SNMP OID for Bandwidth Usage

Unanswered Question
Jul 17th, 2007

Hi!

I'm retrieving the bandwidth usage of our Cat3500XL and Cat2950 using the SNMP OID .iso.org.dod.internet.private.enterprises.cisco.ciscoMgmt.ciscoC2900MIB.c2900MIBObjects.c2900BandwidthUsage.c2900BandwidthUsageCurrent.0 (1.3.6.1.4.1.9.9.87.1.5.1.0).

Does anybody know the Bandwidth Usage SNMP OID for the new Cat2960 switches?

Regards, L. Schieder

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Leonardo Schieder Wed, 07/18/2007 - 03:29

Hi!

Not quite. The TechNote you posted describes how to calculate the bandwidth for ONE interface of any mib-2 compatible device. The OID I'm using for the Cat3500 and Cat2950 switches is Cisco specific and has the following description:

--- BEGIN

Object: c2900BandwidthUsageCurrent

OID: 1.3.6.1.4.1.9.9.87.1.5.1

Description:

The current bandwidth consumed. The measurement unit is in megabits per second (1,000,000 bits/second). This value gives a reasonable estimate of the amount of traffic currently flowing through the switch.

It is calculated as follows:

Octets*8 + Frames*(96 + 64)

---------------------------

Measurement Interval * 1,000,000 * 2

Where:

Measurement Interval is the amount of time over which the Octets and Frames were collected, in seconds.

Measurement Interval is always one second in current implementation.

Octets is the total number of octets transmitted or received by all network interfaces, excluding framing data but including FCS. This includes octets in frames which were partially transmitted or received (due to collisions, for example).

Frames is the total number of frames transmitted or received by all network interfaces, including frames with errors.

The number of frames is multiplied by 96 plus 64 in order to estimate the delay between each frame for Ethernet's IPG and preamble/SFD.

The '2' in the divisor makes this a forwarding bandwidth counter. A frame received on one interface is typically forwarded out another interface. In order to avoid double-counting this frame's bandwidth, once on the receiving interface and once on the transmitting interface, the total bandwidth is divided by two.

Since multicast and broadcast frames can be sent to multiple ports, the above is at best a lower bound.

--- END

Regards, L. Schieder

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