Consider the following quadratic program

$$\begin{align*}&\text{min } \frac{1}{2}b^TDb+d^Tb\\ &\text{s.t. } Ab\le b_0\end{align*}$$ where $D$ is a positive semidefinite matrix.

The Lagrangian dual function is written as follows

$$\begin{align*}L(b,\lambda)&=\frac{1}{2}b^TDb+d^Tb+\lambda^T(Ab-b_0)\\ &=\frac{1}{2}b^TDb+(d^T+\lambda^TA)b-\lambda^Tb_0 \end{align*}$$

Then I got stuck on finding the dual of the problem.

Similar background problem:


Informally, the approach is to swap the $\inf$, $\sup$ to convert from the primal to the dual.

With the Lagrangian above, the primal is $P: \ \inf_b \sup_{\lambda \ge 0} L(b,\lambda)$, hence we go ahead and swap the $\inf$, $\sup$ to get the dual, that is $D:\ \sup_{\lambda \ge 0} \inf_b L(b,\lambda)$.

Note that $\inf_b L(b,\lambda)$ is finite iff the convex function $b \mapsto L(b,\lambda)$ has a (global) minimiser iff there is some $b$ such that $Db + d +A^T \lambda = 0$, hence we can write $D:\ \sup_{\lambda \ge 0} \inf_{\{b | Db + d +A^T \lambda = 0\}} | L(b,\lambda)$.

The expression $L(b,\lambda)$ can be simplified using the pseudo inverse of $D$ and the fact that if $Db + d +A^T \lambda = 0$, then $b = - D^\dagger (A^T \lambda + d)$. This results in semi definite quadratic program.

  • $\begingroup$ how do you then from the dual function formulate the dual problem? I'm working on this problem here: math.stackexchange.com/questions/2550644/… which is looking like it might be the exact same question the OP asked here. $\endgroup$ – ALannister Dec 4 '17 at 15:43
  • $\begingroup$ actually, I just realized that the question I linked to is not regarding the same exact problem. My problem is a little different, and so if you could take a look and offer some help, I would be most appreciative! You always give such instructive answers. $\endgroup$ – ALannister Dec 4 '17 at 15:54

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