IPv4 and IPv6

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Internet Protocol/
Network Layer
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Position of IP in TCP/IP protocol suite
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Internet Protocol (IP)
a) The IP is the host-to-host network layer delivery protocol for Internet.
b) IP is a connectionless datagram protocol for packet switching
network.
c) Best effort delivery service: meaning that IP provides no error control
or flow control.
d) IP uses only an error detection mechanism (Chechsum) and discards
the packet if it is corrupted.
e) IP does its best to deliver a packet to its destination, but with no
guarantees. Needs to rely on TCP layer for reliable delivery.
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Internet Protocol (IP)
 IP connectionless network-layer protocol.
 IP is based on datagram routing.
 IP is unreliable !!
 IP has a two-level (at least, traditionally) hierarchical
addressing mechanism.
 Routing in IP is a mixture of hierarchical, distance vector
and link state routing algorithms.
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Datagram
• Packets in the IP layer are called Datagrams.
• A datagram is a variable-length packet consisting of header and data:
• Header: 20 to 60 bytes in length containing essential information for routing and
addressing
• It is customary to display/show the
header in 4-byte sections.
• VER: defines the IP version (IPv4).
• HLEN: variable-length header; x4 to give
it’s the actual length in bytes
• Total length: of IP datagram in bytes;
• Data length = Total length – header length
• Time to live: to control max no. of hops
visited by a datagram by count decrement.
• Protocol: specifies final dest. protocol to
which IP datagram should be delivered;
TCP, UDP, ICMP, IGMP
• Checksum: covers only header field
for testing & debug5
IP datagram
Differentiate Service types
defines the class of datagram
for quality-of-service purpose.
E.g. min. delay, max.
throughput, max. reliability,
min. cost. (Not look into
detail).
Options: upto 40
bytes long, used
for testing and
debugging.
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IP datagram
Do not fragment !
IP Header Length in
terms of 4-bit words !
More fragments
Pointer to this
offsetting fragment
position in the
datagram.
Total Length of the IP
packet in terms of number of
bytes.
Datagram-ID. Useful in Fragmentation. Identifies each datagram from others.
When a datagram is fragmented, all fragments of a datagram must have the same
unique Identification number. The ID number helps destination in reassembling the datagram.
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Protocol field:
6
1
2
17
89
This field defines the higher level protocol (transport-layer) that uses the
services of IP-layer.
An IP datagram can encapsulate data from several higher-level protocols
such as: TCP=6, UDP=17, ICMP=1, IGMP=2, OSPF=89.
This field specifies the final destination protocol to which the IP datagram
should be delivered. Since IP multiplexes and demultiplexes data from
different higher level protocols, the value of this field helps in the
demultiplexing process when the datagram arrives at its final destination.
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Example 1
An IP packet has arrived with the first 8 bits (leftmost) as shown:
 01000010
The receiver discards the packet. Why?
Solution
There is an error in this packet. The 4 left-most bits (0100)
show the version, which is correct. The next 4 bits (0010)
show the header length, which means (2  4 = 8), which is
wrong. The minimum number of bytes in the header must be
20. The packet has been corrupted in transmission.
Note that only HLEN field uses (x4) to get the header length 9
Example 2
In an IP packet, the value of HLEN is 1000 in binary. How many
bytes of options are being carried by this packet?
Solution
The HLEN field value is 8, which means the total
number of bytes in the header is 8  4 or 32 bytes.
The first 20 bytes are the main header, the next 12
bytes are the options.
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Example 3
In an IP packet, the value of HLEN is 516 (or 0101) and the value of
the total length field is 002816 (or 00101000). How many bytes of
data are being carried by this packet?
Solution
The HLEN value is 5, which means the total number of
bytes in the header is 5  4 or 20 bytes (no options). The
total length is 28(hex) = 40 bytes, which means the packet
is carrying 20 bytes of data.
Data length = total length – header length = (40-20) = 20.
Only HLEN field uses (x4) to get header length
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Example 4
An IP packet has arrived with the first few hexadecimal digits as
shown below:
 45000028000100000102...................
