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Re: dlsw - backup peer / cost question .. posted 03/08/2004
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Hi,

   First of all, there are two kinds of connection will be setup between the
DLSW peers. One is the control connection between two peers (hereafter I
will refert it as peer connection), and one are the circuits formed when
there are real user traffics across the two peers (hereafter I will refer it
as circuits). Circuits will only be setup across two peers which have peer
connections.

   The first solution will only form peer connections between R1 and R2, but
not R1 and R3. However, if R2 is unreachable, DLSW will form peer
connections between R1 and R3. (Which is the solution I will pick as the
answer of the question. I think it depends on how you interpret the word
"use the path". To me, the question is asking to make DLSW use the path, and
it generally means peer connection. If the question explicitly states
circuit/user traffic should use the path, then I will pick either solution 2
or 3).

   The second and third solution are basically the same. However, the peer
cost states in remote-peer command will override the cost advertised by the
remote-peer (i.e. the cost in the local-peer command of the remote peer).
Moreover, in these two cases, both peer connection will formed between R1
and R2, R1 and R3. However, circuits only will be formed using the lowest
cost peer connection (i.e R1 and R2 in this example), unless you use the
load balance.


http://www.cisco.com/en/US/products/sw/iworksw/ps2474/prod_technical_reference09186a008007ce40.html

Best Regards,
William Chen

----- Original Message ----- 
From: "Ramasubramanian Sethuraman" <snrmanian@xxxxxxxxxxx>
To: <ccielab@xxxxxxxxxxxxxx>
Sent: Monday, March 08, 2004 3:20 AM
Subject: dlsw - backup peer / cost question ..


> Hi,
>
> If the question is
>
> R1(192.168.4.4) should use R2 (192.168.5.5) when its available.
> When R2 is unreachable, R1 should use the path thru R3 (192.168.6.6)
>
> Pls let me know if either of the foll 3 solutions can be given.
>
> Solution 1)
> -----------------
>
> R1
> ---
> dlsw local-peer peer-id 192.168.4.4
> dlsw remote-peer 0 tcp 192.168.5.5
> dlsw remote-peer 0 tcp 192.168.6.6 backup-peer 192.168.5.5
>
> R2
> ---
> dlsw local-peer peer-id 192.168.5.5 prom
>
> R3
> ----
> dlsw local-peer peer-id 192.168.6.6 prom
>
>
> Solution 2)
> ----------------
>
> R1
> ---
> dlsw local-peer peer-id 192.168.4.4
> dlsw remote-peer 0 tcp 192.168.5.5 cost 1
> dlsw remote-peer 0 tcp 192.168.6.6 cost 2
>
> R2
> ----
> dlsw local-peer peer-id 192.168.5.5 prom
>
> R3
> ----
> dlsw local-peer peer-id 192.168.6.6 prom
>
>
> Solution 3)
> --------------
> R1
> ---
> dlsw local-peer peer-id 192.168.4.4
> dlsw remote-peer 0 tcp 192.168.5.5
> dlsw remote-peer 0 tcp 192.168.6.6
>
> R2
> ----
> dlsw local-peer peer-id 192.168.5.5 prom cost 1
>
> R3
> ----
> dlsw local-peer peer-id 192.168.6.6 prom cost 2
>
>
> One more question, if i give the cost for R5 locally(in local-peer
> statement) as  well
> in R2 (using the remote-peer statement for R5), which cost will be used
for
> R5 by R2 ?
>
> thanks,
> subbu
>
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