Computer Networks: Distance Vector Routing and Hierarchical Routing

1. EEC-484/584Computer Networks Lecture 12 Wenbing Zhao wenbing@ieee.org (Part of the slides are based on Drs. Kurose & Ross’s slides for their Computer Networking book, and on materials supplied by Dr. Louise Moser at UCSB and Prentice-Hall)

2. Outline • Distance vector routing • Hierarchical routing • Broadcast routing • Internet protocol • Header • Fragmentation EEC-484/584: Computer Networks

3. Distance Vector Routing • Also called Bellman-Ford or Ford-Fulkerson • Each router maintains a table (a vector), giving best known distance to each destination and which line to use to get there • Table is updated by exchanging info with neighbors • Table contains one entry for each router in network with • Preferred outgoing line to that destination • Estimate of time or distance to that destination • Once every T msec, router sends to each neighbor a list of estimated delays to each destination and receives same from those neighbors EEC-484/584: Computer Networks

4. A d(A,Y) Y At router A, for Z Compute d(A,X) + d(X,Z) and d(A,Y) + d(Y,Z), take minimum d(Y,Z) d(A,X) Z X d(X,Z) Distance Vector Routing d(A,Z) = min {d(A,v) + d(v,Z) } where min is taken over all neighbors v of A EEC-484/584: Computer Networks

5. cost to x y z x 0 2 7 y from ∞ ∞ ∞ z ∞ ∞ ∞ 2 1 7 z x y d(x,z) = min{d(x,y) + d(y,z), d(x,z) + d(z,z)} = min{2+1 , 7+0} = 3 d(x,y) = min{d(x,y) + d(y,y), d(x,z) + d(z,y)} = min{2+0 , 7+1} = 2 node x table cost to x y z x 0 2 3 y from 2 0 1 z 7 1 0 node y table cost to x y z x ∞ ∞ ∞ 2 0 1 y from z ∞ ∞ ∞ node z table cost to x y z x ∞ ∞ ∞ y from ∞ ∞ ∞ z 7 1 0 time EEC-484/584: Computer Networks

6. cost to x y z x 0 2 7 y from ∞ ∞ ∞ z ∞ ∞ ∞ 2 1 7 z x y d(x,z) = min{d(x,y) + d(y,z), d(x,z) + d(z,z)} = min{2+1 , 7+0} = 3 d(x,y) = min{d(x,y) + d(y,y), d(x,z) + d(z,y)} = min{2+0 , 7+1} = 2 node x table cost to cost to x y z x y z x 0 2 3 x 0 2 3 y from 2 0 1 y from 2 0 1 z 7 1 0 z 3 1 0 node y table cost to cost to cost to x y z x y z x y z x ∞ ∞ x 0 2 7 ∞ 2 0 1 x 0 2 3 y y from 2 0 1 y from from 2 0 1 z z ∞ ∞ ∞ 7 1 0 z 3 1 0 node z table cost to cost to cost to x y z x y z x y z x 0 2 7 x 0 2 3 x ∞ ∞ ∞ y y 2 0 1 from from y 2 0 1 from ∞ ∞ ∞ z z z 3 1 0 3 1 0 7 1 0 time EEC-484/584: Computer Networks

7. Distance Vector Routing • Delay A to B 12ms, to C 25ms, to D 40ms, to G 18ms • Delay J to A 8ms, to I 10ms, to H 12ms, to K 6ms • Delay J to A to G 8+18 = 26ms to I to G 10+31 = 41ms to H to G 12+6=18ms to K to G 6+31=37ms EEC-484/584: Computer Networks

8. Distance Vector Routing • Good news travels fast • Bad news travels slow • Count to infinity problem: Takes too long to converge upon router failure × Routers’ knowledge about the cost to A EEC-484/584: Computer Networks

9. Distance Vector Routing: Exercise • Consider the subnet shown below. Distance vector routing is used, and the following vectors have just come in to router C: from B: (5, 0, 8, 12, 6, 2); from D: (16, 12, 6, 0, 9, 10); and from E: (7, 6, 3, 9, 0, 4). The measured delays to B, D, and E, are 6, 3, and 5, respectively. What is C's new routing table? Give both the outgoing line to use and the expected delay. EEC-484/584: Computer Networks

10. Hierarchical Routing • Bigger network => bigger routing table • Use hierarchical structure similar to telephone network • Regions: router knows details of how to route packets within its region, does not know internals of other regions • Clusters of regions, zones of clusters, groups of zones • Tradeoff: savings in memory space may result in longer path EEC-484/584: Computer Networks

11. Hierarchical Routing EEC-484/584: Computer Networks

12. duplicate creation/transmission duplicate duplicate in-network duplication sourceduplication R2 R3 R4 R3 R4 R1 R1 R2 Broadcast Routing • Deliver packets from source to all other nodes • Source duplication is inefficient: EEC-484/584: Computer Networks

13. In-network Duplication • Flooding: when node receives broadcast packet, sends copy to all neighbors • Problems: cycles & broadcast storm • Controlled flooding: node only broadcasts packet if it hasn’t broadcast same packet before • Node keeps track of packet ids already broadcasted • Or reverse path forwarding (RPF): only forward packet if it arrived on shortest path between node and source • Spanning tree • No redundant packets received by any node EEC-484/584: Computer Networks

14. Flooding • Idea: every incoming packet is sent out on every outgoing line except the one it arrived on • Advantage: flooding always chooses shortest path, no other algorithm has shorter delay • Disadvantage: generates lots of duplicates EEC-484/584: Computer Networks

