[Tmrg] Queue size - Towards a Common TCP Evaluation Suite

[krasnoj at gmx.at (Stefan Hirschmann)]

Stefan Hirschmann wrote:
> Greeting Andrew and all other readers,
> 
>> Lachlan Andrew wrote:
>>> 2008/7/30 Stefan Hirschmann <krasnoj at gmx.at>:
> 
>> Greetings Stefan,
>>
>> Thanks for your interest in the test suite.  I apologise for the long
>> delay in getting back to you.
> 
> I apologize for this long delay too. But it was not easy to find anyone
> with a 56K POTS modem still in use.
> 
> 
>>> In the "Common TCP Evaluation Suite draft-irtf-tmrg-tests-00" there is the section:
>>> "3.2. Delay/throughput tradeoff as function of queue size"
>>> describing the buffer sizes of the routers, but only for the access link scenario.
>>>
>>> I wanted to extend the values to the other scenarios and noticed a problem with it.
>>> The BDP of the Dial-Up Link scenario is 64Kbps * 0.1 s / 8 = 0.8 KByte -> 0.8 / 1.5 = 0,53 packets.
>>>
>>> So even if I use the BDP the value is much too small. A rounding to one is IMHO also not realistic. What value should be used as a minimum buffer size and why?
>> The Dial-Up scenario is there partly for POTS modems, and partly for
>> GPRS.  You should find out the buffer size used by either one of those
>> (and then it would be great to post it to the list!).
>>
>> If you have access to a dial-up connection, you could try to measure
>> the buffer size:  Ping the next-hop node with an idle link, and then
>> while downloading something large.  The difference in RTTs will give a
>> good estimate of the buffer size.
> 
> 
> OK I have done it. The test were made:
> DATE: 2008/10/01 around 19:30
> Used 56K POTS Provider: Tele2 Austria
> Operating System: Windows XP Media Centre Edition
> Large background traffic: A linux kernel image from ftp://ftp2.kernel.org
> 
> The exact testprotocol is at the end of the email.
> The most important datas are:
> uncongested:
> Minimum = 134ms, Maximum = 148ms, Mean = 141ms
> 
> congested:
>  Minimum = 5963ms, Maximum = 8541ms, Mittelwert = 7177ms
> 
> The correct formula should be:
> max(queuing time) = max(congested) - min(uncongested)
> 8407 ms = 8541 ms - 134 ms
> 
> 56 KBit/s is 7 KByte/s. 6 KByte/s is a realistic value for the real
> usable value. In this case:
> time * bandwidth = amount of data
> 8,541 s * 6 KByte/s = 51,246 KByte
> 
> If you say, that the packetsize is 1,5 KByte than:
> 51,246 KByte / 1.5 KByte =  34,164
> 
> So 35 is the Queuesize in packets.
> 
> 
> 
> Cheers Stefan
> 
> 
> 
> Now the complete console log (was a German Windows vesion):
> ===============================================================================================
> Microsoft Windows XP [Version 5.1.2600]
> (C) Copyright 1985-2001 Microsoft Corp.
> 
> C:\Dokumente und Einstellungen\Leo>ping www.google.at
> 
> Ping www.l.google.com [209.85.129.147] mit 32 Bytes Daten:
> 
> Antwort von 209.85.129.147: Bytes=32 Zeit=148ms TTL=244
> Antwort von 209.85.129.147: Bytes=32 Zeit=146ms TTL=244
> Antwort von 209.85.129.147: Bytes=32 Zeit=136ms TTL=244
> Antwort von 209.85.129.147: Bytes=32 Zeit=134ms TTL=244
> 
> Ping-Statistik f?r 209.85.129.147:
>     Pakete: Gesendet = 4, Empfangen = 4, Verloren = 0 (0% Verlust),
> Ca. Zeitangaben in Millisek.:
>     Minimum = 134ms, Maximum = 148ms, Mittelwert = 141ms
> 
> 
> C:\Dokumente und Einstellungen\Leo>ping -w 9999 www.google.at
> 
> Ping www.l.google.com [209.85.129.104] mit 32 Bytes Daten:
> 
> Antwort von 209.85.129.104: Bytes=32 Zeit=7027ms TTL=244
> Zeit?berschreitung der Anforderung.
> Antwort von 209.85.129.104: Bytes=32 Zeit=8541ms TTL=244
> Antwort von 209.85.129.104: Bytes=32 Zeit=5963ms TTL=244
> 
> Ping-Statistik f?r 209.85.129.104:
>     Pakete: Gesendet = 4, Empfangen = 3, Verloren = 1 (25% Verlust),
> Ca. Zeitangaben in Millisek.:
>     Minimum = 5963ms, Maximum = 8541ms, Mittelwert = 7177ms
> 
> C:\Dokumente und Einstellungen\Leo>
>