# LAN Capacity

Mar 26th, 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getâ commands each second. What percentage of a 100Mbps LAN link's capacity would the resulting response traffic represent?

sK

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## Replies

Giuseppe Larosa Thu, 03/26/2009 - 13:06

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 Thu, 03/26/2009 - 14:03

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 Thu, 03/26/2009 - 14:37

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 Thu, 03/26/2009 - 14:56

Ok... Now it is clear!

Thanks again Giuseppe.

SK

vlatkorun Fri, 03/27/2009 - 04:15

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

40 * 138 * 8 bps = 44160 bps

Giuseppe Larosa Fri, 03/27/2009 - 04:21