From: Chris Lewis (chrlewiscsco@yahoo.com)
Date: Thu Oct 27 2005 - 09:35:21 GMT-3
JC, I think the equation for Bc and statement of default Tc value are correct.
Selias, I do not see what you reference in the solutions guide for lab 1, at least not in the volume 1 workbook, so without the context of the question, it is not possible to answer your question completely. Having said that, you are unlikely to get more information on this than is already supplied in the solutions guide for lab 1, it really is a good explanation of why the values are chosen.
Chris
Jaycee Cockburn - BCX SS <Jaycee.Cockburn@bcx.co.za> wrote:
Hi S,
From what I understand, and help if I'm wrong:
CIR = commited rate in kbps
Bc = commited burst per time interval.
So to get bc, divide cir by 8. Why, because tc is normally 125msec, and
the formula to get all this is
Bc = Cir x Tc
(Which is why I believe your CIR value is wrong, should be 192000)
In effect it means you van send 24k of data per time period (Tc). So in
8 time periods you can send 24k x 8 = 192k which is your CIR...
HTH, in a funny sort of way...
Chris Lewis can probably explain in a lot better/correct...
Cheers
JC
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
selias@uaeu.ac.ae
Sent: 27 October 2005 11:27 AM
To: ccielab@groupstudy.com
Subject: Cir and Bc Task9.1-9.9 Lab1
In this Lab we have this configuration :
map-class frame-relay REMAINING_BW
frame-relay cir 19200
frame-relay bc 24000
How did he find the value for bc ?
He said in the Breakdown:
This means that we need to subtract the CIR from the port speed to
determine what value to shape the remainig trafiic to.
the port speed is 512 kbps,,,could you please some expalin this
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