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Re: [Fwd: [Fwd: I-D ACTION:draft-ietf-rtgwg-microloop-analysis-00.txt]]

Subject: Re: [Fwd: [Fwd: I-D ACTION:draft-ietf-rtgwg-microloop-analysis-00.txt]]
From: Russ White
Date: Thu, 28 Jul 2005 08:29:13 -0400
I'm completely confused about what you are saying here. We are talking about B wanting to send traffic to C, right? In BOTH cases there is a loop. Clearly so in the first case, since there is no other path than through B. But in the second case there is also a loop because of ECMP. If ANY ECMP path may loop then you have to consider it a loop.
Then you suggest changing the B-D cost to 25. Again in BOTH cases there
is loop. I don't know what you are trying to show here.
So, this is part of the problem--what do you mean by "loop?" Do you mean
D has an alternate path that does not go through B, or do you mean D
would not send traffic via B to get to C given the current metrics?
Those are two different definitions, and I've not figured out which one
the draft is using. I _think_ the draft "wants to mean" that the first
is a loop free path, and the second is a "downstream neighbor," but the
formula's don't seem to work this way.
However if you changed the cost of B-D to 51 THEN it becomes interesting. Case 1 is clearly a loop, but case 2 is not.
>> A-(25)-B-(25)-C
>> |      |
>> |     (50)
>> |      |
>> +-(25)-D
>> A-(25)-B-(25)-C
>>         |      |
>>        (50)    |
>>         |      |
>>         D-(75)-+

In neither case is D a downstream neighbor (since the cost from D to C is > cost from B to C).
But consider LFAs

we have an LFA if D_opt(N, D) < D_opt(N, S) + D_opt(S, D)

or in this case if  D_opt(D, C) < D_opt(D, B) + D_opt(B, C)

But in case 1 D is NOT an LFA

D_opt(D, C) = 75 (DABC)
D_opt(D, B) = 50 (DAB)
D_opt(B, C) = 25 (BC)

hence 75 !< 50 + 25

whereas in case 2 D IS an LFA

D_opt(D, C) = 75 (DC)
D_opt(D, B) = 51 (DB)
D_opt(B, C) = 25 (BC)

hence 75 < 51 + 25
I see, you're counting on the looped path showing up as an alternate
lower cost path to the neighbor itself. In other words, this could be
simlified to if there is a path with lower cost to the neighbor through
some other path, then the path through the neighbor must be a loop.
Then why not just say so? :-)

The draft is very confusing, and doesn't really explain the principle being used here very well at all....


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