How do you get the primal solution of an LP from the dual solution?

Question

I am new to Optimization so I think the following question may be very easy, but I'm not sure how to solve it.

The dual of an LP is an LP. If we solve the dual LP, we can get the optimal value for the primal problem. But:

1. How do we get the optimal decision variables for the primal?

2. Does it make a difference if we only relax some of the constraints?

My Work for Finding Dual of LP

\begin{align*} &\text{min } &c^Tx \\ &\text{s.t. } &Ax = b \\ &&Bx \le d \\ \end{align*}

We form the Lagrangian ($\mu \ge 0$): \begin{align*} L(x, \lambda, \mu) &= c^Tx + \lambda^T (Ax - b) + \mu^T (Bx - d) \\ &= (c + A^T \lambda + B^T \mu)^Tx - \lambda^Tb - \mu^Td \end{align*}

Thus we see that the dual function is $$d(\lambda, \mu) = \inf_x L(x, \lambda, \mu) = \begin{cases} - \lambda^Tb - \mu^Td \ \ \ &\text{if } \ \ \ c + A^T \lambda + B^T \mu = 0 \\ - \infty & \text{otherwise} \end{cases}$$ and therefore the dual problem is \begin{align*} &\text{max } &-\lambda^Tb - \mu^Td \\ &\text{s.t. } &c + A^T \lambda + B^T \mu = 0\\ \end{align*}

Answer 1

The duals or shadow prices of the dual model give you a primal solution. See http://yetanothermathprogrammingconsultant.blogspot.com/2022/08/primal-dual-and-equilibrium-format-of.html for a simple example.

Answer 2

I'm also interested in this problem. I think the transformation between the dual solution and the primal solution must be related to KKT conditions, so I give my intuitive solution.