Stewart

I agree that for finding times between routers this is a good approach.

However, you only need this approach when there is asymmetry.
Between two directly connected routers there will not usually be asymmetry,
and simpler approaches are possible (e.g., BFD echo mode).

The exceptions would be routers connected over one-way SONET rings,
routers connected over a complex (not necessarily symmetric) L2 network,
or routers connected with different rates in the two directions.

Y(J)S

-----Original Message-----
From: Stewart Bryant [mailto:[email protected]] 
Sent: Monday, January 25, 2010 19:03
To: Yaakov Stein
Cc: David Mills; [email protected]; [email protected]; Mike Shand
Subject: Re: [TICTOC] interesting article on a global mechanism for one-way 
delay measurement

Yaakov Stein wrote:
> Stewart
>
>
> What you did was to discretize the area of possible solutions
> by looking for them using a brute force method, and to average them.
>
>   

Yaakov, the reason that I took this approach was a practical one - there 
are a lot of triangular loops in a network, particularly and enterprise 
network, so with some very simple network technology we can determine 
the asymmetry in a large number of sub-networks within the network as a 
whole. So we can mechanistically determine these paths and probe them.

I also not that in the second example the information gathered at node 1 
is essentially the next neighbour delays for all nodes plus the three 
shortest indirect paths back to 1 via the next hop, which could be 
approached mechanistically.

In general what you need is the time on your own SPF tree, not the whole 
set of times that exist in the network. So it would be logical for 1 to 
work with the loops 1-2-3-5-6 and 1-10-9-8-6 and for 1 to gather the 
2-4-7-9 times from the measurements taken by one of those nodes.

Also note that if 4-7 did not exist and the clock was at 4 of 7 all 
nodes would have to make the standard symmetry assumption for the 4-2 
and the 9-7 hops.

- Stewart


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