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Given a matrix $A$, each element $A_{i,j} \geq 0$, find the vector $\vec x$ that maximizes the minimum element in $\vec b$ ($\vec b = A \vec x$). Note that this is not a linear equation system as I don't know $\vec b$.

Extra contraints on the solution are $x_i \geq 0$, and $\sum x_i = 1$.

Is this possible to solve, and if so, how? Can it have 0 or more than one solution?

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    $\begingroup$ For A a matrix, what does A >= 0 mean? Do you mean A is non-negative definite? $\endgroup$
    – user5810
    Commented Aug 9, 2010 at 19:37
  • $\begingroup$ I meant simply that all elements in A are >= 0, updated to clarify, thank you. $\endgroup$
    – SoftMemes
    Commented Aug 9, 2010 at 19:47
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    $\begingroup$ I think it certainly has a solution by compactness (in other words, the set of $x_i \geq 0$ with $\sum x_i = 1$ is a compact set). You have a finite set of linear functions, the min of these functions is still continuous so there should be a minimum value. This looks vaguely like something someone in OR should immediately be able to answer. $\endgroup$
    – Karl Schwede
    Commented Aug 9, 2010 at 20:08

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This is a linear program. Put down the contraints $r \leq b_i \ \forall i$, together with all the linear constraints you have above, and maximize $r$. All the constraints are linear, so linear programming will do this. There are tons of linear programming packages (Mathematica and MATLAB have decent ones), and there should be some good introductory material on linear programming on the web that is better than anything I would write here, so I'll let you look for that.

Linear programs can in general have a lots of solutions (although they're all on one face of a polytope), or they may have no solutions. In this case, it certainly has at least one solution (see the compactness argument in the comments), but it mgiht have many.

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  • $\begingroup$ This is exactly what I needed, thank you! $\endgroup$
    – SoftMemes
    Commented Aug 9, 2010 at 21:21

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