1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16

The Network Layer

3 Purpose of Network layer Given a packet, send it across the network to destination 2 key issues: –Portability: connect different technologies –Scalability To the Internet scale network data link physical network data link physical network data link physical network data link physical network data link physical network data link physical network data link physical network data link physical application transport network data link physical application transport network data link physical

4 What does it involve? Two important functions: routing: determine path from source to dest. forwarding: move packets from router’s input to output T1 T3 Sts-1 T3 T1

5 Network service model Q: What service model for “channel” transporting packets from sender to receiver? guaranteed bandwidth? preservation of inter-packet timing (no jitter)? loss-free delivery? in-order delivery? congestion feedback to sender? ? ? ? virtual circuit or datagram? The most important abstraction provided by network layer: service abstraction Which things can be “faked” at the transport layer?

6 Two connection models Connectionless (or “datagram”): –each packet contains enough information that routers can decide how to get it to its final destination Connection-oriented (or “virtual circuit”) –first set up a connection between two nodes –label it (called a virtual circuit identifier (VCI)) –all packets carry label B A bb C B A 11 C 1

7 Virtual circuits: signaling protocols used to setup, maintain teardown VC setup gives opportunity to reserve resources used in ATM, frame-relay, X.25 not used in today’s Internet application transport network data link physical application transport network data link physical 1. Initiate call 2. incoming call 3. Accept call 4. Call connected 5. Data flow begins 6. Receive data

8 Virtual circuit switching Forming a circuit: –send a connection request from A to B. Contains VCI + address of B –rule: VCI must be unique on the link its used on –switch creates an entry mapping input messages with VCI to output port –switch picks a new VCI unique between it and next switch a b c 1 2 1

9 (Input link,VCI) (output link, new VCI) (1, 2) (?, ?) (1, 5) (?, ?) Virtual circuit forwarding For each VCI switch has a table which maps input link to output link and gives the new VCI to use –if a’s messages come into switch 1 on link 2 and go out on link 3 then the table will be: a b c Switch 1 Switch 2 Switch 3

10 Virtual Circuits: Discussion Plusses: easy to associate resources with VC –Easy to provide QoS guarantees (bandwidth, delay) –Very little state in packet Minuses: –Not good in case of crashes Requires explicit connect and teardown phases –What if teardown does not get to all routers? –What if one switch crashes? Will have to teardown and rebuild route

11 Datagram networks no call setup at network layer routers: no state about end-to-end connections –no network-level concept of “connection” packets typically routed using destination host ID –packets between same source-dest pair may take different paths Best effort: data corruption, packet drops, route loops application transport network data link physical application transport network data link physical 1. Send data 2. Receive data

12 Datagrams: Forwarding How does packet get to the destination? switch creates a “forwarding table”, mapping destinations to output port (ignores input ports) when a packet with a destination address in the table arrives, it pushes it out on the appropriate output port when a packet with a destination address not in the table arrives, it must find out more routing information (next problem) a b c 1 d S1 S2 S

13 Datagrams Plusses: –No round trip connection setup time –No explicit route teardown –No resource reservation  each flow could get max bandwidth –Easily handles switch failures; routes around it Minuses –Difficult to provide resource guarantees –Higher per packet overhead Internet uses datagrams: IP (Internet Protocol)

14 Datagrams Forwarding How to build forwarding tables? –Manually enter it What if nodes crashed What about scale? The graph-theoretic routing problem –Given a graph, with vertices (switches), edges (links), and edge costs (cost of sending on that link) –Find the least cost path between any two nodes Path cost =  (cost of edges in path)

15 Simple Routing Algorithm Choose a central node –All nodes send their (nbr, cost) information to this node –Central node uses info to learn entire topology of the network –It then computes shortest paths between all pairs of nodes Using All Pair Shortest Path Algorithm –Sends the new matrix to every node Nice, simple, elegant! What is the problem? –Scalability: centralization hurts scalability –Central node is “crushed” with traffic

16 Link State Routing Basic idea: –Every node propagates its (nbr, cost) information –This information at all nodes is enough to construct topology –Can use a graph algorithm to find the shortest routes Mechanisms required: –Reliable flooding of link information –Method to calculate shortest route (Dijkstra’s algorithm) Example link state update packet: –[node id, (nbr, cost) list, seq. no., ttl] Seq. no. to identify latest updates, ttl specifies when to stop msg.

17 Reliable flooding receive(pkt) If already have a copy of LSP from pkt.ID if pkt’s sequence number <= copy’s discard pkt else decrement pkt.TTL replace copy with pkt forward pkt to all links besides the one that we received it on # done every 10 minutes or so gen_LSP() increment node’s sequence # by one recompute cost vector send created LSP to all neighbors

18 Discussion: Link-State Routing Plusses: –Simple, determines the optimal route most of the time –Used by OSPF Minuses: –Might have oscillations –Avoid using load as cost metric, reduce herding effect A D C B 1 1+e e 0 e A D C B 2+e e 1 A D C B 0 2+e 1+e A D C B 2+e 0 e 0 1+e 1 Initially start with almost equal routes … everyone goes with least loaded … recompute Least loaded => Most loaded … recompute

19 Is our routing algo scalable? Route table size grows with size of network –Because our address structure is flat! Solution: have a hierarchical structure –Used by OSPF –Divide the network into areas, each area has unique number Nodes carry their area number in the address 1.A, 2.B, etc. –Nodes know complete topology in their area –Area border routers (ABR) know how to get to any other area

20 Hierarchical Addressing 1.a 3.b 2.b 1 1.b S1 S2 S a 3 3.a 2 Zone 3 Zone 2 Forwarding table for switch 1 Destination switch port 2.? 3. ? 1.b ? 1.a ?

21 IP has 2-layer addressing Each IP address is 32 bits –Network part: which network the host is on? –Host part: identifies the host. All hosts on same network have the same network part 3 classes of addresses: A, B and C network 32-bits host 0 net host bits 1 0 net host bits 110 net host bits

22 IP addressing The different classes: Problems: inefficient, address space exhaustion 0 network host 10 network host 110 networkhost 1110 multicast address A B C D class to to to to bits Unicast Multicast 1111 reserved E to Reserved

23 IP addressing: CIDR Classless InterDomain Routing –network portion of address of arbitrary length –address format: a.b.c.d/x, where x is # bits in network portion –Examples: Class A: /8 Class B: /16 Class C: / network part host part /23

24 Internet Protocol Datagram ver length 32 bits data (variable length, typically a TCP or UDP segment) 16-bit identifier Internet checksum time to live 32 bit source IP address IP protocol version Number header length max number remaining hops (decremented at each router) for fragmentation/ reassembly total datagram length (bytes) upper layer protocol to deliver payload to head. len type of service “type” of data flgs fragment offset upper layer 32 bit destination IP address Options (if any) E.g. timestamp, record route taken, pecify list of routers to visit.

25 Datagram Portability IP Goal: To create one logical network from multiple physical networks –All intermediate routers should understand IP –IP header information sufficient to carry the packet to destination –Goal: Run over anything! Problem: –Physical networks have different MTUs –“max. transmission unit”: 1500 for Ethernet, 48 for ATM Solution 1: –Fit everything in the MTU (!)

26 IP Fragmentation & Reassembly Solution 2: (the one used) –If packet size > MTU of network, then fragment into pieces Each fragment is less than MTU size Each has IP headers + frag bit set + frag id + offset –Packets may get refragmented on the way to destination –Reassembly only done at the destination –What is a good initial packet size? fragmentation: in: one large datagram out: 3 smaller datagrams reassembly