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Estimating SNMP agents...
qwertyface
Registered Users Posts: 2 ■□□□□□□□□□
in Off-Topic
Hi,
I'm doing some reading on SNMP and have Stallings' book "SNMP, SNMPv2 and RMON" (1996).
He shows how by using some basic math you are able to "determine the maximum number of stations that the management station can handle when engaged in full-time polling" (p. 194, 7.5.4 "Polling Frequency").
He gives the following:
N <= T/V
N = No. of agents
T = desired polling interval (in seconds)
V = average time required to perform a single poll
He gives a worked example:
"The example consists of a single LAN, where each managed device is to be polled every 15 minutes. Assuming processing times on the order of 50ms, and a network delay of about 1ms (packet size of 1,000 bytes, no significant network congestion), then V is approx. 0.202 sec. Then
N <= (15*60) / 0.202 = approx. 4,500."
I find this quite unclear. How do you get 0.202 sec from the information above?
Are there any other ways to determine how many stations can be handled by the management station?
Regards,
QF
I'm doing some reading on SNMP and have Stallings' book "SNMP, SNMPv2 and RMON" (1996).
He shows how by using some basic math you are able to "determine the maximum number of stations that the management station can handle when engaged in full-time polling" (p. 194, 7.5.4 "Polling Frequency").
He gives the following:
N <= T/V
N = No. of agents
T = desired polling interval (in seconds)
V = average time required to perform a single poll
He gives a worked example:
"The example consists of a single LAN, where each managed device is to be polled every 15 minutes. Assuming processing times on the order of 50ms, and a network delay of about 1ms (packet size of 1,000 bytes, no significant network congestion), then V is approx. 0.202 sec. Then
N <= (15*60) / 0.202 = approx. 4,500."
I find this quite unclear. How do you get 0.202 sec from the information above?
Are there any other ways to determine how many stations can be handled by the management station?
Regards,
QF
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Mrock4
Banned Posts: 2,359 ■■■■■■■■□□
Hmm..interesting. Maybe he is assuming 50ms processing at each end + 1ms (one-way)..this results in 101ms..for return traffic + processing (another 101ms) you could see it as 202ms. -
Optionsqwertyface
Registered Users Posts: 2 ■□□□□□□□□□
Hmm..interesting. Maybe he is assuming 50ms processing at each end + 1ms (one-way)..this results in 101ms..for return traffic + processing (another 101ms) you could see it as 202ms.
Thanks for your input; sounds logical.