OK, thanks for the clarification.
However, in the meantime I also conducted another search, and I found two
sources -
1. IEEE 802.1Q (2003), see pg 39 (actual page, including all prefaces etc),
look near the end (specifically at the sentence "Include additional octets
before the FCS field in order for the transmitted frame length for such
frames to be 68 octets. This results in a minimum tagged frame length of 68
octets.")
(the file can be found here: ftp://ftp.ecitele.com/notes/IEEE802.1-Q.-2003
_(VLAN).pdf)
2. According to CISCO - 68 bytes (
http://www.cisco.com/en/US/tech/tk389/tk390/technologies_tech_note09186a0080094665.shtml#frame2
)
Now, my problem is that I got a bet with R&D guys around here, and it seems
my sources say that it's 68 bytes, and you're saying it's 64... or have I
misunderstood the IEEE doc??
TnX
Jaap Keuter
<jaap.keuter@xs4a
ll.nl> To
Sent by: Ethereal user support
ethereal-users-bo <ethereal-users@xxxxxxxxxxxx>
unces@xxxxxxxxxxx cc
m
Subject
Re: [Ethereal-users] (Slightly OT)
18/01/2006 13:34 Ethernet length question
Please respond to
Ethereal user
support
<ethereal-users@e
thereal.com>
Hi,
What you should decouple is the minumum 64 octet Ethernet requirement
from the actual use of these octets.
The 802.1Q spec only tells you that if the Ethernet II header contains a
type value of 0x8100 the next two octets are to be interpreted as VLAN
tag. That defines use.
When a frame is present on the wire it needs to be minimum 64 octets,
whatever the use of these octets.
These are seperate things.
Hope it helps,
Jaap
On Wed, 18 Jan 2006 Ran.Shenhar@xxxxxxxxxxx wrote:
> TnX Jaap,
> I know the CSMA/CD principles behind the 64 byte minimum requirement.
> However, my question is really focused on the 802.1Q change (or no
change)
> in that demand, and I didn't understand the answer for that issue from
your
> answer.
> Could you please try to clarify?
>
> TnX
>
>
>
> Jaap Keuter
> <jaap.keuter@xs4a
> ll.nl>
To
> Sent by: Ethereal user support
> ethereal-users-bo <ethereal-users@xxxxxxxxxxxx>
> unces@xxxxxxxxxxx
cc
> m
>
Subject
> Re: [Ethereal-users] (Slightly OT)
> 18/01/2006 11:54 Ethernet length question
>
>
> Please respond to
> Ethereal user
> support
> <ethereal-users@e
> thereal.com>
>
>
>
>
>
>
> Hi,
>
> The layer protocol model is the best reference you can take in mind.
> The minimum ethernet frame length (that is bits on the wire) is 64*8.
> Together with the wirespeed this gives a minimum amount of transmission
> time. This time is needed for the other endpoint (in a point-to-point
> link) or other stations (on a shared medium) to assure recognition of the
> frame and collision avoidance. All this is Physical layer stuff.
>
> Then comes the meaning of the bits in the frame. If the ethernet header
> says that it's a VLAN tag that is following then the next 2 octets are to
> be interpreted that way. From there on the next protocol layer starts.
>
> That is the true meaning of VLAN, a VIRTUAL LAN. It's not a real one, as
> seen on the wires, but on a layer above that.
>
> Hope it helps,
> Jaap
>
>
> On Wed, 18 Jan 2006 Ran.Shenhar@xxxxxxxxxxx wrote:
>
> > Hi All,
> > I need help with a non-ethereal specific - I know that an ethernet
frame
> is
> > of minimum length = 64 bytes, including FCS.
> > However, what is the minimum length of a VLAN tagged ethernet frame? Is
> it
> > 64+4, or is it still 64?
> >
> > I tried looking in IEEE 802.3q 2003 edition, however I was not able to
> find
> > a definite answer.
> > If you could also point me to a reference about the answer, it'd be
even
> > better.
> >
>
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