From: Swaroop Potdar (swarooppotdar@hotmail.com)
Date: Tue Sep 21 2004 - 14:02:00 GMT-3
HI Venkat,
10 MS is the ideal serialization delay on a frame-relay segment to clock
data.
So the calculation goes like this.
if the cir is 100 kbps it means in 10 ms it will be clocking 1 kb
and in 100 cycles of 10ms interval each the whole 256 kb data is clocked
out.
so in the above analogy teh fragment size would be 125 (* bytes)
fragment value is always in bytes..
So if you calculate
in your case for 256 kbps cir..
2560 bits of data is clocked per 10 ms
and 256000 is sent in 1 second
so ideally if you divide 2560 / 8 to get teh bytes value its 320 bytes.
HTH
Swaroop.
>From: venkat garigipati <garigipati@yahoo.com>
>Reply-To: venkat garigipati <garigipati@yahoo.com>
>To: ccielab@groupstudy.com
>Subject: frame-realy fragment Date: Tue, 21 Sep 2004 08:18:58 -0700 (PDT)
>
>hi,
>
>Referring to QoS section of Lab 10 in IPExpert. Can some one help me how
>the "frame-relay fragment" works ?. The solution states it as "320", given
>CIR 256kbps and Tc as 10ms.
>
>When I left it to default, my BGP network goes down, but when I configure
>it to 320, everything works..I couldnt get to the right documentation on
>how "320" is derived..
>
>thx//venkat
>
>
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Regards,
Swaroop.
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