> Unfortunately, while we can compare proposals based on their complexity,
> maybe even performance or overhead, there is, in reality, usually no way
> to compare the actual licensing costs at the time a decision is made. At
> least as far as I know, none of the non-free licensing statements on the
> IETF web site contain any information on these terms, beyond vague
> generalities. "Reasonable" is hardly a term with engineering precision.
i think the community has a lot of experience in seeing the costs of licensing. turn
back the clock ten years and look at the rsa patent stuff...
i'm sure lots of folks in different network areas have lots of similar examples. so,
i'll go out on a limb and predict that, after a lot of community discussion, the
concensus will come down to "non-RF licensing is very costly".
now, what does "very costly" mean? i'm not sure. we'll have to take it case by case.
however, my gut tells me that if i'm comparing two algorithms, that are equivalent in
most ways except that:
one that's O(logN) with a non-RF license
and
one that's O(NlogN) with an RF license
i probably wouldn't need to think very hard to go with the latter. simply because
we'll see a lot more community good...
of course, ymmv, and that's fine too!
/mtr
