[asterisk-users] G729 License to Bridge calls through VOIP provider?

[Marco Mouta]

Erik,

What a great and detailled explanation! Thank you very much!

Ps. If you know anything about legal issues asked abouta g729 please
post it here:)

Best regards,
Marco Mouta

On 7/26/06, Erik <erik at infopact.nl> wrote:
> Marco Mouta wrote:
>
> > By the way could any one tell me wich is the Bandwith with IP over
> > head for this codec. about 8kb/s?
> >
> Let's do some calculations on that:
>
> g729a 20ms results in 20 bytes RTP payload in each packet, in order to traverse the OSI model there's some headers that need to be added,
> as RTP gets an RTP header, the RTP packet gets an UDP header and the UDP datagram gets an IP header:
> So add a 12 byte RTP header, 8 byte UDP header and a 20 Byte IP header
>
> This results in:
> 20 byte RTP payload + 12 byte RTP header + 8 byte UDP header + 20 byte IP header=60 byte on the ip layer.
> Thats a 40 byte overhead (so 2/3 of the packet is just headers :) and were still only on the IP layer now)
>
> So to transmit just 1 RTP packet you are actualy transmitting 60 bytes on the IP layer, so in order to get the real used bandwidth we
> need to knowhow many packets we are sending and on which medium (DSL/ethernet/slip/smokesignals):
>
> 20 ms results in 50 packets/s so: 50 packet/s *60 bytes/packet=3000 bytes/s
> that's 300*8=24000 bit/s total bandwidth on the IP layer so the overhead is 24000-8000=16000 bit/s.
>
> The fun starts if you are going to send this over DSL, let's continue the calculation:
>
> 50 packets of 60 byte IP, add the 2 byte PPPoA header for DSL= 62 bytes per packet.
> However, DSL operates with 53 bytes ATM cells, in which you can fit 48 bytes payload (and a 5 byte header) so in order to transmit the 62 bytes of
> data you need: 62/48=2 ATM cells.
>
> Why 2 cells you say? Because ATM can't utilize the unused part of cells, so to transmit 62 bytes you use the same amount of bandwith (on dsl) as you
> would use to transmit 96 (48*2) bytes.
>
> So 1 RTP packet uses 96 bytes on the DSL line, as you already know we have 50 packets/s so that's 50 packets/s*2 cells=100 Cells/s
> 100 cells/s * 53 byte = 53000 bytes/s on the DSL line thats 424000 bits/s to transmit a 8 kbit/s stream :)
>
> So the total overhead is 424000-8000=416000 bit/s overhead.
>
>
> If you would use G723 with a 10 ms frequency it gets even worse :)
> G723 on 10 ms produces 8 byte RTP payload per packet, so with headers that's 48 bytes on the IP layer, but now were sending 100 packets/s
> so: 48*100=4800 bytes/s --> 38400 bit/s on the ip layer
> On DSL this would result in 50 byte packets (pppoa header) with won't fit in 1 cell, so you would use 2 cells for each IP packet.
> 100 packets/s * 2 cells = 200 cells/s
> that's 200 cells * 53 bytes/cell = 10600 bytes/s on the DSL line
> 10600*8=84800 bit/s to transmit a 6400 bit/s stream --> 78400 bit/s overhead
>
> If you would use G723 with 20 ms (16 byte RTP payload) you only have 42400 at the DSL layer, so by adjusting the sample frequency you could cut the
> overhead in half :)
>
>
> Erik Versaevel
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-- 
Com os melhores cumprimentos,

Marco Mouta