(New page: The class of the data is written as the second co-ordinate. <math>{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0 </math> **3 inequalities**: <math>1.w+b\leq-1 </math> <math>2.w+b\leq+1 </...)
 
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The class of the data is written as the second co-ordinate.
 
The class of the data is written as the second co-ordinate.
 
<math>{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0 </math>
 
<math>{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0 </math>
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Formulating the Dual Problem:
 
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}]
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<math>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}]
" />
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</math>
  
  
 
Subject to constraints
 
Subject to constraints
  
<img alt="tex:-{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0" />  
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<math>{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=0 </math>
 
and
 
and
  <img alt="tex:{\alpha }_{1}\geq 0; {\alpha }_{2}\geq 0;{\alpha }_{3}\geq 0" />
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  <math>{\alpha }_{1}\geq 0; {\alpha }_{2}\geq 0;{\alpha }_{3}\geq 0</math>
  
Differentiating partially with respect to  <img alt="tex:{\alpha }_{1},{\alpha }_{2}, {\alpha }_{3}" />
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Differentiating partially with respect to  <math>{\alpha }_{1},{\alpha }_{2}, {\alpha }_{3}</math>
  
  
<img alt="tex:1-{\alpha }_{1}+2{\alpha }_{2}+3{\alpha }_{3}=0" />
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<math>{\alpha }_{1}+2{\alpha }_{2}+3{\alpha }_{3}=0 </math>
  
<img alt="tex:1+2{\alpha }_{1}-4{\alpha }_{2}-6{\alpha }_{3}=0" />
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<math>1+2{\alpha }_{1}-4{\alpha }_{2}-6{\alpha }_{3}=0 </math>
  
<img alt="tex:1+3{\alpha }_{1}-6{\alpha }_{2}-9{\alpha }_{3}=0" />
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<math>1+3{\alpha }_{1}-6{\alpha }_{2}-9{\alpha }_{3}=0 </math>
  
 
On Solving, we get
 
On Solving, we get
  
<img alt="tex:{\alpha }_{1}=2" />
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<math>{\alpha }_{1}=2</math>
  
<img alt="tex:{\alpha }_{2}=2" />
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<math>{\alpha }_{2}=2</math>
  
<img alt="tex:{\alpha }_{3}=0" />
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<math>{\alpha }_{3}=0</math>
  
 
This yields w = 2, b = -3.  
 
This yields w = 2, b = -3.  

Revision as of 17:33, 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.

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