[rbridge] Distribution Tree and ECMP

[Suresh Boddapati]

4.3.1 of the base protocol spec says: 

   "When a node RBn has two or more minimal equal cost
paths toward the Root RBi a deterministic tie-breaker
is needed to guarantee that all Rbridges calculate the
same distribution tree. This is obtained by selecting
the path that goes to the parent that has the lower
IS-IS System ID." 

This however does not cover the case when the ECMP
paths are through the same RBridge. For example, 

          
A ----- B ---------- C
          ----------

If B is the designated IS for both the links connected
to C for the tree rooted at A, it is possible that C
might compute the top link to be on the distribution
tree whereas B might compute the bottom one.

It seems to me that the algorithm will be more
deterministic if the LAN ID were to be used for
tie-breaking, when System IDs are the same. In this
case if the top link's LAN ID were to be B.01 and the
bottom link's were to be B.02, both could
deterministically decide that B.01 should be the link
on the distribution tree. 

Comments?

Suresh