CS158a
Chris Pollett
Feb. 21, 2011
Which of the following is true?
64 bits | 48 bits | 48 bits | 16 bits | 32 bits | |
Preamble | Destination Address | Source Address | Type | Body | CRC |
The smarts of Ethernet resides in how the sender sends frames it needs to send. It uses the following transmitter algorithm:
In fast ethernet pseudo-random number generation is done using Linear feedback shift registers.
The worst case for detecting a collision is when the hosts are on opposite ends of the Ethernet. If d is the time for a signal to go the complete length of the Ethernet. Then Host A might transmit for d seconds and Host B might just start transmitting and immediately send a jamming sequence, this takes d second to get back to A. So A might transmit for 2*d before detecting a collision. On a 10Mbps Ethernet limited to 2500m, the delay would be at most 51.2μs and this would correspond to 512 bits = 96 bytes, which is why this is used as the shortest frame length.
NumStations×THT + RingLatency
8 | 8 | 8 | 48 | 48 | 32 | 8 | 8 | |
Start Delimeter | Access Control | Frame Control | Destination Address | Source Address | Body | Checksum | End Delimiter | Frame Status |
The frame is sent using Manchester encoding, except start and end delimiters.
Access Control - contains the frame priority, and reservation bits.
Frame Control - says the higher level protocol.
Frame Status - has the A and C bits.