Re: [MMUSIC] WGLC for draft-ietf-mmusic-sdp-bundle-negotiation-19 - Adam's technical comments

[Paul Kyzivat <[e-mail]>]

Comment at end.

On 4/17/15 5:58 AM, Christer Holmberg wrote:
> Hi Adam,
>
> Most of your editorial comments seem ok. I'll comment them later.
>
> However, at least a couple (please let me know if you think there are more non-editorial ones) of your comments was technical, and I'll focus on those in this reply.
>
> ---------------------------------------------------------------------------
>
>> The procedure described in Section 8.3.2 has a rather unfortunate property that I'm not sure we want. If I were to offer something like:
>>
>> a=group:BUNDLE foo bar
>> m=video 20002 RTP/AVP 97
>> a=rtpmap:97 H261/90000
>> a=mid:foo
>> m=video 0 RTP/AVP 97 98
>> a=rtpmap:97 H261/90000
>> a=rtpmap:98 VP8/90000
>> a=bundle-only
>> a=mid:bar
>>
>> And the remote endpoint wanted to reject the first stream, it would be forced to also reject the second. The historical intention of using "bundle-only" was as a mechanism
>> for UDP port preservation on the offerer's side. This procedure unintentionally turns it into a semantic binding among streams in bundle.
>>
>> I think we could restore the original purpose of the bundle-only handling if we used the port from the rejected 'm=' line containing the BUNDLE-tag for any 'bundle-only' lines.
>
> So, I guess that means we would *always* use the BUNDLE-tag m- line port - even if the m- line is rejected, AND even if there are other non-bundle-only m- lines that are NOT rejected.
>
> Example:
>
> a=group:BUNDLE foo bar zoo
> m=video 20002 RTP/AVP 97
> a=rtpmap:97 H261/90000
> a=mid:foo
> m=video 40002 RTP/AVP 97
> a=rtpmap:99 H264/99999
> a=mid:zoo
> m=video 0 RTP/AVP 97 98
> a=rtpmap:97 H261/90000
> a=rtpmap:98 VP8/90000
> a=bundle-only
> a=mid:bar
>
> Now, if the UAS rejects the "foo" m- line, it would still select port 20002 for the offerer. The UAS would NOT select port 40002, even if it accepts the "zoo" m- line.

There are more implications. If you are going to do this for the port, 
then you need to do it for the address (c=) too. While most likely you 
would see the same address used for each, it is possible that when 
getting a unique address/port pairs that you can't (or at least don't) 
get the same address for each. We don't require them to be the same.

	Thanks,
	Paul