From: Mahaffey, Brian (bmahaffey@pelco.com)
Date: Fri May 09 2008 - 00:58:30 ART
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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