Hi Levente,

From: Levente Csikor <[email protected]<mailto:[email protected]>>
Date: Wednesday, October 28, 2015 at 2:04 PM
To: Pushpasis Sarkar <[email protected]<mailto:[email protected]>>, 
"[email protected]<mailto:[email protected]>" 
<[email protected]<mailto:[email protected]>>, 
"[email protected]<mailto:[email protected]>" 
<[email protected]<mailto:[email protected]>>
Subject: Re: New Version Notification for 
draft-ietf-rtgwg-rlfa-node-protection-04.txt

Dear Pushpasis,
I agree with you with the previous observations, except the following can not 
gracefully leave my mind :)

[Pushpasis]As per RFC 7490.. Ext-P space(S, S-E) is a union of P-space(S, S-E) 
and P-Space(all neighbors of S except E, S-E). In the example you have chosen Y 
(later renamed to N2) is in P-Space(S, S-E) though not in P-Space(N1, S-E). The 
inequality is a generalised version that fits other scenarios. Even though the 
example chosen does not seem to be a good one to demonstrate the use of this 
generalised inequality, the inequality still does work even with this example. 
:)

Tell me if I did not get your point, but according to your inequalities, it 
seems that N2 being evaluated at first gives an error, i.e.,

"D_opt(N1, N2) < D_opt(N1,S) + D_opt(S,N2) is not satisfied
    2              1             1",
so N2 is not in extended P-space (hereafter, always assume that ext.P-space is 
considered w.r.t. the failed link).
However, as a second (or a further) step, when distance from N2 to N2 itself is 
evaluated, then

"D_opt(N2, N2) < D_opt(N2,S) + D_opt(S,N2) is indeed satisfied
     0             1              1 "
So, finally "somehow" N2 becomes the part of the extended P-space.

To me, it's still not clear for the following reasons:
i) from an implementation point of view, why should I evaluate the distance 
from me, to myself (D_opt(N2,N2))? Isn't it an unnecessary step that an 
algorithm should avoid.
[Pushpasis]Like mentioned before an implementation needs to be generic enough 
to handle all kinds of the topology example. If you can suggest a better 
inequality I will be more than happy to accommodate it the document. Please let 
me know if you have a better inequality in your mind.

ii) again, from an implementation point of view, and also according to the 
algorithm proposed in RFC7490, if for each node (w.r.t. the failed link) we 
store only a binary (boolean) variable to indicate whether it is in the 
ext.P-space or not, then who guarantees that in the above example the final 
state of this boolean variable of N2 will be True. It's really typical (at 
least in usual programming languages ) that traveling all the nodes in the 
graph more times, then you definitely won't get the same order at each case.
So, what I want to tell here, that it could be the case that "D_opt(N2, N2) < 
D_opt(N2,S) + D_opt(S,N2)"  will be evaluated first an N2 becomes the part of 
the ext.P-space, but at the end of the loop, N2 won't be in the ext.P-space
[Pushpasis] What you are trying to discuss here is more of like implementation 
details. As a implementor I have implemented this for the company I work for 
and am sure can be implemented using any programming language. How to implement 
this should be outside the scope of this document. However still here is a 
high-level algorithm for your understanding.

For each node ā€˜X’ in the LSDB
-  Set X.InExtPSpace ==> FALSE
 - For each direct neighbor ’N’ in Neighbor-List
    - If D_opt(N, X) < (D_opt(N,S) + D_opt(S,X) then
       - Set X.InExtPspace ==> TRUE
       - Continue to next node in LSDB
  - EndFor
EndFor


What is your opinion about this?
[Pushpasis] Hope the above explanation helps clarifying things.

Thanks
-Pushpasis

Regards,
Levente


On 10/27/2015 06:31 PM, Pushpasis Sarkar wrote:
Or, the statement should emphasize in the beginning that node Y is not in the 
P-space of S w.r.t. the failed link S-E.
Pls tell me if I'm wrong.
Btw, the inequality for node-protecting extended P-space is valid.
[Pushpasis] I think you might have misunderstood it.. Ni is not a single 
neighbor of S.. The term Ni in the above inequality stands for all neighbors of 
S other than E (primary next hop).. Here is the text once more..

"A node Y is in link-protecting extended P-space w.r.t to the link (S-E) being 
protected, if and only if, there exists at least one direct neighbor of S, Ni, 
other than primary nexthop E, that satisfies the following condition.
   D_opt(Ni,Y) < D_opt(Ni,S) + D_opt(S,Y)"

So one possible value for Ni can be Y itself..  So now if you substitute Y for 
Ni in the above inequality it satisfies the inequality..

D_opt(Y, Y) < D_opt(Y,S) + D_opt(S,Y)
              0                     1                     1

However to make it more clear let me rename the Nodes Ni and Y in the above 
diagram as N1 and N2. So the diagram now looks like..

  N2
2/ \
N1--S--x--E
|        /
B       D
\      /
 \    /
  \  /
   A

Ni = {N1, N2}

Now with N2 as the candidate,

substituting Ni = N1

D_opt(N1, N2) < D_opt(N1,S) + D_opt(S,N2) is not satisfied
                  2                         1                          1

However substituting Ni = N2

D_opt(N2, N2) < D_opt(N2,S) + D_opt(S,N2) is indeed satisfied
                  0                        1                          1

So N2 is in link-protecting Ext-P-Space of S, wrt S-E link.

Essentially for Y to be in link-protected Extended P-Space(S, S-E) S (or one of 
the neighbors of S other than E) should be able to reach Y without taversing 
the S-E link.. The above inequality is satisfied by substituting Ni as Y.


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