# Question about the formulation of SVM

The formulation of the SVM optimization problem is:

\begin{aligned} & max_{w,b} \frac{1}{||w||} \\ & \text{ subject to } \\ & y_i(w^{T}x_i+b) \geq 1 \end{aligned}

What I do not understand is why do we use $$w^Tx_i+b=1$$ in the setup. My question is specifically about why 1? I understand that $$w^Tx_i+b$$ is the equation of a hyperplane and multiplying it by binary class labels $$y_i \in \{-1,1\}$$ we get the inequality but why do we initially not use $$w^Tx_i+b = 2$$ or 0 or any number. I am assuming we can adjust for this number since we have b as a hyperparameter.

Thank you!

## 1 Answer

The goal of a hard-margin SVM (you didn't mention slack variables, $$\xi$$, for soft-margin SVMs) is to minimize the Euclidean norm $$||\mathbf{w}||$$ to impose the inequality $$y_i D(\mathbf{x}_i)/||\mathbf{w}||>1$$ (Boser et al, 1992). This is accomplished by using both the margin value $$M$$ and the weight vector $$\mathbf{w}$$ and enforcing the constraint $$M ||\mathbf{w}||=1$$, which when solving for the optimal margin gives $$M=1/||\mathbf{w}||$$.

For hard-margin SVMs, the unconstrained Lagrangian function is also $$$$L({\bf w},b,\boldsymbol{\alpha})=\frac{1}{2}{\bf w}^T{\bf w} - \sum_{i=1}^n \alpha_i [y_i ({\bf w}^T{\bf x}_i + b) - 1 ]\\$$$$

$$s.t.\quad \quad \alpha_i \geq 0, \quad y_i ({\bf w}^T{\bf x}_i + b) - 1=0$$

Reference:

B.E. Boser, I.M. Guyon, V.N. Vapnik. A training algorithm for optimal margin classifiers. $$\textit{Proc. 5th Annual Work. Comp. Learning Theory (COLT'95),}$$ pp. 144-152. New York (NY), ACM Press, 1992.

• thanks for your answer @nxglogic, I am wondering why the hyperplane equation we use is $w^Tx_i + b= 1$ instead of $w^Tx_i + b = 0$ – Kaan Yolsever Apr 21 at 13:28