From: Lee Donald (Lee.Donald@t-systems.co.uk)
Date: Tue Apr 12 2005 - 05:02:56 GMT-3
It's ok, I had an extra 0.
My brains had enough..............
-----Original Message-----
From: Lee Donald
Sent: 12 April 2005 08:49
To: simon hart; Lee Donald; ccielab@groupstudy.com
Subject: RE: Frame relay traffic shaping ?
Simon,
160000 ( 16k) divided by 125 = 1280 ? How do you get 2000 ?
Regards
Lee.
-----Original Message-----
From: simon hart [mailto:simon.hart@btinternet.com]
Sent: 11 April 2005 21:53
To: Lee Donald; ccielab@groupstudy.com
Subject: RE: Frame relay traffic shaping ?
Hi Lee,
I guess from you question you know how to determine bc, then be is a fairly
simple step.
However you need to know what the Tc is for the shaped rate. This could be
explicitly defined or you maybe asked to use the default value.
If asked to use the default value then apply a map-class to the frame port
with just the cir configured and enable traffic shaping.
Then go to show traffic shaping, you will now see what the Tc value is. For
a cir of 32k you should find that this defaults to 125ms.
Therefore bc = cir*tc = 4000 bits
Now you wish to add a burst up to 48k
48k burst minus 32k cir = 16k
Therefore be = 16k/125ms = 2000
Your be value is 2000 and your bc value is 4000
HTH
Simon
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
Lee Donald
Sent: 11 April 2005 21:28
To: ccielab@groupstudy.com
Subject: Frame relay traffic shaping ?
Hi,
Can anyone give me some pointers on frame-relay traffic shaping.
Specifically cir, bc, be.
I know roughly the cir and bc values when asked to configure parameters but
it is calculating the be that I can't get my head around.
For instance, I have a pvc that has a cir of 32k, the router is allowed to
burst up to 75% of the AR 64k. I know that is 48k but what should the "be"
value be and how do I work it out?
Thanks in advance.
Regards
Lee.
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