Lagrangian relaxation

El directorio enciclopédico desde la Wikipedia.

Lagrangian relaxation is a relaxation technique which works by moving hard constraints into the objective so as to exact a penalty on the objective if they are not satisfied.

[edit] Mathematical description

Given an LP problem $x\in \mathbb{R}^n$ and $A\in \mathbb{R}^{m,n}$ on the following form:

max		$c T x$
s.t.
		$Ax \le b$

If we split the constraints in $A$ such that $A_1\in \mathbb{R}^{m_1,n}$ , $A_2\in \mathbb{R}^{m_2,n}$ and $m 1 + m 2 = m$ we may write the system:

max		$c T x$
s.t.
(1)		$A_1 x \le b_1$
(2)		$A_2 x \le b_2$

We may introduce the constraint (2) into the objective:

max		$c T x + λ T (b 2 - A 2 x)$
s.t.
(1)		$A_1 x \le b_1$

If we let $\lambda=(\lambda_1,\ldots,\lambda_{m_2})$ be nonnegative weights, we get penalized if we violate the constraint (2), and we are also rewarded if we satisfy the constraint strictly. The above system is called the Lagrangian Relaxation of our original problem.

Of particular use is the property that for any fixed set of $\tilde{\lambda}$ values, the optimal result to the Lagrangian Relaxation problem will be no smaller than the optimal result to the original problem. To see this, let $\hat{x}$ be the optimal solution to the original problem, and let $\bar{x}$ be the optimal solution to the Lagrangian Relaxation. We can then see that

$c^T \hat{x} \leq c^T \hat{x} +\tilde{\lambda}^T(b_2-A_2 \hat{x} ) \leq c^T \bar{x} +\tilde{\lambda}^T(b_2-A_2 \bar{x} )$

The first inequality is true because $\hat{x}$ is feasible in the original problem and the second inequality is true because $\bar{x}$ is the optimal solution to the Lagrangian Relaxation.

This in turn allows us to take address the original problem by instead solving the partially dualized problem

min

P (λ)

s.t.

$\lambda \geq 0$

where we define $P (λ)$ as

max		$c T x + λ T (b 2 - A 2 x)$
s.t.
(1)		$A_1 x \le b_1$

A Lagrangian Relaxation algorithm thus proceeds to explore the range of feasible $λ$ values while seeking to minimize the result returned by the inner $P$ problem. Each value returned by $P$ is a candidate upper bound to the problem, the smallest of which is kept as the best upper bound. If we additionally employ a heuristic, probably seeded by the $\bar{x}$ values returned by $P$ , to find feasible solutions to the original problem, then we can iterate until the best upper bound and the cost of the best feasible solution converge to a desired tolerance.

Página espejo de la Wikipedia

Directorio de Enlaces Directorio dmoz Directorio espejo dmoz Pedro Bernardo