1

Datagram Networks:Internet Protocol (IPv4)

2

The Internet Network layer: IP• Internet Network Layer Components:

– IP protocol (addressing, datagram format and handling), Routing Protocols and ICMP protocol

forwardingtable

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

Networklayer

3

IP Addressing: Introduction• IP address: 32-bit

identifier for host, router interface

• interface: connection between host/router and physical link– router’s typically

have multiple interfaces

– host may have multiple interfaces

– IP addresses associated with each interface

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

223.1.1.1 = 11011111 00000001 00000001 00000001

223 1 11

4

IP Addressing• IP address:

– network part (high order bits)

– host part (low order bits)

– NetworkID.HostID

• What’s an IP network ? – device interfaces with

same network part of IP address

– Hosts within the same IP network can reach each other without intervening router

network consisting of 3 IP networks:223.1.1.0, 223.1.2.0, 223.1.3.0

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

5

IP Addressing

• IP networks are not restricted to Ethernet segments– Here we have 3

point-to-point links and each have a different IP network defined over them.

223.1.1.1

223.1.1.3

223.1.1.4

223.1.2.2223.1.2.1

223.1.2.6

223.1.3.2223.1.3.1

223.1.3.27

223.1.1.2

223.1.7.0

223.1.7.1223.1.8.0223.1.8.1

223.1.9.1

223.1.9.2

Interconnected system consistingof 6 IP networks

6

IP Addressing

• How to find IP networks?1. Detach each

interface from router, host

2. Create “islands of isolated IP networks

3. Each island defines an IP network

4. Internet consists of millions of such IP networks

223.1.1.1

223.1.1.3

223.1.1.4

223.1.2.2223.1.2.1

223.1.2.6

223.1.3.2223.1.3.1

223.1.3.27

223.1.1.2

223.1.7.0

223.1.7.1223.1.8.0223.1.8.1

223.1.9.1

223.1.9.2

Interconnected system consisting

of 6 IP networks

7

IP Addresses

0network host

10 network host

110 network host

1110 multicast address

A

B

C

D

class1.0.0.0 to127.255.255.255

128.0.0.0 to191.255.255.255

192.0.0.0 to223.255.255.255

224.0.0.0 to239.255.255.255

32 bits

• given notion of “IP network”, let’s re-examine IP addresses: – We have “class-full” addressing– Original Internet design

8

IP addressing: CIDR• Classful addressing:

– inefficient use of address space, address space exhaustion• e.g., class B net allocated enough addresses for 65K hosts,

even if only 2K hosts in that network

– No longer used in the current Internet• Solution? Classless Inter Domain Routing (CIDR)

• CIDR: Classless InterDomain Routing– Standardized in 1993– Network portion of address of arbitrary length– Address format: a.b.c.d/x, where x is # bits in network

portion of address

11001000 00010111 00010000 00000000

networkpart

hostpart

200.23.16.0/23

9

Netmask• With CIDR a new way is needed to determine

the IP network given an IP address:– Solution: Define a netmask– Given an IP address of the form networkID.hostID, the

netmask of the IP address is obtained by putting all “1”s in the networkID portion and all “0”s in the hostID portion

11001000 00010111 00010000 00000000

networkpart

hostpart

200.23.16.0/23

11111111 11111111 11111110 00000000 = 255.255.254.0

Given an IPAddr and a netmask, we bit-wise AND IPAddr and netmask to obtain the IP network. The rest is the hostID. NetworkID = IPAddr & Netmask HostID = IPAddr & ~Netmask

10

IP Layer Broadcast• Recall that a host can send a LL

broadcast message by putting FF-FF-FF-FF-FF-FF in destination MAC address

• How can a host send IP-layer broadcast packet? 2 ways:– Put 255.255.255.255 in destination IP

• Means that all IP hosts within the same LL broadcast domain will receive this IP datagram

– Make <hostID> all 1’s• Means that all hosts within the same IP subnet will

receive this datagram• Example: If IP subnet is 192.169.34.0, then a packet

with a destination IP: 192.169.34.255 will be received by all hosts whose IP subnet is 192.169.34.0

11

IP addresses: how to get one?

