Chapter 4 Network Layer

A note on the use of these ppt slides: We ’ re making these slides freely available to all (faculty, students, readers). They ’ re in PowerPoint form so you see the animations; and can add, modify, and delete slides (including this one) and slide content to suit your needs. They obviously represent a

lot

of work on our part. In return for use, we only ask the following:  If you use these slides (e.g., in a class) that you mention their source (after all, we ’ d like people to use our book!)  If you post any slides on a www site, that you note that they are adapted from (or perhaps identical to) our slides, and note our copyright of this material.

Thanks and enjoy! JFK/KWR All material copyright 1996-2012 J.F Kurose and K.W. Ross, All Rights Reserved

Computer Networking: A Top Down Approach

6 th edition Jim Kurose, Keith Ross Addison-Wesley March 2012 Network Layer 4-1

Network layer

     transport segment from sending to receiving host on sending side encapsulates segments into datagrams on receiving side, delivers segments to transport layer network layer protocols in

every

host, router router examines header fields in all IP datagrams passing through it application transport 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 network data link physical network data link physical data link physical network data link physical application transport network data link physical Network Layer 4-2

Two key network-layer functions



forwarding:

output move packets from router ’ s input to appropriate router 

routing:

determine route taken by packets from source to dest. 

routing algorithms

analogy:



routing:

planning trip from source to dest process of 

forwarding :

getting through single interchange process of Network Layer 4-3

Interplay between routing and forwarding

routing algorithm local forwarding table header value output link 0100 0101 0111 1001 3 2 2 1 value in arriving packet ’ s header 0111 3 2 1 routing algorithm determines end-end-path through network forwarding table determines local forwarding at this router Network Layer 4-4

Datagram networks

   no call setup at network layer routers: no state about end-to-end connections  no network-level concept of “ connection ” packets forwarded using destination host address application transport network data link physical 1. send datagrams 2. receive datagrams application transport network data link physical Network Layer 4-5

Datagram forwarding table

routing algorithm local forwarding table dest address output link address-range 1 address-range 2 address-range 3 address-range 4 3 2 2 1 4 billion IP addresses, so rather than list individual destination address list range of addresses (aggregate table entries) IP destination address in arriving packet ’ s header 3 2 1 Network Layer 4-6

Datagram forwarding table

Destination Address Range 11001000 00010111 00010000 00000000

through

11001000 00010111 00010111 11111111 11001000 00010111 00011000 00000000

through

11001000 00010111 00011000 11111111 11001000 00010111 00011001 00000000

through

11001000 00010111 00011111 11111111

otherwise Link Interface 0 1 2 3

Q:

but what happens if ranges don ’ t divide up so nicely? Network Layer 4-7

Longest prefix matching

longest prefix matching

when looking for forwarding table entry for given destination address, use

longest

address prefix that matches destination address.

Destination Address Range 11001000 00010111 00010*** ********* 11001000 00010111 00011000 ********* 11001000 00010111 00011*** ********* otherwise examples: DA: 11001000 00010111 00010110 10100001 DA: 11001000 00010111 00011000 10101010 Link interface 0 1 2 3 which interface?

which interface?

Network Layer 4-8

Router architecture overview

two key router functions:   run routing algorithms/protocol (RIP, OSPF, BGP) forwarding datagrams from incoming to outgoing link

forwarding tables computed, pushed to input ports

routing processor routing, management control plane (software) forwarding data plane (hardware) high-seed switching fabric router input ports router output ports Network Layer 4-9

Input port functions

physical layer: bit-level reception data link layer: e.g., Ethernet see chapter 5 line termination link layer protocol (receive) lookup, forwarding queueing decentralized switching:    switch fabric given datagram dest., lookup output port using forwarding table in input port memory (“match plus action”) goal: complete input port processing at ‘ line speed ’ queuing: if datagrams arrive faster than forwarding rate into switch fabric Network Layer 4-10

Switching fabrics

   transfer packet from input buffer to appropriate output buffer switching rate: rate at which packets can be transfer from inputs to outputs  often measured as multiple of input/output line rate  N inputs: switching rate N times line rate desirable three types of switching fabrics memory memory bus crossbar Network Layer 4-11

Switching via memory

first generation routers:

   traditional computers with switching under direct control of CPU packet copied to system ’ s memory speed limited by memory bandwidth (2 bus crossings per datagram) input port (e.g., Ethernet) memory output port (e.g., Ethernet) system bus Network Layer 4-12

