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November 19

Mathematical modelling problem

I am a landlord. I am interested in buying a house near a large organisation to rent out to that organisations personnel.

The most important number for me to try to estimate, is the likely time that the house I buy will stay empty before it is let to someone. If it is going to be too long a time, then it is not worth investing.

I have some statistics, and I would like to know if I have sufficient information to estimate this number. There are 360 households who rent houses in the surrounding area. Each household stays in the area for four years before moving on. There is a list of 40 vacant houses that new personnel coming in can choose from.

It is easy to see that there will be 360/4 = 90 new households arriving in the area every year and looking for something to rent.

But is it possible to estimate the average time that a vacant house will take to let please? There may be a simple solution that I am not quite seeing, or it may be more complex and require assumptions about the relevant probability distributions.

Would the answer in fact be 40/90 years, or slightly over five months, or have I got this wrong please?

I arrive at the same result, under the assumption that every house is rented out at least every now and then. My reasoning goes as follows: There are 360+40 = 400 houses for 360 households, so the occupancy rate is 90% and the vacancy rate is 10%. Per "turn" between one household moving in and the next, the average occupancy time is 4 years, which accounts for 90%. The vacancy time, which accounts for the remaining 10% = 10/90th of 90%, is therefore 10/90 × 4 years = 5+ months. Of course, there is no guarantee that any particular single house will see a rate anywhere near the average. --Lambiam ^Talk 16:15, November 19, 2006 (UTC)

It seems to me that the expected time of initial vacancy would be half that, since we don't know where in the ideal 4/9 year of vacancy we are when we start (though obviously we know we're in it as no one's living there!). --Tardis 17:39, 20 November 2006 (UTC)[reply]

I don't think this type of estimation will be very accurate, as it doesn't include whether the housing you are considering is preferable to others. Instead, I suggest you demand rental records from the current owners, to know exactly what portion of the time they were able to rent it out. (Don't trust them, demand to see the records.) StuRat 07:36, 20 November 2006 (UTC)[reply]

While I agree this would be useful, one disadvantage would be your assuming the housing environment in that area hasn't changed. In reality, this is unlikely to be the case. The question is, how has it changed and by how much? Nil Einne 17:27, 20 November 2006 (UTC)[reply]

Quartic equation

for a quadratic x^2+ax+b=0 a solution(s) is found using the substitution x=y-a/2

for a cubic x^3+ax^2+bx+c=0 solutions are found using the substitution x=y-a/3 and then y=z-a'/3z (a' is a new coefficient found in the equation in terms of y)

as I'm sure you will be aware.

My question is:

Can a quartic (x^4+ax^3+bx^2+cx+d=0) be solved by substuting x=y-a/4, then y=z-a'/4z followed by a further substitution of q = fn(z) {as yet unknown to me?} It seems likely that a cubic involving (q^4) terms would be formed just as the solution to the cubic involves a quadratic involving (z^3) terms. If so what is the substitution.. If not can anyone explain why this pattern breaks down. Thank you83.100.138.99 18:45, 19 November 2006 (UTC)[reply]

What is a'? If it is the coefficient of z^3, then a' = 0. If it is the coefficient of z^2, what good does this do? Did you study Ferrari's method (see Quartic equation)? --Lambiam ^Talk 19:03, 19 November 2006 (UTC)[reply]

As per Cubic_equation#Cardano.27s_method then a' would be p in the depressed cubic(in the wiki page it is expressed in terms of t not y. In the quartic I assumed it would be α in the depressed quartic (as shown in ferrari's method - the wikipedia page)

Well, then just do your substitution u = z − α/4 on that depressed quartic and you'll see an undepressed quartic, setting you back. --Lambiam ^Talk 20:47, 19 November 2006 (UTC)[reply]

(For Lambian):The second substitution was u=z-(α/4z) not u=z−(α/4) (don't know if that was a typo) Does that make it any clearer as to what I am trying to find out here?

but I was asking about a third substitution q=fn(z) to give a solution - does anyone know what that substitution is (or of a pattern?)

I read too quickly, but still don't see how this brings you any further. Applying the substitution u := z−(α/4z) to u⁴ + αu² + βu + γ results in Q/(4z)⁴, where

Q = 256z⁸ + 256βz⁵ + (−32α²+256γ)z⁴ − 64αβz³ + α⁴.

This is much messier than what we started out with and does not look promising in any way. Did you actually do the substitution yourself? --Lambiam ^Talk 05:08, 21 November 2006 (UTC)[reply]

Yes I've tried this sub. and knew it didn't make the equation look any better. (I haven't checked to see if you've done it right - assumed you have). However reading my first comment you will notice I was looking for a third substittion that would convert this in to a cubic in powers of 4 eg x^12+ax^8+bx^4+c. (which can the be solved by the cubic method) I was hoping that someone would be aware of a method like this. Please don't trouble yourself too much if you don't want to.

Clearly there's a pattern in the first substitution that works for quadratic and cubics, my guess that the second substitution would follow that pattern was just that - a guess. It's possible that if this method works the second substitution for the quartic will be related to the second substitution for the cubic (but possibly not in the simplistic way I've listed here). Getting the third substitution seem fairly impossible to me.. But I hoped such a method might exist - I'm still hoping that.87.102.16.174 17:38, 21 November 2006 (UTC)[reply]