Let $\left\{\begin{matrix} \operatorname{min}_xc^Tx\\Ax\leq b \\ x\in \mathbb{R}^+ \end{matrix}\right.$ be a LP primal problem and $\left\{\begin{matrix} \operatorname{max}_yb^Ty\\A^Ty\geq c \\ y\in \mathbb{R}^+ \end{matrix}\right.$ its dual. Weak duality theorem says that the value of objective function of primal is bounded above by the value of objective function of dual. In fact:

$$c^Tx\leq (A^Ty)^Tx=y^T(Ax)=y^Tb=b^Ty$$

where $A^Ty\geq c \Rightarrow (A^Ty)^T\geq c^T \Rightarrow c^Tx \leq (A^Ty)^Tx$. But applying the dual function to calculate the dual problem of a Conic primal I obtain the contrary sign. Let $K^*$ be the dual cone of non-negative orthant $K\equiv \mathbb{R}_+^n=\begin{Bmatrix} x\in \mathbb{R}^n:x\geq 0 \end{Bmatrix}$. Thus

$$L:\mathbb{R}^n\times\mathbb{R}^m\times K^* \mapsto \mathbb{R} \space \operatorname{s.t} \space L(x,y,s)=c^Tx+\overset{=y(Ax-b) \operatorname{for} Ax-b=0\Leftrightarrow h(x)=0}{\overbrace{y^T(b-Ax)}}-\overset{=s(x) \operatorname{for} x\geq0\Leftrightarrow f(x)\leq0}{\overbrace{s^Tx}}$$

$$\Rightarrow g(y,s)\doteq \operatorname{min}_xL(x,y,s)=\operatorname{min}_xx^T(c-A^Ty-s)+b^Ty=\left\{\begin{matrix} b^Ty \operatorname{if} c-A^Ty-s=0\\ -\infty \operatorname{otherwise} \end{matrix}\right.$$

and for $L(x,y,s)\leq f_0(x)\Rightarrow g(y,s)=b^Ty\leq c^Tx$ with $x\in \begin{Bmatrix} x\in \mathbb{R}^n:Ax=b \end{Bmatrix}\cap K$.

What am I doing wrong? Thanks in advance.


1 Answer 1


Not sure about your conic problem, but for LP you have the roles of $\min$ and $\max$ reversed.

  • 1
    $\begingroup$ This seems like a comment, not an answer. $\endgroup$
    – LSpice
    Commented Sep 3, 2022 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.