How many hops can this packet travel before being dropped? The
data belong to what upper layer protocol?
Solution
To find the time-to-live field, we should skip 8 bytes (16
hexadecimal digits). 4 bits (binary) = 1 bit (hex); 1 byte = 8 bits binary = 2 bits (hex)
The time-to-live field is the ninth byte, which is 01. This
means the packet can travel only one hop. The protocol field is
the next byte (02), which means that the upper layer protocol
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is IGMP.
Checksum
To create the checksum the sender does
the following:
1. The packet is divided into k sections,
each of n bits.
2. All sections are added together using
one’s complement arithmetic.
3. The final result is complemented
to make the checksum.
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Checksum concept
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Checksum in one’s complement arithmetic
The checksum in the IP covers only the header, not the data.
At sender side, the initial checksum = 0000 0000, then the checksum is
carried out by summing up all the fields and applying one’s compliment
arithmetic to it to obtain the checksum value before it sends out the IP
datagram.
At receiver side, same procedure of checksum is carried out. The final value
of the result must be 0000 0000, otherwise it will be seen as error and
discarded.
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Decimal value representation
00000000 00000000
Example of
checksum
calculation
in binary
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Example of
checksum
calculation
in
hexadecimal
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Fragmentation
a) A datagram can travel through different networks.
b) Each router decapsulates the IP datagram from the frame it receives,
process it and then encapsulates it in another frame.
c) The format and size of the received frame depend on the protocol used
by the physical network through which the frame has just travelled.
d) The format and size of the sent frame depend on the protocol used by
the physical network through which the frame is going to travel.
e) E.g.: if a router connects an Ethernet and ATM networks, it receives a
frame in the Ethernet format and sends a frame in the ATM formats.
f) Fragmentation is the process of slitting the data into smaller chucks of
data (fragment) at the lower layer defined by its maximum transfer
size, and add the header to each of these smaller chucks of fragment.
g) To make the IP independent of the physical network, the packagers
decided to make the max length of IP datagram = largest maximum
transfer unit (MTU) defined so far as 65535 bytes.
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Maximum Transfer Unit (MTU)
Each data-link layer protocol has its own frame format.
One of the fields defined in the format is the maximum size of the data field.
In other words, when a datagram is encapsulated in a frame, the total size of
the datagram must be less then this maximum size.
When a datagram is fragmented, each fragment has its own header with most
of fields repeated, but some changed.
A fragmented datagram may itself be fragmented if it encounters a network
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with even smaller MTU. (Possible several fragmentation b4 final destination).
Fragmentation
a) A datagram can be fragmented by the source host or any router in
the path.
b) The reassembly of the datagram, however, is done only by the
destination host because each fragment becomes an independent
datagram.
c) Whereas the fragmented datagram can travel through different
routes, and we can never control or guarantee which route a
fragmented datagram may take
d) All the fragments belonging to the same datagram should arrive
at the destination host. So it is logical to do the reassembly at the
final destination.
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Fragmentation
The fields that are related to fragmentation and reassembly of an IP datagram
are: the identification, flags and fragmentation offset fields.
Identification: this field identifies a datagram originating from the source host.
When a datagram is fragmented, the value in the ID field is copied into all
fragments. This help the reassembly process at destination host.
Flags: This is a 3-bit field; 1st is reserved, 2nd is do not
fragment bit and 3rd is more fragment bit: if ‘1’ meaning
not the last fragment and more fragment after this one;
if ‘0’ meaning is the last fragment or only fragment.
Fragmentation Offset: The 13-bit field shows the relative position of this
fragment with respect to the whole datagram. It is the offset of the data in the
original datagram measured in units of 8 bytes.
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Fragmentation example
1st fragment
last byte number
2nd fragment
last byte number
3rd fragment
1st byte number
last byte number
This example shows a datagram with a data size of 4000 bytes fragmented
into 3 parts. The bytes in the original datagram are numbered 0 to 3999.
The value of offset is measured in units of 8 bytes because the length of the
offset field is only 13 bits long cannot represent a sequence more than 8192.
