Re: [rtcweb] Lower-overhead protocol variations

On Feb 21, 2013, at 3:54 PM, Randell Jesup wrote:

> On 2/21/2013 8:03 AM, Michael Tuexen wrote:
>> On Feb 20, 2013, at 10:31 PM, Randell Jesup wrote:
>> 
>>> This is a relevant thread to current discussions:
>>> http://www.w3.org/mid/D1737092-F63D-4821-B3FA-D425267E4F23@cisco.com
>>> and continued with subject change in
>>> http://www.w3.org/mid/4F4D6315.9000601@jesup.org
>>> 
>>> I'm re-evaluating if the original decision against even/odd was required,
>>> in order to see if we can collapse the current draft 0-RTT proposal
>>> into a single declarative "open" message on a stream with no response or ack
>>> required. Even/odd (perhaps based on a property of the SCTP association?)
>>> would avoid the need for mismatched channel pairs, and thus avoid the need
>>> for the response/ack and the need to send with the in-order bit.
>> I'm not sure what you mean you even/odd?
>> 
>> I guess you will send the "open" message reliable and ordered. OK.
>> Are you proposing that there is no ACK sent back? What would happen
>> if one side sends an "open" message indicating an unordered data channel,
>> sends a user message on this channel and the user message is delivered
>> first?
> 
> As you infer below, this would be one side uses even channels to initiate, the other uses odd (to avoid glare).
> The reliability question is important; the simplest solution would be to buffer any data that arrives without an Open message until the Open message is received, and then deliver it.  There's no issue in the other direction, as once the open message is received the receiver can then send on that stream/channel with no restrictions. I assume there's some way to reliably choose roles for even/odd selection in SCTP?  If not, we can find other ways to do it (even SDP if we had to), though I think we could also key off the DTLS.
Not sure we can choose based on SCTP. The port numbers can be the same and we have no addresses
available. Maybe we can use the client/server identity from DTLS (the DTLS client uses even,
the DTLS server uses odd).
> 
>>> The only hole I've seen so far is that if the return stream isn't
>>> currently within the range of valid streams (i.e. at the far end) there's
>>> no easy way to return an error.  However, we expect to be able to
>> How are the stream in both directions tied together? How is the allocation
>> done? Can't the sender of the "open" message know that there is no backward
>> channel?
> 
> Actually, it probably can since it knows how many streams the other side can send with.   Some verbiage around how negotiation of the number of streams is done may be all we need here.  I don't see this as a major problem.
My questions are answered by the odd/even rule and buffering received messages before the Open is received.
> 
>> Maybe you are thinking about each side only using even outgoing streams for
>> data channels initiated by its own, and the next higher odd for the ones
>> initiated by the peer. If you are thinking like this, each side knows if
>> the resources are there and, if not, can request more. Do you have something
>> like this in mind?
> 
> Yes.  I don't remember if you can request that the *other* side increase streams, but if we add verbiage that says "when a stream increase is requested by one side, the other side shall request a number of streams that large or larger" I think we're covered - then if the other side wouldn't have a return channel, you just request a stream increase (maybe even an increase of 0 streams, but that would require the other side to equal you).
Yes, you can add incoming and outgoing streams (we thought it might be useful...).
So your idea would work...

Best regards
Michael
> 
> 
> -- 
> Randell Jesup
> randell-ietf@jesup.org
> 
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