Assume that an average SNMP response message is 100 bytes long. Assume that a manager sends 40 SNMP Get commands each second. What percentage of a 128 kbps WAN link would the response message represent?

Answer 1

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

Answer 2

That is correct my friend. I just did mine and aced it. :)