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Re: obsrvtaion on packet numbers

Roberto Peon <fenix@fb.com> Thu, 07 September 2017 21:45 UTC

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From: Roberto Peon <fenix@fb.com>
To: Jana Iyengar <jri@google.com>, Roni Even <roni.even@huawei.com>
CC: joel jaeggli <joelja@bogus.com>, QUIC WG <quic@ietf.org>, Martin Thomson <martin.thomson@gmail.com>
Subject: Re: obsrvtaion on packet numbers
Thread-Topic: obsrvtaion on packet numbers
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Date: Thu, 07 Sep 2017 21:45:24 +0000
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Concretely: 2^63 represents 11,690,840.9 years of packets at 25kpps, or 292 years at 1 Gpps.

Even if it was only 2^32, though, a new connection per day would be reasonable—many sites won’t want to maintain connection state for substantially more than a day (people update binaries/change config, need to shuffle load around to deal with load asymmetries, etc.).

-=R

From: QUIC <quic-bounces@ietf.org> on behalf of Jana Iyengar <jri@google.com>
Date: Thursday, September 7, 2017 at 2:36 PM
To: Roni Even <roni.even@huawei.com>
Cc: joel jaeggli <joelja@bogus.com>, QUIC WG <quic@ietf.org>, Martin Thomson <martin.thomson@gmail.com>
Subject: Re: obsrvtaion on packet numbers

Roni,

> 2^32 isn't half of 2^64, it's half of 2^33. 25kpps is 2 billion odd packets a day
> whereas 2^64 is 1.84 * 10^19.
[Roni Even]I was basing it on the text assuming that packet number does not start at 0  " The initial value for packet number MUST be selected from an uniform
   random distribution between 0 and 2^31-1 "

Even if your initial packet number is the largest possible, that is, 2^32, you still have > 2^63 numbers left.
2^64 = 2^63 + 2^63, which is the total number of possible packet numbers, and 2^63 + 2^32 is the number of packets you can send with this worst case initial packet number.

- jana

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