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
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
Addressing
Packet size
Access mechanism
Timeouts
Error recovery
Flow control
Routing
User access control
Connection based or connectionless



Advantages



Flexibility
Robust
No unnecessary overhead
Unreliable


Not guaranteed delivery
Not guaranteed order of delivery

Packets can take different routes

Reliability is responsibility of next layer up
(e.g. TCP)

Go to Router X
MAC address for Router X
Encapsulated with LAN protocol
Encapsulated with X.25 protocol
IP PDU






Routing
Datagram lifetime
Fragmentation and re-assembly
Error control
Flow control



End systems and routers maintain routing tables

Indicate next router to which datagram should be sent


Static

May contain alternative routes
Dynamic

Flexible response to congestion and errors
Source routing

Source specifies route as sequential list of routers to be followed



Datagrams could loop indefinitely


Consumes resources
Transport protocol may need upper bound on datagram life
Datagram marked with lifetime




Time To Live field in IP
Once lifetime expires, datagram discarded (not forwarded)
Hop count

Decrement time to live on passing through a each router
Time count

Need to know how long since last router



Different packet sizes
When to re-assemble

At destination


Results in packets getting smaller as data traverses internet
Intermediate re-assembly



Need large buffers at routers
Buffers may fill with fragments
All fragments must go through same router
 Inhibits dynamic routing



IP re-assembles at destination only
Uses fields in header


Data Unit Identifier (ID)

Identifies end system originated datagram
 Source and destination address
 Protocol layer generating data (e.g. TCP)
 Identification supplied by that layer
Data length

Length of user data in octets



Offset

Position of fragment of user data in original datagram

In multiples of 64 bits (8 octets)
More flag

Indicates that this is not the last fragment

Application data
IP header is the same as origin
TCP header is not duplicated –
More is the same as original


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Part of TCP/IP

Used by the Internet
Specifies interface with higher layer
 e.g. TCP
Specifies protocol format and mechanisms
IPv4


 addresses are 32 bits wide
Its header is 20 bytes at minimum
Uses doted-decimal notation (e.g. 43.23.43.56)
IPv6




Provides larges address domain; addresses are 128 bits wide
Multiple separate headers are supported
Handles audio and video; providing high quality paths
Supports unicast, multicast, anycast

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Re-assembly may fail if some fragments get lost


Requires buffer
Need to detect failure – but how?
Re-assembly time out


Assigned to first fragment to arrive
If timeout expires before all fragments arrive, discard partial data
Use packet lifetime (time to live in IP)

If time to live runs out, kill partial data


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+
0
32
64
96
128
160
Source address
Destination address
Protocol

Recipient e.g. TCP
Type of Service

Specify treatment of data unit during transmission through networks
0 - 3
Version
4 - 7
Header length
Identification
Time to Live
8 - 15
Type of Service

Protocol
Source Address
Destination Address
Options + padding
Identification

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

Source, destination address and user protocol
Uniquely identifies PDU
Needed for re-assembly and error reporting
Send only
16 - 18 19 - 31
Total Length
Flags Fragment Offset
Header Checksum
192 Data


TTL (time-to-live) refers to the number of router hops the IP packet is allowed before it must be discarded.

Each router that receives a packet subtracts one from the count in the TTL field.

When the count reaches zero, the router detecting it discards the packet and sends an Internet Control
Message Protocol (ICMP) message back to the originating host

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
The IP datagram contains data and IP address
The IP datagram is encapsulated in a frame with physical address
The header changes as the frame goes from one network domain to the next
IP Datagram
Frame
Frame
Address
IP
Address
Data

Ethernet Frame Carrying IP Packet
MAC and
Associated
IP address

Ethernet Frame Carrying IP Packet
An Ethernet frame containing IP information has
08 00 in its type field
IP starting with 45 Hex indicates IPv4 with standard length of 20 bytes
IP starting with 4F Hex indicates IPv4 with standard length of 60 bytes
Remember: 2 4 =16; 45= 0100 0101 = One Byte
0000
Protocol Analyzer Display:
00 00 C0 A0 51 24 00 C0 93 21 88 A7
0020 B8 03 2a B4 DD …..
08 00 45 08
0010 00 5A DC 28 00 00 FF 01 88 08 C0 99 B8 01 C0 99
Example:
99 is one byte
1001 1001

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Two address types


Physical address (the frame has the physical address)


