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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