Introduction to Computer Networks 09/23OutlineReviewExtending LANsBridges = LAN switchesLearning BridgesSpanning TreeEach LAN segment can have many bridgesSlide 9Spanning Tree Protocol OverviewSpanning Tree AlgorithmSlide 12Slide 13Slide 14Introduction to Computer Networks09/23Presenter:Fatemah PanahiOutline•Client-Server code•Project overview•Review from last session•Spanning Tree AlgorithmReview•Shared medium vs. Point to point links•Ethernet as a shared medium link•LAN is a Local Area Network, Usually based on the Ethernet technology•Ethernet limitation: Cable length, Host numberExtending LANs•Repeaters: A device that forwards signals, just like an amplifier•Hub: A multi-way repeater; Usually used when you cannot attach a host directly to a thin cable.•No notion of frames•Broadcast: Everybody can get the message that someone sends.Bridges = LAN switches•Switch: multi-input, multi-output device which transfers packets from an input to one or more outputs•Using repeaters to extend a LAN has its own limits:–Total number of repeaters–Total length•Alternative: Have a node between 2 Ethernet segments to forward frames–Isolates collision domains if used carefully–Increases total bandwidth of the networkLearning Bridges•It is inefficient to always broadcast message to all Ethernet segments•How do we know on which port a host is?–Manual table–Learning mechanism•Learning:–Inspect the source address of each frame–Each table entry has a timeout–Not sure what to do? BroadcastCS 640 7Spanning TreeHow to handle forwarding in complex Layer 2 topologies?Each LAN segment can have many bridges•More complex topologies can provide redundancy.–But can also create loops.•E.g. What happens when there is no table entry?–Multiple copies of data Could crash the network has happened often!host host host host hosthost host host host hosthosthostBridge BridgeWhat is a Spanning Tree?•Reduce our topology graph to a tree:–Make sure there are no loops in the topology–All LAN segments are still connected to the LAN and can receive messages•Main idea: Bridges choose the ports over which they have to forward frames.Spanning Tree Protocol OverviewEmbed a tree that provides a single unique default path to each destination:Bridges designate ports over which they will or will not forward framesBy removing ports, extended LAN is reduced to a treeAddresses the crashing problem; but tree is not resilientWhen switch/link fails, rerun protocol to converge to new treeSpanning Tree Algorithm•Root of the spanning tree is elected first the bridge with the lowest identifier.–All ports are part of tree•Each bridge finds shortest path to the root.–Remembers port that is on the shortest path–Used to forward packets•Select for each LAN a designated bridge that will forward frames to root–Has the shortest path to the root.–Identifier as tie-breakerB3B7B5B2B1B4B612111 1Spanning Tree Algorithm•Each node sends configuration message to all neighbors.–Identifier of the sender–Id of the presumed root–Distance to the presumed root•Initially each bridge thinks it is the root.–B5 sends (B5, B5, 0)•When B receive a message, it decide whether the solution is better than their local solution.–A root with a lower identifier?–Same root but lower distance?–Same root, distance but sender has lower identifier?•Message from bridge with smaller root ID–Not root; stop generating config messages, but can forward•Message from bridge closer to root –Not designated bridge; stop sending any config messages on the portB3B7B5B2B1B4B612111 1Spanning Tree Algorithm•Each bridge B can now select which of its ports make up the spanning tree:–B’s root port–All ports for which B is the designated bridge on the LAN•States for ports on bridges–Forward state or blocked state, depending on whether the port is part of the spanning tree•Root periodically sends configuration messages and bridges forward them over LANs they are responsible for•Any bridge failure => Start overB3B7B5B2B1B4B612111 1Spanning Tree AlgorithmExample–B3 receives (B2,B2,0)–Since 2<3 B3 accepts B2 as a root–B3 adds one to the distance advertised by B2(0) and thus sends (B3,B2,1) toward B5–Meanwhile B2 accepts B1 as the root and sends (B2,B1,1)–B5 accepts B1 as the root and sends (B5,B1,1)–B3 accepts B1as the root and figures that B1 and B2 are closer to the root. So stops forwarding on both interfaces.B3B7B5B2B1B4B612111
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