Network Layer UNIT 3

8/18/2019 Network Layer UNIT 3

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McGraw-Hill © The McGraw-Hill Companies, Inc., 2000

Network Layer


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!osition o" the Network

Layer

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19.3

Issues at the Network Layer

•Addressing

IP addresses on the Internet

• SwitchingDatagram packet switching

Virtual circuit packet switching

•


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19.6


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How Networks

#i$er

%ome o" the many ways networks can &i$er.


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IPv4 ADDRESSESIPv4 ADDRESSES

An An IPv4 address IPv4 address is ais a 32-bit 32-bit address that uniquely andaddress that uniquely and

universally defines the connection of a device (foruniversally defines the connection of a device (for

example, a computer or a router to the Internet!example, a computer or a router to the Internet!

Address Space

Notations

!lassful Addressing!lassless Addressing

Network Address "ranslation #NA"$

"opics discussed in this section#"o pics discussed in this section#


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An IPv4 address is 32 bits long.

$ote


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The IPv4 addresses are uniueand universal.

$ote


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The address s!a"e o# IPv4 is232 or 4$294$967$296.

$ote


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%otted-decimal notation and binary notation for an IPv4 address


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Class"'l

(&&ressin)

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In "lass#ul addressing$ the addresss!a"e is divided into #ive "lasses%

A$ &$ '$ ($ and ).

$ote


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&indin' the classes in binary and dotted-decimal notation


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$umber of blocs and bloc si)e in classful IPv4 addressin'


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In "lass#ul addressing$ a large !art o#the available addresses *ere *asted.

$ote


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%efault mass for classful addressin'


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19.20

'lass#ul addressing$ *hi"h is al+ostobsolete$ is re!la"ed *ith "lassless

addressing.

$ote


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Classless (&&ressin)

To overcome address depletion, classlessaddressing was designed.

No classes, but addresses are granted in

blocks. Rules

Addressees in blocks must be contiguous

The number of addresses in block must be power of 2

(1,2,,!,1",#$ The first address must be evenl% divisible b% the

number of addresses.

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A bloc of *+ addresses 'ranted to a small or'ani)ation


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Mask

A mask is 32 bit number in which n leftmost bits are 1s and the 32-n rightmost bits are 0s.

Range from & to '2.

A block of addresses can be defined as .%.).t *n , in which .%.).t defines one of the

addresses and the *n defines the mask .

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19.24

The #irst address in the blo", "an be#ound b- setting the right+ost

32 n bits to 0s.

$ote


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19.25

The last address in the blo", "an be#ound b- setting the right+ost

32 n bits to 1s.

$ote

l


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19.26

x! A bloc of addresses is 'ranted to small or'ani)ation!

ne of the address is 2./!*+!30!312

&ind the last address for the bloc!

olution"he binary representation of the 'iven address is

**..**.* ...*.... ..*..*.* ..*..***

If 5e set 32 6 2 ri'htmost bits to *, 5e 'et

**..**.* ...*.... ..*..*.* ..*.**** or

2./!*+!30!40

"his is actually the bloc sho5n in &i'ure *1!3!

xample


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19.27

The nu+ber o# addresses in the blo","an be #ound b- using the #or+ula

232n.

$ote

l *1 1


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19.28

Another 5ay to find the first address, the last address, and

the number of addresses is to represent the mas as a 32-bit binary (or -di'it hexadecimal number! "his is

particularly useful 5hen 5e are 5ritin' a pro'ram to find

these pieces of information! In xample *1!/ the 2 can

be represented as******** ******** ******** ****....

(t5enty-ei'ht *s and four .s!

&ind

a! "he first address

b! "he last address

c! "he number of addresses!

xample *1!1

l *1 1 ( ti d


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olutiona! "he first address can be found by A$%in' the 'iven

addresses 5ith the mas! A$%in' here is done bit by

bit! "he result of A$%in' 2 bits is * if both bits are *s7

the result is . other5ise!

xample *1!1 (continued

l *1 1 ( ti d


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19.30

b! "he last address can be found by 8in' the 'iven

addresses 5ith the complement of the mas! 8in'

here is done bit by bit! "he result of 8in' 2 bits is . if

both bits are .s7 the result is * other5ise! "he

complement of a number is found by chan'in' each * to . and each . to *!


l *1 1 ( ti d


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c! "he number of addresses can be found by complementin' the mas, interpretin' it as a decimal

number, and addin' * to it!



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19.32

A net5or confi'uration for the bloc 2./!*+!30!322


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19.33

The #irst address in a blo", isnor+all- not assigned to an- devi"e/

it is used as the net*or, address that

re!resents the organiation

to the rest o# the *orld.

