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CChhaapptteerr 44 
IIPP AAddddrreesssseess:: 
CCllaassssffuull AAddddrreessssiinngg 
Objectives 
Upon completion you will be able to: 
• Understand IPv4 addresses and classes 
• Identify the class of an IP address 
• Find the network address given an IP address 
• Understand masks and how to use them 
• Understand subnets and supernets 
TCP/IP Protocol Suite 1
44..11 IINNTTRROODDUUCCTTIIOONN 
TThhee iiddeennttiiffiieerr uusseedd iinn tthhee IIPP llaayyeerr ooff tthhee TTCCPP//IIPP pprroottooccooll ssuuiittee ttoo 
iiddeennttiiffyy eeaacchh ddeevviiccee ccoonnnneecctteedd ttoo tthhee IInntteerrnneett iiss ccaalllleedd tthhee IInntteerrnneett 
aaddddrreessss oorr IIPP aaddddrreessss.. AAnn IIPP aaddddrreessss iiss aa 3322--bbiitt aaddddrreessss tthhaatt uunniiqquueellyy 
aanndd uunniivveerrssaallllyy ddeeffiinneess tthhee ccoonnnneeccttiioonn ooff aa hhoosstt oorr aa rroouutteerr ttoo tthhee 
IInntteerrnneett.. IIPP aaddddrreesssseess aarree uunniiqquuee.. TThheeyy aarree uunniiqquuee iinn tthhee sseennssee tthhaatt 
eeaacchh aaddddrreessss ddeeffiinneess oonnee,, aanndd oonnllyy oonnee,, ccoonnnneeccttiioonn ttoo tthhee IInntteerrnneett.. TTwwoo 
ddeevviicceess oonn tthhee IInntteerrnneett ccaann nneevveerr hhaavvee tthhee ssaammee aaddddrreessss.. 
TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: 
AAddddrreessss SSppaaccee 
NNoottaattiioonn 
TCP/IP Protocol Suite 2
NNoottee:: 
An IP address is a 32-bit address. 
TCP/IP Protocol Suite 3
NNoottee:: 
The IP addresses are unique. 
TCP/IP Protocol Suite 4
NNoottee:: 
The address space of IPv4 is 
232 or 4,294,967,296. 
TCP/IP Protocol Suite 5
Figure 4.1 Dotted-decimal notation 
TCP/IP Protocol Suite 6
NNoottee:: 
The binary, decimal, and hexadecimal 
number systems are reviewed in 
Appendix B. 
TCP/IP Protocol Suite 7
ExamplE 1 
Change the following IP addresses from binary notation to 
dotted-decimal notation. 
a. 10000001 00001011 00001011 11101111 
b. 11000001 10000011 00011011 11111111 
c. 11100111 11011011 10001011 01101111 
d. 11111001 10011011 11111011 00001111 
Solution 
We replace each group of 8 bits with its equivalent decimal 
number (see Appendix B) and add dots for separation: 
a. 129.11.11.239 b. 193.131.27.255 
c. 231.219.139.111 d. 249.155.251.15 
TCP/IP Protocol Suite 8
ExamplE 2 
Change the following IP addresses from dotted-decimal 
notation to binary notation. 
a. 111.56.45.78 b. 221.34.7.82 
c. 241.8.56.12 d. 75.45.34.78 
Solution 
We replace each decimal number with its binary equivalent: 
a. 01101111 00111000 00101101 01001110 
b. 11011101 00100010 00000111 01010010 
c. 11110001 00001000 00111000 00001100 
d. 01001011 00101101 00100010 01001110 
TCP/IP Protocol Suite 9
ExamplE 3 
Find the error, if any, in the following IP addresses: 
a. 111.56.045.78 b. 221.34.7.8.20 
c. 75.45.301.14 d. 11100010.23.14.67 
Solution 
a. There are no leading zeroes in dotted-decimal notation (045). 
b. We may not have more than four numbers in an IP address. 
c. In dotted-decimal notation, each number is less than or equal 
to 255; 301 is outside this range. 
d. A mixture of binary notation and dotted-decimal notation is 
not 
allowed. 
