This page contains some example calculations for the operation of 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) | 12 Bytes |
| MAC Preamble (+ SFD) | 8 Bytes |
| MAC Destination Address | 6 Bytes |
| MAC Source Address | 6 Bytes |
| MAC Type (or Length) | 2 Bytes |
| Payload (Network PDU) | 46 Bytes |
| Check Sequence (CRC) | 4 Bytes |
| Total Frame Physical Size | 84 Bytes |
Calculation of number of bit periods occupied by smallest size of Ethernet frame
The maximum number of frames per second is:
Ethernet Data Rate (bits per second) / Total Frame Physical Size (bits)
= 10 000 0000 / (84 x 8 )
= 14 880 frames per second.
In practice, this exceeds the forwarding capacity of many routers and bridges (typically 1000's of frames per second). This is however not usually a concern, since most Ethernet networks carry packets with a range of lengths and usually transmit a significant proportion of maximum sized frames (the maximum rate of transmission of maximum sized frames is only 812 frames per second). If the forwarding capacity is momentarilly exceeded due to a large number of small frames, the bridge/router will simply discard the excess frames which it can not process (this is allowed, since Ethernet provides only a best effort link layer service).
The maximum frame payload is 1500 Bytes, this will offer the highest throughput, when the frames are transmitted by only one node on the network (i.e. there are no collisions) To calculate the throughput provided by the link layer, one must first calculate the maximum frame rate for this size of frame.
| Frame Part | Maximum Size Frame |
| Inter Frame Gap (9.6µs) | 12 Bytes |
| MAC Preamble (+ SFD) | 8 Bytes |
| MAC Destination Address | 6 Bytes |
| MAC Source Address | 6 Bytes |
| MAC Type (or Length) | 2 Bytes |
| Payload (Network PDU) | 1500 Bytes |
| Check Sequence (CRC) | 4 Bytes |
| Total Frame Physical Size | 1538 Bytes |
Calculation of number of bit periods occupied by largest size of Ethernet frame
The largest frame consists of 1526 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) therefore corresponds to1538 Bytes.
The maximum frame rate is:
Ethernet Data Rate (bits per second) / Total Frame Physical Size (bits)
= 812.74 frames per second
The link layer throughput (i.e. number of payload bits transferred per second) is:
Frame Rate x Size of Frame Payload (bits)
= 812.74 x (1500 x 8)
= 9 752 880 bps.
This represents a throughput efficiency of 97.5 %.
The utilisation is the percentage of time the physical link is transmitting data. This calculation will assume that the transmissions do not collide. (Thisd may need to be reviewed if the utilisation is greater than about 10%).
The maximum frame payload is 1500 Bytes, this will offer the highest throughput. To calculate the throughput provided by the link layer, one must first calculate the maximum frame rate for this size of frame.
| Frame Part | Frame 1 | Frame 2 |
| Inter Frame Gap (9.6µs) | 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) | 100 Bytes | 1000 Bytes |
| Check Sequence (CRC) | 4 Bytes | 4 Bytes |
| Total Frame Physical Size | 138 Bytes | 1038 Bytes |
Calculation of number of bit periods occupied by largest size of Ethernet frame
The two frames occupy respectively a period corresponding to 138 Bytes and 1038 Bytes (including the 9.6 µs inter-frame gap). The total number of bits which are utilised in 1 second are therefore be:
(Frame Size 1 x Frame Rate 1) + (Frame Size 2 x Frame Rate 2)
= (138 x 8 x 10) + (1038 x 8 x 2)
= 27648 bits
This represents a utilisation of :
(27648 / 10 000 0000) x 100
= 0.28 %
(N.B. 0.28 % << 10%, therefore it was permissable in this calculation to neglect the effect of collisions.)