$ote


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19.34

"5o levels of heirachy in an IPv4 Address


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19.35

)a"h address in the blo", "an be

"onsidered as a t*olevel

hierar"hi"al stru"ture%

the le#t+ost n bits !re#i de#ine

the net*or,/

the right+ost 32 n bits de#inethe host.

$ote


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%'*nettin)+

hat are the maor reasons "or s'*nettin) orse)mentin) yo'r network

. To &i/i&e a lar)e network into smaller se)ments tore&'ce trac an& spee& 'p the sections o" yo'r

network.2. To connect networks across )eo)raphical areas.

1. To connect &i$erent topolo)ies s'ch as thernet, Token3in), an& 4##I to)ether /ia ro'ters.

5. To a/oi& physical limitations s'ch as ma6im'm ca*le

len)ths or e6cee&in) the ma6im'm n'm*er o"comp'ters on a se)ment.

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The ABCD University is planning to deploy an IPnetwork in their main campus .Its main campus

location includes only a single physical building with 4 floors and 8 total departments.

Each of the different departments needs to beseparate and have their own IP address space.

The address space that has been allocated to theUniversity by their Internet Service Provider (ISP) is172.16.0.0/23.

Each of the different departments requires at least

40 different usable addresses and at least 10 extraaddresses allocated for future growth.

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)a+!le et*or,

(esign the IP address range #or the et*or,


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*elow shows the allocate& space )i/en *y the I%! an& how it is&isplaye& in *inary.

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The re7'irements state& that each o" the 8 &epartments nee&e& at aminim'm 50 a&&resses with an a&&itional 0 allocate& "or "'t're)rowth "or a total o" 90 re7'ire& a&&resses assi)ne& per &epartment.Gi/en this re7'irement, what is the smallest s'*net that wo'l& *e

re7'ire& 4i)'re *elow shows that the smallest s'*net a/aila*lewo'l& *e allocatin) each network ;5 total a&&resses.

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The ne6t thin) to 4i)'re o't iswhether the n'm*er o" the a&&ressesallocate& *y the I%! is eno')h to

meet the re7'irements o" theor)ani


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Network Range Calculation

4irst %'*network 3an)e19.41

Now that it has been calculated that the ISPs allocation of addresses is enough to meet the

requirements of the organization, the next task is to come up with the different ranges that will be used

to allocate to each department.shows the calculation of the first range using the !!.!!.!!."# $%&' subnet mask calculated


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The secon& s'*network will *e)in where the rst le"t o$ at =2.;.0.;5 an& )o'p to =2.;.0.2=> this ran)e is shown.


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The thir& &epartment+ =2.;.0.28 thro')h=2.;.0.D

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(n& so onE..

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h i h h & &


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The i)hth &epartment+ =2.;..D2 an& )o 'p to=2.;..299

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"hree le%el hierarchy

+uppose an organi)ation has three offices and

needs to divide the addresses into ' subblocks

of &'()*()* addresses. rgani)ation is given

the block )+,)',)-,./'*

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"h l l hi h


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19.49

"hree le%el hierarchy1. +uppose the mask for 1st subnet is n1 then 2'2-n1 must be

'2, which means n12/

2. 0ask for 2nd subnet is n2 then 2'2-n2 must be 1", hence

n22!

'. 0ask for 'rd subnet is n' then 2'2-n' must be 1", hence

n'2!


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Note0 1e ha%e the masks '+( '2( '2

for su3nets with the organi4ation

mask 3eing '*

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19.51

5igure )6,2 "hree-level hierarchy in an IPv4 address

Su3netting


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9onfi'uration and addresses in a subnetted net5or


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7ore le%els of hierarchy

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xample *1!*.


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An IP is 'ranted a bloc of addresses startin' 5ith

*1.!*..!.!.*+ (+/,/3+ addresses! "he IP needs to

distribute these addresses to three 'roups of customers as

follo5s#

a! "he first 'roup has +4 customers7 each needs 2/+ addresses!

b! "he second 'roup has *2 customers7 each needs *2

addresses!

c! "he third 'roup has *2 customers7 each needs +4 addresses!

%esi'n the subblocs and find out ho5 many addresses

are still available after these allocations!

xample *1!*.

xample *1!*. (continued


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19.55

olution

&i'ure *1!1 sho5s the situation!

xample *1!*. (continued

:roup *

&or this 'roup, each customer needs 2/+ addresses! "his

means that (lo'2 2/+ bits are needed to define eachhost! "he prefix len'th is then 32 6 ; 24! "he addresses

are


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19.58

5igure )6,6 An example of address allocation and distribution by an IP

N t k (&& T l ti


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Network (&&ress Translation

N'm*er o" Internet 'sers areincreasin).