Switching via a bus

   datagram from input port memory to output port memory via a shared bus

bus contention:

switching speed limited by bus bandwidth 32 Gbps bus, Cisco 5600: sufficient speed for access and enterprise routers bus Network Layer 4-13

Switching via interconnection network

    overcome bus bandwidth limitations banyan networks, crossbar, other interconnection nets initially developed to connect processors in multiprocessor advanced design: fragmenting datagram into fixed length cells, switch cells through the fabric. Cisco 12000: switches 60 Gbps through the interconnection network crossbar Network Layer 4-14

Output ports

switch fabric datagram buffer link layer protocol (send) queueing line termination  

buffering

required when datagrams arrive from fabric faster than the transmission rate

scheduling discipline

chooses among queued datagrams for transmission Network Layer 4-15

Output port queueing

switch fabric switch fabric at

t,

packets more from input to output one packet time later   buffering when arrival rate via switch exceeds output line speed

queueing (delay) and loss due to output port buffer overflow!

Network Layer 4-16

Input port queuing

  fabric slower than input ports combined -> queueing may occur at input queues 

queueing delay and loss due to input buffer overflow!

Head-of-the-Line (HOL) blocking: queued datagram at front of queue prevents others in queue from moving forward switch fabric output port contention: only one red datagram can be transferred.

lower red packet is blocked

switch fabric one packet time later: green packet experiences HOL blocking Network Layer 4-17

The Internet network layer

host, router network layer functions: network layer transport layer: TCP, UDP

routing protocols

• path selection • RIP, OSPF, BGP

IP protocol

• addressing conventions • datagram format • packet handling conventions forwarding table link layer

ICMP protocol

• error reporting • router “ signaling ” physical layer Network Layer 4-18

IP datagram format

IP protocol version number header length (bytes) “ type ” of data max number remaining hops (decremented at each router) upper layer protocol to deliver payload to

how much overhead?

 20 bytes of TCP   20 bytes of IP = 40 bytes + app layer overhead 32 bits ver head.

len type of service 16-bit identifier time to live upper layer flgs length fragment offset header checksum 32 bit source IP address 32 bit destination IP address options (if any) data (variable length, typically a TCP or UDP segment) total datagram length (bytes) for fragmentation/ reassembly e.g. timestamp, record route taken, specify list of routers to visit.

Network Layer 4-19

 

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 ” final destination only at IP header bits used to identify, order related fragments

reassembly fragmentation:

in:

one large datagram

out:

3 smaller datagrams Network Layer 4-20

IP fragmentation, reassembly

example:

  4000 byte datagram MTU = 1500 bytes 1480 bytes in data field offset = 1480/8 length =4000 ID =x fragflag =0 offset =0

one large datagram becomes several smaller datagrams

length =1500 ID =x fragflag =1 offset =0 length =1500 ID =x fragflag =1 offset =185 length =1040 ID =x fragflag =0 offset =370 Network Layer 4-21

IP addressing: introduction

  

IP address:

32-bit identifier for host, router

interface

223.1.1.2

interface:

 router ’ connection between host/router and physical link s typically have multiple interfaces  host typically has one or two interfaces (e.g., wired Ethernet, wireless 802.11)

IP addresses associated with each interface

223.1.1.1

223.1.1.3

223.1.1.4

223.1.3.27

223.1.3.1

223.1.2.1

223.1.2.9

223.1.2.2

223.1.3.2

223.1.1.1 = 11011111 00000001 00000001 00000001 223 1 1 1 Network Layer 4-22

IP address format

   The 32-bit IP address is broken up into 4 octets, which are arranged into a dotted-decimal notation scheme.

An octet is a set of 8 bits & not a musical instrument.

Example of an IP version 4: 172.64.126.52

Network Layer 4-23

Decimal to binary conversion

2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 1 1 1 1 1 1 1 1 128 64 32 16 8 4 2 1

The binary number 1111 1111 converts into the decimal number:

128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255

The largest decimal number that can be stored in an IP address octet is 255 Network Layer 4-24

IP address classes

Class

A B C D E

Leftmost bits

0xxx 10xx 110x 1110 1111

Start address

0.0.0.0

128.0.0.0

192.0.0.0

224.0.0.0

240.0.0.0

Finish address

127.255.255.255

191.255.255.255

223.255.255.255

239.255.255.255

255.255.255.255

Class D addresses are used for multicasting and class E addresses Reserved for experimental uses Network Layer 4-25

Network and host parts

 The 32 bits of the IP address are divided into Network & Host portions, with the octets assigned as a part of one or the other.

