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

lachlan.andrew at gmail.com (Lachlan Andrew) Wed, 01 October 2008 18:42 UTC

From: "lachlan.andrew at gmail.com"
Date: Thu, 02 Oct 2008 04:42:16 +1000
Subject: [Tmrg] Queue size - Towards a Common TCP Evaluation Suite
In-Reply-To: <48E3C1F5.40906@gmx.at>
References: <20080730103251.299310@gmx.net> <aa7d2c6d0808311521l2fb03761l350017c02548382a@mail.gmail.com> <48E3C1F5.40906@gmx.at>
Message-ID: <aa7d2c6d0810011142q15db75fct588b62f467365141@mail.gmail.com>

Thanks Stefan.

Those numbers are interesting.  I'm surprised that there was 8s delay
when congested.  I'm wondering if  ping  packets are treated
differently.  (Many systems give ICMP packets lower priority.) Still,
35 packets sounds a reasonable buffer size.

Does anyone else on the list have any data to support or contradict
this?  My parents-in-law use dial-up, so I'll try to check their
connection soon.

Cheers,
Lachlan

2008/10/2 Stefan Hirschmann <krasnoj at gmx.at>:
> 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>
>



-- 
Lachlan Andrew  Dept of Computer Science, Caltech
1200 E California Blvd, Mail Code 256-80, Pasadena CA 91125, USA
Ph: +1 (626) 395-8820    Fax: +1 (626) 568-3603
http://netlab.caltech.edu/lachlan