Embedded in the hardware (NIC, e.g., 00 00 11 00 11 33)
Also called the Media Address Control (MAC) address
Logical

IP datagram contains the logical IP address
To transport IP packets both physical and IP addresses must be known


Static address resolution
Dynamic address resolution
How to map physical and IP addresses



Local tables of IP addresses
Centralized directory
Address resolution mechanism
Address resolution mechanisms


ARP (address resolution protocol) – IP  48 bit Ethernet address
RARP (reverse) address resolution protocol are used to convert
MAC to IP address and vice versa



A network address is divided into Netid and Hostid
IP Address classification

( number of hosts per network )
Class
Class A
Class B
Class C
Class D (multicast)
Class E (reserved)
Leading bits
0
10
110
1110
1111
Network
Address (Netid)
7 bit ( 125 ) *
14 bit ( 16,368 )
21 bit ( 2,096,896 )
Host
Address (Hostid)
24 bit ( 16,777,151,750 )
16 bit ( 65,534 )
8 bit ( 254 )
* Some values are reserved!

Network
Address
Host Address
Multicast Address
Network Address Host Address
Reserved for Internet research
Network
Address
Host
Address

Q1: Determine the network address for the following IP addresses:
1- 84.42.58.11 (84 = 54 Hex = 0101 0100)
 Netid=84.0.0.0
 Class A
 Hostid=0.42.58.11
2- 144.54.67.5 (144 = 90 Hex = 1001 0000)
 Netid=144.62.0.0
 Class B
 Hostid=0.0.67.5
Q2: What type of IP address classification will a large organization with 1000 individual users in 150 dispersed buildings use?  Class B



Routing packets requires having knowledge about the network


Partial (know your own neighbors)
Full (know the entire network elements)
Retrieving network information (network discovery protocols)


RIP (routing information protocol): routing based on the least number of hops
OSPF (open shortest path first): Routing based on number of hops, link speed, congestion

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Bridge

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
IS used to connect two LANs using similar LAN protocols
Address filter passing on packets to the required network only
OSI layer 2 (Data Link)
Router



Connects two (possibly dissimilar) networks
Uses internet protocol present in each router and end system
OSI Layer 3 (Network)

00 10 00 00
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Each LAN assigned subnet number
Site looks to rest of internet like single network
Host portion of address partitioned into subnet number and host number
Local routers route within subnetted network
Subnet mask indicates which bits are subnet number and which are host number
Allow arbitrary complexity of internetworked
LANs within organization
Insulate overall internet from growth of network numbers and routing complexity

IP Address of B: 192.228.17.57
IP Address of A: 192.228.17.33
IP Address of X: 192.228.17.32
00 1 | 1 10 01
00 1 | 0 00 01
00 1 | 0 00 00
3-bit subnet
5-bit host
Subnet mask: 255.255.255.224
11 10 00 00
Note: if we XOR IP Address of B & Subnet Mask
We will have:
00 11 10 01 XOR 00 10 00 00 = 00 01 10 01
Host number is 25  00 01 10 01;
192  1100 0000, hence, Class C network (8-bit host/subnet) !
NetId SubNetId Host

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Address space exhaustion

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Two level addressing (network and host) wastes space
Network addresses used even if not connected to Internet
Growth of networks and the Internet
Extended use of TCP/IP
Single address per host
Requirements for new types of service

Note:
IPv5 used for Stream
Protocol- IP-layer protocol that provides end-to-end guaranteed service across a network.
Features:
Extended address space
Improved option mechanism
Dynamic address assignment
Multicasting and anycasting
Flow routing
128 bits
128 bits


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
Over half million networks are connected to the
Internet
Network numbers are managed by ICNN
(Internet Corporation for Assigned Names and
Numbers) http://www.icann.org/

Delegates part of address assignments to regional authorities

IP addresses are given to ISPs and companies
IP addresses are based on dotted decimal notation: 192.41.7.32



IP address 0.0.0.0 refer to machine’s own network when it is being booted (This host)
255.255.255.255 broadcast on the LAN
127.x.y.z reserved for loopback testing




Routers can be connected to multiple LANs
LANS are divided into subnets each identified by a subnet mask: 255.255.252.0 (… 1111 1100 0000 0000)  32-
10=22 to identify the subnet!


