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Re: [AVT] Frame Interleaving in AMR profile->RFC 3267
Hi,
When it comes to meeting the limitations the sender side has the freedom
to select any combination of ILL and number of frames that satisfies the
restrictions. Another restriction then I that may be used is "maxptime"
which puts a limit on the number of frames per packet.
In your example with I=30 the sender has the choice to go from ILL=0 (No
interleaving, with 30 frames in a single RTP packet, up to ILL=14 and
only two frames per packet. It is assumed that the sender will use the
combination most suitable for the current network performance and
congestion state.
There is no need for the payload format to put in more limitations. The
RTP payload format has great flexibility and for any receiver
implementing the interleaving mode can decode these payloads independent
of which ratio between ILL and N that are used. The I restriction
value is there to put a limit on buffering and interleaving induced
delay. However using many AMR frames per packet is generally not a good
idea. If one would use the 130 frames per packet that you calculate, I
hope you are aware that this represent 2.6 seconds. That can be a whole
sentence lost if this single packet is lost. So filling voice packet to
the limit is not suitable in environments where losses occur. However if
one has retransmission or uses a reliable channel more frames per
packets may give satisfactory performance. My opinion is that more than
5-10 frames per packet is to much. And if one is going as high as 10
frames per packet one should definitely consider using interleaving with
fairly decent length.
[Chairs hat on]
RFC 3267 is a payload format NOT a profile. A RTP profile is something
else and must not be confused with payload formats. I have seen this
mistake being done also by others. The current available profiles are
RFC 1890 (AVP) or rather draft-ietf-avt-profile-new-13. Two new profiles
are with the IESG, SRTP (SAVP) draft-ietf-avt-srtp-09 and Extended RTCP
feedback (AVPF) draft-ietf-avt-rtcp-feedback-07.
[chairs hat off]
Best Regards
Magnus
B Prashanth wrote:
Dear All
The frame interleaving in the AMR profile (RFC 3267) states that
" The sender of the payload MUST only apply interleaving if the
receiver has signalled its use through out-of-band means. Since
interleaving will increase buffering requirements at the receiver,
the receiver uses MIME parameter "interleaving=I" to set the maximum
number of frame-blocks allowed in an interleaving group to I.
When performing interleaving the sender MUST use a proper number of
frame-blocks per payload (N) and ILL so that the resulting size of an
interleave group is less or equal to I, i.e., N*(L+1)<=I. "
ILL is the interleaving Length i.e. number of RTP packets across which
interleaving happens.
NOTE: The value of ILL starts from 0. Hence, one needs to do ILL +1 in
the above formula.
Nmax, the maximum value N (number of frame-blocks per payload) can take,
is calculated by the formula ->
Nmax = (RTP MTU SIZE)/(The frame size of a frame block)
Assuming,
RTP MTU SIZE = 1500 bytes, and
The frame size of a AMR frame block = 13 octets for AMR frame type=0
The above assumption gives:
Nmax = 1500/13 = 115 packets.
The information obtained from out-of-band means, say, SDP is
"interleaving=30", which means that an interleaving group can have a
maximum of 30 frames per group such that:
(ILL+1) * N less than or equal to I
where N varies from 0<N<Nmax.
ILL is the interleaving Length i.e. number of RTP
packets across which interleaving happens.
Now, in the case taken by me above, Nmax > I (i.e. 115>30). So, the
condition (ILL+1)*N less than or equal to I with N=Nmax doesn't hold good.
So, how does one select the value of N and ILL such that it is less than
I, in the above case?
Right now, I can think of only one option of handling this:
My option: Assume the minimum value of ILL, say 1. A value of 2 makes
sense because one needs at least 2 RTP packets to do interleaving. Based
on this value calculate N
N = (30/2)-1
N = 14
With this value of N we satisfy both the conditions, namely,
(ILL+1) * N less than or equal to I -> (2+1)
and
N varies from 0<N<Nmax.
I wanted to ask the others on this list whether this solution is the
only solution? Or there are other ways of handling this?
Shouldn't the values of N and ILL be very apparent rather than AMR
profile designer making such assumptions about it?
--
Prashanth B.P.
Sr. Design Engr.
NeoMagic Design Center.
Polyplex Building, First Floor
B-37 Sector 1
Noida
INDIA 201301
E-Mail: bprashanth@neomagic.com
Tel: +91-120-2536208
Fax: +91-120-2536207
--
Magnus Westerlund
Multimedia Technologies, Ericsson Research EAB/TVA/A
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Ericsson AB | Phone +46 8 4048287
Torshamsgatan 23 | Fax +46 8 7575550
S-164 80 Stockholm, Sweden | mailto: magnus.westerlund@ericsson.com
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