In soft-margin SVM, is it guaranteed that some points will lie on the margin?

In soft-margin SVM, once we solve the dual-form optimization problem (using quadratic programming, which is guaranteed to converge on a global optimum because the problem is convex), we get a vector $\alpha$ of coefficients that we can use to recover the hyperplane parameters $\beta$ and $\beta_0$. The additional constraint in soft-margin (compared to hard-margin) is $0 \leq \alpha_i \leq C$ for some positive constant $C$. We say that:

1. If $\alpha_i = 0$, then the corresponding $x_i$ is outside the margin
2. If $\alpha_i = C$, then the corresponding $x_i$ is on the wrong side of the margin
3. If $0 \lt \alpha_i \lt C$, then the corresponding $x_i$ is on the margin

We recover $\beta$ through the KKT condition $\beta = \sum_{i}{\alpha_i y_i x_i}$, but we recover $\beta_0$ by selecting any $x_i$ whose corresponding $\alpha_i$ satisfies the third condition above and solving $y_i(\beta^T x_i + \beta_0)=1$ (we should get the same value for $\beta_0$ using any such point satisfying the third condition above).

Here's my question: given an optimal $\alpha$, is it possible for the third condition to not be satisfied for any $\alpha_i$? For example, I actually have a dataset that's linearly separable for which I'm trying to train a soft-margin SVM with $C=0.5$, but I'm finding that every value in $\alpha$ is either $0$ or $0.5$ and that there are no values in between, hence I cannot recover $\beta_0$. Am I doing something wrong if my $\alpha$ doesn't satisfy the third condition above for any $i$? My intuition tells me that there should always be at least two points on the margin because the primal optimization problem maximizes the size of the margin; but the $\alpha$ for my dataset is contradicting my intuition.

If it helps, I'm doing this in MATLAB with the following (pseudo)code:

X = 2 by n matrix of linearly separable data points
y = n by 1 matrix of class labels (-1 or +1)
X_hat = X with each column signed to its label
H = X_hat' * X_hat;
f = -1 * ones(n, 1);
Aeq = y';
beq = 0;
lb = zeros(n, 1);
ub = 0.5 * ones(n, 1);
alpha = quadprog(H, f, [], [], Aeq, beq, lb, ub); % alpha is always 0 or 0.5

• Set an upper bound higher than 0.5 . If your upper bound is 0.5, you will never get an alpha higher than 0.5 – user73812 Apr 17 '15 at 15:22

But I suggest you change the value of $C$ first and see if there is still no free SV.