Fiber Optical Networks and Data Communication
FarhanGhafoor7
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Jul 03, 2024
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About This Presentation
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Slide Content
Software Engineering Departments
COE (College of Engineering
Lecture 11
Network Layer:
Logical Addressing
Computer Communication and Networks
Engr. Farhan Ghafoor
COE (College of Engineering
Lecture 11
Network Layer:
Logical Addressing
Computer Communication and Networks
Engr. Farhan Ghafoor
19.2
19-1 IPv4 ADDRESSES
AnIPv4addressisa32-bitaddressthatuniquelyand
universallydefinestheconnectionofadevice(for
example,acomputerorarouter)totheInternet.
Address Space
Notations
Classful Addressing
Classless Addressing
Network Address Translation (NAT)
Topics discussed in this section:
19-1 IPv4 ADDRESSES
AnIPv4addressisa32-bitaddressthatuniquelyand
universallydefinestheconnectionofadevice(for
example,acomputerorarouter)totheInternet.
Address Space
Notations
Classful Addressing
Classless Addressing
Network Address Translation (NAT)
Topics discussed in this section:
19.3
An IPv4 address is 32 bits long.
Note
An IPv4 address is 32 bits long.
Note
19.4
The IPv4 addresses are unique
and universal.
Note
The IPv4 addresses are unique
and universal.
Note
19.5
The address space of IPv4 is
2
32
or 4,294,967,296.
Note
The address space of IPv4 is
2
32
or 4,294,967,296.
Note
19.6
Figure 19.1 Dotted-decimal notation and binary notation for an IPv4 address
Figure 19.1 Dotted-decimal notation and binary notation for an IPv4 address
19.7
Numbering systems are reviewed in
Appendix B.
Note
Numbering systems are reviewed in
Appendix B.
Note
19.8
ChangethefollowingIPv4addressesfrombinary
notationtodotted-decimalnotation.
Example 19.1
Solution
Wereplaceeachgroupof8bitswithitsequivalent
decimalnumber(seeAppendixB)andadddotsfor
separation.
ChangethefollowingIPv4addressesfrombinary
notationtodotted-decimalnotation.
Example 19.1
Solution
Wereplaceeachgroupof8bitswithitsequivalent
decimalnumber(seeAppendixB)andadddotsfor
separation.
19.9
ChangethefollowingIPv4addressesfromdotted-decimal
notationtobinarynotation.
Example 19.2
Solution
Wereplaceeachdecimalnumberwithitsbinary
equivalent(seeAppendixB).
ChangethefollowingIPv4addressesfromdotted-decimal
notationtobinarynotation.
Example 19.2
Solution
Wereplaceeachdecimalnumberwithitsbinary
equivalent(seeAppendixB).
19.10
Findtheerror,ifany,inthefollowingIPv4addresses.
Example 19.3
Solution
a.There must be no leading zero (045).
b.There can be no more than four numbers.
c.Each number needs to be less than or equal to 255.
d.A mixture of binary notation and dotted-decimal
notation is not allowed.
Findtheerror,ifany,inthefollowingIPv4addresses.
Example 19.3
Solution
a.There must be no leading zero (045).
b.There can be no more than four numbers.
c.Each number needs to be less than or equal to 255.
d.A mixture of binary notation and dotted-decimal
notation is not allowed.
19.11
In classful addressing, the address
space is divided into five classes:
A, B, C, D, and E.
Note
In classful addressing, the address
space is divided into five classes:
A, B, C, D, and E.
Note
19.12
Figure 19.2 Finding the classes in binary and dotted-decimal notation
Figure 19.2 Finding the classes in binary and dotted-decimal notation
19.13
Findtheclassofeachaddress.
a.00000001000010110000101111101111
b.11000001100000110001101111111111
c.14.23.120.8
d.252.5.15.111
Example 19.4
Solution
a.The first bit is 0. This is a class A address.
b.The first 2 bits are 1; the third bit is 0. This is a class C
address.
c.The first byte is 14; the class is A.
d.The first byte is 252; the class is E.
Findtheclassofeachaddress.
a.00000001000010110000101111101111
b.11000001100000110001101111111111
c.14.23.120.8
d.252.5.15.111
Example 19.4
Solution
a.The first bit is 0. This is a class A address.
b.The first 2 bits are 1; the third bit is 0. This is a class C
address.
c.The first byte is 14; the class is A.
d.The first byte is 252; the class is E.
19.14
Table 19.1 Number of blocks and block size in classful IPv4 addressing
Table 19.1 Number of blocks and block size in classful IPv4 addressing
19.15
In classful addressing, a large part of the
available addresses were wasted.