TCP/IP Protocol Suite 10
ExamplE 4 
Change the following IP addresses from binary notation to 
hexadecimal notation. 
a. 10000001 00001011 00001011 11101111 
b. 11000001 10000011 00011011 11111111 
Solution 
We replace each group of 4 bits with its hexadecimal 
equivalent (see Appendix B). Note that hexadecimal notation 
normally has no added spaces or dots; however, 0X (or 0x) is 
added at the beginning or the subscript 16 at the end to show 
that the number is in hexadecimal. 
a. 0X810B0BEF or 810B0BEF16 
b. 0XC1831BFF or C1831BFF16 
TCP/IP Protocol Suite 11
4.2 CLASSFUL ADDRESSING 
a IP addresses, when started a few decades aggoo,, uusseedd tthhee ccoonncceepptt ooff 
ccllaasssseess.. TThhiiss aarrcchhiitteeccttuurree iiss ccaalllleedd ccllaassssffuull aaddddrreessssiinngg.. IInn tthhee mmiidd--11999900ss,, 
aa nneeww aarrcchhiitteeccttuurree,, ccaalllleedd ccllaasssslleessss aaddddrreessssiinngg,, wwaass iinnttrroodduucceedd aanndd wwiillll 
eevveennttuuaallllyy ssuuppeerrsseeddee tthhee oorriiggiinnaall aarrcchhiitteeccttuurree.. HHoowweevveerr,, ppaarrtt ooff tthhee 
IInntteerrnneett iiss ssttiillll uussiinngg ccllaassssffuull aaddddrreessssiinngg,, bbuutt tthhee mmiiggrraattiioonn iiss vveerryy ffaasstt.. 
TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: 
RReeccooggnniizziinngg CCllaasssseess 
NNeettiidd aanndd HHoossttiidd 
CCllaasssseess aanndd BBlloocckkss 
NNeettwwoorrkk AAddddrreesssseess 
SSuuffffiicciieenntt IInnffoorrmmaattiioonn 
MMaasskk 
CCIIDDRR NNoottaattiioonn 
AAddddrreessss DDeepplleettiioonn 
TCP/IP Protocol Suite 12
Figure 4.2 Occupation of the address space 
TCP/IP Protocol Suite 13
TTaabbllee 44..11 AAddddrreesssseess ppeerr ccllaassss 
TCP/IP Protocol Suite 14
Figure 4.3 Finding the class in binary notation 
TCP/IP Protocol Suite 15
Figure 4.4 Finding the address class 
TCP/IP Protocol Suite 16
ExamplE 5 
How can we prove that we have 2,147,483,648 addresses in 
class A? 
Solution 
In class A, only 1 bit defines the class. The remaining 31 bits 
are available for the address. With 31 bits, we can have 231 
or 2,147,483,648 addresses. 
TCP/IP Protocol Suite 17
ExamplE 6 
Find the class of each address: 
a. 00000001 00001011 00001011 11101111 
b. 11000001 10000011 00011011 11111111 
c. 10100111 11011011 10001011 01101111 
d. 11110011 10011011 11111011 00001111 
Solution 
See the procedure in Figure 4.4. 
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 bit is 1; the second bit is 0. This is a class B address. 
d. The first 4 bits are 1s. This is a class E address.. 
TCP/IP Protocol Suite 18
Figure 4.5 Finding the class in decimal notation 
TCP/IP Protocol Suite 19
ExamplE 7 
Find the class of each address: 
a. 227.12.14.87 b.193.14.56.22 c.14.23.120.8 
d. 252.5.15.111 e.134.11.78.56 
Solution 
a. The first byte is 227 (between 224 and 239); the class is D. 
b. The first byte is 193 (between 192 and 223); the class is C. 
c. The first byte is 14 (between 0 and 127); the class is A. 
d. The first byte is 252 (between 240 and 255); the class is E. 
e. The first byte is 134 (between 128 and 191); the class is B. 
TCP/IP Protocol Suite 20
ExamplE 8 
In Example 5 we showed that class A has 231 (2,147,483,648) 
addresses. How can we prove this same fact using dotted-decimal 
notation? 
Solution 
The addresses in class A range from 0.0.0.0 to 
127.255.255.255. We need to show that the difference between 
these two numbers is 2,147,483,648. This is a good exercise 
because it shows us how to define the range of addresses 
between two addresses. We notice that we are dealing with base 
256 numbers here. Each byte in the notation has a weight. The 
weights are as follows (see Appendix B): 
See Next Slide 
TCP/IP Protocol Suite 21
ExamplE 8 (continuEd) 
2563, 2562, 2561, 2560 
Now to find the integer value of each number, we multiply 
each byte by its weight: 
Last address: 127 × 2563 + 255 × 2562 + 
255 × 2561 + 255 × 2560 = 2,147,483,647 
First address: = 0 
If we subtract the first from the last and add 1 to the result 
(remember we always add 1 to get the range), we get 
2,147,483,648 or 231. 
