## Assignment 3 - Problem 1B: What if C=0?

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Notschko
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### Assignment 3 - Problem 1B: What if C=0?

Hi,

by definition C is the weight for $$\xi_i$$. Does this mean, if C=0 the result will be the same as in Problem1A? Still, I'm getting some different values for w and b compared to 1A even if C is 0. Also, my values for $$\xi_i$$ are greater than 0 for C=0 and I don't know if this is right and what to do with these values.

Notschko

hymGo
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### Re: Assignment 3 - Problem 1B: What if C=0?

C determines the degree of punishment for the slack variables. To achieve results as in a) (or rather no slack) C must be choosen ...
Sorry can't give the full answer. If you don't get it, just try it out. There are only two possibilities
Zuletzt geändert von hymGo am 1. Jul 2013 18:41, insgesamt 1-mal geändert.

lustiz
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### Re: Assignment 3 - Problem 1B: What if C=0?

Notschko hat geschrieben:by definition C is the weight for $$\xi_i$$. Does this mean, if C=0 the result will be the same as in Problem1A? (...) Still, I'm getting some different values for w and b compared to 1A even if C is 0
You should not consider C to be a weight. It just describes the trade-off between the penalties of the slack variables and the margin which holds for all variables. I can't give you the answer right away, but you should probably try some different settings and then take the resulting change of margins into consideration. Varying C you will get a feeling about its influence.
If you find the right C you will end up with something pretty similar to the results of a).

Notschko hat geschrieben:Also, my values for $$\xi_i$$ are greater than 0 for C=0
The slack variables are $$\xi_i \ge 0$$ per definition. If you have $$\xi_i = 0$$, it means the associated point is on the correct side of the margin. On the other hand, if $$\xi_i > 0$$ then $$\xi_i = |t_i - y(x_i)|$$ holds and the associated point is not on the correct side. In this case there are two options: Either the slack variable is > 1 or the slack variable is < 1 and both cases correspond to the sides of the decision boundary, respectively. In other words: The result of your optimization should never give you negative slack variables! If it does, there is something going wrong..
Notschko hat geschrieben:... and I don't know if this is right and what to do with these values.
Easy answer: Nothing! They are just help variables required for the optimization, you can discard them afterwards..

EDIT: ahhhh too late