Hi,
I am trying to have a clearer understanding of ATM VBR-NRT calculation on this article posted on Cisco web site:
http://www.cisco.com/univercd/cc/td/doc/cisintwk/itg_v1/tr1922.htm
In ATM, traffic shaping works by inserting equal times between the cells. For example, if an OC-3/STM-1 connection is 155 Mbps, only about 149 Mbps can be used for forwarding ATM cells. (SONET/SDH has approximately 3 percent of overhead.) As a result, the max rate is 353.208 cells (353.208 ¥ 53 ¥ 8 bits can fit in the OC-3c/STM-1 frames payload in a second). If a user requests a connection of 74.5 Mbps (half the line rate), equal spaces of 2.83 usec will be inserted between each cell. 2.83 usec is the time needed to send one cell at OC3c/STM-1 (1/353.208 sec). Because we requested half the line rate, we can send one cell, wait an equal amount of time, and then start over again.
I sat down and did some maths and did not understand how the author came up with 353.208 cells that are able to fit into a OC3 frame payload per second. I did my own calculation and found it to be 353207.208 cells that are able to fit into a OC3 frame payload per second.
1 ATM cell --> 53 bytes = (53*8) bits
1 ATM cell --> 424 bits
OC3 = 155Mbps = 149.760 Mbps (due to OC3/Sonet overheads) = 149760000 bit per second
Hence,
# of ATM cell per second
= 149760000 / 424 = 353207.54
I am not sure if I did overlook something or should have included some additional factors in my calculation.
Does anyone know the author comes up with the # of cells to be 353.208 cells per second that may be sent/received over OC3 ATM link?
Thank you,
Ben
