Short and slick proof of the strong duality theorem for linear programming

Question

Consider the linear programs

\begin{array}{|ccc|} \hline Primal: & A\vec{x} \leq \vec{b} \hspace{.5cm} & \max \vec{c}^T\vec{x} \\ \hline \end{array} \begin{array}{|ccc|} \hline Dual: & \vec{c} \leq \vec{y}^TA \hspace{.5cm} & \min \vec{y}^T\vec{b} \\ \hline \end{array}

The weak duality theorem states that if $\vec{x}$ and $\vec{y}$ satisfy the constraints then $\vec{c}^T\vec{x} \leq \vec{y}^T\vec{b}$. It has a short and slick proof using linear algebra: $\vec{c}^T\vec{x} \leq \vec{y}^T A \vec{x} \leq \vec{y}^T\vec{b}$.

The strong duality theorem states that if the $\vec{x}$ is an optimal solution for the primal then there is $\vec{y}$ which is a solution for the dual and $\vec{c}^T\vec{x} = \vec{y}^T\vec{b}$.

Is there a similarly short and slick proof for the strong duality theorem?

Answer 1

Probably not. Here is a conceptual argument based on

Farkas Lemma: Exactly one of the following alternatives has a solution:

1. $Ax \le b$ and $x \ge 0$
2. $y^TA\ge 0$ and $y^Tb < 0$

Now let $\delta$ be the optimal objective value of the primal. Let $\epsilon > 0$ be arbitrary. Let $A'$ to be $A$ with an additional $-c^T$ as the last row. Let $b'$ to be $b$ with an additional $-\delta - \epsilon$ as the last value.

The system $A'x'\le b'$ has no solution. By Farkas, there is a $y' = (y,\alpha)$ such that:

$y^TA\ge \alpha c$ and $y^Tb < \alpha (\delta + \epsilon)$.

Note that if $\epsilon = 0$ we are in the other alternative of Farkas. Therefore $\alpha > 0$.

Scale $y'$ so that $\alpha = 1$. $y$ is dual feasible. The weak duality implies $\delta \le y^Tb < \delta + \epsilon$.