TCP/IP Protocol Suite 22
Figure 4.6 Netid and hostid 
TCP/IP Protocol Suite 23
NNoottee:: 
Millions of class A addresses are 
wasted. 
TCP/IP Protocol Suite 24
Figure 4.7 Blocks in class A 
TCP/IP Protocol Suite 25
Figure 4.8 Blocks in class B 
TCP/IP Protocol Suite 26
NNoottee:: 
Many class B addresses are wasted. 
TCP/IP Protocol Suite 27
Figure 4.9 Blocks in class C 
TCP/IP Protocol Suite 28
NNoottee:: 
The number of addresses in class C is 
smaller than the needs of most 
organizations. 
TCP/IP Protocol Suite 29
NNoottee:: 
Class D addresses are used for 
multicasting; there is only one block in 
this class. 
TCP/IP Protocol Suite 30
NNoottee:: 
Class E addresses are reserved for 
future purposes; most of the block is 
wasted. 
TCP/IP Protocol Suite 31
NNoottee:: 
In classful addressing, the network 
address (the first address in the block) 
is the one that is assigned to the 
organization. The range of addresses 
can automatically be inferred from the 
network address. 
TCP/IP Protocol Suite 32
ExamplE 9 
Given the network address 17.0.0.0, find the class, the block, 
and the range of the addresses. 
Solution 
The class is A because the first byte is between 0 and 127. The 
block has a netid of 17. The addresses range from 17.0.0.0 to 
17.255.255.255. 
TCP/IP Protocol Suite 33
ExamplE 
10 
Given the network address 132.21.0.0, find the class, the block, 
and the range of the addresses. 
Solution 
The class is B because the first byte is between 128 and 191. 
The block has a netid of 132.21. The addresses range from 
132.21.0.0 to 132.21.255.255. 
TCP/IP Protocol Suite 34
ExamplE 
11 
Given the network address 220.34.76.0, find the class, the 
block, and the range of the addresses. 
Solution 
The class is C because the first byte is between 192 
and 223. The block has a netid of 220.34.76. The 
addresses range from 220.34.76.0 to 220.34.76.255. 
TCP/IP Protocol Suite 35
Figure 4.10 Masking concept 
TCP/IP Protocol Suite 36
Figure 4.11 AND operation 
TCP/IP Protocol Suite 37
TTaabbllee 44..22 DDeeffaauulltt mmaasskkss 
TCP/IP Protocol Suite 38
NNoottee:: 
The network address is the beginning 
address of each block. It can be found 
by applying the default mask to any of 
the addresses in the block (including 
itself). It retains the netid of the block 
and sets the hostid to zero. 
TCP/IP Protocol Suite 39
ExamplE 
12 
Given the address 23.56.7.91, find the beginning address 
(network address). 
Solution 
The default mask is 255.0.0.0, which means that only the first 
byte is preserved and the other 3 bytes are set to 0s. The 
network address is 23.0.0.0. 
TCP/IP Protocol Suite 40
ExamplE 
13 
Given the address 132.6.17.85, find the beginning address 
(network address). 
Solution 
The default mask is 255.255.0.0, which means that the first 2 
bytes are preserved and the other 2 bytes are set to 0s. The 
network address is 132.6.0.0. 
TCP/IP Protocol Suite 41
ExamplE 
14 
Given the address 201.180.56.5, find the beginning address 
(network address). 
Solution 
The default mask is 255.255.255.0, which means that the first 3 
bytes are preserved and the last byte is set to 0. The network 
address is 201.180.56.0. 
TCP/IP Protocol Suite 42
NNoottee:: 
Note that we must not apply the 
default mask of one class to an address 
belonging to another class. 
TCP/IP Protocol Suite 43
4.3 OTHER ISSUES 
i In this section, we discuss some other issssuueess tthhaatt aarree rreellaatteedd ttoo 
aaddddrreessssiinngg iinn ggeenneerraall aanndd ccllaassssffuull aaddddrreessssiinngg iinn ppaarrttiiccuullaarr.. 
TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: 
MMuullttiihhoommeedd DDeevviicceess 
LLooccaattiioonn,, NNoott NNaammeess 
SSppeecciiaall AAddddrreesssseess 
PPrriivvaattee AAddddrreesssseess 
UUnniiccaasstt,, MMuullttiiccaasstt,, aanndd BBrrooaaddccaasstt AAddddrreesssseess 
TCP/IP Protocol Suite 44
Figure 4.12 Multihomed devices 
TCP/IP Protocol Suite 45
TTaabbllee 44..33 SSppeecciiaall aaddddrreesssseess 
TCP/IP Protocol Suite 46
Figure 4.13 Network address 
TCP/IP Protocol Suite 47
Figure 4.14 Example of direct broadcast address 
TCP/IP Protocol Suite 48
Figure 4.15 Example of limited broadcast address 
TCP/IP Protocol Suite 49
Figure 4.16 Examples of “this host on this network” 
TCP/IP Protocol Suite 50
Figure 4.17 Example of “specific host on this network” 
TCP/IP Protocol Suite 51
Figure 4.18 Example of loopback address 
TCP/IP Protocol Suite 52
Table 44..55 AAddddrreesssseess ffoorr pprriivvaattee nneettwwoorrkkss 
TCP/IP Protocol Suite 53
NNoottee:: 
Multicast delivery will be discussed in 
depth in Chapter 15. 
TCP/IP Protocol Suite 54
TTaabbllee 44..55 CCaatteeggoorryy aaddddrreesssseess 
TCP/IP Protocol Suite 55
Table 44..66 AAddddrreesssseess ffoorr ccoonnffeerreenncciinngg 
TCP/IP Protocol Suite 56
Figure 4.19 Sample internet 
TCP/IP Protocol Suite 57
4.4 SUBNETTING AND 
SUPERNETTING 
In the previous sections we discussed the pprroobblleemmss aassssoocciiaatteedd wwiitthh 
ccllaassssffuull aaddddrreessssiinngg.. SSppeecciiffiiccaallllyy,, tthhee nneettwwoorrkk aaddddrreesssseess aavvaaiillaabbllee ffoorr 
aassssiiggnnmmeenntt ttoo oorrggaanniizzaattiioonnss aarree cclloossee ttoo ddeepplleettiioonn.. TThhiiss iiss ccoouupplleedd wwiitthh 
tthhee eevveerr--iinnccrreeaassiinngg ddeemmaanndd ffoorr aaddddrreesssseess ffrroomm oorrggaanniizzaattiioonnss tthhaatt wwaanntt 
ccoonnnneeccttiioonn ttoo tthhee IInntteerrnneett.. IInn tthhiiss sseeccttiioonn wwee bbrriieeffllyy ddiissccuussss ttwwoo 
ssoolluuttiioonnss:: ssuubbnneettttiinngg aanndd ssuuppeerrnneettttiinngg.. 
TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: 
SSuubbnneettttiinngg 
SSuuppeerrnneettttiinngg 
SSuuppeerrnneett MMaasskk 
OObbssoolleesscceennccee 
TCP/IP Protocol Suite 58
NNoottee:: 
IP addresses are designed with two 
levels of hierarchy. 
TCP/IP Protocol Suite 59
Figure 4.20 A network with two levels of hierarchy (not subnetted) 
TCP/IP Protocol Suite 60
Figure 4.21 A network with three levels of hierarchy (subnetted) 
TCP/IP Protocol Suite 61
Figure 4.22 Addresses in a network with and without subnetting 
TCP/IP Protocol Suite 62
Figure 4.23 Hierarchy concept in a telephone number 
TCP/IP Protocol Suite 63
Figure 4.24 Default mask and subnet mask 
TCP/IP Protocol Suite 64
ExamplE 
15 
What is the subnetwork address if the destination address is 
200.45.34.56 and the subnet mask is 255.255.240.0? 
Solution 
We apply the AND operation on the address and the subnet 
mask. 
Address ➡ 11001000 00101101 00100010 00111000 
Subnet Mask ➡ 11111111 11111111 11110000 00000000 
Subnetwork Address ➡ 11001000 00101101 00100000 00000000. 
TCP/IP Protocol Suite 65
Figure 4.25 Comparison of a default mask and a subnet mask 
TCP/IP Protocol Suite 66
Figure 4.26 A supernetwork 
TCP/IP Protocol Suite 67
NNoottee:: 
In subnetting, we need the first 
address of the subnet and the subnet 
mask to define the range of addresses. 
In supernetting, we need the first 
address of the supernet and the 
supernet mask to define the range of 
addresses. 