There is short "all o" a&&resses.

%ol'tion Network (&&ress Translation

Lar)e a&&resses internally %in)le or small set o" a&&resses )lo*ally.

Fni7'e insi&e , *'t not )lo*ally.

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19.60

"a3le Addresses for private net5ors


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$A" implementation


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19.62

5igure )6,)) Addresses in a $A"


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5igure )6,)' $A" address translation


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"a3le )6,- &ive-column translation table

5i )6 )&


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19.65

5igure )6,)& An IP and $A"


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20.66

INTERNETWORKINGINTERNETWORKING

In this section, 5e discuss internet5orin', connectin' net5ors In this section, 5e discuss internet5orin', connectin' net5ors

to'ether to mae an internet5or or an internet!to'ether to mae an internet5or or an internet!

Need for Network Layer

Internet as a Datagram Network

Internet as a !onnectionless Network



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5igure '.,)


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$et5or layer in an internet5or


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$et5or layer at the source, router, and destination


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*it"hing at the net*or, la-er in the Internet uses the datagra+

a!!roa"h to !a",et s*it"hing.

$ote


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'o++uni"ation at the net*or, la-er in the Internet is "onne"tionless.

$ote


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20.72

IPv4 IPv4

The Internet Protocol version 4 ( IPv4 ) is the delivery mechanism

used by the TCPIP !rotocols"

Datagram

5ragmentation

!hecksum

8ptions



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5igure '.,- Position of IPv4 in "9PIP protocol suite


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IPv4 data'ram format


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5igure '.,* ervice type or differentiated services


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The !re"eden"e sub#ield *as !art o# version 4$ but never used.

$ote


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"a3le '.,) "ypes of service

"a3le '.,' %efault types of service


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"a3le '.,& =alues for codepoints


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The total length #ield de#ines the total length o# the datagra+ in"luding

the header.

$ote


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5igure '.,+ ncapsulation of a small data'ram in an thernet frame


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Protocol field and encapsulated data


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Protocol values

xample 2.!*


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An IPv4 pacet has arrived 5ith the first bits as sho5n#

.*....*.

"he receiver discards the pacet! >hy?

olution

There is an error in this packet. The 4 leftmost bits (0100) show the version, which is correct. The

next 4 bits (0010) show an invalid header length (2 × 4 = 8). The minimum number of bytes in the

header must be 20 . The packet has been corrupted in transmission.

xample 2.!2


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20.85

In an IPv4 pacet, the value of @


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xample 2.!4


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An IPv4 pacet has arrived 5ith the first fe5 hexadecimal di'its as sho5n!

.x4/....2...*.....*.2 ! ! !

@o5 many hops can this pacet travel before bein' dropped? "he data belon' to 5hat upper-layer

protocol?

olution

To find the time-to-live field, we skip 8 bytes. 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 is IGMP.


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5igure '.,6 Baximum transfer unit (B"C


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"a3le '.,9 B"Cs for some net5ors


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dentification


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dentification

dentifies the datagram originating from the

source

ombination of identification and 3 address

uni4uel% identifies the datagram.

5ses a counter to label.

6hen a datagram is fragmented, value in the

identification field is copied in all the fragments.

7elps in re-assembl%

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20.93


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RE-ASSEMBLE?

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19.95

1 2

34

3e-assem*le


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3e assem*le

8irst fragment has offset e4ual to 9ero :ivide the length of first fragment b% !, to get

2nd fragment offset.

:ivide the total length of 1st

and 2nd

b% ! to get'rd fragment offset.

ontinue the process, last fragment has more

value set to &.

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19.97

xample 2.!/


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20.98

A pacet has arrived 5ith an B bit value of .! Is this the first fra'ment, the last fra'ment, or a

middle fra'ment? %o 5e no5 if the pacet 5as fra'mented?

olution

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 non-fragmented packet is considered

the last fragment.

xample 2.!+


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20.99

A pacet has arrived 5ith an B bit value of *! Is this the first fra'ment, the last fra'ment, or a

middle fra'ment? %o 5e no5 if the pacet 5as fra'mented?

olution

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).

xample 2.!0


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A pacet has arrived 5ith an B bit value of * and a fra'mentation offset value of .! Is this the first

fra'ment, the last fra'ment, or a middle fra'ment?

olution

Because the M bit is 1, it is either the first fragment or a middle one. Because the offset value is 0, it

is the first fragment.

xample 2.!


