I don’t understand
If is “each of 2 participants receives 2 streams” then Server outbound bandwidth must be : 176Kbit/s (88 Kbit/s x 2) ……….So the result shown on the site is wrong. But I think this: SERVER INBOUND : 44Kbit/s for each partecipant => 44Kbit/s * 2 => 88Kbit/s SERVER OUTBOUND : 44Kbit/s for each partecipant => 44Kbit/s * 2 * 1 => 88Kbit/s (TOTAL) CLIENT INBOUND : 44Kbit/s (44Kbit/s because I receive only the stream of the other …my stream is local from my webcam) CLIENT OUTBOUND : 44Kbit/s (my stream) Bye Da: Maxim Solodovnik [mailto:solomax...@gmail.com] Inviato: giovedì 26 febbraio 2015 15:50 A: Openmeetings user-list Oggetto: Re: Bandwidth calculation each of 2 participants receives 2 streams, 1 for him/herself and one for other party On Thu, Feb 26, 2015 at 8:13 PM, Dario Guida <produzion...@infotelsistemi.com> wrote: Why for each Partecipant inbound bandwidth is 88 Kbit/s ….. If server send 88Kbit/s (total) Partecipant 1 receive : 44Kbit/s Partecipant 2 receive : 44Kbit/s So Partecipant inbound bandwidth is 44 Kbit/s -- WBR Maxim aka solomax