Mask: netID + SunnetID or /22 (subnet mask is 22 bit long)
Subnets are not visible outside the network
Example: Assume subnet mask is 255.255.252.0/22

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
Subnet 1: 130.50.4.1
Subnet 2: 130.50.8.1

Subnet 3: 130.50.12.1

…000010|00 00000001
…000011|00 00000001
Assume a packet’s destination is 130.50.15.6
 which subnet does it belong to?
Mask: … 1111 11|00 0000 0000
Adrs: … 0000 11|11 0000 0110
Hence: the packet must go to Subnet 3





Classful routing is not very efficient
Having IP address classes creates issues

Addresses can be under utilized (Class A)

Addresses can be over utilized (Class C)


Management of addresses may be difficult
Organizations can grow!
Classless InterDomain Routing (CIDR)

Allocate remaining IP addresses in variable-sized block; no regard to class!


Use 32-bit mask!
Uses a single routing table

What happens if a packet has an address of 194.24.17.4
?
Where does it go?
O: ….1111 0000 0000 0000 AND 0001 0001 0000 0100 
…. 0001 0000 0000 0000 (194.24.16.0)
How do you represent class B using CIDR?
16-bit NetID + 16-bit HostID  /16




Addresses are growing! What is the solution?


Use IPV6
Use NAT
NAT:


Allows using one IP address per company
Internally new nodes can be added!
How?




IP reserved addresses



10.x.y.z
172.16.x.y
192.168.x.y
Receiving a packet from the
Internet


Sender



Add IP address
TCP will have the destination port
(0-1023)
The port determines which server on the remote (destination) side to process the packet
NAT box:

Using the PORT address in TCP, change the IP address to a designated address (10.0.01)
Sending a packet into the Internet

NAT box:

Changes 10.0.0.1 to 198.60.42.12

198.60.42.12






Addresses are not unique: many 10.0.0.1!
NAT controls are the incoming and outgoing packets – reliability!
NAT accesses TCP and IP layers – layers should work independent of one another
NAT only allows TCP/IP or UDP/IP
NAT does not support applications which insert the IP address in the body (FTP or
H.323)



IP protocol only deal with the data transfer
We need control protocol to check the network

ICMP, ARP, RARP, BOOTP, HDCP


ICMP error messages are used by routers and hosts to tell a device that sent a datagram about problems encountered in delivering it



Used to test the network
Messages are encapsulated in the IP packet
Has many message types
Code field is used for subtypes





Exploits broadcast property of a LAN
Each host on LAN maintains a a table of IP subnetwork addresses
If the address can not be found ARP broadcasts a request

Shouting: Who know about this IP address?
Other hosts listen and reply

The reply includes IP address and MAC

Any interested host can learn about the new information


Assume 1 is sending a message to 2 ( rose@sonoma.edu
)





Sonoma.edu is the host
Host 1 sends a message to Domain Name System (DNS): what is the IP address for Sonoma.edu?  192.31.65.5
What is the MAC address for 192.31.65.5? Use ARP broadcast!

Host 2 respond: it is E2
Host 1 maps IP and MAC; encapsulate the IP message in the Ethernet frame and send it
Cashing can enhance ARP operation


Assume 1 is sending a message to 4 ( rose@sonoma.edu
)



Sonoma.edu is the host
Host 1 sends a message to Domain Name System (DNS): what is the IP address for Sonoma.edu?  192.31.65.5 What is the MAC address for 192.31.65.5? ARP cannot pass through the router!
Two choices:

Reconfigure routers to response to ARP (Proxy ARP)

Send the message to the LAN router (E3)  F1  F3  4 – Each router looks are the IP address and passes it to the next node using the routing table





Reverse ARP translates the Ethernet address to
IP address

A diskless machine when it is booking can ask: My MAC is 12.03.23.43.23.23; what is my IP?
RARP broadcasts the question (destination address is all one)

Not passed through the router!
Each LAN needs a RARP server!
Bootstrap protocol uses UDP and forwards over routers

Mapping must be done manually in each router!

Uses one server but harder to implement!





DHCP allows many and automatic configuration
Replaces BOOTP and RARP
DHCP sends a DISCOVER Packet

What is the IP address for this MAC?
Even when an IP address is assigned, how long is it good for?

Before the IP address is removed find another IP address….called Leasing




This is My MAC; what is my IP address?
RARP / DHCP
This is the destination host name, what it is IP address? DNS Server
This is the IP address, what is your` MAC address? ARP




Tanenbaum
Tomasi Text Book
Comer Text book