Ah, Jerry, you may be awarded an honorary degree in math with this work!!
Ferritus emeritus?  My own working theory for ferret math was simply that
the number of ferrets would grow to fill all available cage space, much in
the same fashion as computer programs and data expand to fill all available
disk space.  However your formula takes accounts for some significant
environment variables.  I commend you also for assigning set values that
assure there will not be a divide by zero error in the formula.
 
Jerry wrote:
 
>C = Cost factor (1=low, 2=mid, 3=high)
>F = Financial factor (1=bad, 2=ok, 3=great)
>v = available ferrets in the area (0=no, 1= yes)
>a = Number of family members who approve (not including yourself)
>d = Number of family members who disapprove
>p = Persuasion factor (How easy you can persuade someone) (1=high, 2=low)
>q = Odd number quotient (Do the number of f you own 1=even, 2=odd)
>y = Number ferrets you own
>l = How many ferrets will fit in the current living quarters?
 
>n =e(F/C)(a-(d/p))(q+(l-y)
>Ferret new ownership percentage (n*10)
 
Now it happens that in the 13 months I have owned ferrets (I have two), I
have acquired no less than three extra cages (all small).  So I currently
have two ferrets and five cages.  (Mine live in two which are connected with
a short length of drainage pipe; one serves as a litterbox area.)  I could
easily fit an extra ferret or two into the cage, and have ready access to
pet store ferrets, and occaisionaly there are free ferrets offered in this
area.  (Even yesterday's FML I think had something for the Bronx, which
isn't too far.)
 
Yet, oddly, I've managed to maintain some resistance, at least so far, from
acquiring more ferrets.
 
I maintain my resistance levels through regular exposure to the kits at
alocal pet store (reminding myself as I play with them that this way I can
*always* play with kits, if I actually *got* one, it would outgrow the kit
stage quickly), reminding myself that with extra ferrets comes extra vet
bills.  But I have yearned for a more quantitative explanation, so I plugged
some values into the formula you provided.
 
(Somebody check my math here!)
 
Ah ha!  While working with the formula I discovered that the middle section
is critical.  If you can maintain that section to work out to zero, you
won't get hit by ferret math, not matter what the other numbers are!!
 
However, I am puzzled by three things.  The first is an apparent omisisonof
the v ( available ferrets in the area) from the formula itself.  The second
is a lack of definition as to what e is.  (In the absence of a definition,
I assigned a value of 1 to e.) The last is a missing left bracket from the
end of the equation, which I think probably should read (q+(l-y)).  which
would be equivilent to (q+l-y)
 
I feel that v could be worked either into the middle portion of the formula,
or in the end.  Alternatively, it could be left as a stand alone factor, as
in n=v*(f/c)(a-(d/p))(q+l-y)
 
The problem with leaving it as a stand alone factor, is that it does not
account for the possibility that in an area devoid of ferrets, one could
have them brought in from another area.  I propose the following for that
section of the equation:
 
>C = Cost factor (1=low, 2=mid, 3=high)
>F = Financial factor (1=bad, 2=ok, 3=great)
>v = available ferrets in the area (0=no, 1= yes)
 
R=available ferrets outside the area that could be shipped to your location
 (0=no, 1= yes)
S=cost of shipping  (1=low, 2=mid, 3=high)
X=general availability of ferrets (0=no, 1+= yes)
P=Practicality of getting a ferret shipped to you from another area
(R*(S-F+1), (If result is negative, round up to zero)
 
X=V +P
 
The modified equation would read:
 
n =X * e(F/C)(a-(d/p))(q+(l-y)
 
Note that X becomes a critical factor in the equation; if there are no
ferrets to be had, it will be 0, and the calculation always works out that
you can't get another ferret.
 
Let us know about the e value!!
 
-Ilena Ayala
[Posted in FML issue 2282]