Message-Id: <200211190816.gAJ8Gf217919@masaka.cs.ohiou.edu> From: ronny.tittoto@exgate.tek.com Subject: RE: Does tcptrace measure RTT for every ACK? Date: Tue, 19 Nov 2002 09:08:07 +0100
a good article could be found here
http://citeseer.nj.nec.com/jiang02passive.html
Ronny
> -----Original Message-----
> From: uaca@alumni.uv.es [mailto:uaca@alumni.uv.es]
> Sent: lunedì 18 novembre 2002 20.34
> To: TCPTRACE Mailing List
> Cc: Rogelio Montañana; StasK@narus.com
> Subject: Re: Does tcptrace measure RTT for every ACK?
>
>
> On Sat, Nov 16, 2002 at 09:10:09PM -0800, Stas Khirman wrote:
> > Gentlemen,
> >
> > Did you considered to use three-way handshake as a good
> estimation for RTT?
>
> Yes:
>
> What do you think about the following?
>
>
> suposse the following scenario:
>
> Monitor
> [Host A]------------------------------------[Host B]
>
> SYN
> t0 ····················t1···················t2
>
> SYN+ACK
> t5 ····················t4···················t3
>
> ACK
> t6 ····················t7···················t8
>
>
> then
>
> t4-t1= two way delay from monitor to Host B
> t7-t4= two way delay from monitor to Host A
>
> so when we have the following situation
>
> Monitor
> [Host A]------------------------------------[Host B]
>
> A sends packet i
> ))))))))))))))))))))))))>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>
> B acks packet i
> ((((((((((((((((((((((((<<<<<<<<<<<<<<<<<<<<<<<<<<<<
>
>
> tcptrace actualy meassures rtt as
>
> sum('>') +sum('<')
>
> in this case making the above calculations we do instead
>
> sum('>') +sum('<') +(t7 -t4)
>
>
> Note: this makes tha asumtion that t3-t2 and t6-t5 is near zero
>
> The same can be done in the other way
>
> This is based on a white paper I read from packeteer.com
>
> I have just made a patch for tcptrace that does this wich is at
>
> http://despistao.cjb.net:8000/tcptrace/symetric_rtt.diff
>
> Please note that the code is very untested
>
> any comment will be greatly apprecited
>
> regards
>
> Ulisses
>
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