CENG415 – Communication Networks

CENG415 – Communication Networks Lectures 20 Data Link Layer - Ethernet

Ethernet Ethernet is the most commonly used protocol for local networks Created in 1978 by DEC, Intel, and Xerox Standardized by the IEEE (RFC 802.3) • Read the history from the book Ethernet uses BUS technology (one single cable to which nodes (computers) attach

Physical properties Versions: 10Mbps, 100Mbps (fast Ethernet), 1000Mbps (Gigabit Ethernet). • Cable: Coax, up to 500m. • Taps: At least 2.5m apart. • Transceiver: Detect idle line, drive signal on TX, receive signal on RX. • Repeater:Amplifies and forwards electric signals. No more than 4 between any two hosts determines a maximum reach of 2500m.(5 cables of 500m connected with 4 repeaters) • Terminators:Placed at the end of a segment to avoid signals from bouncing back. • Medium:Broadcast. • Maximum number of hosts:1024.(2500 / 2.5) ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Host Repeater

Ethernet - characteristics • While one computer transmits info, all other computers must wait • The most interesting aspect of Ethernet is the mechanism used to coordinate transmission • An Ethernet network doesn’t have a centralized controller that tells each computer how to take turns using the shared cable • Distributed coordination scheme called Carrier Sense Multiple Access (CSMA)* * Will be described later

Ethernet Frame • When a frame is put on the wire, it is received by all NICs • The NIC picks up only the frames addressed to itself • Destination address matches with his MAC address • Broadcast address: all bits 1 • Multicast address: first bit 1, but not all • The adaptor only passes to hosts the frames it accepts unless it is in promiscuous mode. * Destination address first to allow the NIC to know ASAP if the frame need to be delivered or discarded

Preamble Fact: hardware can detect a change in voltage more easily then a fixed one Manchester Encoding: uses rising and falling edges to encode data • Ethernet uses Manchester Encoding • It is important so that sender and receiver agree on the bit slot (transmission clock rate)

Preamble • 8 bytes • Each of the first seven bytes of the preamble is 10101010 • The last byte is 10101011 • The first seven bytes of the preamble serve to "wake up" the receiving adapters and to synchronize their clocks to that of the sender's clock • At the receiver side: • NIC senses the cable for current • Current is there: synchronize clock according to the sender clock (preamble) • Data received 10101011  End of preamble  Start of Data • No current  End of frame

Ethernet – Sender side 1-persistent protocol • Sender has frame to transmit • Sender listens to the medium • If line is busy, wait until free • Transmit frame immediately p-persistent protocol • Sender has frame to transmit • Sender listens to the medium • If line is busy, wait until free • Transmit frame immediately with probability p, defer to another host with probability q=1-p. Question: What happens when more than one host decides to transmit at the same time?

MAC protocols - Recall Three broad classes: • Channel Partitioning • divide channel into smaller “pieces” (time slots, frequency, code) • allocate piece to node for exclusive use • Random Access • channel not divided, allow collisions • “recover” from collisions • “Taking turns” • Nodes take turns, but nodes with more to send can take longer turns

CSMA / CD • CSMA Cannot prevent all possible conflicts • Interference between two signals is called collision • Collision does not harm the hardware • Collision is allowed • Ethernet standard requires the sender to monitor signals on the cable (collision detection CD) • When collision is detected • Stop and wait for the Ether to idle • After the Ether is idle • Select a random number between 0 and d (delay) • Wait (back off) • When random number time out, restart transmission • If collision occurs again, the delay is doubled (exponential back off)

CSMA / CD Host A listens and finds the bus idle. Host A starts TX. Host B starts TX. Host A detects collision. Host B completes TX. Host A backs off. COLLISION! Host B backs off. time Host B listens and finds the bus idle. Host B detects collision. Host B listens and finds the bus idle. Host B starts TX.

