30 Jun
2004
30 Jun
'04
1:52 a.m.
Andrei Pelinescu-Onciul wrote:
[snip]
So all the packets comming from the same ip will be
sent to the same
fron end SER? (hashing after src ip)?
Yes, using ciscos "Sticky IP" which I admit, I do not know about, but
I'm told it will do this job properly.
Anyway there are some problems related to the nat
traversal:
1. nat ping - nat ping needs to access usrloc, so that it would know
which users to ping. However on your setup the front-end servers have no
ideea about this, so they wouldn't be able to nat ping. The "main"
server (User accounts) knows who to ping but its ping won't traverse a
symmetric nat (the nat will have an open binding only with the outbound
proxy, which would be one of the load balanced front-ends).
I do realize this now, so I'm considering running a non-persistent
usr_loc (no mysql back end) on all the front end servers, and using
t_replicate between all of them. I admit I have not verified if this is
possible, so please forgive me if I'm talking non-sense here at this
stage. My concern here, as I mentioned in my reply to Klaus's post, is
that if I use t_replicate will all my front end ser servers, will they
all spit udp at a single natted client when the client has only one udp
session with one front end server?
2. consider user A calling user B, where at least B is
behind a nat.
The invite would reach the "main" server which will look up B and will
try to send the message to B's address. Unfortunately B's nat will drop
the packet, because it has an open binding only between B and the load
balanced ip. (this will work only if B has a full cone nat which is very
very unlikely)
I'm not sure on the solution here. I will need to make the call go via
the front end ser server that has the active udp session with the
client. I'm going to sleep on this!
3. assuming the above stuff will work somehow, you
still have to be very
carefull to open only one rtp proxy session (since each front end has
its own rtp proxy you should make sure you use force_rtp_proxy on only
one of them, for the same call)
I agree, and I realize that I'm making some challenging issues for
myself :)
Thank you Andrei for your comments!
-Jev