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Subject: Re: optimization
Date: Mon, 5 Jul 2010 00:33:05 +0000 (UTC)
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"Hobst Friedli" <tobias.friedli@access.uzh.ch> wrote in message <i0q8lh$5v8$1@fred.mathworks.com>...
> > "Hobst Friedli" <tobias.friedli@access.uzh.ch> wrote in message <i0nnmc$cp2$1@fred.mathworks.com>...
> > > .......
> > > (ni. - nii)/(1-pi) +(n.i - nii)/(pi) +lamda = 0 für alle i = 1,...,20
> > > 
> > > and Sum (pi) =1
> > > 
> > > n.i = sum row
> > > ni. = sum column
> > > nii = diagonal entries
> > > Pii = cell-percentage of row-sum
> > > 
> > > I now want to solve this for the Pi's, probably with fmincon. But i just dont get how to specifie the commands.
> > > 
> > > For every kind of help i would be very grateful... I need to do this estimation for my bachelor thesis and i cant find help anywhere... Thank you very much in advance
> > > 
> > > Here i add the part of the paper i want to replicate :
> > > 
> > > http://img340.imageshack.us/img340/9861/paperu.png
> .......
> clear all
> A = [.....]
> B = [.....]
> C = [.....]
> x0 = [0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1]
> 
> for ind =1:15
>      nii=A(ind)
>      n_i=B(ind)
>      ni_=C(ind)
>     fun = @(x) [(nii-ni_)/(1-x) + (nii-n_i)/x - L, sum(x)-1]
> 
> x(ind,: ) = fsolve(fun,x0)
>  
> I dont know if any of this code is useable. And there are obviously some parts missing.
> 
> Now my problem is still, how can i  specifie the L to be a constant for every i ? and it seems that the fsolve does only solve for one variable. But i have 2 Variables x and L. 
> 
> Any tipps would be appreciated. Thank you
- - - - - - - - -
  There are at least three things wrong with your anonymous function fun.  First the expression in brackets must be a vector expression, not something that is done-one-at-a-time in a for-loop.  Second, the division operators need a dot so as to avoid matrix division.  Finally, you need to include the unknown L (which I presume is your -lambda) as one of the elements of vector x to be evaluated, say the last one.  That means if k is the k of the paper you referenced, then x should have a length of k+1.

  Let A and B be row vectors of length k and defined for each i as

 A(i) = ni. - nii
 B(i) = n.i - nii

in your earlier notation.  Then do:

 fun = @(x) [A./(1-x(1:k))+B./x(1:k)-x(k+1),sum(x(1:k))-1]

 x = fsolve(fun,x0);

Also be sure to include an initial estimate for L in the k+1 element of x0.


  However, I would predict that fsolve might have difficulty successfully solving the equations in this particular problem.  My reasoning goes something like this.  For each particular value of L, the i-th equation

 A(i)/(1-x(i))+B(i)/x(i) = L

is equivalent to a quadratic equation in x(i) (where I am assuming A(i) and B(i) are always positive quantities since they derive from "n-counts".)  There are two real roots, one root, or none, to this quadratic depending on whether

 sqrt(A(i))+sqrt(B(i))

is less than, equal to, or greater than L, respectively.  If L is greater than the maximum of this sum of two square roots for all i, then all k equations would have two roots, so there would be 2^k = 32,768 possible cases which fsolve would have to "investigate" for such L values to make sure it didn't overlook a possible solution to sum(x(1:k))-1=0.  It looks like a very ugly problem when looked at from that point of view, and I can conceive of fsolve getting hopelessly stuck in these numerous blind alleys.

  I can envision a process by which a single solution can certainly be found, but I see no guarantee that it would be the only possible one.  It can easily be seen that as L approaches infinity, one of the roots of each of the above quadratics must approach zero while the other root must approach one.  Start with a very high value of L and choose in each case the smaller of the two possible roots for x(i).  If L is sufficiently large the sum of these x(i) would be less than 1.  Now imagine L decreasing continuously unless reaching the greatest

 sqrt(A(i))+sqrt(B(i))

at an i=i0, which would be the lowest value that L can be allowed to have.  The corresponding sum(x(1:k)) would continuously increase as the roots increase.  If it passes through one somewhere along the line, there is your solution.  If not, then let L commence increasing again from this minimum value but choosing the greater of the two possible roots for the i0 case while the other roots remain at their lesser values.  This would be a continuous process for sum(x(1:k)) with no discontinuous jumps, since you would have passed through the one-root point at i0 when making the switch in roots.  Since x(i0) would now approach one as L again approaches infinity, there surely must come a point where sum(x(1:k)) crosses one, and there would be your solution.

  In other words, there must always exist a solution in which either all the x(i) roots are chosen as the lesser of a pair, or else there is a solution in which just one of the x(i) is the greater of a pair and all the other roots are still at their lesser values.

  However, making the transition for any other of the root pairs from the lower to the upper one would in general involve a discontinuous jump in sum(x(1:k)), so we are not sure any other solutions can be found in the L-varying process.  If they do exist I foresee fsolve having a difficult time finding them.

  As for the single "sure-fire" solution above, it should be possible to translate that process to something fzero, rather than fsolve, could solve in terms of the single unknown L.  It can be determined in advance whether sum(x(1:k)) is above or below one for L chosen at sqrt(A(i0))+sqrt(B(i0)), and this would allow you to determine which of the above two alternatives to select for the x(i0) root in starting fzero with a pair of surrounding estimates for L.

  My apologies for not going into greater detail in the previous paragraph with respect to fzero.  I could not be sure you would be interested in such an approach.

Roger Stafford