Re: my broken TCP is faster on broken networks [Re: Very poor TCP/SACK performance]

Savochkin Andrey Vladimirovich (saw@msu.ru)
Fri, 11 Sep 1998 08:38:04 +0400


On Thu, Sep 10, 1998 at 10:24:59PM +0200, Andrea Arcangeli wrote:
> On Thu, 10 Sep 1998, Savochkin Andrey Vladimirovich wrote:
[...]
> >giving a very low probability of considering a traveling packet as lost.
> >As far as I remember the value of the timeout is set to
> >round trip time plus its mean deviation multiplied by 4.
>
> I don' t know what is mean deviation. Can you give me the whole formula so
> I' ll do the calc by hand and I' ll see the value calculated by the
> kernel, just to be sure?

The kernel calculates mean deviation as an approximation for
M | rtt - M rtt |,
where M denotes mean value. In other words it's the mean of
the absolute values of the difference between RTT of a single packet
and the mean RTT. The calculation is coded and supplied by good comments in
tcp_rtt_estimator() in tcp_input.c.

[...]
> Now I' ll only verify that the rto got calculated right from the kernel.

Well, I've done the check of the RTO evaluation. I added a debug output
if the kernel increases RTO in more than two or three times at once
(I don't remember) and didn't find anything unexpected.
The only strange place in the RTO evaluation code
was the initialization of SRTT and MDEV values.

Please report if your verification shows a bug in the code.

Regards
Andrey V.
Savochkin

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