(65536/8 = 213). This force the hosts or routers that fragment datagrams to
choose the size of each fragment so that the 1st byte number is divisible by 22
8.
Fragmentation example
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Example 5
A packet has arrived with an M bit value of 0. Is this the first
fragment, the last fragment, or a middle fragment? Do we know if
the packet was fragmented?
Solution
If the M bit is 0, it means that there are no more
fragments; the fragment is the last one. However, we
cannot say if the original packet was fragmented or not. A
nonfragmented packet is considered the last fragment.
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Example 6
A packet has arrived with an M bit value of 1. Is this the first
fragment, the last fragment, or a middle fragment? Do we know if
the packet was fragmented?
Solution
If the M bit is 1, it means that there is at least one more
fragment. This fragment can be the first one or a middle
one, but not the last one. We don’t know if it is the first
one or a middle one; we need more information (the value
of the fragmentation offset). However, we can definitely
say the original packet has been fragmented because the M
bit value is 1.
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Example 7
A packet has arrived with an M bit value of 1 and a fragmentation
offset value of zero. Is this the first fragment, the last fragment, or a
middle fragment?
Solution
Because the M bit is 1, it is either the first fragment or a
middle one. Because the offset value is 0, it is definitely
the first fragment.
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Example 8
A fragment packet has arrived in which the offset value is 100.
What is the number of the first byte? Do we know the number of
the last byte?
Solution
To find the number of the first byte, we multiply the offset
value by 8. (100  8 = 800) This means that the first byte
number is 800. We cannot determine the number of the
last byte unless we know the length of the data.
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Example 9
A fragment packet has arrived in which the offset value is 100, the
value of HLEN is 5 and the value of the total length field is 100.
What is the number of the first byte and the last byte?
Fragment Offset
Solution
1st byte number
last byte number
100
The first byte number is 100  8 = 800.
Given that the total length = 100 bytes & the header length = 20 bytes
(5  4), hence the data length = 80 bytes in this datagram.
(Data length = total length – header length = 100 – 20 = 80)
If the first byte number is 800, the last byte number must 879.
(= 800 + 80 – 1 ; since it started from ‘0’ not ‘1’)
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IP components
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Internet Protocol version 6:
IPv6
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IPv6
a) Although IPv4 was well designed, it is clear that IPv4 is becoming
insufficient for the fast growing Internet and the needs:
• Due to growing demand of internet usage, the use of addressing space is
becoming insufficient for conventional netid & hostid two-level
addressing structure in Classful addressing.
• Real-time application i.e. video transmission requires minimum delay
strategies and reservation of resource not provided in IPv4
• Encryption and authentication of data and security mechanism not
provided in IPv4.
• Limitation of flexibility and mobility supported in IPv4.
b) While CIDR may buy a few years’ time, the days of IPv4 has already
been numbered.
c) In IPv6, the internet protocol was extensively modified to
accommodate the unforeseen growth of Internet.
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Improvement in IPv6
a) IPv6 has some advantages over IPv4:
• Longer addresses than IPv4: 128 bits long > 32 bits
currently in IPv4. That means a huge increase of
additional 296 addressing space.
• Simplification in Header: contains only 8 vital fields
< 13 in IPv4. (speeds up routing process). – 5 fields
becomes options.
• New options: to support additional functionalities
• Security: Authentication & privacy are key features
of new IP
• More Services: to handle special/real-time
multimedia traffic.
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IPv4 datagram
Totally remove bcs
fixed 40 bytes
header is used
Replaced by Hop
limit in IPv6
Replaced by
Flow label in
IPv6
Taken out bcs Next header in
IPv6 tells what follows the last
IP header (e.g. a UDP or TCP)
All replaced by functions in
both Next header & payload
extension header:
fragmentation in IPv6
Replaced by
payload length in
IPv6
Totally removed bcs calculating this greatly
reduce performances. Data-link and
Transport layer have their own checksum for
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better support of reliability,
Format of an IPv6 datagram
32 bits
Base Header:
Occupies 40 bytes
Payload:
Extension header
+
Data from upper layer
Occupies up to 65535
bytes
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IPv6 packet field
a) VER: IP version; 4 for IPv4 and 6 for IPv6
b) PRI: priority field defines priority of packets w.r.t. traffic congestion;
0-7: for packets that can slow down in event of congestion; 8-15 for
real-time traffic with constant sending rate – Audio & Video.
