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
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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
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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
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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?
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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
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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