Consider the optimization problem \begin{align}\label{opt-lp}\tag{Primal} \begin{array}{cl} \underset{x \in \mathbb{R}^n}{\text{minimize}} & c^\top x \\ \text{subject to} & Ax = a \\ & Bx = b \end{array} \end{align} where $A \in \mathbb{R}^{m \times n}, a \in \mathbb{R}^m, B \in \mathbb{R}^{q \times n}, b \in \mathbb{R}^q$ are the problem data, and the problem has a nonempty feasible set.

I would like to introduce a partial Lagrange relaxation to only $Ax = a$ constrains, so that the partial Lagrange function is $$L(x, \lambda) =c^\top x + \lambda^\top ( Ax - a).$$ As this is a "partial" Lagrange relaxation, I define the Lagrange dual function as $$ g(\lambda) = \underset{x : Bx = b}\inf L(x, \lambda)$$ that is, I add the constraint of $Bx = b$ already. It is clear that $g(\lambda)$ lower bounds \eqref{opt-lp}.

I think the Lagrange dual problem becomes: \begin{align}\label{dual}\tag{Dual} \sup_\lambda \inf_{x : Bx =b} c^\top x + \lambda^\top (Ax - a). \end{align} My question is whether what I am doing here has a name and whether I can simply say strong duality holds between problems \eqref{opt-lp} and \eqref{dual} in a paper without going much in detail.


2 Answers 2


Based on the mentioned references, suppose the primal problem is: \begin{align} \begin{array}{cl} \underset{}{\text{minimize}} & c x \\ \text{subject to} & Ax = a \\ & Dx \leq e \\ & x \geq \text{0} \end{array} \end{align}

The idea behind Lagrangian relaxation is to relax the complicating constraints to produce an easier problem by adding this constraint into the objective function with a penalty so-called Lagrange multipliers. The Lagrangian subproblem is:

\begin{align} \begin{array}{cl} \underset{}{\text{minimize}} & cx \ + \ \lambda(b \ - \ Ax)\\ \text{subject to} & Dx \leq e \\ & x \geq \text{0} \end{array} \end{align}

Solving the Lagrangian subproblem can produce a lower bound on the optimal objective value of the primal problem. Let $Z^*$ be the optimal objective value of primal and let $Z_{LR}(\lambda)$ be the optimal objective value of The Lagrangian subproblem. Then we have the following result:

For any $\lambda \in \mathbb{R}^m$ ($m$ be the number of rows in $A$) $$Z_{LR}(\lambda) \leq Z^*$$

Since primal is a minimization problem, we want lower bounds that are as large as possible; these are the most accurate and useful bounds. Different values of $\lambda$ will give different values of $Z_{LR}(\lambda)$, and hence different bounds. We’d like to find $\lambda$ that gives the largest possible bounds. That is, we want to solve:

\begin{align} \begin{array}{cl} \underset{\lambda}{\text{maximize}} & Z_{LR}(\lambda)\\ \end{array} \end{align}

According to the primal-dual relation: $Z_{LP} \leq Z_{LR}$.

$$\begin{aligned} z_{\mathrm{LR}} &\geq \max _{\lambda}\left\{\min _{x} c x+\lambda(b-A x) \mid D x \leq e, x \geq 0\right\} \\ &=\max _{\lambda}\left\{\min _{x}(c-\lambda A) x+\lambda b \mid D x \leq e, x \geq 0\right\} \\ &=\max _{\lambda}\left\{\max _{\mu} \mu e+\lambda b \mid \mu D \leq c-\lambda A, \mu \leq 0\right\} \\ &=\max _{\lambda, \mu}\{\mu e+\lambda b \mid \mu D \leq c-\lambda A, \mu \leq 0\} \\ &=\max _{\lambda, \mu}\{\mu e+\lambda b \mid \mu D+\lambda A \leq c, \mu \leq 0\} \\ &=\min _{y}\{c y \mid A y=b, D y \leq e, y \geq 0\} \\ &=z_{\mathrm{L} P} \end{aligned}$$

where the last one is LP dual of the entire problem.



This is called Lagrangian relaxation, no matter what subset of constraints you choose to dualize.

  • $\begingroup$ Thank you for your answer. Is strong duality as clear as the typical case where we dualize all constraints? $\endgroup$ Commented Dec 4, 2021 at 23:56
  • 1
    $\begingroup$ @independentvariable I would say yes, it is as clear. You just squuece what you call partial LR between the original LP and the full LR. $\endgroup$
    – Sune
    Commented Dec 5, 2021 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.