# Formulating the dual of an exponential cone optimization problem

I have an optimization problem that is written \begin{align} \min_{\{s_\kappa\} }& \frac{1}{2}\sum_{\kappa} \eta_\kappa \left( \exp(r_\kappa s_\kappa) + \exp\left( \frac{s^2_\kappa}{2} \right) \right) \\ \text{s.t.} & \sum_{\kappa} \exp\left( \frac{s_\kappa}{2} \right) \leq C \end{align} where $$r_\kappa < 0$$, $$C> 0$$ and $$\{ s_\kappa\}$$ is a finite set of decision variables.

I believe then that the Lagrangian is given by \begin{align} \mathcal{L}(s_\kappa,\lambda) = \frac{1}{2}\sum_{\kappa} \eta_\kappa \left( \exp(r_\kappa s_\kappa) + \exp\left( \frac{s^2_\kappa}{2} \right) \right) + \lambda\left( \sum_{\kappa} exp\left( \frac{s_\kappa}{2} \right) - C\right) \end{align} and the KKT conditions then give $$\lambda \geq 0$$ (dual feasibility), $$0 = \lambda \left( \sum_{\kappa} exp\left( \frac{s_\kappa}{2} \right) - C \right)$$ (complimentary slackness), and \begin{align} 0 &= \frac{1}{2} \eta_\kappa \left( r_\kappa \exp(r_\kappa s_\kappa) + s_\kappa \exp\left(\frac{s^2_\kappa}{2}\right) \right) + \frac{\lambda \exp(\frac{s_\kappa}{2})}{2},\ \forall \kappa \end{align}

I would like to write the dual problem for this, which should be $$g(\lambda) = \inf_{\{s_\kappa\} } \mathcal{L}(s_\kappa,\lambda)$$. I would like to this because I have from other sources some properties on $$\lambda$$ which I would like to translate into properties on the optimal objective.

My problem is that though I can write $$\lambda$$ in terms of $$s_\kappa$$ for a given $$s_\kappa$$ \begin{align} \lambda = - \eta_\kappa \left( r_\kappa \exp(r_\kappa s_\kappa) + s_\kappa \exp\left( \frac{s^2_\kappa}{2} \right) \right) \end{align} I am struggling to write $$s_\kappa$$ in terms of $$\lambda$$ in order to substitute back into the Lagrangian to get the dual objective function. Any suggestions?

• Given the presence of the term $s_{\kappa}\exp\left(\frac{{s_{\kappa}}^2}{2}\right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{\kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual. – nathan.j.mcdougall Jan 15 at 0:49
• An alternative to that term is to introduce a constraint of the form $$s^2_\kappa \leq \alpha, \ \forall \kappa$$ and then the $\eta_\kappa$ term becomes $\eta_\kappa r_\kappa \exp(r_\kappa s_\kappa)$. The problem here is then that there would be a new lagrange multiplier $\mu_\kappa$ associated with each of these constraints. Would something like this be more ameanable? – NeedsToKnowMoreMaths Jan 15 at 1:53
• I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$\frac{1}{2}\eta_{\kappa}\left[r_{\kappa}\exp(r_{\kappa}s_{\kappa})+s_{\kappa}\exp\left(\frac{{s_{\kappa}}^2}{2}\right)\right]+\frac{1}{2}\lambda\exp\left(\frac{s_{\kappa}}{2}\right)+2s_{\kappa}\mu_{\kappa}=0$$ for $s_{\kappa}$ in terms of both $\lambda$ and $\mu_{\kappa}$, which seems more difficult rather than easier. I don't see how the $\eta_{\kappa}$ term changes. – nathan.j.mcdougall Jan 15 at 2:11
• Sorry, I meant that I could change the objective to $$\min \sum_\kappa \eta_\kappa \exp(r_\kappa s_\kappa)$$ by adding in the additional constraints. The $s^2_\kappa$ term came from trying to regularize to avoid the need for the constraints. – NeedsToKnowMoreMaths Jan 15 at 15:05

\begin{align*} \min_{\{s_{\kappa}\}}\quad&\sum_{\kappa}\eta_{\kappa}\exp(r_{\kappa}s_{\kappa})\\ \text{s.t.}\quad&\sum_{\kappa}\exp\left(\frac{1}{2}s_{\kappa}\right)\leq C\\ &s_{\kappa}^2\leq \alpha. \end{align*} We can make the transformation $$x_{\kappa}=\exp(\frac{1}{2}s_{\kappa})$$ to render this more linear. Especially, we can change the $$s_{\kappa}^2\leq \alpha$$ constriants into pairs of linear constraints $$x_{\kappa}\leq \exp(\sqrt{\alpha})$$, and $$x_{\kappa}\geq \exp(-\sqrt{\alpha})$$. This gives a new problem \begin{align*} \min_{\{x_{\kappa}\}}\quad&\sum_{\kappa}\eta_{\kappa}(x_{\kappa})^{2r_{\kappa}}\\ \text{s.t.}\quad&\sum_{\kappa}x_{\kappa}\leq C\\ &x_{\kappa}\leq \exp(\sqrt{\alpha})\\ &x_{\kappa}\geq \exp(-\sqrt{\alpha})\\ \end{align*}
The KKT stationary-point constraint for the transformed problem is given by $$2\eta_{\kappa}r_{\kappa}(x_{\kappa})^{2r_{\kappa}-1}+\lambda+\mu_{\kappa}-\nu_{\kappa}=0.$$
Then, it is a simple rearrangement for $$x_{\kappa}$$ to give $$x_{\kappa}^*=\left[\frac{\lambda+\mu_{\kappa}-\nu_{\kappa}}{2\eta_{\kappa}(-r_{\kappa})}\right]^{\frac{1}{2r_{\kappa}-1}}.$$
The Lagrangian for the transformed problem is then $$\mathcal{L}(x_{\kappa},\lambda,\mu,\nu)=\sum_{\kappa}\eta_{\kappa}(x_{\kappa})^{2r_{\kappa}}+\lambda\left[\sum_{\kappa}x_{\kappa}-C\right]+\sum_{\kappa}\mu_{k}[x_{\kappa}-\exp(\sqrt{\alpha})]-\sum_{\kappa}\nu_{k}[x_{\kappa}-\exp(-\sqrt{\alpha})].$$
I will leave the cumbersome exercise of substituing $$x_{\kappa}^*$$ into the Lagrangian, but hopefully this approach will help you progress.