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?

Beitrag von Notschko » 1. Jul 2013 15:56

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.

Thanks for your help.

Notschko

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

Beitrag von hymGo » 1. Jul 2013 18:05

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 :wink:
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?

Beitrag von lustiz » 1. Jul 2013 18:08

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 :P

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