LAN Capacity

skhirbash · ‎03-26-2009

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 â€œgetâ€ commands each second. What percentage of a 100Mbps LAN link's capacity would the resulting response traffic represent?

Thanks in advance,

sK

Giuseppe Larosa · ‎03-26-2009

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

skhirbash · ‎03-26-2009

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

Giuseppe Larosa · ‎03-26-2009

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

skhirbash · ‎03-26-2009

Ok... Now it is clear!

Thanks again Giuseppe.

SK

vlatkorun · ‎03-27-2009

May I ask why is the multiplication by 8bps in the equatation:

40 * 138 * 8 bps = 44160 bps

Giuseppe Larosa · ‎03-27-2009

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