduced...

;)

 
Neutron:

Once again.

At k=0 you will have zero in your pocket, not the maximum! Is that clear?

We maximize the amount of money withdrawn and do not consider (do not touch) the value of the deposit. That is how the condition was set.



Sergei, don't get too hot... Read my post, I corrected it and just do the math on your fingers, no need for high-handed rhetoric, I'm not your enemy.
 
Sorento:

From an 'economic' point of view, the depreciation of money over time should also be introduced...

;)


We can't swallow an idealised condition here yet. Let alone find a solution to the problem. And you, Sorento, about inflation...


Sergei, take it easy... Read my post, I corrected it and just do the math on your fingers, don't make grandiloquent pronouncements, I'm not your enemy.

Sorry Lord_Shadows, I'm getting a kick out of Jurin's way of talking. I'll have a look.

 
Neutron:

We can't swallow an idealised condition here yet. Let alone find a solution to the problem. And you, Sorento, about inflation...

Discounting is the basis of financial mathematics...

;)