Collision detection • collisions detected within short time • colliding transmissions aborted, reducing channel wastage • easy in wired LANs: measure signal strengths, compare transmitted, received signals • When a host detects a collision, it first transmits a 32-bit jamming sequence (plus preamble) and then stops the transmission (a total of 96 bits) • The objective of Jam sequence is to make sure that all other transmitters are aware of collision • If collision is far, more than 32-bit are needed • Ethernet’s shortest frame size is chosen to be 512 bits…

Ethernet Efficiency The efficiency of an Ethernet can be determined using a relatively simple formula Example: • cable is 300 meters long, so τ is 300/(300*106) • Packet is 1000 bytes long, bandwidth 10 Mbps • a = 0.00125 • ρ = 1/(1+6.44*0.00125) = 0.99 • If bandwidth  Efficiency  where τ: end-to-end propagation time (length / Speed) T: transmission time of a packet (Size / bandwidth)

Ethernet Type Field • Two bytes • Permits Ethernet to "multiplex" network-layer protocols • hosts can use other network-layer protocols besides IP • A host may support multiple network layer protocols, and use different protocols for different applications • When the Ethernet frame arrives at adapter B, it needs to know to which network-layer protocol it should pass the contents of the data field • Example: the ARP protocol has its own type number (0x806)

Ethernet Star Topology • Bus topology popular through mid 90s • Now star topology prevails • Connection choices: hub or switch hub or switch

twisted pair hub Ethernet Technologies • 10Base2: thin coaxial cable in a bus topology. Transmission rate is 10Mbps (2 for 200m the maximum distance between two nodes without repeaters) • 10BaseT: which uses twisted-pair cooper wire in a star topology and has a transmission rate of 10 Mbps (T for twisted 10 for the speed) • 100 m max distance between nodes and hub

Gigabit Ethernet • Gigabit Ethernet is an extension to the highly successful 10 Mbps and 100 Mbps Ethernet standards • Offers a raw data rate of 1000 Mbps • Full compatibility with the installed base of Ethernet equipment. • The standard for Gigabit Ethernet is IEEE 802.3z • Uses the standard Ethernet frame format • Allows for point-to-point links as well as shared broadcast channels. • Point-to-point links use switches • broadcast channels use hubs "buffered distributors" • Uses CSMA/CD for shared broadcast channels. In order to have acceptable efficiency, the maximum distance between nodes must be severely restricted. • Allows for full-duplex operation at 1000 Mbps in both directions for point-to-point channels.

Interconnectors • Repeater: to connected two Ether cables • bits coming from one link go out to the other links • at the same rate • no frame buffering • Collision transmitted also • Bridge: electronic device (processor, memory …) • Bridges are repeaters and more • Help isolate the problems • Discard frames with errors • Frame filtering • Bridge knows if he needs to send a frame from one segment to another

Bridge – Self learning Example: consider the following LAN Segment 1 Segment 2 X Y Z Bridge A B C

Planning a Bridged Network • Performance on a bridged network can be maximized by attaching a set of computers that interact frequently to the same segment • In addition to filtering, bridges perform buffering (frames might arrive faster than they can be processed) • Cycles in a bridged network • If all bridges are functional, a broadcasting frame will cycle forever in the network Bridge Bridge Bridge Bridge

Distributed Spanning Tree • To prevent the problem of infinite loops. A bridged network must not allow both of the following conditions to occur simultaneously • All bridges forward all frames • The bridged network contains a cycle of bridged segments At boot time: bridges perform a computation know as DST (distributed spanning tree) algorithm to determine which bridges will not forward frames DST Algorithm Applies on a network with N nodes (bridges) and M edges Objective is to select N-1 edge such that: • There is a path between each pair of nodes

x z w u y v DST algorithm • Step 1: • Set S={a node selected Randomly} , U=N/S • Step 2: • Select any edge (a, b) such that a  S and b  U • Add b to S • Step 3: • Repeat step 2 until the number of chosen edges = the number of nodes -1 Example: S= {u} U={v, w, x, y, z} S={u, w} U={v, x, y, z} S={u, w, y} U={v, x, z} S={u, w, y, z} U={v, x} S={u, w, x, y, z} U={v} S={u, v, w, x, y, z} U=

Interconnectors • 4 ports Hub or Switch • Hub: connectors are repeaters • Switch: connectors are bridges C C C C C C connector C

LAB: Bridge learning • Consider the following network • The objective is to see how bridgesswitch learn how to route frames • When starting this network • ARP tables on hosts are empty arp –a to verify • Learning in Switch is empty (except for the router)

LAB: Bridge learning • From PC0 ping PC3 • From PC3 ping PC0 • Check the ARP table of PC0 and PC3 • Check the mac-address-table of the Switch

LAB: Bridge learning • Switch to simulation mode • From PC0 ping PC3 • Notice the ICMP message goes directly from PC0 to PC3 • Complete the ping (realtime mode) • From the client screen of the switch clean the mac-address-table clear mac-address-table dynamic • Back to simulation mode • From PC0 ping PC3 • Notice the behavior of the Bridge

CENG415 – Communication Networks