Subject: Re: [PATCH 1/1] tcp: Wrong timeout for SYN segments

From: Alex Bergmann <[email-masked]>
Date: Wed, 22 Aug 2012 11:29:28 +0200

On 08/22/2012 10:58 AM, Eric Dumazet wrote:
> On Wed, 2012-08-22 at 10:48 +0200, Alex Bergmann wrote:
>> On 08/22/2012 10:06 AM, Eric Dumazet wrote:
>>>> Prior to 9ad7c049 the timeout was defined with 189secs. Now we have only
>>>> a timeout of 63secs.
>>>>
>>>>           ((2 << 5) - 1) * 3 secs = 189 secs
>>>>           ((2 << 5) - 1) * 1 secs = 63 secs
>>>
>>> Strange maths ... here I have :
>>>
>>> (1+2+4+8+16) * 3 = 93 secs
>>> vs
>>> (1+2+4+8+16) * 1 = 31 secs
>>>
>>> So even before said commit, we were not rfc1122 compliant.
>>>
>>> Using 7 retries would give 127 seconds, still not rfc compliant.
>>
>> You're missing the timeout after the 5th SYN packet was sent. This
>> would result in another 32 seconds (96 seconds).
>>
>> The timeout is calculated here:
>>
>> net/ipv4/tcp_timer.c(146:150)
>>
>> 	if (boundary <= linear_backoff_thresh)
>> 		timeout = ((2 << boundary) - 1) * rto_base;
>> 	else
>> 		timeout = ((2 << linear_backoff_thresh) - 1) * rto_base +
>> 			(boundary - linear_backoff_thresh) * TCP_RTO_MAX;
>
> Thats the code yes but you miss the fact that last occurence of the
> timer doesnt send a frame on the _network_
>
> R2 is derived from the last frame sent.
>
> Fact that the connect() is a bit long to return to user space is not
> relevant. We could block the task for 2 hours and still be non RFC
> compliant.
>
> Actual 5 frames are sent, so the effective global timeout is the one I
> quoted.
>
> 1 + 2 + 4 + 8 + 16   and its 31
>
> Just do a tcpdump and you can see it.

Actual 6 SYN frames are sent. The initial one and 5 retries.

The kernel is waiting another 32 seconds for a SYN+ACK and then gives 
the ETIMEDOUT back to userspace.

Do you mean that we have to send another SYN packet after the 3 minutes?