Q: How does a host get an IP address?– hard-coded by system admin in a file

• Wintel: control-panel->network->configuration->tcp/ip->properties

• UNIX: /etc/rc.config

– DHCP: Dynamic Host Configuration Protocol: dynamically get address from as server• “plug-and-play” (more shortly)

12

Q: How does network get network part of IP addr?

A: gets allocated portion of its provider ISP’s address space

ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20

Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/23 Organization 1 11001000 00010111 00010010 00000000 200.23.18.0/23 Organization 2 11001000 00010111 00010100 00000000 200.23.20.0/23 ... ….. …. ….

Organization 7 11001000 00010111 00011110 00000000 200.23.30.0/23

IP addresses: how to get one?

13

Hierarchical addressing: route aggregation

“Send me anythingwith addresses beginning 200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7Internet

Organization 1

ISPs-R-Us“Send me anythingwith addresses beginning 199.31.0.0/16”

200.23.20.0/23Organization 2

...

...

• CIDRized addresses facilitate hierarchical routing• Fly-By-Night-ISP advertises that any IP datagram whose addresses

begin with 200.23.16.0/20 should be sent to it. The rest of the world need not that there are 8 other organizations each with its own IP network. This is called route aggregation

• Dividing an IP network into smaller IP networks as done in here is called subnetting. Each organization can further divide their IP address range into smaller IP subnets

14

Hierarchical addressing: more specific routes

ISPs-R-Us has a more specific route to Organization 1. By the longest prefix matching rule, packets with destination addresses beginning with 200.23.18.0/23 are sent to ISPs-R-Us

“Send me anythingwith addresses beginning 200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7Internet

Organization 1

ISPs-R-Us“Send me anythingwith addresses beginning 199.31.0.0/16or 200.23.18.0/23”

200.23.20.0/23Organization 2

...

...

15

IP addressing: the last word...Q: How does an ISP get block of

addresses?A: ICANN: Internet Corporation for Assigned

Names and Numbers– allocates addresses– manages DNS root servers– assigns domain names, resolves disputes

16

IP datagram format

ver length

32 bits

data (variable length,typically a TCP

or UDP segment)

16-bit identifier

Header checksum

time tolive

32 bit source IP address

IP protocol versionnumber

header length (bytes)

max numberremaining hops

(decremented at each router)

forfragmentation/reassembly

total datagramlength (bytes)

upper layer protocolto deliver payload to

head.len

type ofservice

“type” of data flgsfragment

offsetupper layer

32 bit destination IP address

Options (if any) E.g. timestamp,record routetaken, specifylist of routers to visit.

how much overhead with TCP?

• 20 bytes of TCP• 20 bytes of IP• = 40 bytes +

app layer overhead

17

Forwarding: Getting a datagram from source to dest.

IP datagram:

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A

BE

miscfields

sourceIP addr

destIP addr data

• datagram remains unchanged, as it travels from source to destination

Dest. Net. Next Hop Nhops

223.1.1 1223.1.2 223.1.1.4 2223.1.3 223.1.1.4 2

forwarding table in A

18

Starting at A, send IP datagram addressed to B:

• look up network address of B in forwarding table

• find B is on same net. as A• B and A are directly

connected– link layer will send

datagram directly to B inside link-layer frame. How?

Dest. Net. next router Nhops

223.1.1 1223.1.2 223.1.1.4 2223.1.3 223.1.1.4 2

miscfields223.1.1.1223.1.1.3data

forwarding table in A

Forwarding: Getting a datagram from source to dest.

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A

BE

19

Delivering the packet from A to B

• Starting at A, given IP datagram addressed to B:– look up net. address of B,

find B on same net. as A– link layer send datagram to B

inside link-layer frame– How does A know the MAC

address of B? ARP protocol

B’s MACaddr

A’s MACaddr

A’s IPaddr

B’s IPaddr

IP payload

datagramframe

frame source,dest address

datagram source,dest address

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A

BE

20

ARP: Address Resolution Protocol

• Each IP node (Host, Router) on LAN has ARP module, table

• ARP Table: IP/MAC address mappings for some LAN nodes

< IP address; MAC address; TTL>

< ………………………….. >

– TTL (Time To Live): time after which address mapping will be forgotten (typically 20 min)

Question: how to determineMAC address of Bgiven B’s IP address?