Note
In classful addressing, a large part of the
available addresses were wasted.
Note
19.16
Table 19.2 Default masks for classful addressing
Table 19.2 Default masks for classful addressing
19.17
Classful addressing, which is almost
obsolete, is replaced with classless
addressing.
Note
Classful addressing, which is almost
obsolete, is replaced with classless
addressing.
Note
19.18
Figure19.3showsablockofaddresses,inbothbinary
anddotted-decimalnotation,grantedtoasmallbusiness
thatneeds16addresses.
Wecanseethattherestrictionsareappliedtothisblock.
Theaddressesarecontiguous.Thenumberofaddresses
isapowerof2(16=2
4
),andthefirstaddressisdivisible
by16.Thefirstaddress,whenconvertedtoadecimal
number,is3,440,387,360,whichwhendividedby16
resultsin215,024,210.
Example 19.5
Figure19.3showsablockofaddresses,inbothbinary
anddotted-decimalnotation,grantedtoasmallbusiness
thatneeds16addresses.
Wecanseethattherestrictionsareappliedtothisblock.
Theaddressesarecontiguous.Thenumberofaddresses
isapowerof2(16=2
4
),andthefirstaddressisdivisible
by16.Thefirstaddress,whenconvertedtoadecimal
number,is3,440,387,360,whichwhendividedby16
resultsin215,024,210.
Example 19.5
19.19
Figure 19.3 A block of 16 addresses granted to a small organization
Figure 19.3 A block of 16 addresses granted to a small organization
19.20
In IPv4 addressing, a block of
addresses can be defined as
x.y.z.t /n
in which x.y.z.t defines one of the
addresses and the /ndefines the mask.
Note
In IPv4 addressing, a block of
addresses can be defined as
x.y.z.t /n
in which x.y.z.t defines one of the
addresses and the /ndefines the mask.
Note
19.21
The first address in the block can be
found by setting the rightmost
32 − nbits to 0s.
Note
The first address in the block can be
found by setting the rightmost
32 − nbits to 0s.
Note
19.22
Ablockofaddressesisgrantedtoasmallorganization.
Weknowthatoneoftheaddressesis205.16.37.39/28.
Whatisthefirstaddressintheblock?
Solution
Thebinaryrepresentationofthegivenaddressis
11001101 00010000 00100101 00100111
If we set 32−28 rightmost bits to 0, we get
11001101 00010000 00100101 0010000
or
205.16.37.32.
This is actually the block shown in Figure 19.3.
Example 19.6
Ablockofaddressesisgrantedtoasmallorganization.
Weknowthatoneoftheaddressesis205.16.37.39/28.
Whatisthefirstaddressintheblock?
Solution
Thebinaryrepresentationofthegivenaddressis
11001101 00010000 00100101 00100111
If we set 32−28 rightmost bits to 0, we get
11001101 00010000 00100101 0010000
or
205.16.37.32.
This is actually the block shown in Figure 19.3.
Example 19.6
19.23
The last address in the block can be
found by setting the rightmost
32 − n bits to 1s.
Note
The last address in the block can be
found by setting the rightmost
32 − n bits to 1s.
Note
19.24
FindthelastaddressfortheblockinExample19.6.
Solution
The binary representation of the given address is
11001101 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 1, we get
11001101 00010000 00100101 00101111
or
205.16.37.47
This is actually the block shown in Figure 19.3.
Example 19.7
FindthelastaddressfortheblockinExample19.6.
Solution
The binary representation of the given address is
11001101 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 1, we get
11001101 00010000 00100101 00101111
or
205.16.37.47
This is actually the block shown in Figure 19.3.
Example 19.7
19.25
The number of addresses in the block
can be found by using the formula
2
32−n
.
Note
The number of addresses in the block
can be found by using the formula
2
32−n
.
Note
19.26
FindthenumberofaddressesinExample19.6.
Example 19.8
Solution
Thevalueofnis28,whichmeansthatnumber
ofaddressesis2
32−28
or16.
FindthenumberofaddressesinExample19.6.
Example 19.8
Solution
Thevalueofnis28,whichmeansthatnumber
ofaddressesis2
32−28
or16.
19.27
Anotherwaytofindthefirstaddress,thelastaddress,and
thenumberofaddressesistorepresentthemaskasa32-
bitbinary(or8-digithexadecimal)number.Thisis
particularlyusefulwhenwearewritingaprogramtofind
thesepiecesofinformation.InExample19.5the/28can
berepresentedas
11111111 11111111 11111111 11110000
(twenty-eight1sandfour0s).
Find
a.Thefirstaddress
b.Thelastaddress
c.Thenumberofaddresses.