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20.101

A pacet has arrived in 5hich the offset value is *..! >hat is the number of the first byte? %o 5e

no5 the number of the last byte?

olution

"o find the number of the first byte, 5e multiply the offset value by ! "his means that the first byte

number is ..! >e cannot determine the number of the last byte unless 5e no5 the len'th!


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xample 2.!*.


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20.105

&i'ure 2.!*3 sho5s an example of a checsum calculation for an IPv4 header 5ithout options! "he

header is divided into *+-bit sections! All the sections are added and the sum is complemented! "he

result is inserted in the checsum field!

xample of checsum calculation in IPv4


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20.106

xample of checsum calculation in IPv4


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20.107

"axonomy of options in IPv48ptions field is used for network testing and de3ugging

A record route option is used to record the nternet routers that

handle the datagram.

t can list up to nine router addresses. t can be used for debuggingand management purposes.


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20.108


A stri"t sour"e route o!tion is used b- the sour"e to

!redeter+ine a route #or the datagra+ as it travelsthrough the Internet


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20.109


)a"h router in the list +ust be visited$ but the datagra+

"an visit other routers as *ell.


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A ti+esta+! o!tion is used to re"ord the ti+e o#

datagra+ !ro"essing b- a router.

Limitations o" I!/5


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IPv6 ADDRESSESIPv6 ADDRESSES


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19.112

%espite all short-term solutions, address depletion is %espite all short-term solutions, address depletion isstill a lon'-term problem for the Internet! "his andstill a lon'-term problem for the Internet! "his and

other problems in the IP protocol itself have been theother problems in the IP protocol itself have been the

motivation for IPv+!motivation for IPv+!

Structure

Address Space


Advantages of 3v"


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g

Larger Address Space :etter Header 5ormat

New 8ptions

Allowance for e;tensions

Support for resource Allocation

Support for more Security

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19.114

An IPv6 address is 128 bits long.

$ote

5igure )6,)- IPv+ address in binary and hexadecimal colon notation


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19.115

5igure )6,)9 Abbreviated IPv+ addresses


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19.116

xample *1!**


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19.117

xpand the address .#*/##*#*2#*2*3 to its ori'inal!

olution

>e first need to ali'n the left side of the double colon to

the left of the ori'inal pattern and the ri'ht side of the

double colon to the ri'ht of the ori'inal pattern to findho5 many .s 5e need to replace the double colon!

"his means that the ori'inal address is!

(&&ress %pace


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(&&ress %pace

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228

1.5 6018


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19.119

"ype prefixes for IPv+ addresses

"ype prefixes for IPv+ addresses# %efines the cate'ory!

"a3le )6,9 "ype prefixes for IPv+ addresses (continued


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19.120

Cnicast Address#Provider based address# used for


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19.121

'lobal communication over the Internet!

North America

"&( ) (*

T-!e "ode 3 &it

010 Provider based identi#ier

Cnicast Address#Provider based address


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19.122


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Anycast Address# Pacet is delivered to any one of the

ibl d i h


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'roup! Possibly nearest node in that 'roup !

8esearved Address

5igure )6,)6


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19.125

Used when an organization wants to use these addresses

without connecting to the internet.


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5i '. )9 IP + d t h d d l d


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20.127

5igure '.,)9 IPv+ data'ram header and payload

&ormat of an IPv+ data'ram


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$ext header codes for IPv+


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5low La3el


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xtension header types

9omparison bet5een IPv4 and IPv+ pacet headers


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9omparison bet5een IPv4 options and IPv+ extension headers


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TRANSITION FROM IPv4 TO IPv6TRANSITION FROM IPv4 TO IPv6


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20.135

Decause of the hu'e number of systems on the Internet, the transition Decause of the hu'e number of systems on the Internet, the transition

from IPv4 to IPv+ cannot happen suddenly! It taes a considerable from IPv4 to IPv+ cannot happen suddenly! It taes a considerable

amount of time before every system in the Internet can move from IPv4amount of time before every system in the Internet can move from IPv4

to IPv+! "he transition must be smooth to prevent any problems bet5eento IPv+! "he transition must be smooth to prevent any problems bet5een

IPv4 and IPv+ systems! IPv4 and IPv+ systems!

Dual Stack

"unneling

Header "ranslation



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"hree transition strate'ies


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%ual stac


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"unnelin' strate'y


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@eader translation


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Network Layer UNIT 3

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