On Thu, 25 Oct 2007, in the Usenet newsgroup comp.os.linux.networking, in
article <(E-Mail Removed) .com>,
Philippe Signoret wrote:
>> NOTE: Posting from groups.google.com (or some web-forums) dramatically
>> reduces the chance of your post being seen. Find a real news server.
>Which one can I use for free?
Some people have been using 'teranews.com' but this seems to be poorly
administered, and is subject to substantial delays (and may be in a lot
of killfiles as well - look at
http://www.teranews.com). Another used is
'aioe.org' with apparently better results (sorry - don't have a URL), and
still another is motzarella.org (again - no URL). I offer no opinions
either way. There is an alternative Usenet newsgroup "alt.free.newsservers"
and another "alt.usenet.news-server-comparison" you may want to look at,
but be well aware that they are infested with trolls.
>> What network? What is on this network?
>My home wireless network.
If this is _only_ your own network, then yes - this is to high. What I
would do would be to run a packet sniffer and see who is ARPing for who.
I've never bothered using Ethereal (now called Wireshark), as it puts
to MUCH information in the "User Friendly" display, which is quite useless
for me. If you have 'tcpdump' installed, a suitable command would be
/usr/sbin/tcpdump -n -i eth0 -x arp
(though you'll probably have to run that as root). The output will look
something like
20:36:01.250000 arp who-has 192.168.1.102 tell 192.168.1.17
0001 0800 0604 0001 0020 af57 d129 c0a8
0111 0000 0000 0000 c0a8 0166
The last four double-octets in the middle line (0020 af57 d129 c0a8)
is the MAC address of the source (00:20:AF:57

1:29) and the first
two octets of the IP address (c0a8 = 192.168). On the last line is
the other two octets of the source IP address (0111 = 1.17 which makes
the source address 192.168.1.17), the next three pairs are zeros,
because this is the desired information (the MAC address of 192.168.1.102)
and the last two pairs (c0a8 0166) are the IP address we are searching
for "who is 192.168.1.102").
20:36:01.260000 arp reply 192.168.1.102 is-at 08:0:20:c2:e3:14
0001 0800 0604 0002 0800 20c2 e314 c0a8
0166 0020 af57 d129 c0a8 0111
There is the reply. Note that in the second and third lines, the source
and destination MAC and IP addresses are swapped, because the reply is
coming from 192.168.1.102 at 08:0:20:c2:e3:14, and is being sent to
192.168.1.17 at 00:20:af:57:d1:29.
>>> Is this a normal ARP packet percentage? It seems a bit high to me.
>> Not enough information. The other question is what operating system
>> are the hosts running? That may also have impact on the traffic.
>Most hosts (5 of them) are running Windows XP, one Ubuntu Linux. Linux
>and two Windows XP are wired, others are wireless.
I don't use windoze, but windoze is EXTREMELY talkative, and wants to
talk to every address it's ever heard of. What you may be seeing is
windoze looking for hosts listed in shares.
Old guy