The Relationship of Bandwidth and Packet Forwarding Rate

Network 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 Part	Minimum Frame Size	Maximum 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)
Using the computed maximum and minimum frame rate values above of 1,488,096 f/s and 81,274 f/s for a 1 Gb/s Ethernet link, the computed maximum and minimum frame rate values of 1,488,096 f/s and 81,274 f/s for a 1-Gb/s Ethernet link can be plotted as shown in Figure 1. Figure 1 also displays other common link speeds, such as 10 Mb/s, 100 Mb/s, and 10 Gb/s Ethernet, for comparison purposes. A constant rate line is also shown at 100 Kp/s. Whether hardware- or software-based, network devices have a maximum rate at which they can forward packets. Thus, graphing this maximum forwarding rate can provide an indication of the equivalent bandwidths that a device may be capable of handling for various packet sizes.

Ref: Bandwidth, Packets Per Second, and Other Network Performance Metrics