From: Edison Ortiz (edisonmortiz@gmail.com)
Date: Sun Jan 14 2007 - 13:14:33 ART
#1
128k or 384k now becomes your AR and the Tc is given for the circuit on that
case.
Be = (128-64)x32/1000
Be = (384-64)x64/1000
Forget about the 512k AR, it's there to confuse you.
#2
Other signs to look for, if you have voice on the circuit. Voice latency
needs the minimum
Tc value which is 10ms. Then you need to fragmented, example:
Bc = (CIR*Tc)/1000
Bc = (512000*10)/1000 which is Bc 5120 (640 bytes) - passes fragmentation
rule
Whereas if you have
Bc = (512000*32)/1000 which is Bc 16384 (2048 bytes) - fails fragmentation
rule and this
packet must be fragmented.
----- Original Message -----
From: "prashant shukla" <shukla_cisco@yahoo.co.in>
To: <ccielab@groupstudy.com>
Sent: Sunday, January 14, 2007 2:31 AM
Subject: IEWB--- Be calculations.
> Gurus,
>
> This is what i understand and need clarification on.
> AR= 512Kbps ; CIR = 64Kbps.
>
> Bc= CIR x Tc / 1000
> Be= (AR-CIR) x Tc / 1000
>
> #1. So im able to the above calculation if the question mentions " allow
> DLCi 103 to burst upto port speed " then the Be formula works with
> (512-64)kbps x TC/1000, my doubt comes when the question says, " Allow the
> DLCI to burst upto 128 /384 Kbps for 32ms or 64ms etc etc." here how to
> look for the Be, as the burst is not upto port speed, so im struggling to
> find a "Universal rule".
>
> #2.Secondly; in questions where its not mentioned any Tc value I would
> like to
> go for 125ms as default, what other "Signs" i should look for where the
> Tc value can be influenced, e.g. In one IEWB it mentions " The packet
> above 960 bytes should be fragmented". Is this a catch!! as the the Tc
> chosen made .
>
> hope i was clear enough... :-)
>
> Shukla.
>
>
>
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