$$\begin{array}{ll} \text{minimize} & \frac{1}{2} x^\top P x + q^\top x\\ \text{subject to} & A x = b\\ & x \in \mathbb{R}^n {\geq 0}\end{array}$$

where $P \succ 0$. Let the minimizer be $x^*$.

Without the non-negativity constraint $x \geq 0$, the optimizer $y^*$ is the solution of the KKT system

$$\left[ \begin{matrix} P & A^\top \\ A & 0\end{matrix} \right] \left( \begin{matrix} x \\ \nu\end{matrix}\right) = \left( \begin{matrix} -q \\ b \end{matrix}\right)$$

Suppose that the inverse of $\left[ \begin{matrix} P & A^\top \\ A & 0\end{matrix} \right]$ exists and is known analytically, so that $y^*$ is known analytically as well.

Now with the non-negativity constraint $x \geq 0$, does $x^*$ have a known analytical expression (as a function of $y^*$)?

Not especially. Even in two dimensions, the nature of the problem is adequately presented. Either $y^* \geq 0$ so $y^*$ is the constrained solution or it is not. If it is, great. If not, on the new boundary imposed by the constraint, the solution can be found by solving a reduced dimensionality problem on each facet. In two dimensions, this means optimizing the quadratic form on the positive $x$-axis and then again on the positive $y$-axis. You may find that the solution is actually at the intersection of several facets, i.e. at the origin in two dimensions.
• No. Any version can be transformed to the unit simplex by rescaling the variables. ($Ax = b$ is already linear...) – Eric Towers Apr 27 '14 at 20:09
• Ok, thanks. But, just out of curiosity, if $A$ has more than $1$ row how is it possible to rescale the variables to get to $1^\top x = 1$? – user693 Apr 27 '14 at 20:11
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