c) Flow label: (~ Service in IPv4) Special handling for a particular flow
of data; pseudo-connection btw source and dest. for reserved
bandwidth and stringent delay requirement. (Still experimental).
d) Payload length: tells how many payload bytes follow the 40-bytes
header; excluding the count for header based – 40 bytes not included
e) Next header: 8-bit field defines the subsequent headers that follow the
base header to allow extension; either optional extension headers used
by IP or header for an upper-layer protocol, e.g. UDP & TCP.
f) Hop limit: serves the same purpose as the TtL in IPv4.
g) Source and Destination address field: both occupy 16 bytes (128 bits)
h) Extension Headers: 6 extension headers which is optional in IPv4.
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Comparison between IPv4 and IPv6 packet headers
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Table 20.7 Priorities for congestion-controlled traffic
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Table 20.8 Priorities for noncongestion-controlled traffic
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IPv6 datagram
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Extension header format
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Table 20.6 Next header codes for IPv6
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Extension header types
Miscellaneous info for routers
Full or partial route to follow
Manage datagram fragment
Verification of sender’s ID
Info about encrypted contents
Additional info for destination
Same concept of fragmentation as IPv4. However, in IPv6, only original source can
fragment the packet using a path MTU discovery method to find the smallest MTU
supported by any network on the path. If not, then fragment to a size of 576 bytes, the
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minimum size of MTU required for each network connected to Internet.
Table 20.10 Comparison between IPv4 options and IPv6 extension headers
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Hop-by-hop option header format
Used for information that all routers along the path must examine.
Start by telling what kind of header come next and then the length and
size of datagram.
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Source routing
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Source routing example
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Fragmentation
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Authentication
Provide a mechanism for the receiver to confirm and be sure of the sender
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Encrypted security payload
When sender and receiver wish to communicate securely.
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IPv6 address
IPv4
4 bytes
32 bits
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Abbreviated address
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Abbreviated address with consecutive zeros
Two or more groups of 16 zeros can
be replaced by a pair of colons
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CIDR address
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Transition from IPv4 to IPv6
a) Because of huge number of systems on the Internet,
the transition from IPv4 to IPv6 cannot happen
suddenly.
b) It will take considerable amount of time before every
system in the Internet can move from IPv4 to IPv6
c) The transition should be a smooth process to prevent
any problems due to their differences.
d) 3 transition strategies:
• Dual stack
• Tunnelling
• Header translation
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Three transition strategies
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Dual stack
A station should simultaneously run both IPv4 & IPv6 protocols until all
Internet uses IPv6.
To determine which version to use, the source host queries the DNS and
send whichever version of IP packet the DNS returns.
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Tunneling
Tunnelling is a strategy used when two hosts using IPv6 want to connect with
each other when the packet must pass through a region that uses IPv4.
To pass thru the region, packet must have an IPv4 address. So, the IPv6
packet is encapsulate in an IPv4 packet when it enters the region, and IPv6
leaves its capsule when it exits the region.
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Header translation
Header translation is necessary when majority of the Internet has moved to
IPv6 but some systems still use IPv4.
The sender wants to use IPv6 but the receiver does not understand IPv6.
Tunnelling does not work because packet must be in the IPv4 format to be
understood by the receiver.
In this case, the header format must be changed totally through header
translation to down-convert from IPv6 to IPv4 header.
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Table 20.11 Header translation
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Further Reading
1- “Computer Networks”, Andrew Tanenbaum, 4th Ed. to learn more
about the generic network layer.
2- “Internetworking with TCP/IP vol.1”, Douglas Comer, 4th Ed.,
provides a detailed and comprehensive presentation of TCP/IP.
3- “Data Communications and Networking”, Behrouz Forouzan, 4th
Ed.,
Copyright Information : Some figures used in this presentation have been
either directly copied or adapted from several books.
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