Re: [rbridge] Parent node ID selection for multi destination trees

The only problem with making the change below is that it is very
important that all RBridges calculate exactly the same distribution
trees. Otherwise multi-destinationn traffic get dropped due to the
Reverse Path Forwarding Check. The existing specification, while a bit
odd, works; but an implementation of the existing specification and
the proposed changed specification would have interoperability
problems for multi-destination TRILL Data frame.

I'd really like to hear from other implementors as to whether they
have implemented it as specified in RFC 6325 and what their feelings
are about the proposed change.

Thanks,
Donald
=============================
 Donald E. Eastlake 3rd   +1-508-333-2270 (cell)
 155 Beaver Street, Milford, MA 01757 USA
 d3e3e3@gmail.com

On Sat, Mar 17, 2012 at 2:24 PM, ayabaner <ayabaner@cisco.com> wrote:
> Varun,
>
> Agree, that was the original intention. I believe that if kept as-is we will
> have more unnecessary changes in the SPF outputs when topology changes.
>
> I believe that we should just add this into the
> “draft-ietf-trill-clear-correct-01”.
>
> Thanks,
> Ayan
>
>
>
>
> On 3/16/12 5:30 PM, "varshah" <varshah@cisco.com> wrote:
>
> Hi,
>
> Reading section 4.5 of RFC 6325
> (http://tools.ietf.org/html/rfc6325#section-4.5) it seems that the selection
> of a parent node  (7 byte- node ID) for a node N on a multi destination tree
> j; should be based on j % p, where p is the number of equal cost parent
> nodes to reach node N from root node R.
>
> The intent seems to use lowest parent node ID for tree 1, next for tree 2
> and so on.... And circle through if we have more possible trees than parent
> nodes.
>
> However, the trees are numbered starting 1, and the parent node IDs are
> indexed starting 0 to (p – 1).  So, for a node N, when the network supports
> 2 trees and with two possible parent nodes,  we would use
>
> parent node indexed at 1 for tree 1, j % p --> 1 % 2 = 1 and
> Parent node indexed at 0 for tree 2, j % p --> 2 % 2 = 0.
>
> To stay with the real intention, it would help if we had (j - 1) %p.
>
> In that case,
> parent node indexed at 0 for tree 1, (j  - 1)% p --> 0 % 2 = 0 and
> Parent node indexed at 1 for tree 2,( j – 1) % p --> 1 % 2 = 1.
>
> Thanks,
> -Varun
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