# Where does the definition of the hyperplane in a simple SVM come from?

I'm trying to figure out support vector machines using this resource. On page 2 it is stated that for linearly separable data the SVM problem is to select a hyperplane such that $\vec{x}_i\vec{w} + b \geq 1$ for $y_i \in 1$ and $\vec{x}_i\vec{w} + b \leq -1$ for $y_i \in -1$. I'm having trouble to understand where the right-hand side of the constraints come from?

P.S The next question would be how to show that the SVM's margin is equal to $\frac{1}{||\vec{w}||}$.

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See if you find this useful - stats.stackexchange.com/questions/43727/… – TenaliRaman Jan 2 '13 at 22:58
yes, the answer for this question is part of the answer given here stats.stackexchange.com/a/43811/16837 – entropy Jan 3 '13 at 1:05

Essentially these two constraints basically require the training data to be correctly classified, and at least a certain distance from the decision threshold 0. The hyperplane that fulfils these constraints with the smallest norm of the weights will have the maximal margin. The value $\pm 1$ is essentially arbitrary, you could replace it with $\pm$ any value you like and it would merely rescale the coefficients of the hyper-plane, but without changing the decision boundary. A value of 1 is used just to keep the maths neat.

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wx + b = -1 and wx + b = 1; These equations represent two parallel hyperplanes that are formed based on samples class (-1, +1). These two hyperplanes are used to optimize the distance between classes and to get optimal hyperplane. For optimization problem Lagrange multipliers are used.

You can find brief description on resources listed below.

Online course on Machine Learning by Andrew Ng is a great place to understand SVM and other ML algorithms: Machine Learning - Andrew Ng Hyperplane is thoroughly explained.

In order to better understand math behind the SVM, learning Optimization is the right choice. There is a great free ebook by S.Boyd: Optimization - Boyd

Here you can find SVM papers: svms.org/tutorials/

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