03-26-2009 09:14 AM - edited 03-06-2019 04:50 AM
I am doing an analysis on the network and have the following question:
Assume that an average SNMP response message is 100 bytes long. Assume that a manager sends 40 SNMP âgetâ commands each second. What percentage of a 100Mbps LAN link's capacity would the resulting response traffic represent?
Thanks in advance,
sK
03-26-2009 01:06 PM
Hello Sadik,
I would use the following calculation:
100 byte IP packet takes 18 byte ethernet overhead
+
20 byte of "silence" inter frame gap
40 * 138 * 8 bps = 44160 bps
0,4% of BW of a 100 Mbps link
Hope to help
Giuseppe
03-26-2009 02:03 PM
Thank you Giuseppe for the quick response. I am not clear on how you get the 4%?
Also, since the 44160 is in bps don't you have to convert the 100Mbps to bps to get the bps percentage out of the 100Mbps?
Thanks,
sK
03-26-2009 02:37 PM
Hello Sadik,
sorry I was meaning 0.04 % in my country we use , instead of . for decimal separation
take 44160 bps
44160 / 100 10^6 = 0,0004416
to express a percentage you need to multiply by 100
so you get 0.04%
Edit:
(I had also made an error dividing by 10^7 instead of 10^8)
I hope you got the process ..
Hope to help
Giuseppe
03-26-2009 02:56 PM
Ok... Now it is clear!
Thanks again Giuseppe.
SK
03-27-2009 04:15 AM
May I ask why is the multiplication by 8bps in the equatation:
40 * 138 * 8 bps = 44160 bps
03-27-2009 04:21 AM
Hello Vladimir,
the 8 is for 8 bit/byte the packet size is expressed in bytes (100 byte ip plus the other overheads )
Hope to help
Giuseppe
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