I understand that SVM is about solving the constrained optimization such that

$$\min_{\mathbf{w}} \dfrac{1}{2} \mathbf{w}^T\mathbf{w}$$ subject to $$y_i(\mathbf{w}^T\mathbf{x_i}+b)\geq{1}, i=1, 2, ...,n$$

And this is handled using nonlinear optimization method Karush–Kuhn–Tucker approach where one step is based on the necessary complementary slackness condition such that

$${\alpha}_i\left(y_i(\mathbf{w}^T\mathbf{x_i}+b)-1\right)=0, i=1, 2, ...,n$$ has to be satified.

Because for non-gutter dots (i.e., the points not on the edge of the separating hyperplane), we have

$$y_i(\mathbf{w}^T\mathbf{x_i}+b)-1 > 0$$

the corresponding multiplier $\alpha_i$ then must be $0$. But my question is for gutter points, because

$$y_i(\mathbf{w}^T\mathbf{x_i}+b)-1 = 0$$

we know that the corresponding multiplier $\alpha_i$ should be non-negative, but are they necessarily positive? In other words, if I define support vector as any $\mathbf{x_i}$ on the gutter, then is this the same as if I define support vector as any $\mathbf{x_i}$ whose multiplier is positive?

  • 2
    $\begingroup$ I think I know what you mean, but I don't think "gutter point" is standard terminology? (When I google for "gutter point" and "svm" I only get this page.) $\endgroup$ Commented Nov 5, 2016 at 22:02

1 Answer 1


It's mathematically possible to have cases where the constraint $y_i(\mathbf{w}^T\mathbf{x_i}+b)\geq{1}$ is satisfied with equality and where Lagrange multiplier $a_i = 0$ if the constraint isn't active in the sense that without the constraint, you would still have the same solution $\mathbf{w}$. In practice, this is probably a knife-edge case.

Intuitively, multipliers are penalties for violating a constraint

The intuitive interpretation of Lagrangian multipliers is that they are the price of violating a constraint.

Imagine that we have the problem:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $\mathbf{x}$)} & f(\mathbf{x}) \\ \mbox{subject to} & g(\mathbf{x}) \leq 0 \end{array} \end{equation}

The Lagrangian is:

$$ \mathcal{L}\left(\mathbf{x}, \lambda \right) = f(\mathbf{x}) + \lambda g(\mathbf{x}) $$

$\lambda$ is a penalty for having a positive $g$. The intuitive idea is that there exists a price $\lambda^*$ for violating the constraint such that the optimizer is better off not violating the constraint.

(Side note: it's worth the time to understand KKT sufficient vs. necessary conditions, the primal vs. dual problem, weak duality, strong duality, and the saddle point property of the Lagrangian.)

Simple example:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $x$)} & x^2 \\ \mbox{subject to} & -x \leq 0 \end{array} \end{equation}

Lagrangian is:

$$ \mathcal{L} = x^2 - \lambda x $$ First order condition implies:

$$ 2x = \lambda $$

Solution is $x=0$, $\lambda = 0$. The solution to the unconstrained problem $\min x^2$ is $x = 0$. Basically, the point is on the boundary, but the constraint isn't doing any work.

If the unconstrained solution to the problem would have placed the point $x$ on the boundary of the constraint anyway, the penalty for violating the constraint may be zero. In practice, for many problems, this will be a knife-edge case.

  • $\begingroup$ X=0 gives the solution to minimize x^2 and it is in the constraint -x <=0. In such cases, do we need to formulate Lagrangian? I am wondering we need to form the Lagrangian only when the the optimal point for the convex cost function does not satisfy the constraint (which I am struggling to understand now). $\endgroup$
    – mon
    Commented Jan 12, 2017 at 9:23

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.