Dear Pushpasis,
I agree with you with the previous observations, except the following
can not gracefully leave my mind :)
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.
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
What is your opinion about this?
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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