1. 4. Interconnecting Networks: Routers

2. © Tallal Elshabrawy 2 Bridges Vs Routers BRIDGES DO WELL IN SMALL (FEW HUNDRED HOSTS) WHILE ROUTERS USED IN LARGE NETWORKS (THOUSANDS OF HOSTS) Both store and forward devices Routers: Network layer devices Bridges: Datalink layer devices Routers maintain routing tables, implement routing algorithms Bridges maintain forwarding tables, implement filtering, learning and spanning tree algorithms

3. © Tallal Elshabrawy 3 Bridges Characteristics Advantages: Bridge operation is simpler requiring less processing bandwidth Limitations Homogeneous link layer (e.g., all Ethernet) is desirable for transparent bridging Topologies are restricted with bridges: a spanning tree must be built to avoid cycles As a result of the spanning tree algorithm the root bridge may represent a bottleneck Bridges do not offer protection from broadcast storms (endless broadcasting by a host will be forwarded by a bridge)

4. © Tallal Elshabrawy 4 Routers Characteristics Goal: Scalable interconnection of a large numbers of networks of different types Advantages Arbitrary topologies can be supported, cycling is limited by TTL (time-to-live counters) and good routing protocols. Provides firewall protection against broadcast storms DISADVANTAGES: Require IP address configuration (not plug and play) Require higher processing bandwidth

5. © Tallal Elshabrawy 5 Router Functions Forwarding Move packet from input link to the appropriate output link Purely local computation Must go be very fast (executed for ever packet) Routing Keep track of network topology so you know where to forward packets Queuing Buffer (i.e., store) packets if router can’t forward it right now Buffer Management Drop packet if you run out of buffer space

6. © Tallal Elshabrawy 6 Routing Algorithm classification Global or decentralized information? Global: all routers have complete topology, link cost info “link state” algorithms Decentralized router knows physically-connected neighbors, link costs to neighbors iterative process of computation, exchange of info with neighbors “distance vector” algorithms Static or dynamic? Static: routes change slowly over time Dynamic: routes change more quickly periodic update in response to link cost changes

7. © Tallal Elshabrawy 7 Routing Approaches Static Type in the right answers and hope they are always true Distance vector Tell your neighbors what you know about everyone Link state Tell everyone what you know about your neighbors

8. © Tallal Elshabrawy 8 Assumption Each router knows own address & cost to reach neighbors Objective Calculate routing table containing next-hop information for every destination at each router Distributed Bellman-Ford algorithm Each router maintains a vector of costs to all destinations Initialize neighbors with known cost, others with infinity Periodically send copy of distance vector to neighbors On reception of a vector, If neighbor’s path to a destination is shorter, switch to it Bellman-Ford Distance Vector Routing Algorithm

9. © Tallal Elshabrawy 9 Bellman-Ford Algorithm: Shortest Path to Node 6 Step (Iteration) Node 1Node 2Node 3Node 4Node 5 Initial(-1,∞) 1 (6,1)(-1,∞)(6,2) 2(3,3)(5,6)(6,1)(3,3)(6,2) 3(3,3)(4,4)(6,1)(3,3)(6,2) 4(3,3)(4,4)(6,1)(3,3)(6,2) Note: (a,b)=(next hop, total cost to destination)

10. © Tallal Elshabrawy 10 Bellman-Ford Algorithm (From Communicataion Networks A. Garcia) IterationNode 1Node 2Node 3Node 4Node 5 Initial (-1,  ) 1 2 3 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 San Jose Table entry @ node 1 for dest SJ Table entry @ node 3 for dest SJ

11. © Tallal Elshabrawy 11 IterationNode 1Node 2Node 3Node 4Node 5 Initial (-1,  ) 1 (6,1) (-1,  ) (6,2) 2 3 San Jose D 6 =0 D 3 =D 6 +1 n 3 =6 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 D 6 =0 D 5 =D 6 +2 n 5 =6 0 2 1 Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

