ECE/CS 372 introduction to computer networks

Transcription

1 ECE/CS 372 introduction to computer s Lecture 9 Announcements: Chapter 4 Credit for lecture slides to Professor Bechir Hamdaoui Adapted from Jim Kurose & Keith Ross (original copyright) Chapter 4, slide: 1

2 Chapter 3: Recap What we covered in chapter 3: Transport layer services: required vs. available Reliability/Bandwidth/Timing/Security Connectionless vs. connection-oriented TCP vs. UDP Reliable data transfer ACK/Retransmit/RTT and Timeout Flow control Sender should not overwhelm receiver s buffer Congestion control TCP Tahoe/Reno: Fast retransmit, slow start, congestion avoidance Chapter 4, slide: 2

3 Chapter 4: Network Layer Chapter goals: understand principles behind layer services: layer service models forwarding versus routing subnetting and IP addressing routing algorithms (path selection) Chapter 4, slide: 3

4 Network layer layer protocols run at end systems & routers Sender side: get segments from transport layer encapsulates segments into IP datagrams router examines header fields in all IP datagrams Receiver side: delivers segments to transport layer application transport application transport Chapter 4, slide: 4

5 Interplay between routing and forwarding forwarding table: a lookup table for figuring out output port for each input pkt routing algorithm local forwarding table header value output link routing algorithm: constructs routing tables forwarding process: move pkts from input to output value in arriving packet s header Source routing process: find route taken by packets from source to dest. 3 2 Destination Chapter 4, slide: 5

6 Two Key Network-Layer Functions forwarding: move packets from router s input to appropriate router output routing: determine route taken by packets from source to dest. analogy: routing: process of planning trip from source to dest forwarding: process of getting through single interchange routing algorithms Chapter 4, slide: 6

7 Network service model Q: What services are needed/offered to deliver datagrams from sender to receiver? Example services for individual datagrams: Reliability Guaranteed delivery End-to-end delay guaranteed delivery within 40 msec delay Example services for a flow of datagrams: In-order in-order datagram delivery Throughput guaranteed minimum bandwidth to flow Jitter delay restrictions on changes in inter-packet spacing Chapter 4, slide: 7

8 Network layer: connection and connection-less services Network-layer versus transport-layer services Transport layer Network layer Service Process to process Host to host Choice Reliable (TCP) and unreliable (UDP) Unreliable only (Best effort) Implementation Edge (Hosts) Core (routers) datagram provides -layer connectionless service Chapter 4, slide: 8

9 Datagram s no call setup at layer no state about end-to-end connections is kept in routers no -level concept of connection packets forwarded using dest. host address packets (same source-dest pair) may take different paths application transport 1. Send data 2. Receive data application transport Chapter 4, slide: 9

10 Forwarding table 4 billion possible entries Destination Address Range Link Interface through through through otherwise 3 Chapter 4, slide: 10

11 Longest prefix matching Prefix Match Link Interface otherwise 3 Examples DA: Which interface? DA: Which interface? Chapter 4, slide: 11

12 Chapter 4: Network Layer Introduction IP: Internet Protocol IPv4 addressing NAT, IPv6 Routing algorithms Link state Distance Vector Routing in the Internet RIP, OSPF, BGP Chapter 4, slide: 12

13 The Internet Network layer Host, router layer functions: Transport layer: TCP, UDP Network layer Routing protocols path selection RIP, OSPF, BGP forwarding table IP protocol addressing conventions datagram format packet handling conventions ICMP protocol error reporting router signaling Link layer layer Chapter 4, slide: 13

14 IP Fragmentation & Reassembly links have MTU (max.transfer size) - largest possible link-level frame. different link types, different MTUs fragmentation: in: one large datagram out: 3 smaller datagrams 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 reassembly Chapter 4, slide: 14

15 IP Fragmentation & Reassembly (ctd) Example 4000 byte datagram = 20 (header) (data) MTU = 1500 bytes length =4000 ID =x fragflag =0 offset =0 One large datagram becomes several smaller datagrams Chapter 4, slide: 15

16 IP Fragmentation & Reassembly (ctd) Example 4000 byte datagram = 20 (header) (data) MTU = 1500 bytes 1480 bytes in data field offset = 1480/8 length =4000 ID =x length =1500 length =1500 fragflag =0 ID =x ID =x ID =x offset =0 One large datagram becomes several smaller datagrams length =1040 fragflag =1 fragflag =1 fragflag =0 offset =0 offset =185 offset = = 20 (header) (data) 1020 (data) = Chapter 4, slide: 16

17 IP Addressing: introduction IP address: 32-bit identifier for host, router interface = interface: connection between host/router and link multiple interfaces per router one interface per host one IP address per interface Chapter 4, slide: 17

18 Subnets IP address: subnet part (higher bits) host part (lower bits) subnet part /23 What s a subnet? host part device interfaces with same subnet part of IP address can ly reach each other without intervening router subnet consisting of 3 subnets Chapter 4, slide: 18

19 Subnets / /24 Recipe To determine the subnets, detach each interface from its host or router, creating islands of isolated s. Each isolated is called a subnet /24 Subnet mask: /24 Chapter 4, slide: 19

20 Subnets How many? Chapter 4, slide: 20

21 IP addressing: CIDR Classful addressing: A, B, C A: /8 B: /16 C: /24 (only 2 8 subnets, but 2 24 hosts per subnet) (2 16 subnets, and 2 16 hosts per subnet) (2 24 subnets, but only 2 8 hosts per subnet) Problem: see which class is needed for 300 hosts? Issue? CIDR: Classless InterDomain Routing subnet portion of address of arbitrary length address format: a.b.c.d/x, where x is # bits in subnet portion of address subnet part host part /23 Chapter 4, slide: 21

22 IP addresses: how to get one? Q: How does host get IP address? hard-coded by system admin in a file DHCP: Dynamic Host Configuration Protocol: dynamically get IP address from a server when joining the IP address can be reused by other hosts if released Can renew IP addresses if stayed connected Chapter 4, slide: 22

23 DHCP client-server scenario A DHCP server B E arriving DHCP client needs address in this Chapter 4, slide: 23

24 IP addresses: how to get one? Q: How does get subnet part of IP addr? A: gets allocated portion of its provider ISP s address space ISP's block /20 Organization /23 Organization /23 Organization / Organization /23 Chapter 4, slide: 24

25 Example Three subnets All interfaces in all these subnets are required to have prefix: /24 Subnet 1 is required to support 125 interfaces Subnet 2 & 3 are each required to support 60 interfaces Subnet 1 Subnet 2 Question: Provide 3 addresses in the form: a.b.c.d/x for subnets 1, 2, and 3 Subnet 3 Answer: see board (board notes will be posted) Chapter 4, slide: 25