TCP/IP Protocol Suite 68
Figure 4.27 Comparison of subnet, default, and supernet masks 
TCP/IP Protocol Suite 69
NNoottee:: 
The idea of subnetting and 
supernetting of classful addresses is 
almost obsolete. 
TCP/IP Protocol Suite 70

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Chap 04 ip addresses classful

  • 1. CChhaapptteerr 44 IIPP AAddddrreesssseess:: CCllaassssffuull AAddddrreessssiinngg Objectives Upon completion you will be able to: • Understand IPv4 addresses and classes • Identify the class of an IP address • Find the network address given an IP address • Understand masks and how to use them • Understand subnets and supernets TCP/IP Protocol Suite 1
  • 2. 44..11 IINNTTRROODDUUCCTTIIOONN TThhee iiddeennttiiffiieerr uusseedd iinn tthhee IIPP llaayyeerr ooff tthhee TTCCPP//IIPP pprroottooccooll ssuuiittee ttoo iiddeennttiiffyy eeaacchh ddeevviiccee ccoonnnneecctteedd ttoo tthhee IInntteerrnneett iiss ccaalllleedd tthhee IInntteerrnneett aaddddrreessss oorr IIPP aaddddrreessss.. AAnn IIPP aaddddrreessss iiss aa 3322--bbiitt aaddddrreessss tthhaatt uunniiqquueellyy aanndd uunniivveerrssaallllyy ddeeffiinneess tthhee ccoonnnneeccttiioonn ooff aa hhoosstt oorr aa rroouutteerr ttoo tthhee IInntteerrnneett.. IIPP aaddddrreesssseess aarree uunniiqquuee.. TThheeyy aarree uunniiqquuee iinn tthhee sseennssee tthhaatt eeaacchh aaddddrreessss ddeeffiinneess oonnee,, aanndd oonnllyy oonnee,, ccoonnnneeccttiioonn ttoo tthhee IInntteerrnneett.. TTwwoo ddeevviicceess oonn tthhee IInntteerrnneett ccaann nneevveerr hhaavvee tthhee ssaammee aaddddrreessss.. TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: AAddddrreessss SSppaaccee NNoottaattiioonn TCP/IP Protocol Suite 2
  • 3. NNoottee:: An IP address is a 32-bit address. TCP/IP Protocol Suite 3
  • 4. NNoottee:: The IP addresses are unique. TCP/IP Protocol Suite 4
  • 5. NNoottee:: The address space of IPv4 is 232 or 4,294,967,296. TCP/IP Protocol Suite 5
  • 6. Figure 4.1 Dotted-decimal notation TCP/IP Protocol Suite 6
  • 7. NNoottee:: The binary, decimal, and hexadecimal number systems are reviewed in Appendix B. TCP/IP Protocol Suite 7
  • 8. ExamplE 1 Change the following IP addresses from binary notation to dotted-decimal notation. a. 10000001 00001011 00001011 11101111 b. 11000001 10000011 00011011 11111111 c. 11100111 11011011 10001011 01101111 d. 11111001 10011011 11111011 00001111 Solution We replace each group of 8 bits with its equivalent decimal number (see Appendix B) and add dots for separation: a. 129.11.11.239 b. 193.131.27.255 c. 231.219.139.111 d. 249.155.251.15 TCP/IP Protocol Suite 8
  • 9. ExamplE 2 Change the following IP addresses from dotted-decimal notation to binary notation. a. 111.56.45.78 b. 221.34.7.82 c. 241.8.56.12 d. 75.45.34.78 Solution We replace each decimal number with its binary equivalent: a. 01101111 00111000 00101101 01001110 b. 11011101 00100010 00000111 01010010 c. 11110001 00001000 00111000 00001100 d. 01001011 00101101 00100010 01001110 TCP/IP Protocol Suite 9
  • 10. ExamplE 3 Find the error, if any, in the following IP addresses: a. 111.56.045.78 b. 221.34.7.8.20 c. 75.45.301.14 d. 11100010.23.14.67 Solution a. There are no leading zeroes in dotted-decimal notation (045). b. We may not have more than four numbers in an IP address. c. In dotted-decimal notation, each number is less than or equal to 255; 301 is outside this range. d. A mixture of binary notation and dotted-decimal notation is not allowed. TCP/IP Protocol Suite 10
