Re: [Rift] Clarification on North/South bound SPF

"Nagendra Kumar Nainar (naikumar)" <naikumar@cisco.com> Thu, 03 May 2018 16:23 UTC

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From: "Nagendra Kumar Nainar (naikumar)" <naikumar@cisco.com>
To: Tony Przygienda <tonysietf@gmail.com>
CC: "draft-przygienda-rift@ietf.org" <draft-przygienda-rift@ietf.org>, "rift@ietf.org" <rift@ietf.org>
Thread-Topic: [Rift] Clarification on North/South bound SPF
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Date: Thu, 03 May 2018 16:23:18 +0000
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Subject: Re: [Rift] Clarification on North/South bound SPF
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Hi Tony,

Thanks.  I think I get it now ☺

Just confirming my understanding – While computing Northbound SPF, a node will consider a) “locally” originated Node N-TIE, b) Node N-TIE from same level c) Node/Prefix S-TIE. But it does not consider any other N-TIE(ex: the one from downstream or lower level nodes). On the same line, while computing southbound SPF, it will consider a)Entire Prefix/Node N-TIE b) (obviously) locally originated Node S-TIE.

Is that a fair understanding?

Thanks,
Nagendra

From: Tony Przygienda <tonysietf@gmail.com>
Date: Wednesday, April 25, 2018 at 12:34 AM
To: Nagendra Kumar Nainar <naikumar@cisco.com>
Cc: "draft-przygienda-rift@ietf.org" <draft-przygienda-rift@ietf.org>, "rift@ietf.org" <rift@ietf.org>
Subject: Re: [Rift] Clarification on North/South bound SPF

Hey Nagendra, good drilling, good observations, let me expound a bit ;-)
a. "SPF" per se is a lie in the draft to seem more familiar to existing stuff  than it is ;-), RIFT is loop free & with that RIFT CAN compute all feasible paths, i.e. you could use a "modified kind of SPF" if you will where you have a candidate list per level independent of distance (i.e. you get all paths with all distances & you can use all of those). Draft doesn't mention that, will probably do in the future but when you implement stuff it becomes blatantly obvious what I mean ;-}
b. Your point 1: It's a bit more complicated albeit you are right.
i) when you're a leaf you do not strictly speaking have a S node TIE (since it's not necessary) so you'd use your own (leaf's N node-tie & then use the upper level's S node-tie [as you say]) to double-check the bi-directionality.
c. as meta-point:
observe that N- and S-node-TIEs are fully symmetric right now (in terms of adjacencies) for many reasons so it really doesn't matter which one you use ;-)
d. as meta-meta-point:
N- and S-node-TIEs are NOT fully symmetric  (if you read the schema you see mysterious unexplained stuff in NodeTIEElement, this has to do with something we call "transitive disaggregation" and will be added to the draft. I'm discussing the math with Pascal and we try to find a simple representation of it but in terms of SPF it plays no role).
I hope I clarified more than I confused here, otherwise just keep poking ;-)
--- tony


On Tue, Apr 24, 2018 at 2:40 PM, Nagendra Kumar Nainar (naikumar) <naikumar@cisco.com<mailto:naikumar@cisco.com>> wrote:
Hi Authors,

While I am still reading, I thought of quickly clarifying a couple of points to better understand the remaining part of the draft ☺


  1.  In Section 4.2.5.1, N-SPF is defined as below

“N-SPF uses northbound and east-west adjacencies in North Node TIEs
   when progressing Dijkstra.”

Based on my reading so far, N-SPF is triggered to compute the path towards upper level (Level0 to Level1 and so on) and I believe this requires N-SPF to be run on Node S-TIE and not Node N-TIE right?. Should it not be “N-SPF uses northbound and east-west adjacencies in Node S-TIE when progressing Dijkstra”?.


  1.  Similar one in Section 4.2.5.2 – S-SPF should use soutbound adjacencies in node N-TIE right?.

Am I missing something?.

Thanks,
Nagendra

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