Subject: [trill] RPF entries ..RFC 6325 section 4.5.2

From: gururam m <[email-masked]>
Date: Tue, 22 May 2012 15:44:31 +0530

Hi,
      I have a query related to the RPF check from RFC 6325.

Let us say a TRILL domain setup has about 6 RBridges each with a single
nickname and 6 multicast trees are computed by each RBridge in the TRILL
domain and the trill domain setup is as shown below.

RB1-------RB2--------RB3
  |             |              |
  |             |              |
  RB4------RB5----------RB6

As per my understanding each RB will have RPF entries that are equal to the
ingress interfaces on that RB. For e.g RB1 has two interfaces hence it has
two RPF entries and RB2,RB5 will have 3 RPF entries since it has 3
interfaces or to state the number of RPF entries on each RBridges depends
on the number of ports that are part of the multicast trees on that
particular RBridge.

Please correct me if my understanding is not correct.

Regards
Ram

---

From: Donald Eastlake <[email-masked]>
Date: Tue, 22 May 2012 09:22:40 -0400

Hi,

On Tue, May 22, 2012 at 6:14 AM, gururam m <[email-masked]> wrote:

> RB4-------RB5--------RB6
> As per my understanding each RB will have RPF entries that are equal
> to the ingress interfaces on that RB.

I think you are talking about TRILL Data traffic so I would avoid
using the word "ingress" here. In TRILL "ingress" normally refers to
receiving native traffic, not TRILL Data frames.

> For e.g RB1 has two interfaces hence it has two RPF entries and RB2,
> RB5 will have 3 RPF entries since it has 3 interfaces or to state
> the number of RPF entries on each RBridges depends on the number of
> ports that are part of the multicast trees on that particular
> RBridge.

TRILL does not constrain how the RPF (Reverse Path Forwarding) Check
entries are organized.

If you think about it as a lookup under the triple { tree number;
port; ingress RBridge nickname } that returns a boolean, then there
are a lot more entries than you are talking about. In this case
(number of ports)*(number of trees)*(number of possible ingress
RBridges) or (number of ports)*30. If you think about it as a lookup
under the double { tree number; ingress RBridge nicknme } that returns
the expected receiving port that you then compare with the actul
receiving port for the check, then there are 30 entries at each
RBridge in your example. (This assumes that no RBridge is announcing
the trees it is using as a subset of the computed trees. If one does,
it reduces the RPF Check information.)

Other methods are possible.

Thanks,
Donald
=============================
 Donald E. Eastlake 3rd   +1-508-333-2270 (cell)
 155 Beaver Street, Milford, MA 01757 USA
 [email-masked]