  • 11. ExamplE 4 Change the following IP addresses from binary notation to hexadecimal notation. a. 10000001 00001011 00001011 11101111 b. 11000001 10000011 00011011 11111111 Solution We replace each group of 4 bits with its hexadecimal equivalent (see Appendix B). Note that hexadecimal notation normally has no added spaces or dots; however, 0X (or 0x) is added at the beginning or the subscript 16 at the end to show that the number is in hexadecimal. a. 0X810B0BEF or 810B0BEF16 b. 0XC1831BFF or C1831BFF16 TCP/IP Protocol Suite 11
  • 12. 4.2 CLASSFUL ADDRESSING a IP addresses, when started a few decades aggoo,, uusseedd tthhee ccoonncceepptt ooff ccllaasssseess.. TThhiiss aarrcchhiitteeccttuurree iiss ccaalllleedd ccllaassssffuull aaddddrreessssiinngg.. IInn tthhee mmiidd--11999900ss,, aa nneeww aarrcchhiitteeccttuurree,, ccaalllleedd ccllaasssslleessss aaddddrreessssiinngg,, wwaass iinnttrroodduucceedd aanndd wwiillll eevveennttuuaallllyy ssuuppeerrsseeddee tthhee oorriiggiinnaall aarrcchhiitteeccttuurree.. HHoowweevveerr,, ppaarrtt ooff tthhee IInntteerrnneett iiss ssttiillll uussiinngg ccllaassssffuull aaddddrreessssiinngg,, bbuutt tthhee mmiiggrraattiioonn iiss vveerryy ffaasstt.. TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: RReeccooggnniizziinngg CCllaasssseess NNeettiidd aanndd HHoossttiidd CCllaasssseess aanndd BBlloocckkss NNeettwwoorrkk AAddddrreesssseess SSuuffffiicciieenntt IInnffoorrmmaattiioonn MMaasskk CCIIDDRR NNoottaattiioonn AAddddrreessss DDeepplleettiioonn TCP/IP Protocol Suite 12
  • 13. Figure 4.2 Occupation of the address space TCP/IP Protocol Suite 13
  • 14. TTaabbllee 44..11 AAddddrreesssseess ppeerr ccllaassss TCP/IP Protocol Suite 14
  • 15. Figure 4.3 Finding the class in binary notation TCP/IP Protocol Suite 15
  • 16. Figure 4.4 Finding the address class TCP/IP Protocol Suite 16
  • 17. ExamplE 5 How can we prove that we have 2,147,483,648 addresses in class A? Solution In class A, only 1 bit defines the class. The remaining 31 bits are available for the address. With 31 bits, we can have 231 or 2,147,483,648 addresses. TCP/IP Protocol Suite 17
  • 18. ExamplE 6 Find the class of each address: a. 00000001 00001011 00001011 11101111 b. 11000001 10000011 00011011 11111111 c. 10100111 11011011 10001011 01101111 d. 11110011 10011011 11111011 00001111 Solution See the procedure in Figure 4.4. 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 bit is 1; the second bit is 0. This is a class B address. d. The first 4 bits are 1s. This is a class E address.. TCP/IP Protocol Suite 18
  • 19. Figure 4.5 Finding the class in decimal notation TCP/IP Protocol Suite 19
  • 20. ExamplE 7 Find the class of each address: a. 227.12.14.87 b.193.14.56.22 c.14.23.120.8 d. 252.5.15.111 e.134.11.78.56 Solution a. The first byte is 227 (between 224 and 239); the class is D. b. The first byte is 193 (between 192 and 223); the class is C. c. The first byte is 14 (between 0 and 127); the class is A. d. The first byte is 252 (between 240 and 255); the class is E. e. The first byte is 134 (between 128 and 191); the class is B. TCP/IP Protocol Suite 20
  • 21. ExamplE 8 In Example 5 we showed that class A has 231 (2,147,483,648) addresses. How can we prove this same fact using dotted-decimal notation? Solution The addresses in class A range from 0.0.0.0 to 127.255.255.255. We need to show that the difference between these two numbers is 2,147,483,648. This is a good exercise because it shows us how to define the range of addresses between two addresses. We notice that we are dealing with base 256 numbers here. Each byte in the notation has a weight. The weights are as follows (see Appendix B): See Next Slide TCP/IP Protocol Suite 21
  • 22. ExamplE 8 (continuEd) 2563, 2562, 2561, 2560 Now to find the integer value of each number, we multiply each byte by its weight: Last address: 127 × 2563 + 255 × 2562 + 255 × 2561 + 255 × 2560 = 2,147,483,647 First address: = 0 If we subtract the first from the last and add 1 to the result (remember we always add 1 to get the range), we get 2,147,483,648 or 231. TCP/IP Protocol Suite 22
