From: Nadeem Ansari (nadeem.ansari574@gmail.com)
Date: Thu Feb 26 2009 - 06:20:59 ARST
Hi Dale,
Continuing to my previous reply I think Wendel Odom is refering 1/16 value
for increased in shaping rate and (Bc+Be)/16 value for increase in tokens
every Tc, so there is two things which increases one is shaping rate by 1/16
and other is numbers of token replenished (Bc+Be)/16, correct me if I am
wrong, And I think we should thank to Wendell Odom for his in depth
explaination
Regards
Nadeem
On Thu, Feb 26, 2009 at 1:36 PM, Nadeem Ansari
<nadeem.ansari574@gmail.com>wrote:
> Hi Dale,
>
> Cisco Doc simply says that rate is exceeded to CIR whenever Queue depth
> decreased from its max depth value with no BECN received, Also I did not
> find any information about consecutive intervals what is that ?
>
> Regards
> Nadeem
> On Thu, Feb 26, 2009 at 12:02 PM, Dale Shaw <dale.shaw@gmail.com> wrote:
>
>> Hi all,
>>
>> Regarding 'shape adaptive' ..
>>
>> I've read the docs on this and it seems pretty straight-forward, but
>> something I read in Wendell Odom's QoS book is a bit confusing.
>>
>> When talking about the way adaptive shaping brings the actual shaping
>> rate back up to the configured shaping rate, he says:
>>
>> "As a reminder, after 16 consecutive intervals with no received BECNs,
>> this router would start increasing the Shaping rate by 1/16 of the
>> original 96-kbps shaping rate teach Tc. To do so, CB Shaping actually
>> starts replenishing the token bucket with a little more each Tc,
>> specifically by (Bc + Be)/16 each Tc. As a result, the rate will
>> likely increase more slowly than it decreased in the presence of
>> BECNs."
>>
>> It's the second sentence that's got me stuck. Does adaptive shaping
>> cause the actual shaping rate to go up by 1/16 of the original rate
>> each Tc, or by the amount specified in the formula (Bc + Be)/16 each
>> Tc? Because they are not necessarily the same.
>>
>> Example:
>>
>> Configured shaping rate: 4,000,000bps (4Mbps)
>> Tc: 25ms (default)
>> Bc: 100,000 bits (derived from Tc)
>> Be: 100,000 bits (same as Bc)
>>
>> 1/16 of 4,000,000 = 250,000, implying that the actual shaping rate
>> increased by 250Kbps every 25ms.
>>
>> The second sentence implies that it would actually increase by 12,500
>> (12.5Kbps) every 25ms.
>>
>> Which is it? Or am I just confused :-)
>>
>> cheers,
>> Dale
>>
>>
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