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1 Ethernet “dominant” LAN technology: cheap $20 for 100Mbs! first widely used LAN technology Simpler, cheaper than token LANs and ATM Kept up with speed.

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Presentation on theme: "1 Ethernet “dominant” LAN technology: cheap $20 for 100Mbs! first widely used LAN technology Simpler, cheaper than token LANs and ATM Kept up with speed."— Presentation transcript:

1 1 Ethernet “dominant” LAN technology: cheap $20 for 100Mbs! first widely used LAN technology Simpler, cheaper than token LANs and ATM Kept up with speed race: 10, 100, 1000 Mbps Uses Manchester encoding or 4B/5B Encoding Exports a connectionless, unreliable service interface Metcalfe’s Ethernet sketch

2 2 Ethernet Service Interface Connectionless: No handshaking between sending and receiving adapter. Unreliable: Does not implement any reliable delivery protocol: Receiving adapter doesn’t send acks or nacks to sending adapter –stream of datagrams passed to network layer can have gaps. –reliability must be implemented by upper layer protocols if desired –otherwise, application will see the gaps

3 3 Ethernet Frame Structure Preamble: –7 bytes with pattern 10101010 followed by one byte with pattern 10101011 (start of frame flag) –used to synchronize receiver, sender clock rates & to have the receiver detect the beginning of the frame Addresses: 6 bytes Type: Mux/Demux key –indicates the higher layer protocol CRC: Uses CRC-32 as EDC No Length Field! End of frame is detected by the lack of current on the link – LL & PL are strictly tied to each other 64 48 16 32 46-1500 bytes

4 4 Ethernet CSMA/CD algorithm CS – Carrier Sense TxF – Transmit Frame LBT – Listen Before Talking LWT – Listen While Talking Ready to send Backoff (2 n *RTT) CS (LBT) Tx JAM Signal TxF CS (LWT) Tx Done Busy Not Busy Collision No Collision New Frame after the n^th collision, adapter chooses a K at random from {0,1,2,…,2^ n -1}. Adapter waits K*512 bit times (random wait) Jam Signal: make sure all other transmitters are aware of collision; 48 bits;

5 5 Collision Detection (CD) For A to detect a collision, we have to assure that TxT > RTT. How? –Small RTT (short link), long packet, or slow link Suppose distance <= 2.5km –Speed of Light C = 2*10^5 km/sec –Then RTT < (2*2.5)/C = 25us –25us@10Mbps = 250 bits, which is about 30 bytes –Thus the required minimum size, 46 bytes, for the payload of an Ethernet frame is sufficient to detect collision for 10 Mbps shared-medium Also means that shared medium Ethernets of speeds 100Mbps or more are not practical. That’s why they use Point-to-Point dedicated physical layer A B Time B detects collision, sends JAM signal B starts transmission here A detects collision here TxT LBT


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