Ripe 181

  > cengiz at ISI.EDU (Cengiz Alaettinoglu) writes:
  > 
  > > So the partial ordering is
  > > 
  > > L1R1 = L1R3 < L1R2 = L1R4
  > > 
  > > Easy to derive too.
  > > Is this clear now?
  > 
  > Not quite. This only works when you have two to choose from.
  > If you have more:
  > 
  > interas-in: from AS2 L1R1 1 accept AS100
  > interas-in: from AS2 L1R2 2 accept AS100
  > interas-in: from AS3 L1R3 1 accept AS100
  > interas-in: from AS3 L1R4 2 accept AS100
  > interas-in: from AS3 L1R5 3 accept AS100
  > 
  > i.e.
  > 	L1R1 < L1R2        and           L1R3 < L1R4 < L1R5
  > hence there are two possible orderings:
  > 	1. L1R1 = L1R3 < L1R2 = L1R4 < L1R5
  > 	2. L1R1 = L1R3 < L1R4 < L1R2 = L1R5
  > 
  > Choosing either of them means some comparison of interas-in
  > preferences across AS's. 
  > 
  > After some local discussion here, we decided to go with 1. 
  > Since it is easier to compute:
  > 	o  renumber interas-in preferences so that the smallest
  >            preference value is now 1, next smallest is 2, ...
  > 	o  compare interas-in preferences across AS's as
  >            well to get the final ordering.
  > 		as-in preference * Range + interas-in preference
  > 
  > No need to maintain ordered lists per route. (Actually with equal
  > preferences ordered lists in your algorithm are actually ordered
  > graphs).
  > 
  > Folks, please let me know if there are problems with this approach.

This is how I would do it.
Since I do not expect this to be exercised much 
in practise, much less that it will really matter at all,
I do not care much either.

Daniel

Daniel



-------- Logged at  Tue Oct 18 17:51:24 MET 1994  ---------