Example 1: Calculate the maximum frame rate of a node on an Ethernet LAN.
The minimum frame payload is 46 Bytes (dictated by the slot time of the Ethernet LAN architecture). The maximum frame rate is achieved by a single transmitting node which does not therefore suffer any collisions. This implies a frame consisting of 72 Bytes (see table above) with a 9.6 µs inter-frame gap (corresponding to 12 Bytes at 10 Mbps). The total utilised period (measured in bits) corresponds to 84 Bytes.
| Frame Part | Minimum Size Frame |
| Inter Frame Gap (9.6µs) | |
| MAC Preamble (+ SFD) | |
| MAC Destination Address | |
| MAC Source Address | |
| MAC Type (or Length) | |
| Payload (Network PDU) | |
| Check Sequence (CRC) | |
| Total Frame Physical Size |
Example 2: Calculate the maximum throughput of the link service provided by Ethernet
| Frame Part | Maximum Size Frame |
| Inter Frame Gap (9.6µs) | |
| MAC Preamble (+ SFD) | |
| MAC Destination Address | |
| MAC Source Address | |
| MAC Type (or Length) | |
| Payload (Network PDU) | |
| Check Sequence (CRC) | |
| Total Frame Physical Size |
Example 3: One node transmits 100 B frames at 10 frames per second, another transmits 1000 B frames at 2 frames per second, calculate the utilisation of the Ethernet LAN.
| Frame Part | Frame 1 | Frame 2 |
| Inter Frame Gap (9.6µs) | ||
| MAC Preamble (+ SFD) | ||
| MAC Destination Address | ||
| MAC Source Address | ||
| MAC Type (or Length) | ||
| Payload (Network PDU) | ||
| Check Sequence (CRC) | ||
| Total Frame Physical Size |