Consider the SVM dual, i.e., \begin{align} &\text{maximize} \sum_{i=1}^n \alpha_i-\frac{1}{2\lambda} \sum_{i,j=1}^n \alpha_i \alpha_j y_i y_j K(x_i,x_j)\cr &\text{subject to, } 0\leq \alpha_i \leq 1 \end{align} where $K$ is the kernel matrix and $\lambda$ is the regularization parameter.

My questions are:

  • What is the meaning of $\sum_{i=1}^n \alpha_i$?
  • When this value is equal to $n$, Can we say that model is
  • Is it meaningful to add this constraint $\sum_{i=1}^n \alpha_i<n-t$
    with $t$ a parameter, to the above problem?
  • If yes, what is this constraint in the primal?
  • $\begingroup$ why subject to $0\le \alpha_i \le 1$? $\endgroup$
    – Kuo
    Feb 21 at 4:08

1 Answer 1


From strong duality, it's easy to see that $$ \sum_{i=1}^n \alpha_i = \frac{\lambda^2 + 1}{2\lambda}||w||^2 + L(w) $$ where $L(w)$ in your formulation is $\sum_{i=1}^n \max\{0, 1- y_i w^T x_i\}$.

If this value is equal to $n$, then you are no better than the case where $w=0$. So this is basically the null-case where you haven't really learned anything (and underfitted would seem to be apt).

Adding such a constraint is odd - one already has $\lambda$ in place to handle the tradeoff between model complexity and error. This constraint may be effective in finding a sparser solution (i.e., fewer support vectors) - of course, there is also the problem that the problem may not be solvable (i.e., for $t$ which is too large you have no solution). But there exist a large body of work already on finding sparse solutions to SVM which are well motivated - this constraint is not. Also adding in this constraint may alter the expression for $w$ itself, so it's not entirely clear what the end result would be.

Adding an additional constraint in the dual will result in a new variable in the primal, and the primal objective will look different. It's largely a mechanical process to transform from dual -> primal, but someone else (or perhaps OP) can do that.


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