1. Communication Networks Recitation 3 Bridges & Spanning trees

2. Bridges  Link layer device  stores and forwards Ethernet frames  examines frame header and selectively forwards frame based on MAC dest address  when frame is to be forwarded on segment, uses CSMA/CD to access segment  transparent  hosts are unaware of presence of bridges  plug-and-play, self-learning  bridges do not need to be configured

3. Some bridge features  Isolates collision domains resulting in higher total max throughput  limitless number of nodes and geographical coverage  Can connect different Ethernet types  Transparent (“plug-and-play”): no configuration necessary

4. Bridges: traffic isolation  Bridge installation breaks LAN into LAN segments  bridges filter packets:  same-LAN-segment frames not usually forwarded onto other LAN segments  segments become separate collision domains bridge collision domain collision domain = hub = host LAN (IP network) LAN segment

5. Forwarding How do determine to which LAN segment to forward frame? Looks like a routing problem...

6. Self learning  A bridge has a bridge table  entry in bridge table:  (Node LAN Address, Bridge Interface, Time Stamp)  stale entries in table dropped (TTL can be 60 min)  bridges learn which hosts can be reached through which interfaces  when frame received, bridge “learns” location of sender: incoming LAN segment  records sender/location pair in bridge table

7. Filtering/Forwarding When bridge receives a frame: index bridge table using MAC dest address if entry found for destination then{ if dest on segment from which frame arrived then drop the frame else forward the frame on interface indicated } else flood forward on all but the interface on which the frame arrived

8. Question Bridge B1 = {1,a};{1,f};{2,b} abc 1 2 Bridge B2 = {1,f};{1,c};{2,b} de 1 2 fg

9. Find all errors in the table and explain why? Bridge nameError in tableExplain

10. All Errors Bridge B1 = {1,a}; {1,f} ; {2,b} abc 1 2 Bridge B2 = {1,f} ;{1,c}; {2,b} de 1 2 fg

11. Does a message reaches destination? From C to G From A to F From F to A What will happen to the tables?

12. From C to G Bridge B1 = {1,a};{1,f};{2,b}{1,c} abc 1 2 Bridge B2 = {1,f};{1,c};{2,b} de 1 2 fg

13. From A to F Bridge B1 = {1,a};{1,f};{2,b} abc 1 2 Bridge B2 = {1,f};{1,c};{2,b} de 1 2 fg

14. From F to A Bridge B1 = {1,a};{,f};{2,b} abc 1 2 Bridge B2 = {,f};{1,c};{2,b} de 1 2 fg 2 1 1 2 12

15. Loop Resolving  The simple learning mechanism described fails in presence of loops in the LAN  Loops may be present by mistake, or deliberately provided for redundency  This problem is resolved by running a distributed spanning tree algorithm

16. Spanning Tree Algorithm  Creates a logical, or “active” topology that behaves like a spanning tree  Makes alternate bridges redundant  Is run periodically, so will discover failures and use alternate bridges if necessary

17. Spanning tree  Think of the LAN as a graph that possibly has loops (LAN segments as nodes, bridges as edges)  The spanning tree is a sub graph of this graph that covers all vertices (LAN segments), but contains no cycles.

18. Spanning tree algorithm  Spanning tree algorithm is a protocol used by a set of bridges to agree upon a spanning tree for a particular extended LAN.  Essentially, this means that each bridge decides the ports over which it is and is not willing to forward packets.  Some ports (or even entire bridges) may not participate in a spanning tree  How does the bridge select the ports to include (/exclude)?

19. Spanning Tree Algorithm  Working: Bridges regularly exchange frames known as Bridge Protocol Data Units (BPDUs). This exchange does the following: 1. Each bridge has a unique Identifier 2. Bridge with highest priority and smallest ID is selected as root bridge. 3. Each bridge determines for each port, the least cost path from root bridge to this port. This is the Root Path Cost (RPC) for this port. a)Select the port which has the least RPC and designate it as the Root Port (RP). This is the port which will be used for communicating with the root.

20. Algorithm... 1. Once root port is determined, one bridge port is selected for each LAN segment as the designated bridge port (DP) over which frames will be sent for that LAN segment. a)This is a port (which is NOT a root port) which has the least path cost to the root b)The ports of the root bridge are always DPs for the LAN segments connected to the root bridge 2. The state of the bridge ports can be set either to forwarding or blocking. a)All ports that are either RPs or DPs are forwarding, the rest are blocking.

21. Example: A C E D B K F H J G I B5 B2 B3 B7 B4 B1 B6  B1 is the root bridge  B3 and B5 are both connected to LAN A, but B5 is the designated port since it's closer to root  B5 and B7 are both connected to LAN B, but B5 is the designated port due to smaller ID (equal distance).

22. Topology Initialization  BPDUs are sent to a broadcast MAC address of all bridges on the LAN  All bridges initially assume they are the root bridge  Each BPDU contains (self ID, root ID, transmitting port ID, RPC of this port)  A bridge updates its own info if it receives an update which  identifies a root with smaller id or  identifies a root with equal id but with shorter distance  the root id and distance are equal, but the sending bridge has a smaller id  The bridge adds 1 to the received RPC in the above update and saves this info.

23. Designated port / Root Port A C E D B K F H J G I B5 B2 B3 B7 B4 B1 B6 What are these And these And this one And

24. STP Run – Find Root A C E D B K F H J G I B3 B7 B4 B2 B5 B1 B6 B8 L M B9 3 5 7 4 6 1 2 9 8 B3 sends BPDU 3 B2 sends BPDU 2 2 B1 sends BPDU 21 1 1 1 B4, B2 sends BPDU B8 sends BPDU 1 1 1 1

25. Proof sketch  if there is a bridge which has a different value THEN There is a segment on which one bridge has the correct minimum and the other a larger value.  When the minimum will broadcast, the other bridge would update, and we have one more correct bridge

26. STP Run – Block Ports A C E D B K F H J G I B3 B7 B4 B2 B5 B1 B6 B8 L M B9 B5: 5, 0, 1 B2: 2, 1, 1 B3: BLOCK B7: 7, 0, 1 B5: 5, 0, 1 B7: BLOCK

27. Data A C E D B K F H J G I B3 B7 B4 B2 B5 B1 B6 B8 L M B9 Laptop A Laptop B Message A to B Message B to A