15. How to Reduce Duplicates • Include a hop counter in the header of each packet • Decrement the counter at each hop • Packet discarded when counter reaches 0 EEC-484/584: Computer Networks

16. How to Reduce Duplicates • The router assigns a sequence number in each packet it receives from its hosts • Each router has a list for each source router telling which sequence numbers originating at that source have already been seen • Each list augmented with counter k, meaning all sequence numbers up through k have been seen • If packet is duplicate, router discards it Source router k EEC-484/584: Computer Networks

17. Reverse Path Forwarding • Rely on router’s knowledge of unicast shortest path from it to sender • Each router has simple forwarding behavior: if (datagram received on incoming link on shortest path back to center) then flood datagram onto all outgoing links else ignore datagram

18. Reverse Path Forwarding • Assumption: • I is the sender • The cost for each hop is 1 • Optimal route is that gives the min hop count A subnet EEC-484/584: Computer Networks

19. 1 2 3 4 5 Reverse Path Forwarding • Hop 1: I sends packets to F, H, J, N; each arrives on preferred path to I • Hop 2: F sends packets to A,D; H to E,K; J to G,O; N to M,O • All are previously unvisited routers; All but E,K,O arrive on preferred path • Hop3: A sends packet to E; D to C,G; G to D; O to N; M to K; Only E,C,K arrive on preferred path EEC-484/584: Computer Networks

20. Exercise – Broadcast Routing • Looking at the subnet shown below, how many packets are generated by a broadcast from B, using (a) flooding with duplicate control through sequence number? (b) reverse path forwarding? EEC-484/584: Computer Networks

21. Design Principles for Internet • Make sure it works • Build prototypes first • Keep it simple • When in doubt, use the simplest solution • Make clear choices • If there are several ways of doing the same thing, choose one • Exploit modularity • Use protocol stacks, each of whose layers is independent of all the other ones EEC-484/584: Computer Networks

22. Design Principles for Internet • Expect heterogeneity • Different types of hardware, transmission facilities, and applications will occur on any large network • Avoid static options and parameters • Have the sender and receiver negotiate a value • Look for a good design; it need not be perfect • Be strict when sending and tolerant when receiving • Think about scalability • Consider performance and cost EEC-484/584: Computer Networks

23. Collection of Subnetworks The Internet is an interconnected collection of many networks, or Autonomous Systems (ASes) EEC-484/584: Computer Networks

24. Host, router network layer functions: • ICMP protocol • error reporting • router “signaling” • IP protocol • addressing conventions • datagram format • packet handling conventions • Routing protocols • path selection • RIP, OSPF, BGP forwarding table The Network Layer in Internet Transport layer: TCP, UDP Network layer Link layer physical layer EEC-484/584: Computer Networks

25. IP protocol version number 32 bits total datagram length (bytes) header length (bytes) type of service head. len ver length for fragmentation/ reassembly fragment offset “type” of data flgs 16-bit identifier max number remaining hops (decremented at each router) upper layer time to live header checksum 32 bit source IP address 32 bit destination IP address upper layer protocol to deliver payload to E.g. timestamp, record route taken, specify list of routers to visit. Options (if any) data (variable length, typically a TCP or UDP segment) IP Datagram Format How much overhead with TCP? • 20 bytes of TCP • 20 bytes of IP • = 40 bytes + app layer overhead EEC-484/584: Computer Networks

26. The IPv4 Header • Version– 4 • IHL– length of header in 32-bit words • Min 5, max 15 – i.e., 60 bytes • Type of service - to distinguish different classes of service • To accommodate differentiated services (which class this packet belongs to) • Total length– header and data  65,535 (216-1) bytes • Identification– allows destination to determine which datagram a fragment belongs to EEC-484/584: Computer Networks

27. The IPv4 Header • Time to live– counter to limit packet lifetimes • Max lifetime 255sec • Packet is destroyed when counter becomes 0 • Protocol– which transport layer protocols being used • Header checksum– verifies header EEC-484/584: Computer Networks

28. The IPv4 Header • Options– security, error reporting, etc. • Some of the IP options EEC-484/584: Computer Networks

29. IP Fragmentation • Fragmentation Flags • DF – tells routers “Don’t Fragment” • MF – More Fragments. All fragments except last have this set. Used as check against total length • Fragment offset– where in datagram this fragment belongs • All fragments (payload in the IP packet) except last must be multiples of 8 bytes • The number of 8 byte blocks is called Number of Fragment Blocks (NFB) • The unit of the offset is NFB EEC-484/584: Computer Networks

30. Network links have MTU (max.transfer size) - largest possible link-level frame. different link types, different MTUs Large IP datagram divided (“fragmented”) within net one datagram becomes several datagrams “reassembled” only at final destination IP header bits used to identify, order related fragments IP Fragmentation & Reassembly fragmentation: in: one large datagram out: 3 smaller datagrams reassembly EEC-484/584: Computer Networks

31. length =1500 length =1040 length =1500 length =4000 ID =x ID =x ID =x ID =x MF =0 MF =0 MF =1 MF =1 offset =0 offset =185 offset =0 offset =370 IP Fragmentation and Reassembly Example • 4000 byte datagram • MTU = 1500 bytes One large datagram becomes several smaller datagrams 1480 bytes in data field offset = 1480/8 EEC-484/584: Computer Networks

32. Exercise: IP Fragmentation • Suppose that host A is connected to a router R 1, R 1 is connected to another router, R 2, and R 2 is connected to host B. Suppose that a TCP message that contains 900 bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Total length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header, and link R2-B can support a maximum frame size of 512 bytes including a 12-byte frame header. EEC-484/584: Computer Networks