Difference between revisions of "Support Vector Machine Examples OldKiwi" - Rhea

Latest revision as of 16:35, 30 March 2008

The class of the data is written as the second co-ordinate. $ {\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0 $

• • 3 inequalities**:

$ 1.w+b\leq-1 $

$ 2.w+b\leq+1 $

$ 3.w+b\leq+1 $

$ J=\frac{{w}^{2}}{2}-{\alpha }_{1}\left(-w-b-1 \right)-{\alpha }_{2}(2w+b-1)-{\alpha }_{3}(3w+b-1) $

Formulating the Dual Problem:

$ Q(\alpha )= {\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}-[0.5{{\alpha }_{1}}^{2}+2{{\alpha }_{2}}^{2}+4.5{{\alpha }_{3}}^{2}-2{\alpha }_{1}{\alpha }_{2}-3{\alpha }_{1}{\alpha }_{3}-6{\alpha }_{2}{\alpha }_{3}] $

Subject to constraints

$ {\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0 $ and

$ {\alpha }_{1}\geq 0; {\alpha }_{2}\geq 0;{\alpha }_{3}\geq 0 $

Differentiating partially with respect to $ {\alpha }_{1},{\alpha }_{2}, {\alpha }_{3} $

$ {\alpha }_{1}+2{\alpha }_{2}+3{\alpha }_{3}=0 $

$ 1+2{\alpha }_{1}-4{\alpha }_{2}-6{\alpha }_{3}=0 $

$ 1+3{\alpha }_{1}-6{\alpha }_{2}-9{\alpha }_{3}=0 $

On Solving, we get

$ {\alpha }_{1}=2 $

$ {\alpha }_{2}=2 $

$ {\alpha }_{3}=0 $

This yields w = 2, b = -3. Hence the solution of decision boundary is: 2x - 3 = 0. or x = 1.5 This is shown as the dash line in above figure.