Example 19.9
Anotherwaytofindthefirstaddress,thelastaddress,and
thenumberofaddressesistorepresentthemaskasa32-
bitbinary(or8-digithexadecimal)number.Thisis
particularlyusefulwhenwearewritingaprogramtofind
thesepiecesofinformation.InExample19.5the/28can
berepresentedas
11111111 11111111 11111111 11110000
(twenty-eight1sandfour0s).
Find
a.Thefirstaddress
b.Thelastaddress
c.Thenumberofaddresses.
Example 19.9
19.28
Solution
a.ThefirstaddresscanbefoundbyANDingthegiven
addresseswiththemask.ANDinghereisdonebitby
bit.TheresultofANDing2bitsis1ifbothbitsare1s;
theresultis0otherwise.
Example 19.9 (continued)
Solution
a.ThefirstaddresscanbefoundbyANDingthegiven
addresseswiththemask.ANDinghereisdonebitby
bit.TheresultofANDing2bitsis1ifbothbitsare1s;
theresultis0otherwise.
Example 19.9 (continued)
19.29
b.ThelastaddresscanbefoundbyORingthegiven
addresseswiththecomplementofthemask.ORing
hereisdonebitbybit.TheresultofORing2bitsis0if
bothbitsare0s;theresultis1otherwise.The
complementofanumberisfoundbychangingeach1
to0andeach0to1.
Example 19.9 (continued)
b.ThelastaddresscanbefoundbyORingthegiven
addresseswiththecomplementofthemask.ORing
hereisdonebitbybit.TheresultofORing2bitsis0if
bothbitsare0s;theresultis1otherwise.The
complementofanumberisfoundbychangingeach1
to0andeach0to1.
Example 19.9 (continued)
19.30
c.Thenumberofaddressescanbefoundby
complementingthemask,interpretingitasadecimal
number,andadding1toit.
Example 19.9 (continued)
c.Thenumberofaddressescanbefoundby
complementingthemask,interpretingitasadecimal
number,andadding1toit.
Example 19.9 (continued)
19.31
Figure 19.4 A network configuration for the block 205.16.37.32/28
Figure 19.4 A network configuration for the block 205.16.37.32/28
19.32
The first address in a block is
normally not assigned to any device;
it is used as the network address that
represents the organization
to the rest of the world.
Note
The first address in a block is
normally not assigned to any device;
it is used as the network address that
represents the organization
to the rest of the world.
Note
19.33
Figure 19.5 Two levels of hierarchy in an IPv4 address
Figure 19.5 Two levels of hierarchy in an IPv4 address
19.34
Figure 19.6 A frame in a character-oriented protocol
Figure 19.6 A frame in a character-oriented protocol
19.35
Each address in the block can be
considered as a two-level
hierarchical structure:
the leftmost nbits (prefix) define
the network;
the rightmost 32 − n bits define
the host.
Note
Each address in the block can be
considered as a two-level
hierarchical structure:
the leftmost nbits (prefix) define
the network;
the rightmost 32 − n bits define
the host.
Note
19.36
Figure 19.7 Configuration and addresses in a subnetted network
Figure 19.7 Configuration and addresses in a subnetted network
19.37
Figure 19.8 Three-level hierarchy in an IPv4 address
Figure 19.8 Three-level hierarchy in an IPv4 address
19.38
AnISPisgrantedablockofaddressesstartingwith
190.100.0.0/16(65,536addresses).TheISPneedsto
distributetheseaddressestothreegroupsofcustomersas
follows:
a.Thefirstgrouphas64customers;eachneeds256
addresses.
b.Thesecondgrouphas128customers;eachneeds128
addresses.
c.Thethirdgrouphas128customers;eachneeds64
addresses.
Designthesubblocksandfindouthowmanyaddresses
arestillavailableaftertheseallocations.
Example 19.10
AnISPisgrantedablockofaddressesstartingwith
190.100.0.0/16(65,536addresses).TheISPneedsto
distributetheseaddressestothreegroupsofcustomersas
follows:
a.Thefirstgrouphas64customers;eachneeds256
addresses.
b.Thesecondgrouphas128customers;eachneeds128
addresses.
c.Thethirdgrouphas128customers;eachneeds64
addresses.
Designthesubblocksandfindouthowmanyaddresses
arestillavailableaftertheseallocations.
Example 19.10
19.39
Solution
Figure19.9showsthesituation.
Example 19.10 (continued)
Group1
Forthisgroup,eachcustomerneeds256addresses.This
meansthat8(log2256)bitsareneededtodefineeach
host.Theprefixlengthisthen32−8=24.Theaddresses
are
Solution
Figure19.9showsthesituation.