Class

Class A Class B Class C Network & Host Representation By IP Address Class

Octet1 Octet2 Octet3

Network Host Host Network Network Network Network Host Network

Octet4

Host Host Host

Network Layer 4-26

Network mask

   Class A, 255.0.0.0

Class B, 255.255.0.0

Class C, 255.255.255.0

 An AND of an IP address with the network mask results into the network address

IP Address Mask N/W Address 138.

10001010 11111111

138. 101.

01100101 11111111

101. 114.

01110010 00000000

0. 250

11111010 00000000

0

Network Layer 4-27

Special addresses

Network address

0’s 1’s N/W address 127

Host address

0’s 1’s

Description Example

Default Cisco Route Broadcast to local network 0.0.0.0

255.255.255.255

1’s Any Broadcast to N/W address 192.21.12.255

Loopback testing 127.0.0.1

Network Layer 4-28

Private addresses

Class

A B C

Private start address

10.0.0.0

172.16.0.0

192.168.0.0

Private last address

10.255.255.255

172.31.255.255

192.168.255.255

Private addresses can be freely used on networks which are not Connected to the Internet or they are behind Gateways or Firewalls That use NAT(Network Address Translation) Network Layer 4-29

IP addresses: how to get one?

Q: How does a host get IP address?

  hard-coded by system admin in a file  Windows: control-panel->network->configuration >tcp/ip->properties  UNIX: /etc/rc.config

DHCP: D ynamic H ost C onfiguration P rotocol: dynamically get address from as server  “ plug-and-play ” Network Layer 4-30

DHCP: Dynamic Host Configuration Protocol

goal:

server when it joins network  allow host to dynamically obtain its IP address from network can renew its lease on address in use   allows reuse of addresses (only hold address while connected/ “ on ” ) support for mobile users who want to join network (more shortly)

DHCP overview:

  host broadcasts “ DHCP discover ” msg [optional] DHCP server responds with “ DHCP offer ” msg [optional]   host requests IP address: “ DHCP request ” DHCP server sends address: “ DHCP ack ” msg msg Network Layer 4-31

DHCP client-server scenario

DHCP server: 223.1.2.5

DHCP discover

src : 0.0.0.0, 68 dest.: 255.255.255.255,67 yiaddr: 0.0.0.0

transaction ID: 654

DHCP offer

src: 223.1.2.5, 67 dest: 255.255.255.255, 68 yiaddrr: 223.1.2.4

transaction ID: 654 lifetime: 3600 secs

DHCP request

src: 0.0.0.0, 68 dest:: 255.255.255.255, 67 yiaddrr: 223.1.2.4

transaction ID: 655 lifetime: 3600 secs

DHCP ACK

src: 223.1.2.5, 67 dest: 255.255.255.255, 68 yiaddrr: 223.1.2.4

transaction ID: 655 lifetime: 3600 secs arriving client Network Layer 4-32

IP addressing: the last word...

Q: A:

how does an ISP get block of addresses?

ICANN : I nternet C orporation for A ssigned  N ames and N umbers http://www.icann.org/ allocates addresses   manages DNS assigns domain names, resolves disputes Network Layer 4-33

NAT: network address translation

rest of Internet local network (e.g., home network) 10.0.0/24 10.0.0.4

10.0.0.1

10.0.0.2

138.76.29.7

10.0.0.3

all

datagrams network have

leaving same

local single source NAT IP address: 138.76.29.7,different source port numbers datagrams with source or destination in this network have 10.0.0/24 address for source, destination (as usual) Network Layer 4-34

NAT: network address translation

motivation:

as outside world is concerned:  local network uses just one IP address as far range of addresses not needed from ISP: just one IP address for all devices    can change addresses of devices in local network without notifying outside world can change ISP without changing addresses of devices in local network devices inside local net not explicitly addressable, visible by outside world (a security plus) Network Layer 4-35

NAT: network address translation

implementation: NAT router must: 

outgoing datagrams: replace

(source IP address, port #) of every outgoing datagram to (NAT IP address, new port #) . . . remote clients/servers will respond using (NAT IP address, new port #) as destination addr 

remember (in NAT translation table)

every (source IP address, port #) to (NAT IP address, new port #) translation pair 

incoming datagrams: replace

(NAT IP address, new port #) in dest fields of every incoming datagram with corresponding (source IP address, port #) stored in NAT table Network Layer 4-36

NAT: network address translation

2:

NAT router changes datagram source addr from 10.0.0.1, 3345 to 138.76.29.7, 5001, updates table 2 NAT translation table WAN side addr LAN side addr 138.76.29.7, 5001 10.0.0.1, 3345 …… …… S: 138.76.29.7, 5001 D: 128.119.40.186, 80 138.76.29.7

