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:
<img src="svm1.bmp" />
<img alt="tex: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
<img alt="tex:-{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0" /> and
<img alt="tex:{\alpha }_{1}\geq 0; {\alpha }_{2}\geq 0;{\alpha }_{3}\geq 0" />
Differentiating partially with respect to <img alt="tex:{\alpha }_{1},{\alpha }_{2}, {\alpha }_{3}" />
<img alt="tex:1-{\alpha }_{1}+2{\alpha }_{2}+3{\alpha }_{3}=0" />
<img alt="tex:1+2{\alpha }_{1}-4{\alpha }_{2}-6{\alpha }_{3}=0" />
<img alt="tex:1+3{\alpha }_{1}-6{\alpha }_{2}-9{\alpha }_{3}=0" />
On Solving, we get
<img alt="tex:{\alpha }_{1}=2" />
<img alt="tex:{\alpha }_{2}=2" />
<img alt="tex:{\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.