21

ARP protocol

• A knows B's IP address, wants to learn MAC address of B

• A broadcasts ARP query pkt, containing B's IP address – all machines on LAN receive ARP query

• B receives ARP packet, replies to A with its (B's) MAC address

• A caches (saves) IP-to-MAC address pairs until information becomes old (times out) – soft state: information that times out

(goes away) unless refreshed

22

Dest. Net. next router Nhops

223.1.1 1223.1.2 223.1.1.4 2223.1.3 223.1.1.4 2

Starting at A, dest. E:• look up network address of

E in forwarding table• E on different network

– A, E not directly attached

• routing table: next hop router to E is 223.1.1.4

• link layer sends datagram to router 223.1.1.4 inside link-layer frame

• datagram arrives at 223.1.1.4

• continued…..

miscfields223.1.1.1223.1.2.3 data

forwarding table in A

Forwarding: Getting a datagram from source to dest.

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A

BE

23

Arriving at 223.1.4, destined for 223.1.2.2

• look up network address of E in router’s forwarding table

• E on same network as router’s interface 223.1.2.9 – router, E directly

attached• link layer sends datagram

to 223.1.2.2 inside link-layer frame via interface 223.1.2.9

• datagram arrives at 223.1.2.2

miscfields223.1.1.1223.1.2.3 data Dest. Net router Nhops interface

223.1.1 - 1 223.1.1.4 223.1.2 - 1 223.1.2.9

223.1.3 - 1 223.1.3.27

forwarding table in router

Forwarding: Getting a datagram from source to dest.

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A

BE

24

– A creates IP packet with source A, destination E – A uses ARP to get R’s MAC address for 111.111.111.110– A creates Ethernet frame with R's MAC as dest, Ethernet

frame contains A-to-E IP datagram– A’s data link layer sends Ethernet frame to R – R’s data link layer receives Ethernet frame – R removes IP datagram from Ethernet frame, sees its

destined to E– R uses ARP to get E’s MAC address – R creates frame containing A-to-E IP datagram sends to E

A

RE

Another IP Packet Forwarding Example

25

IP packet forwarding algorithm

D = destination IP addressBool found = false;For each forwarding table entry (SubnetNumber, SubnetMask, NextHop)

do {D1 = SubnetMask & D;

if (D1 == SubnetNumber) { if (NextHop is an Interface) { Deliver the datagram directly to the destination within a LL frame } else { Deliver the datagram to NextHop (a router) } //end-else found = true; break; } //end-if} //end-for

If (!found) { if (there is a <default> router) Deliver the datagram to the <default> router else “Report: Destination unreachable”} //end-if

26

IP packet forwarding: Example

Subnet Number SubnetMask NextHop

128.96.34.0 255.255.255.128 Interface0128.96.34.128 255.255.255.128 Interface1128.96.33.0 255.255.255.0 R2

<default> R3

• Assume Destination IP = 128.96.34.68– D1 = 128.96.34.68 & 255.255.255.128 = 128.96.34.0 Deliver the

datagram to the destination over Interface0

• Assume Destination IP = 128.96.34.150– D1 = 128.96.34.150 & 255.255.255.128 = 128.96.34.128 Deliver

datagram to the destination over Interface1

• Assume Destination IP = 128.96.35.44– IP subnet will not match any of the known IP subnetsDeliver the

packet to the <default> router, R3

27

IP Fragmentation & Reassembly

• network links have MTU (max.transfer unit) - 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 datagramout: 3 smaller datagrams

reassembly

28

IP Fragmentation & Reassembly

ID=x

offset=0

fragflag=0

length=4000

ID=x

offset=0

fragflag=1

length=1500

ID=x

offset=1480

fragflag=1

length=1500

ID=x

offset=2960

fragflag=0

length=1040

One large datagram becomesseveral smaller datagrams

Example• 4000 byte datagram

– 3980 byte payload

• MTU = 1500 bytes– 1st packet payload:

1480 bytes– 2nd packet payload:

1480 bytes– 3rd packet payload:

1020 bytes

29

ICMP: Internet Control Message Protocol

• Used by hosts, routers, gateways to communication network-level information– error reporting:

unreachable host, network, port, protocol

– echo request/reply (used by ping)

• ICMP runs over IP:– ICMP msgs carried in IP

datagrams

• ICMP message: type, code plus first 8 bytes of IP datagram causing error

Type Code description0 0 echo reply (ping)3 0 dest. network unreachable3 1 dest host unreachable3 2 dest protocol unreachable3 3 dest port unreachable3 6 dest network unknown3 7 dest host unknown4 0 source quench (congestion control - not used)8 0 echo request (ping)9 0 route advertisement10 0 router discovery11 0 TTL expired12 0 bad IP header

30

Host IP Address Configuration

Q: How does a host get an IP address?– hard-coded by system admin in a file

• Wintel: control-panel->network->configuration->tcp/ip->properties

• UNIX: /etc/rc.config

– DHCP: Dynamic Host Configuration Protocol: dynamically get address from as server• “plug-and-play” (more shortly)

31

DHCP: Dynamic Host Configuration Protocol

Goal: allow host to dynamically obtain its IP address from network server when it joins networkCan renew its lease on address in useAllows reuse of addresses (only hold address while

connected an “on”Support for mobile users who want to join network (more

shortly)

DHCP overview:– host broadcasts “DHCP discover” msg– DHCP server responds with “DHCP offer” msg– host requests IP address: “DHCP request” msg– DHCP server sends address: “DHCP ack” msg

32

DHCP client-server scenario

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A

BE

DHCP server

arriving DHCP client needsaddress in thisnetwork

33

DHCP client-server scenarioDHCP server: 223.1.2.5 arriving

client

time

DHCP discover

src : 0.0.0.0, 68 dest.: 255.255.255.255,67yiaddr: 0.0.0.0transaction ID: 654

DHCP offer

src: 223.1.2.5, 67 dest: 255.255.255.255, 68yiaddrr: 223.1.2.4transaction ID: 654Lifetime: 3600 secs

DHCP request

src: 0.0.0.0, 68 dest:: 255.255.255.255, 67yiaddrr: 223.1.2.4transaction ID: 655Lifetime: 3600 secs

DHCP ACK

src: 223.1.2.5, 67 dest: 255.255.255.255, 68yiaddrr: 223.1.2.4transaction ID: 655Lifetime: 3600 secs

34

NAT: Network Address Translation

10.0.0.1

10.0.0.2

10.0.0.3

10.0.0.4

138.76.29.7

local network(e.g., home network)

10.0.0/24

rest ofInternet

Datagrams with source or destination in this networkhave 10.0.0/24 address for

source, destination (as usual)

All datagrams leaving localnetwork have same single source

NAT IP address: 138.76.29.7,different source port numbers

35

NAT: Network Address Translation

• Motivation: local network uses just one IP address as far as outside word is concerned:– no need to be allocated range of addresses from

ISP: - just one IP address is used 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).

36

NAT: Network Address Translation

10.0.0.1

10.0.0.2

10.0.0.3

S: 10.0.0.1, 3345D: 128.119.40.186, 80

1

10.0.0.4

138.76.29.7

1: host 10.0.0.1 sends datagram to 128.119.40, 80

NAT translation tableWAN side addr LAN side addr

138.76.29.7, 5001 10.0.0.1, 3345…… ……

S: 128.119.40.186, 80 D: 10.0.0.1, 3345

4

S: 138.76.29.7, 5001D: 128.119.40.186, 80

2

2: NAT routerchanges datagramsource addr from10.0.0.1, 3345 to138.76.29.7, 5001,updates table

S: 128.119.40.186, 80 D: 138.76.29.7, 5001

3

3: Reply arrives dest. address: 138.76.29.7, 5001

4: NAT routerchanges datagramdest addr from138.76.29.7, 5001 to 10.0.0.1, 3345

37

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

38

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, eg, P2P applications

– What about embedded IP addresses?• FTP, SMTP, SIP etc.

– address shortage should instead be solved by IPv6