Consider the following dual linear programs: $$ \min \mathbf{c^T x} ~~ \text{s.t.} ~~ A \mathbf{x} \geq \mathbf{b}, \mathbf{x}\geq 0; \\ \max \mathbf{b^T y} ~~ \text{s.t.} ~~ A^T \mathbf{y} \leq \mathbf{c}, \mathbf{y}\geq 0. $$ By the strong duality theorem, if there is an optimal solution $\mathbf{y^*}$ to the dual problem, then the primal problem has an optimal solution $\mathbf{x^*}$ with $\mathbf{c^T x^*} = \mathbf{b^T y^*}$.
Suppose now that the dual problem is very large, and I can solve it only approximately. Specifically, I can find a vector $\mathbf{y'}$ such that:
- $A^T \mathbf{y'} \leq (1+\epsilon)\cdot \mathbf{c}$ - the constraints are only approximately satisfied;
- $\mathbf{b^T y'} \geq \mathbf{b^T y}$ for any vector $\mathbf{y}$ that satisfies the original constraints $\mathbf{y} \leq \mathbf{c}$.
So $\mathbf{y'}$ is optimal for the original constraints, but it may violate these constraints a bit.
QUESTION: Can I draw from this, any conclusion regarding the optimal (or approximately-optimal) solution of the primal LP?
NOTE: It is easy to prove a variant of the weak duality theorem for this setting. If $\mathbf{x}$ is any feasible solution to the primal, then:
$$ \mathbf{c^T x} = \mathbf{x^T c} \geq \\ \geq \mathbf{x^T}\cdot A^T\mathbf{y'}/(1+\epsilon) \\ = (A\mathbf{x})^T\cdot \mathbf{y'}/(1+\epsilon) \\ \geq \mathbf{b^T y'}/(1+\epsilon). $$
So a candidate for a variant of the strong duality theorem is: the primal LP has a solution $\mathbf{x^*}$ for which: $$ \mathbf{b^T y'} \cdot (1+\epsilon) \geq \mathbf{c^T x^*} \geq b^T \mathbf{y'}/(1+\epsilon) $$ Is this true?
If so, is there an algorithm for finding such $\mathbf{x^*}$, given the vector $\mathbf{y'}$?