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.
Best Answer
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.