I'm going through an article lately and there is one point which is very confusing. So, we have the following original constrained binary quadratic problem as the following. The pre-assumption of certain parameters are $Q\in Z^{n*n}$, $A\in Z^{n*n}$, and $b\in Z^{n*1}$.

$\text{min}$ $x^{T}Qx$,

$s.t.,$ $Ax\leq b$ $\text{and}$ $x\in \{0,1\}^{n}$. Let's call this optimization problem the problem $P$.

Now, by the classical lagrangian reduction, this can be relaxed as

$\text{min}$ $x^{T}Qx+\lambda^{T}(Ax-b)$,

$s.t.,$ $x\in \{0,1\}^{n}$ and $\lambda >0$. Let's call this optimization problem the problem $L_{\lambda}$.

Since the relaxation serves as a lower bound, we ideally plan to make it as maximum as possible. So, let $d(\lambda) = x^{T}Qx + \lambda^{T}(Ax-b)$, we have the other optimization,

$\text{max}$ $d(\lambda)$

$s.t.,$ $\lambda \geq 0$. Let's call this optimization problem the problem $L$.

So far so good. Now, here is the confusing point.

The article mentioned that the problem $L$ can be rewritten as the following.

$max$ $\mu$

$s.t.,$ $\mu \leq x^{T}Qx + \lambda^{T}(Ax-b) $ for any $x\in \{0,1\}^{n}$ and $\lambda \geq 0$.

Let's call the last optimization problem the problem $W$.

I do not quite understand why this is a rewritten. So, suppose $\mu_{0}$ is the optimal solution of problem $W$, then it follows that $\mu_{0}\leq x^{T}Qx+\lambda^{T}(Ax-b)$ for any $x\in \{0,1\}^{n}$ and $\lambda \geq 0$. While on the other hand, for the problem $L$, what I understand it is like this: for each given $\lambda$, there is one optimization problem $L_{\lambda}$, and one optimal solution (which we take a minimum by the definition of $L_{\lambda}$). Then, finally, among all the optimum solutions, one maximum value is chosen. Somehow, I can not see that in the formulation of the optimization problem $W$.

Or did I overthink? Any comments would be greatly appreciated.


1 Answer 1


You have not defined $d(\lambda)$ (and, therefore, problem $L$) correctly. It should be $$ d(\lambda) = \min_{x \in \{0,1\}^n} \left(x^TQx+\lambda^T(Ax-b)\right). $$ Now note that any problem of the form $$ \begin{align} \text{maximize }&\min_zf(z)\\ \text{subject to }& \mathcal{C}, \end{align} $$ where $z$ is the (vector) optimization variable and $\mathcal{C}$ is a set of constraints, can be equivalently written as $$ \begin{align} \text{maximize }&t\\ \text{subject to }& t\leq f(z) \text{ for all } z, \\ & \mathcal{C}. \end{align} $$ where the optimaziation variables are the vector $z$ and the scalar $t$.

  • $\begingroup$ Could you be more specific why I have not defined $d(\lambda)$ correctly? $\endgroup$ Feb 14, 2017 at 2:07
  • $\begingroup$ You left the minimization with respect to $x$ out of your definition of $d(\lambda)$. Computing $d(\lambda)$ requires minimizing the expression with respect to $x$. $\endgroup$ Feb 14, 2017 at 2:31
  • $\begingroup$ I see. So, when an optimization problem is considered, the minimization or maximization is always with respect to a certain variable which needs to be emphasized, right? In this case, it need to mention that $x$ is the variable, not $\lambda$. I think I simply mis-treat $\lambda$ as a variable also, which is a mistake. $\endgroup$ Feb 14, 2017 at 2:36

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