S: 128.119.40.186, 80 D: 138.76.29.7, 5001

3:

reply arrives dest. address: 138.76.29.7, 5001 3

1:

host 10.0.0.1 sends datagram to 128.119.40.186, 80 S: 10.0.0.1, 3345 D: 128.119.40.186, 80 1 10.0.0.1

10.0.0.4

10.0.0.2

S: 128.119.40.186, 80 D: 10.0.0.1, 3345 4

4:

NAT router changes datagram dest addr from 10.0.0.3

138.76.29.7, 5001 to 10.0.0.1, 3345 Network Layer 4-37

NAT: network address translation

  16-bit port-number field:  60,000 simultaneous connections with a single LAN-side address!

NAT is controversial:  routers should only process up to layer 3  violates end-to-end argument  • NAT possibility must be taken into account by app designers, e.g., P2P applications address shortage should instead be solved by IPv6 Network Layer 4-38

Subnetting

 Enables to divide one physical network into several logically divided sub-networks  Smaller broadcast domain   Traffic reduction Security Network Layer 4-39

A network with two levels of hierarchy (not subnetted)

Network Layer 4-40

A network with three levels of hierarchy (subnetted)

Network Layer 4-41

Addresses in a network with and without subnetting

Network Layer 4-42

Default mask and subnet mask

Network Layer 4-43

Finding the Subnet Address

 Given an IP address, the subnet address can be found in the same way as for the network address  Apply the AND operation to the binary of IP address and the subnet mask Network Layer 4-44

Example

 What is the subnetwork address if the host address is 172.16.13.29 and the subnet mask is 255.255.255.0?

1 0 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Hence the subnet address is 172.16.13.0

Network Layer 4-45

Creating the subnets

Class B add.

0 subnet Subnet 1 Subnet 2 ….

Subnet 254 Broadcast subnet

172.

16.

1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0

0.

0

1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 Number of subnets=2 Number of subnet bits -2 Subnet 0 and the broadcast subnets should not be used Network Layer 4-46

Valid IP address in subnet 1

Subnet 1

1 st IP 2 nd IP ….

Last IP Bcast IP 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 . . . .

1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 # valid IP addresses or # hosts=2 Number of host bits - 2 1 2 st nd IP IP Last IP Broadcast IP

In dotted decimal

172.16.1.1

172.16.1.1

172.16.1.254

172.16.1.255

Network Layer 4-47

Example

  Subnet the class C IP, 192.168.1.0, using three bits from the host part subnet mask 255.255.255.224

Network Layer 4-48

Example

1 st Subnet 192.168.1.0

10100000.1001000.00000001

.

000 00000 10100000.1001000.00000001

.

001 00000 10100000.1001000.00000001

.

010 00000 10100000.1001000.00000001

.

100 00000 10100000.1001000.00000001

.

101 00000 Last Subnet 10100000.1001000.00000001

.

011 00000 10100000.1001000.00000001

.

110 00000 10100000.1001000.00000001

.

111 00000 Write the valid IP address of subnet 2 Network Layer 4-49

Solution

      Conversion from binary to decimal Subnet number is 192.168.1.64

First address: 192.168.1.65

Broadcast address: 192.168.1.95 (turn all the host bits into 1) Last address: 192.168.1.94 (subtract 1 from the broadcast address) Other addresses are in the range from 1 st to last Network Layer 4-50

Prefix notation

     Subnet masks have consecutive numbers of 1s 255.255.0.0 = 11111111.11111111.00000000.00000000

Can be written as /16 Means that the first 16 bits represent the network part Example, 192.168.1.70/27 has a 27 bits prefix Network Layer 4-51

Finding the subnet number

   Given a subnetted IP address, 192.168.1.70/27, find the subnet number, and range of IP addresses in the subnet Conversion into binary, AND operation Can we do it without using binary conversion?

Network Layer 4-52

Finding the subnet number

     Find the magic number MN=256 – mask’s interesting octet value of the MN=256-224=32 Find the closest multiple to MN that is not greater than the address’ interesting octet (i.e. 70) 32*2=64 put this value in the interesting octet i.e. 192.168.1.64 (subnet number) The first address: 192.168.1.65

Network Layer 4-53

Finding the broadcast address

     Use the magic number again The broadcast address= MN+ Subnet number-1 MN+ Subnet number-1=32+64-1 = 95 The broadcast address=192.168.1.95

The range of valid IP addresses: 192.168.1.65 to 192.168.1.94

Network Layer 4-54