1. EEC-484/584 Computer Networks Lecture 10 Wenbing Zhao wenbingz@gmail.com (Part of the slides are based on Drs. Kurose & Ross ’ s slides for their Computer Networking book)

2. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao Outline Reminder  Time to start working on the project! Routing algorithms  Link state routing (done)  Distance vector routing Internet protocol v4  Header  Fragmentation Internet Protocol v6

3. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao Distance Vector Routing Also called Bellman-Ford or Ford-Fulkerson Each router maintains a table, 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

4. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao Distance Vector Routing: How each entry is updated d(A,X) d(A,Y) A X Z d(Y,Z) d(X,Z) At router A, for Z Compute d(A,X) + d(X,Z) and d(A,Y) + d(Y,Z), take minimum Y d(A,Z) = min {d(A,v) + d(v,Z) } where min is taken over all neighbors v of A

5. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao x y z x y z 0 2 7 ∞∞∞ ∞∞∞ from cost to from x y z x y z 0 from cost to x y z x y z ∞∞ ∞∞∞ cost to x y z x y z ∞∞∞ 710 cost to ∞ 2 0 1 ∞ ∞ ∞ 2 0 1 7 1 0 time x z 1 2 7 y node x table node y table node z table d(x,y) = min{d(x,y) + d(y,y), d(x,z) + d(z,y)} = min{2+0, 7+1} = 2 d(x,z) = min{d(x,y) + d(y,z), d(x,z) + d(z,z)} = min{2+1, 7+0} = 3 32 Each node keeps track of the following info: 1.Its own distance vector: least-cost to each of other routers 2.Each of its neighbor’s distance vector received most recently If there is a change in distance vector, a node sends the update to all its neighbors

6. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao x y z x y z 0 2 7 ∞∞∞ ∞∞∞ from cost to from x y z x y z 0 2 3 from cost to x y z x y z 0 2 3 from cost to x y z x y z ∞∞ ∞∞∞ cost to x y z x y z 0 2 7 from cost to x y z x y z 0 2 3 from cost to x y z x y z 0 2 3 from cost to x y z x y z 0 2 7 from cost to x y z x y z ∞∞∞ 710 cost to ∞ 2 0 1 ∞ ∞ ∞ 2 0 1 7 1 0 2 0 1 7 1 0 2 0 1 3 1 0 2 0 1 3 1 0 2 0 1 3 1 0 2 0 1 3 1 0 time x z 1 2 7 y node x table node y table node z table d(x,y) = min{d(x,y) + d(y,y), d(x,z) + d(z,y)} = min{2+0, 7+1} = 2 d(x,z) = min{d(x,y) + d(y,z), d(x,z) + d(z,z)} = min{2+1, 7+0} = 3

7. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao Distance Vector Routing Distance from A to B 12ms, to C 25ms, to D 40ms, to G 18ms Distance from J to A 8ms, to I 10ms, to H 12ms, to K 6ms Distance from 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

8. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao 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

9. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao The Network Layer in Internet forwarding table Host, router network layer functions: Routing protocols path selection RIP, OSPF, BGP IP protocol addressing conventions datagram format packet handling conventions ICMP protocol error reporting router “signaling” Transport layer: TCP, UDP Link layer physical layer Network layer

10. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao IPv4 Datagram Format ver Total length 32 bits data (variable length, typically a TCP or UDP segment) 16-bit identifier header checksum time to live 32 bit source IP address IP protocol version number header length (bytes) max number remaining hops (decremented at each router) for fragmentation/ reassembly total datagram length (bytes) upper layer protocol to deliver payload to IHL type of service “type” of data flgs fragment offset protocol 32 bit destination IP address Options (if any) E.g. timestamp, record route taken, specify list of routers to visit. How much overhead with TCP? 20 bytes of TCP 20 bytes of IP = 40 bytes + app layer overhead

11. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao 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 (2 16 -1) bytes Identification – allows destination to determine which datagram a fragment belongs to

12. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao 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

13. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao The IPv4 Header Options – security, error reporting, etc.  Some of the IP options

14. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao IPv4 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

15. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao IPv4 Fragmentation & Reassembly 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 fragmentation: in: one large datagram out: 3 smaller datagrams reassembly

16. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao IPv4 Fragmentation and Reassembly ID =x offset =0 MF =0 length =4000 ID =x offset =0 MF =1 length =1500 ID =x offset =185 MF =1 length =1500 ID =x offset =370 MF =0 length =1040 One large datagram becomes several smaller datagrams Example 4000 byte datagram MTU = 1500 bytes 1480 bytes in data field offset = 1480/8 Fragment should be as large as possible

17. Network Layer 4-17 IPv6: motivation initial motivation: 32-bit address space soon to be completely allocated. additional motivation:  header format helps speed processing/forwarding  header changes to facilitate QoS IPv6 datagram format:  fixed-length 40 byte header  no fragmentation allowed

18. Network Layer 4-18 IPv6 datagram format priority: identify priority among datagrams in flow flow Label: identify datagrams in same “flow.” (concept of“flow” not well defined). next header: identify upper layer protocol for data data destination address (128 bits) source address (128 bits) payload len next hdr hop limit flow label pri ver 32 bits

19. Network Layer 4-19 Other changes from IPv4 checksum: removed entirely to reduce processing time at each hop options: allowed, but outside of header, indicated by “Next Header” field ICMPv6: new version of ICMP  additional message types, e.g. “Packet Too Big”  multicast group management functions

20. Network Layer 4-20 Transition from IPv4 to IPv6 not all routers can be upgraded simultaneously  no “flag days”  how will network operate with mixed IPv4 and IPv6 routers? tunneling: IPv6 datagram carried as payload in IPv4 datagram among IPv4 routers IPv4 source, dest addr IPv4 header fields IPv4 datagram IPv6 datagram IPv4 payload UDP/TCP payload IPv6 source dest addr IPv6 header fields

21. Network Layer 4-21 Tunneling physical view: IPv4 A B IPv6 E F C D logical view: IPv4 tunnel connecting IPv6 routers E IPv6 F A B

22. Network Layer 4-22 flow: X src: A dest: F data A-to-B: IPv6 Flow: X Src: A Dest: F data src:B dest: E B-to-C: IPv6 inside IPv4 E-to-F: IPv6 flow: X src: A dest: F data B-to-C: IPv6 inside IPv4 Flow: X Src: A Dest: F data src:B dest: E physical view: A B IPv6 E F C D logical view: IPv4 tunnel connecting IPv6 routers E IPv6 F A B IPv4 Tunneling

23. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao Dijkstra ’ s Algorithm : Exercise Given the subnet shown below, using the Dijkstra ’ s Algorithm, determine the shortest path tree from node u and its routing table u y x wv z 2 2 1 3 1 1 2 5 3 5

24. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao 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.

25. 11/26/2015 EEC-484/584: Computer Networks Wenbing Zhao 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.