How to solve the optimization using Matlab's optimization toolbox

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Sobin cc
Sobin cc 2016-1-18
评论： John D'Errico 2016-1-18
Can somebody help to solve following optimization using Matlab's optimization toolbox
Max r == [ 1 + [(1-pab)/p(ab) ]* pint ] ^ 1/k
subject to 0 < pinit <= 1 ;k>0, and 0 < pab <= 1
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Torsten
Torsten 2016-1-18
A minimum on [0:1]x[0:1] does not exist.
Hint:
See what happens for p arbitrary, q=eps a very small positive number.
Best wishes
Torsten.
John D'Errico
John D'Errico 2016-1-18
Read my answer. You CANNOT find a maximum of this function. It is unbounded.
The minimum is TRIVIAL to find. Again, use common sense!
Simply set pab == 1. r will take on the value of 1. It cannot be any lower. The other parameters are meaningless. Pick any random numbers.
USE COMMON SENSE and high school mathematics.

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John D'Errico
John D'Errico 2016-1-18
编辑：John D'Errico 2016-1-18
IF I assume that p(ab) and pab are the same, and that pinit and pint are the same thing, then...
No, fmincon will be of no real value. Common sense is sufficient.
With pint and pab both positive and no larger than 1, the expression
[(1-pab)/p(ab) ]* pint
is ALWAYS a non-negative number, only ever exactly zero based on your constraints when pab is exactly 1.
Therefore, the expression
[ 1 + [(1-pab)/p(ab) ]* pint ]
is ALWAYS at least 1, and generally larger than 1. Since you then raise it to a power (1/k) we therefore need to maximize 1/k. So make k as small as possible. Zero would be nice, but difficult to achieve, since you said k>0. So make k approach zero in a limit, and you can make r as large as you wish. r is unbounded. Choose any values for pab and pint, as long as pab is not exactly 1. As long as we can make k arbitrarily small, we can make r approach infinity as k approaches zero in the limit.
As I said, fmincon is totally unnecessary, and in fact ANY optimizer would be a complete waste of time. Just use common sense and basic mathematics.

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