Subject: [rbridge] seeking input on abstract for problem and applicabilitystatement

From: petera at nortel.com (Peter Ashwood-Smith)
Date: Wed, 12 Oct 2005 11:23:01 -0400

> I still think the rbridge MUST participate in order to avoid loops, 
> but if that's contentious it can't be part of the abstract yet...)

Let me see if I can help. Radia gave a good description of why the
Rbridge does not need to participate in STP/FSTP but here is another way
to think of it and perhaps it will strike different chords:

Take two node disjoint trees A and B.

Now, connect the trees A and B with only ONE link. The result is still a
tree.

This is "sort of" what the combination of STP and Rbridge protocols are
doing.

STP computes one tree. Rbridge computes the other and finally Rbridge
connects the two with only ONE link using the designated Rbridge
concept.

Since we can apply this same mechanism over and over again, i.e.
induction we are guaranteed the end result is indeed one big tree and
therefore is loop free, ... at least in the steady state. 

Cheers,

Peter Ashwood-Smith

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From: touch at ISI.EDU (Joe Touch)
Date: Thu, 13 Oct 2005 20:28:09 -0700

> therefore is loop free, ... at least in the steady state. 

But isn't this just what STP does?

I.e., a bridge:
	
	accepts BPDUs as input
	determines an internal spanning tree based on port selection
	emits appropriate BPDUs on those ports

Why isn't this exactly what an rbridge needs to do? I.e., neither a
bridge nor an rbridge 'transparently' forward BPDUs, but both are nodes
in the resulting spanning tree.

This still provides a benefit for an rbridge - that it is just ONE node
in the spanning tree, not O(#rbridges) or even O(#egress rbridges).

I.e., it still helps the remaining spanning tree converge faster and
react better. AND rbridges 'inside' the campus (acting as transits) can
move around without affecting the spanning tree at all.

Joe
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