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

[quetchen at caltech.edu (Tom Quetchenbach)]

My ISP gives me dial-up access as a backup to my DSL, so I'll try to
play around with it at some point.

-Tom

Lachlan Andrew wrote:
> 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>
>>
> 
> 
> 

-- 
/* Tom Quetchenbach
 * WAN-in-Lab / Netlab, Dept of Computer Science, Caltech
 * 1200 E California Blvd, MC 256-80, Pasadena CA 91125
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 */