12. © Tallal Elshabrawy 12 IterationNode 1Node 2Node 3Node 4Node 5 Initial (-1,  ) 1 (6, 1) (-1,  ) (6,2) 2(3,3)(5,6)(6, 1)(3,3)(6,2) 3 San Jose 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 0 1 2 3 3 6 Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

13. © Tallal Elshabrawy 13 IterationNode 1Node 2Node 3Node 4Node 5 Initial (-1,  ) 1 (6, 1) (-1,  ) (6,2) 2(3,3)(5,6)(6, 1)(3,3)(6,2) 3(3,3)(4,4)(6, 1)(3,3)(6,2) San Jose 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 0 1 2 6 3 3 4 Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

14. © Tallal Elshabrawy 14 IterationNode 1Node 2Node 3Node 4Node 5 Initial(3,3)(4,4)(6, 1)(3,3)(6,2) 1(3,3)(4,4)(4, 5)(3,3)(6,2) 2 3 San Jose 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 0 1 2 3 3 4 Network disconnected; Loop created between nodes 3 and 4 5 Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

15. © Tallal Elshabrawy 15 IterationNode 1Node 2Node 3Node 4Node 5 Initial(3,3)(4,4)(6, 1)(3,3)(6,2) 1(3,3)(4,4)(4, 5)(3,3)(6,2) 2(3,7)(4,4)(4, 5)(5,5)(6,2) 3 San Jose 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 0 2 5 3 3 4 7 5 Node 4 could have chosen 2 as next node because of tie Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

16. © Tallal Elshabrawy 16 IterationNode 1Node 2Node 3Node 4Node 5 Initial(3,3)(4,4)(6, 1)(3,3)(6,2) 1(3,3)(4,4)(4, 5)(3,3)(6,2) 2(3,7)(4,4)(4, 5)(5,5)(6,2) 3(3,7)(4,6)(4, 7)(5,5)(6,2) San Jose 3 1 5 4 6 2 2 3 4 2 1 1 2 3 5 0 2 5 5 7 4 7 6 Node 2 could have chosen 5 as next node because of tie Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

17. © Tallal Elshabrawy 17 3 5 4 6 2 2 3 4 2 1 1 2 3 5 1 IterationNode 1Node 2Node 3Node 4Node 5 1(3,3)(4,4)(4, 5)(3,3)(6,2) 2(3,7)(4,4)(4, 5)(2,5)(6,2) 3(3,7)(4,6)(4, 7)(5,5)(6,2) 4(2,9)(4,6)(4, 7)(5,5)(6,2) San Jose 0 7 7 5 6 9 2 Node 1 could have chose 3 as next node because of tie Bellman-Ford Algorithm (From Communicataion Networks A. Garcia)

18. © Tallal Elshabrawy 18 3124 111 3124 11 X (a) (b) UpdateNode 1Node 2Node 3 Before break (2,3)(3,2)(4, 1) After break (2,3)(3,2)(2,3) 1 (3,4)(2,3) 2(2,5)(3,4)(2,5) 3 (3,6)(2,5) 4(2,7)(3,6)(2,7) 5 (3,8)(2,7) ………… Counting to Infinity Problem Nodes believe best path is through each other (Destination is node 4)

19. © Tallal Elshabrawy 19 Remedies Split Horizon Do not report route to a destination to the neighbor from which route was learned Poisoned Reverse Report route to a destination to the neighbor from which route was learned, but with infinite distance Breaks erroneous direct loops immediately Does not work on some indirect loops Problem: Bad News Travels Slowly

20. © Tallal Elshabrawy 20 3124 111 3124 11 X (a) (b) Nodes believe best path is through each other UpdateNode 1Node 2Node 3 Before break(2, 3)(3, 2)(4, 1) After break(2, 3)(3, 2) (-1,  ) Node 2 advertizes its route to 4 to node 3 as having distance infinity; node 3 finds there is no route to 4 1(2, 3) (-1,  ) Node 1 advertizes its route to 4 to node 2 as having distance infinity; node 2 finds there is no route to 4 2 (-1,  ) Node 1 finds there is no route to 4 Split Horizon with Poison Reverse

21. © Tallal Elshabrawy Dijkstra Algorithm 21