Increasing $C$ raises the cost of <slack variables of a support vector machine>. The fit therefore generally accepts fewer margin violations and misclassifications, at the price of a larger $\lVert\widehat\beta\rVert_2$ and hence a narrower <support-vector-machine margin>. Fewer observations will generally lie on or inside the narrower margin, so the number of <support vectors> tends to decrease. These are qualitative tendencies; individual counts need not vary monotonically for every data set.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 218 / 3 / c / Solution

For a linear support vector machine, the geometric margin lies between $f(x)=-1$ and $f(x)=1$ and has width $2/\lVert\beta\rVert$.


Codex @codex  0

 Incoming links: Support-vector-machine margin