Example 19.10 (continued)
Group1
Forthisgroup,eachcustomerneeds256addresses.This
meansthat8(log2256)bitsareneededtodefineeach
host.Theprefixlengthisthen32−8=24.Theaddresses
are
19.40
Example 19.10 (continued)
Group2
Forthisgroup,eachcustomerneeds128addresses.This
meansthat7(log2128)bitsareneededtodefineeach
host.Theprefixlengthisthen32−7=25.Theaddresses
are
Example 19.10 (continued)
Group2
Forthisgroup,eachcustomerneeds128addresses.This
meansthat7(log2128)bitsareneededtodefineeach
host.Theprefixlengthisthen32−7=25.Theaddresses
are
19.41
Example 19.10 (continued)
Group3
Forthisgroup,eachcustomerneeds64addresses.This
meansthat6(log264)bitsareneededtoeachhost.The
prefixlengthisthen32−6=26.Theaddressesare
NumberofgrantedaddressestotheISP:65,536
NumberofallocatedaddressesbytheISP:40,960
Numberofavailableaddresses:24,576
Example 19.10 (continued)
Group3
Forthisgroup,eachcustomerneeds64addresses.This
meansthat6(log264)bitsareneededtoeachhost.The
prefixlengthisthen32−6=26.Theaddressesare
NumberofgrantedaddressestotheISP:65,536
NumberofallocatedaddressesbytheISP:40,960
Numberofavailableaddresses:24,576
19.42
Figure 19.9 An example of address allocation and distribution by an ISP
Figure 19.9 An example of address allocation and distribution by an ISP
19.43
Table 19.3 Addresses for private networks
Table 19.3 Addresses for private networks
19.44
Figure 19.10 A NAT implementation
Figure 19.10 A NAT implementation
19.45
Figure 19.11 Addresses in a NAT
Figure 19.11 Addresses in a NAT
19.46
Figure 19.12 NAT address translation
Figure 19.12 NAT address translation
19.47
Table 19.4 Five-column translation table
Table 19.4 Five-column translation table
19.48
Figure 19.13 An ISP and NAT
Figure 19.13 An ISP and NAT
19.49
19-2 IPv6 ADDRESSES
Despiteallshort-termsolutions,addressdepletionis
stillalong-termproblemfortheInternet.Thisand
otherproblemsintheIPprotocolitselfhavebeenthe
motivationforIPv6.
Structure
Address Space
Topics discussed in this section:
19-2 IPv6 ADDRESSES
Despiteallshort-termsolutions,addressdepletionis
stillalong-termproblemfortheInternet.Thisand
otherproblemsintheIPprotocolitselfhavebeenthe
motivationforIPv6.
Structure
Address Space
Topics discussed in this section:
19.50
An IPv6 address is 128 bits long.
Note
An IPv6 address is 128 bits long.
Note
19.51
Figure 19.14 IPv6 address in binary and hexadecimal colon notation
Figure 19.14 IPv6 address in binary and hexadecimal colon notation
19.52
Figure 19.15 Abbreviated IPv6 addresses
Figure 19.15 Abbreviated IPv6 addresses
19.53
Expandtheaddress0:15::1:12:1213toitsoriginal.
Example 19.11
Solution
Wefirstneedtoaligntheleftsideofthedoublecolonto
theleftoftheoriginalpatternandtherightsideofthe
doublecolontotherightoftheoriginalpatterntofind
howmany0sweneedtoreplacethedoublecolon.
Thismeansthattheoriginaladdressis.
Expandtheaddress0:15::1:12:1213toitsoriginal.
Example 19.11
Solution
Wefirstneedtoaligntheleftsideofthedoublecolonto
theleftoftheoriginalpatternandtherightsideofthe
doublecolontotherightoftheoriginalpatterntofind
howmany0sweneedtoreplacethedoublecolon.
Thismeansthattheoriginaladdressis.
19.54
Table 19.5 Type prefixes for IPv6 addresses
Table 19.5 Type prefixes for IPv6 addresses
19.55
Table 19.5 Type prefixes for IPv6 addresses (continued)
Table 19.5 Type prefixes for IPv6 addresses (continued)
19.56
Figure 19.16 Prefixes for provider-based unicast address
Figure 19.16 Prefixes for provider-based unicast address
19.57
Figure 19.17 Multicast address in IPv6
Figure 19.17 Multicast address in IPv6
19.58
Figure 19.18 Reserved addresses in IPv6
Figure 19.18 Reserved addresses in IPv6
19.59
Figure 19.19 Local addresses in IPv6
Figure 19.19 Local addresses in IPv6
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