[asterisk-users] PRI TBCT - Practical Experience, Anybody?

[Jay R. Ashworth]

On Fri, Aug 15, 2008 at 11:56:30PM -0500, Tilghman Lesher wrote:
> > There's no hairpin involved: the point of TBCT is that you tie up *0*
> > timeslots instead of 2, to forward a call.
> 
> There is a hairpin involved.  The call (for several milliseconds at least) is
> using two channels on the PRI before the 2BCT succeeds, and then the call
> no longer takes up any channels.  It is only when the PRI detects the hairpin,
> through the native bridge code that it is able to detect that the call is
> eligible for 2BCT.

I'll assume you've watched it on a PRI, so I'll defer, but I wouldn't
expect that myself; I would expect that when you tell the switch to
transfer it, you go immediately from one B channel to 0.

> > Why would an Asterisk instance call itself on the same span?
> 
> Very simple.  Call your main number, and if you don't have special logic
> in your internal dialplan context to handle that, the call will go out to the
> telco and dutifully come right back in on the same circuit.

Sure, but that's not a target for TBCT anyway.

> > > Similarly, Asterisk cannot complete a 2BCT request, if Asterisk is on the
> > > NET side of the PRI circuit.  That might could be added in the future,
> > > but it is not supported now.
> > >
> > > So in summary, Asterisk can request 2BCT, but it cannot perform a 2BCT if
> > > requested from the other side.
> >
> > Nothing can perform a TBCT unless it's a PRI server, not a client; it's
> > function of 5ESS's and DMSen; you have to be an SS7 speaker to do it in
> > the first case.
> 
> I don't think that's the case.  Matt would know more, and his reply suggests
> that it certainly would be possible for Asterisk to do this.  SS7 is not at
> all required here.

Allow me to phrase it differently:

TBCT is a feature of LEC/IXC edge switches; there isn't much use for it
in any other context.  I don't care if you're using Asterisk to be an
edge switch, but it's a *carrier* feature, by and large.

Certainly in the specific instance I'm discussing, it is.

Cheers,
-- jra
-- 
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