  • 23. Figure 4.6 Netid and hostid TCP/IP Protocol Suite 23
  • 24. NNoottee:: Millions of class A addresses are wasted. TCP/IP Protocol Suite 24
  • 25. Figure 4.7 Blocks in class A TCP/IP Protocol Suite 25
  • 26. Figure 4.8 Blocks in class B TCP/IP Protocol Suite 26
  • 27. NNoottee:: Many class B addresses are wasted. TCP/IP Protocol Suite 27
  • 28. Figure 4.9 Blocks in class C TCP/IP Protocol Suite 28
  • 29. NNoottee:: The number of addresses in class C is smaller than the needs of most organizations. TCP/IP Protocol Suite 29
  • 30. NNoottee:: Class D addresses are used for multicasting; there is only one block in this class. TCP/IP Protocol Suite 30
  • 31. NNoottee:: Class E addresses are reserved for future purposes; most of the block is wasted. TCP/IP Protocol Suite 31
  • 32. NNoottee:: In classful addressing, the network address (the first address in the block) is the one that is assigned to the organization. The range of addresses can automatically be inferred from the network address. TCP/IP Protocol Suite 32
  • 33. ExamplE 9 Given the network address 17.0.0.0, find the class, the block, and the range of the addresses. Solution The class is A because the first byte is between 0 and 127. The block has a netid of 17. The addresses range from 17.0.0.0 to 17.255.255.255. TCP/IP Protocol Suite 33
  • 34. ExamplE 10 Given the network address 132.21.0.0, find the class, the block, and the range of the addresses. Solution The class is B because the first byte is between 128 and 191. The block has a netid of 132.21. The addresses range from 132.21.0.0 to 132.21.255.255. TCP/IP Protocol Suite 34
  • 35. ExamplE 11 Given the network address 220.34.76.0, find the class, the block, and the range of the addresses. Solution The class is C because the first byte is between 192 and 223. The block has a netid of 220.34.76. The addresses range from 220.34.76.0 to 220.34.76.255. TCP/IP Protocol Suite 35
  • 36. Figure 4.10 Masking concept TCP/IP Protocol Suite 36
  • 37. Figure 4.11 AND operation TCP/IP Protocol Suite 37
  • 38. TTaabbllee 44..22 DDeeffaauulltt mmaasskkss TCP/IP Protocol Suite 38
  • 39. NNoottee:: The network address is the beginning address of each block. It can be found by applying the default mask to any of the addresses in the block (including itself). It retains the netid of the block and sets the hostid to zero. TCP/IP Protocol Suite 39
  • 40. ExamplE 12 Given the address 23.56.7.91, find the beginning address (network address). Solution The default mask is 255.0.0.0, which means that only the first byte is preserved and the other 3 bytes are set to 0s. The network address is 23.0.0.0. TCP/IP Protocol Suite 40
  • 41. ExamplE 13 Given the address 132.6.17.85, find the beginning address (network address). Solution The default mask is 255.255.0.0, which means that the first 2 bytes are preserved and the other 2 bytes are set to 0s. The network address is 132.6.0.0. TCP/IP Protocol Suite 41
  • 42. ExamplE 14 Given the address 201.180.56.5, find the beginning address (network address). Solution The default mask is 255.255.255.0, which means that the first 3 bytes are preserved and the last byte is set to 0. The network address is 201.180.56.0. TCP/IP Protocol Suite 42
  • 43. NNoottee:: Note that we must not apply the default mask of one class to an address belonging to another class. TCP/IP Protocol Suite 43
  • 44. 4.3 OTHER ISSUES i In this section, we discuss some other issssuueess tthhaatt aarree rreellaatteedd ttoo aaddddrreessssiinngg iinn ggeenneerraall aanndd ccllaassssffuull aaddddrreessssiinngg iinn ppaarrttiiccuullaarr.. TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: MMuullttiihhoommeedd DDeevviicceess LLooccaattiioonn,, NNoott NNaammeess SSppeecciiaall AAddddrreesssseess PPrriivvaattee AAddddrreesssseess UUnniiccaasstt,, MMuullttiiccaasstt,, aanndd BBrrooaaddccaasstt AAddddrreesssseess TCP/IP Protocol Suite 44
