From: Mahaffey, Brian (bmahaffey@pelco.com)
Date: Fri May 09 2008 - 13:30:01 ART
Thank you Shine for taking the time to break this down for me. This
makes total sense now!
-----Original Message-----
From: Shine [mailto:shinepjoseph@iprimus.com.au]
Sent: Thursday, May 08, 2008 9:38 PM
To: Mahaffey, Brian; ccielab@groupstudy.com
Subject: RE: ATM Technology Question
Brian,
Hopefully the following would shed some light!
Each ATM cell is 53 bytes including over head of 5 bytes. Payload is 48
bytes and overhead of 5 bytes makes total of 53 bytes.
While using the codec of G.729, the transmission normally occurs at
8Kbps
and at 50 packets per packet, the packet size is 20 bytes. (8Kbps = 8000
bps, 8000/8 = 1000 bytes, 1000bytes/50 packets = 20 bytes per packet)
For every voice packet, the header size is 40 bytes (IP+RTP+UDP), makes
the
total size of a voice packet to 60 bytes (20 bytes payload + 40 bytes
overhead - when you use header compression this is reduced to 2-4 bytes)
when G.729 codec is employed.
But, as we have seen earlier, an ATM cell can carry only 48 bytes and
each
voice packet is 60 bytes. So, in order to carry a single voice packet
with
the default packet size, you need at least 2 cells or 2 cells x 52 bytes
=
106 bytes. In other words, to transport 60 bytes of voice packet + 8
bytes
of ATM header, you are using 106 bytes cells or two ATM cells - one uses
48
bytes and the other uses 68-48 = 20 bytes, and thus wasting 28 bytes.
Hope this makes sense!
Regards,
Shine
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Mahaffey, Brian
Sent: Friday, 9 May 2008 1:59 PM
To: ccielab@groupstudy.com
Subject: ATM Technology Question
Quick Question which for some reason has me asking myself other
questions.
hmm...
I am reading Cisco QoS Exam Certification Guide which a few of you have
recommended to me (great book by the way) to better understand the blank
spots
I am having on the QoS section of the CCIE.
Although the IE lab for RS doesn't touch ATM I am seriously having
trouble
calculating the math on this simple calculation.
Quoted from the book Page 72
"ATM can add a significant amount of data-link overhead to a voice
packet.
Each ATM cell has 5 bytes of overhead; in addition, the last cell
holding
parts of the voice packet may have a lot of wasted space. For instance,
a
voice call using G.729 will have a packet size of 60 bytes. ATM adds
about 8
bytes of framing overhead, and then segments the 68-byte frame into two
cells
- one using the full 48 bytes of ATM cell payload, and the other only
using
20
bytes of the cell payload area - with 28 bytes "wasted". Therefore, to
send
one voice "packet" of 60 bytes, two cells, in total 106 bytes, must be
sent
over ATM. One option to lessen the overhead is to change3 the payload
size
to
contain 30 ms of voice with a G.729 codec - which interestingly also
only
takes two ATM cells."
End Quote
So if I understand this correctly. Laugh at me if not.
The voice call 60 bytes
ATM overhead per cell has 5 bytes (regardless) per payload
ATM overhead "framing" 8 bytes (regardless) per/after payload???
ATM cell can only contain 48 bytes per payload total.
48 bytes payload - 5 bytes overhead - 8 bytes framing = 35 bytes of
actual
data contained in each cell? But 2 cells in total = 106 bytes as stated
above
is it?
48 bytes payload - 8 bytes framing = 40 bytes of actual data contained
in
each
cell + 5 bytes of overhead........ which 2 cells then would equal 106
bytes....
Can anyone shed some light here? Maybe I should keep reading then maybe
I
will
understand it :)
I appreciate your time.
-
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