From: Devender Singh (devender.singh@xxxxxxxxxxxxxx)
Date: Sun Aug 12 2001 - 00:29:13 GMT-3
I am sorry, we are talking about dlsw here( Normally we will terminate rif)
and ethernet does not have rif any way. O' well now I am really confused.
Devender Singh
BE(Hons), CCNP
IP Solution Specialist
-----Original Message-----
From: Devender Singh [mailto:devender.singh@cmc.cwo.net.au]
Sent: Sunday, 12 August 2001 12:17
To: ElleJf(Yahoo); ccielab@groupstudy.com
Subject: RE: lf in DLSw when token to ether
When a device ( say in token ring ) attempts to connect to its partner, it
sends explorer to locate the partner. The originator put maximum supported
frame size in the explorer by setting appropriate bits in RIF. As it
traverses diffrent bridges they set this value to their maximum supported
frame size.
So in your example it should work without any problems because R2 should set
bits as per its MTU. But I am not too sure of this, I would like someone to
confirm this.
Devender Singh
BE(Hons), CCNP
IP Solution Specialist
-----Original Message-----
From: ElleJf(Yahoo) [mailto:ellejf@yahoo.com]
Sent: Sunday, 12 August 2001 12:08
To: ccielab@groupstudy.com
Subject: lf in DLSw when token to ether
Is lf required to set lf when connect token to ether via DLSw?
For example, token MTU 4096-->R1--DLSw (TCP MTU 17800)--R2-->ether MTU 1500.
It mean R2 should segment frame to 1500 then send to ethernet. Will R2 do
it without problem? Or we "need" to set lf on TCP connection to 1500?
In RFC1795, "This should be used to ensure that the two end-stations always
negotiate a frame size to be used on a circuit that does not require the
Origin and Target DLSw partners to re-segment frames."
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