  • 45. Figure 4.12 Multihomed devices TCP/IP Protocol Suite 45
  • 46. TTaabbllee 44..33 SSppeecciiaall aaddddrreesssseess TCP/IP Protocol Suite 46
  • 47. Figure 4.13 Network address TCP/IP Protocol Suite 47
  • 48. Figure 4.14 Example of direct broadcast address TCP/IP Protocol Suite 48
  • 49. Figure 4.15 Example of limited broadcast address TCP/IP Protocol Suite 49
  • 50. Figure 4.16 Examples of “this host on this network” TCP/IP Protocol Suite 50
  • 51. Figure 4.17 Example of “specific host on this network” TCP/IP Protocol Suite 51
  • 52. Figure 4.18 Example of loopback address TCP/IP Protocol Suite 52
  • 53. Table 44..55 AAddddrreesssseess ffoorr pprriivvaattee nneettwwoorrkkss TCP/IP Protocol Suite 53
  • 54. NNoottee:: Multicast delivery will be discussed in depth in Chapter 15. TCP/IP Protocol Suite 54
  • 55. TTaabbllee 44..55 CCaatteeggoorryy aaddddrreesssseess TCP/IP Protocol Suite 55
  • 56. Table 44..66 AAddddrreesssseess ffoorr ccoonnffeerreenncciinngg TCP/IP Protocol Suite 56
  • 57. Figure 4.19 Sample internet TCP/IP Protocol Suite 57
  • 58. 4.4 SUBNETTING AND SUPERNETTING In the previous sections we discussed the pprroobblleemmss aassssoocciiaatteedd wwiitthh ccllaassssffuull aaddddrreessssiinngg.. SSppeecciiffiiccaallllyy,, tthhee nneettwwoorrkk aaddddrreesssseess aavvaaiillaabbllee ffoorr aassssiiggnnmmeenntt ttoo oorrggaanniizzaattiioonnss aarree cclloossee ttoo ddeepplleettiioonn.. TThhiiss iiss ccoouupplleedd wwiitthh tthhee eevveerr--iinnccrreeaassiinngg ddeemmaanndd ffoorr aaddddrreesssseess ffrroomm oorrggaanniizzaattiioonnss tthhaatt wwaanntt ccoonnnneeccttiioonn ttoo tthhee IInntteerrnneett.. IInn tthhiiss sseeccttiioonn wwee bbrriieeffllyy ddiissccuussss ttwwoo ssoolluuttiioonnss:: ssuubbnneettttiinngg aanndd ssuuppeerrnneettttiinngg.. TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee:: SSuubbnneettttiinngg SSuuppeerrnneettttiinngg SSuuppeerrnneett MMaasskk OObbssoolleesscceennccee TCP/IP Protocol Suite 58
  • 59. NNoottee:: IP addresses are designed with two levels of hierarchy. TCP/IP Protocol Suite 59
  • 60. Figure 4.20 A network with two levels of hierarchy (not subnetted) TCP/IP Protocol Suite 60
  • 61. Figure 4.21 A network with three levels of hierarchy (subnetted) TCP/IP Protocol Suite 61
  • 62. Figure 4.22 Addresses in a network with and without subnetting TCP/IP Protocol Suite 62
  • 63. Figure 4.23 Hierarchy concept in a telephone number TCP/IP Protocol Suite 63
  • 64. Figure 4.24 Default mask and subnet mask TCP/IP Protocol Suite 64
  • 65. ExamplE 15 What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0? Solution We apply the AND operation on the address and the subnet mask. Address ➡ 11001000 00101101 00100010 00111000 Subnet Mask ➡ 11111111 11111111 11110000 00000000 Subnetwork Address ➡ 11001000 00101101 00100000 00000000. TCP/IP Protocol Suite 65
  • 66. Figure 4.25 Comparison of a default mask and a subnet mask TCP/IP Protocol Suite 66
  • 67. Figure 4.26 A supernetwork TCP/IP Protocol Suite 67
  • 68. NNoottee:: In subnetting, we need the first address of the subnet and the subnet mask to define the range of addresses. In supernetting, we need the first address of the supernet and the supernet mask to define the range of addresses. TCP/IP Protocol Suite 68
  • 69. Figure 4.27 Comparison of subnet, default, and supernet masks TCP/IP Protocol Suite 69
  • 70. NNoottee:: The idea of subnetting and supernetting of classful addresses is almost obsolete. TCP/IP Protocol Suite 70