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Consider minimizing $f(x)$ over a specified domain with inequalities $g_i(x)\le 0$. Assume a primal minimizer exists and strong duality holds with an optimal nonnegative multiplier ${\lambda }^{\ast }$. Let $v(x)=\max (0,g_1(x),\dots ,g_m(x))$. For every $M>\Vert {\lambda }^{\ast }{\Vert }_1$, the exact maximum-violation penaltyhas the same minimum, and every penalized minimizer is feasible and primal optimal.

Indeed the optimal Lagrange dual function bound gives $f(x)+{\sum }_i{\lambda }_i^{\ast }{g}_i(x)\ge p^{\ast }$. Since ${\sum }_i{\lambda }_i^{\ast }{g}_i(x)\le \Vert {\lambda }^{\ast }{\Vert }_{1}v(x)$,A primal optimizer has $v=0$ and reaches $p^{\ast }$. Any infeasible point has strictly larger penalized value, proving the claim. The strict multiplier margin and zero-gap hypotheses explain why existence of an arbitrary optimal dual vector alone is insufficient.

Let ${\lambda }^{\ast }$ be an optimal dual multiplier and define the maximum violationDual stationarity and strong duality imply, for every $x$,Since every component of $Ax-b$ is at most $v(x)$ and ${\lambda }^{\ast }\ge 0$,It follows that the exact maximum-violation penalty obeysChoose any $M>\Vert {\lambda }^{\ast }{\Vert }_1$. A primal optimum has $v(x)=0$ and penalized value $p^{\ast }$, whereas every infeasible point has $v(x)>0$ and penalized value strictly greater than $p^{\ast }$. Thus the penalized problem and the original linear program have exactly the same minimizers.