Re: Custom Queueing and packet size

From: Alexei Monastyrnyi (alexeim@orcsoftware.com)
Date: Tue Oct 17 2006 - 04:46:33 ART


Hi.

Possible way to Rome as well, not very straight though. :-)

With these byte counts Q1 and Q3 will get in fact the same percentage
since both byte counts are less than average packet size. So only one
packet will be transmitted in every turn for Q1 and Q3, whereas 2
packets (or maybe 3) in average will be transmitted for Q2. This will
affect overall percentage layout.

Total bytes transmitted per average round will be
1500+1200+1500=4200
and real percentage 35% 30% 35%
or.
1500+1800+1500=4800
and real percentage 30% 40% 30%

As you can see, the real percentage slightly differs from desirable.

A.

Victor Cappuccio wrote:
> Very nice post Alexei,
>
> Ok, CQ does not remember how many bytes of packets it was send out in the
> first Round per each Q, and also CQ uses the byte-count to adjust the
> bandwidth allocation the determined queue.
>
> I think that 0,25 is ok for Queue 1, But I think that there is another
> solution for this and more simpler, I would love to read your opinion about
> this one.
>
> 1500 600 1500 => Sum == 3600
> 20 40 40 => Bandwidth Allocation per Queue
> 720 1440 1440 => Byte Count Ex: 720 came from 20 * 3600 / 100
> 0,2 0,4 0,4 => Check == (Sum ByteCount)/n
>
> Just another way to get to Rome
>
> Thanks it was very illustrative
>
> Saludos,
> Victor.-
>
> -----Mensaje original-----
> De: nobody@groupstudy.com [mailto:nobody@groupstudy.com] En nombre de Alexei
> Monastyrnyi
> Enviado el: Lunes, 16 de Octubre de 2006 06:40 p.m.
> Para: Michael Zuo
> CC: ccielab@groupstudy.com
> Asunto: Re: Custom Queueing and packet size
>
> Hi.
>
> For CQ in terms of packet sizes it seems to be enough understanding what
> the maximum packet size is.
>
> As per your example, assume that "the rest" is 1500 byte
>
> pkt_size 1500 600 1500
> %q 20%=0.2 etc 0.2 0.4 0.4
> ratio max_pkt/pkt 1 2,5 1
> norm [%q*ratio*10] 2 10 4
> byte_count ptk_s*norm 3000 6000 6000
> check byte_c/summ 3/15 6/15 6/15
> 0.2 0.4 0.4
>
> It matches your ratio pretty much. Ratio 1500 3000 3000 differs in delay
> only, meaning that packets in Q1 (the one with 1500) will wait
> serialization time for 3000+3000 bytes (Q2+Q3). In my case Q1 would wait
> serialization time for 6000+6000 bytes before getting served again.
>
> But it is just a part of the fun. :-)
>
> Let's have "the rest" as z in calculations above.
>
> pkt_size 1500 600 z
> %q 20%=0.2 etc 0.2 0.4 0.4
> ratio max_pkt/pkt 1 2,5 1500/z
> norm [%q*ratio*10] 2 10 6000/z
> byte_count ptk_s*norm 3000 6000 z*(6000/z)=6000 !
> check byte_c/summ 3/15 6/15 6/15
> 0.2 0.4 0.4
>
> Getting into more abstract, assume we have x,y,z for packet sizes.
> max = maximum{x,y,z}
>
> pkt_size x y z
> %q 20%=0.2 etc 0.2 0.4 0.4
> ratio max_pkt/pkt max/x max/y max/z
> norm [%q*ratio*10] 2*(max/x) 4*(max/y) 4*(max/z)
> byte_count ptk_s*norm x*2*(max/x) y*4*(max/y) z*4*(max/z)
> 2*max 4*max 4*max
> check byte_c/summ x/10max y/10max z/10max
>
> So, byte count for Qn is [%qn*10*max]
>
> [] is round-up to the nearest integer.
>
> Much easier than classic books offer. :-)
>
> HTH
> A.
>
>
> Michael Zuo wrote:
>
>> Hi Group,
>>
>>
>>
>> I have a question re: custom queueing. When calculating the byte-count
>> for each queue, does the packet size in each queue affect the
>> calculation?
>>
>>
>>
>> Example:
>>
>>
>>
>> I want to allocation 20% to 1500 byte packets, 40% to 600 byte packets
>> and 40% to the rest. To me, the byte-count would be 1500, 3000, 3000
>> respectively just from normalizing the percentage allocation. But I
>> have read from NetMaster library that the packet size for each queue is
>> part of the calculation and there are various steps that need to be
>> performed to arrive at an approximate percentage that comes close to the
>> desired percentage.
>>
>>
>>
>> Can anyone shed some light on this?
>>
>>
>>
>> Thanks in advance...
>>
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>
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