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

2. 2 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao Outline Reminder (for graduate students only) –Wiki-Page Project #1 due on today midnight! Distance vector routing Hierarchical routing Internet protocol –Header –Fragmentation

3. 3 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao 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

4. 4 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao Distance Vector Routing 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. 5 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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

6. 6 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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. 7 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao 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

8. 8 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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. 9 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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.

10. 10 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao 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

11. 11 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao Hierarchical Routing

12. 12 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao 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

13. 13 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao 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

14. 14 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao Collection of Subnetworks The Internet is an interconnected collection of many networks, or Autonomous Systems (ASes)

15. 15 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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

16. 16 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao IP 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

17. 17 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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

18. 18 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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

19. 19 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao The IPv4 Header Options – security, error reporting, etc. –Some of the IP options

20. 20 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao 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

21. 21 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao IP 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

22. 22 Fall Semester 2007EEC-484/584: Computer NetworksWenbing Zhao IP 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

23. 23 Fall Semester 2007EEC-484/584: Computer NetworksWenbing 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.