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?
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
That is correct my friend. I just did mine and aced it. :)
To calculate the percentage of a 128 kbps WAN link that the response message represents, we need to determine the size of the response message in bits and divide it by the WAN link capacity in bits per second.
First, let's convert the size of the response message from bytes to bits:
Size of the response message: 100 bytes
1 byte = 8 bits
Size of the response message in bits: 100 bytes * 8 bits/byte = 800 bits
Now let's calculate the percentage:
Response message size: 800 bits
WAN link capacity: 128 kbps = 128,000 bits per second
Percentage of WAN link used: (Response message size / WAN link capacity) * 100
= (800 bits / 128,000 bits per second) * 100
≈ 0.625%
Therefore, the response message represents approximately 0.625% of a 128 kbps WAN link.
To find the percentage of a 128 kbps (kilobits per second) WAN link that the response message represents, we need to calculate the amount of bandwidth that the response message requires.
First, let's calculate the bandwidth required for one SNMP response message:
100 bytes * 8 bits/byte = 800 bits
Next, we need to calculate the amount of bandwidth required for the manager to send 40 SNMP Get commands per second:
40 commands/second * 800 bits/command = 32,000 bits/second
Now, we can calculate the percentage of the 128 kbps WAN link that the response message represents:
32,000 bits/second / 128,000 bits/second = 0.25 or 25%
Therefore, the response message would represent 25% of a 128 kbps WAN link.