I edited this question based on information I got from comments. Assume we have an optimization problem (primal problem). we solve it's dual using some kind of primal-dual interior point solver. So, we have multipliers of constraints of the dual or any other relevant information. Knowing that dual of the dual is equal or related to primal, how can we this information to find the primal variable as efficient at possible? Is there any method that can use this information to find the primal variable cheaper than directly solving it or no? to be more clear consider the next few lines. Consider the following optimization problem \begin{align} \min_{\alpha} &f(\alpha)\\ \textrm{s.t.} &-p \leq \alpha \leq r\notag\\ \end{align} where $f(\alpha)$ is a convex function. It's lagrangian is \begin{align} &\mathcal{L}=f(\alpha)-\eta^\mathsf{T} (\alpha+p)+\beta^\mathsf{T} (\alpha-r)\notag\\ \end{align} Dual of this problem is a problem in the form \begin{align} \min_{\eta,\beta} &g(\eta,\beta) \\ \textrm{s.t.} & \eta >= 0 \\ & \beta >=0 \end{align} Now assume we have a solver which solves this problem. If we assume that solver gives as output $\eta^*$,$\beta^*$ and also multipliers of the constriants $\eta>=0$ and $\beta>=0$ say $\nu$ and $\mu$ (at the optimality), how can we compute $\alpha^*$ if strong duality holds using those values efficiently?

To make it clearer, consider a simple quadratic (convex) optimization problem like \begin{align} \min_{\alpha} &c \alpha^\mathsf{T} A \alpha-d^\mathsf{T} \alpha\\ \textrm{s.t.} &-p \leq \alpha \leq r\notag\\ \end{align} where $A$ is a positive semi-definite matrix. It's lagrangian is \begin{align} &\mathcal{L}=c \alpha^\mathsf{T} A \alpha-d^\mathsf{T} \alpha -\eta^\mathsf{T} (\alpha+p)+\beta^\mathsf{T} (\alpha-r)\notag\\ \end{align} Now assume we have answers of the dual optimization problem which is an optimization problem with respect to variables $\eta>=0$ and $\beta>=0$,i.e. \begin{align} \min_{\beta,\eta} &\frac{1}{2c} (d+\eta-\beta)^\mathsf{T}A^{\dagger}(d+\eta-\beta)+\eta^\mathsf{T} p +\beta^\mathsf{T} r\notag\\ \textrm{s.t.} & \beta \geq 0,\\&\eta \geq 0\notag\\ \end{align} If we have $\eta$ ,$\beta$ and the lagrange multipliers for constraints $\eta >=0$ and $\beta>=0$, say $\nu$ and $\mu$ respectively, How to compute primal variable $\alpha$ based on $\eta,\beta,\nu,\mu$, when strong duality holds? Note: For some reasons, I don't want to use pseudo inverse of $A$ in this case .

Comment: According to an answer by @Michael, the above dual is not completely correct, please see his answer.

  • $\begingroup$ Just fix the values of $\eta$ and $\beta$ in the Lagrangian and apply unconstrained minimization. (No, you can't do it any more cheaply in the general case.) $\endgroup$ – Michael Grant May 14 '15 at 12:29
  • $\begingroup$ Thank you Michael Grant. Even if we have the multipliers of the constrained in the dual problem?. In the quadratic case, I have to use $A^\dagger (d+\eta-\beta)$, when I test the code the results are not correct. May be because the pseudo inverse has values like 3 e^5 which results in some numeric difficulties (May be the matrix conditional number is so high). But when I solve the primal problem the strong duality holds and $\alpha$ value is correct. Becuase of the I assume that I must find a way to compute primal problem without having to compute pseudo inverse of $A$. $\endgroup$ – user85361 May 14 '15 at 13:29
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    $\begingroup$ Ah, I see what you're saying. There may be a way to recover $\alpha$ from those. I'm not sure. $\endgroup$ – Michael Grant May 15 '15 at 3:04
  • $\begingroup$ Thank you, @MichaelGrant. Can you suggest a book or literature to work on? $\endgroup$ – user85361 May 17 '15 at 1:59

I assume $A$ is symmetric.

