Bandwidth usage or throughput using counters related to IpInterface MO

[ivanbrij]

Dear experts,

I want to calculate Bandwidth usage or Throughput using counters related to IpInterface, like counter below, but the counter units is "1000 Number of frames". How to convert them to octets, bytes or bps..?

pmDot1qTpVlanPortInFrames The number of valid frames received on this port, belonging to this VLAN and IP interface. If there are more than one IP interface belonging to the same VLAN, the total number of frames for a specific VLAN is the sum of the pmDot1qTpVlanPortInFrames counter for all IP interfaces belonging to the VLAN.
See RFC 2674.

Condition: 1000 ethernet frames are received on this port with this VLAN and a protocol which is supported by the forwarding process.
A frame received on this port is counted by this object if and only if it is for a protocol being processed by the local forwarding process for this VLAN.
This object includes received bridge management frames classified as belonging to this VLAN (e.g. GMRP, but not GVRP or STP).

Counter is reset after measurement period: Yes

Counter type: PEG

Unit: 1000 Number of frames



How to change from frames to bytes or Kbps?

Thanks in advance for your support.

Ivan..!

[consultant]

from cisco:
The Relationship of Bandwidth and Packet Forwarding RateNetwork devices receive and forward packets through physical interfaces that employ Layer 2 technologies, such as Ethernet and Packet Over SONET (POS) framing. The description for these network links always includes bandwidth that is expressed in terms of b/s. By performing simple mathematical manipulations, it is possible to determine the potential range of p/s, or more correctly, frames per second (f/s) that a network link can support.
For example, the very common 1-Gb/s Ethernet interface is capable of transmitting up to 1,000,000,000 b/s. To determine p/s, first convert bits to bytes. (There are eight bits in one byte.) Then consider how many bytes exist in each packet. The size of the packet does not have to be a fixed value, but administrators can bound the problem by recognizing that there are both minimum and maximum packet sizes. The minimum size is based on both the IP-defined minimum IP packet size and the Layer 2-defined minimum frame size. The maximum IP packet size is based on the link maximum transmission unit (MTU) for the Layer 2 technology. Based on these factors, and using Ethernet as an example, the following two calculations can be considered:

Maximum Frame Rate (Minimum Frame Size)The maximum Ethernet frame rate is achieved by a single transmitting node that does not suffer any collisions when Ethernet frames are at their smallest size. The minimum Ethernet frame payload is 46 bytes (based on the slot time of Ethernet), which yields a frame that consists of 72 bytes (see Table 1) plus a 12-byte inter-frame gap, for a total Minimum Frame size of 84 bytes.
Maximum Throughput (Maximum Frame Size)The maximum Ethernet throughput is achieved by a single transmitting node that does not suffer any collisions when the Ethernet frames are at their maximum size. The maximum Ethernet frame payload is 1500 bytes (not considering Jumbo frames), which yields a frame that consists of 1526 bytes (see Table 1) plus a 12-byte inter-frame gap, for a total Maximum Frame size of 1538 bytes. (This calculation provides the lower bound on frame rate.)

Table 1. Maximum Frame Rate and Throughput Calculations For a 1-Gb/s Ethernet Link


Frame PartMinimum Frame SizeMaximum Frame Size

Inter Frame Gap (9.6 ms)

12 bytes


12 bytes



MAC Preamble (+ SFD)

8 bytes


8 bytes



MAC Destination Address

6 bytes


6 bytes



MAC Source Address

6 bytes


6 bytes



MAC Type (or length)

2 bytes


2 bytes



Payload (Network PDU)

46 bytes


1,500 bytes



Check Sequence (CRC)

4 bytes


4 bytes



Total Frame Physical Size

84 bytes


1, 538 bytes




[1,000,000,000 b/s / (84 B * 8 b/B)] == 1,488,096 f/s (maximum rate)
[1,000,000,000 b/s / (1,538 B * 8 b/B)] == 81,274 f/s (minimum rate)