Subject: Fragmentation in RFC1981

From: Giuseppe Federico Rossi <[email-masked]>
Date: Tue, 28 Dec 1999 16:11:19 +0100

Suppose there is an IPv6 host receiving an ICMPv6 "Packet Too Big"
message with a next-hop MTU that is less than the IPv6 minimum link MTU
(1280). According to RFC1981 the host must include the "Fragment Header"
in the successive datagrams. In this case I do not understand if:
- the fragmentation is performed by the sending host (as explained in
RFC2460)
  or
- the fragmentation is performed by the node generating the "Packet Too
Big" message (where the next MTU is < 1280).
In any case, does the fragmenting node generate IPv6 datagrams less than
1280 byte long ? If yes, what is the minimum (allowed) length for an
IPv6 fragment ?

Many thanks to all members of ipng community
Best Regards

Giuseppe

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From: Erik Nordmark <[email-masked]>
Date: Mon, 3 Jan 2000 09:56:20 -0800 (PST)

> 1280 byte long ? If yes, what is the minimum (allowed) length for an

The support for less than 1280 in the packet too big is there to make
life a lot easier for boxes that translate between IPv6 and IPv4.
Examples of such boxes/algorithms are specified in draft-ietf-ngtrans-siit-*
and draft-ietf-ngtrans-natpt-*.

In that case the fragmentation to less than 1280 bytes will occur on
the IPv4 side of the translator. The sole purpose of the sending IPv6
node including the fragment header is to have that node assign an
unique Identification value (in the fragment header) 
which will be carried end-to-end i.e. "translated" to the 16-bit
IPv4 ID value.

   Erik

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From: Erik Nordmark <[email-masked]>
Date: Mon, 3 Jan 2000 09:56:20 -0800 (PST)

Erik