Your dual manipulation is incorrect when $A$ does not have full rank. To see why, let's try to compute the dual function (using your notation):

\begin{align} g(\eta, \beta) &= \sup_{\alpha \in \mathbb{R}^k} [-c\alpha^TA\alpha + d^T\alpha+\eta^T(\alpha + p) - \beta^T(\alpha -r)] \\ &= \sup_{\alpha \in \mathbb{R}^k} [ -c\alpha^TA\alpha + (d+\eta-\beta)^T\alpha ] + \eta^Tp + \beta^Tr \end{align}

Suppose $A$ does not have full rank. Take any nonzero vector $z \in Null(A)$. It follows that: $$ -cz^TAz +(d+\eta-\beta)^Tz = (d+\eta-\beta)^Tz$$ Then if $(d+\eta-\beta)$ is not orthogonal to $z$, the vector $z$ can be scaled by either $M$ or $-M$ (for arbitrarily large real numbers $M$) to ensure the above expression $\rightarrow\infty$. Thus: $$ g(\eta, \beta) = \infty \:\: \mbox{if $(d+\eta-\beta) \notin Null(A)^{\perp}$} $$ But $Null(A)^{\perp} = Col(A^T) = Col(A)$ (by symmetry of $A$). Thus, the $g(\eta, \beta)$ function is only finite when $(d+\eta-\beta) \in Col(A)$. Minimizing the $g(\eta, \beta)$ function thus requires the additional constraint $(d+\eta-\beta) \in Col(A)$, which is not considered in your dual problem. Your dual problem only considers the constraints $\eta \geq 0$, $\beta \geq 0$, which is not enough.

The column space constraint can be enforced by adding two constraints to the dual formulation, as follows: \begin{align} &d + \eta - \beta = Aw \\ & w \in \mathbb{R}^k \end{align}

Now suppose you obtain an optimal solution $\eta^*$ and $\beta^*$ for the dual. Then an optimal primal solution $\alpha^*$ must be a solution to the primal maximization $\sup_{\alpha\in\mathbb{R}^k}[-c\alpha^TA\alpha + (d+\eta^*-\beta^*)^T\alpha]$. Thus, it is a solution to: $$ A\alpha = \frac{1}{c}(d+\eta^* - \beta^*)$$ This has a solution because we already know $d+\eta^*-\beta^*\in Col(A)$ (this was enforced by the dual problem). Unfortunately, there are an infinite number of solutions to the above equation. A pseudo-inverse can give you a particular solution. This particular solution is not guaranteed to satisfy the desired constraints of the primal. For any particular solution $\alpha^p$, the set of all solutions is $\alpha^p + Null(A)$. So, armed only with this information, you can only say that an optimal primal $\alpha^*$ satisfies: $$ \alpha^* \in \alpha^p + Null(A) $$

  • $\begingroup$ Thank you very much. Your answer helped me much. Could you please help me to find a cheaper method to use this information for computing primal variable(s)? $\endgroup$ – user85361 May 17 '15 at 11:25
  • $\begingroup$ If you also have the Lagrange multipliers of the dual, that might help, I will think about that later. The "dual of dual" idea is kind of interesting. $\endgroup$ – Michael May 17 '15 at 13:01
  • $\begingroup$ I'm really thankful. I'm waiting to hearing from you soon. $\endgroup$ – user85361 May 17 '15 at 13:22
  • $\begingroup$ Well I haven't thought much about it, but I do not see an obvious way of using that information. $\endgroup$ – Michael May 22 '15 at 14:29

The process of minimizing $\mathcal{L}$ to construct the dual results in a formula for $\alpha$ as a function of $\beta$ and $\eta$. This is exactly the connection between the optimal primal and dual variables.

  • $\begingroup$ Unfortunately, when I do this method, because matrix $A$ is not full rank we have to use pesudo inverse. In using pesudo inverse, some times the result is wrong, may be becuase of rounding error. Also, we need to compute pesudo inverse which task $O(n^2)$ time. So, I need a method to use available data, that is multpliers of $\eta>=0, \beta>=0$, to compute $\alpha$. $\endgroup$ – user85361 May 13 '15 at 17:14
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    $\begingroup$ Unfortunately, when $A$ is not invertible, I believe that there is no method to compute $\alpha$ which is not as hard as solving the primal problem. $\endgroup$ – Alex Shtof May 14 '15 at 11:05
  • $\begingroup$ The question is that assume we solved the dual problem and it' dual (dual of the dual!, using a primal-dual interior point solver), which is obviously related to the primal problem, so we have it's multipliers, how can we find the primal problem's answer? $\endgroup$ – user85361 May 14 '15 at 15:42
  • $\begingroup$ Still, I don't know of any way you can exploit this information. $\endgroup$ – Alex Shtof May 16 '15 at 9:08

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