Chapter 19
Network Layer:
Logical Addressing
19.1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-1 IPv4 ADDRESSES
An IPv4 address is a 32-bit address that uniquely and
universally defines the connection of a device (for
example, a computer or a router) to the Internet.
Topics discussed in this section:
Address Space
Notations
Classful Addressing
Classless Addressing
Network Address Translation (NAT)
19.2
Note
An IPv4 address is 32 bits long.
19.3
Note
The IPv4 addresses are unique
and universal.
19.4
Note
The address space of IPv4 is
232 or 4,294,967,296.
19.5
Figure 19.1 Dotted-decimal notation and binary notation for an IPv4 address
19.6
Note
Numbering systems are reviewed in
Appendix B.
19.7
Example 19.1
Change the following IPv4 addresses from binary
notation to dotted-decimal notation.
Solution
We replace each group of 8 bits with its equivalent
decimal number (see Appendix B) and add dots for
separation.
19.8
Example 19.2
Change the following IPv4 addresses from dotted-decimal
notation to binary notation.
Solution
We replace each decimal number with its binary
equivalent (see Appendix B).
19.9
Example 19.3
Find the error, if any, in the following IPv4 addresses.
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.
19.10
Note
In classful addressing, the address
space is divided into five classes:
A, B, C, D, and E.
19.11
Figure 19.2 Finding the classes in binary and dotted-decimal notation
19.12
Example 19.4
Find the class of each address.
a. 00000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 14.23.120.8
d. 252.5.15.111
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.
19.13
Table 19.1 Number of blocks and block size in classful IPv4 addressing
19.14
Table 19.3 Addresses for private networks
19.15
Note
In classful addressing, a large part of the
available addresses were wasted.
19.16
Table 19.2 Default masks for classful addressing
Classless Interdomain Routing (CIDR)
19.17
Note
Classful addressing, which is almost
obsolete, is replaced with classless
addressing.
19.18
Figure 19.3 A block of 16 addresses granted to a small organization
19.19
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 /n defines the mask.
19.20
Note
The first address in the block can be
found by setting the rightmost
32 − n bits to 0s.
19.21
Example 19.6
A block of addresses is granted to a small organization.
We know that one of the addresses is 205.16.37.39/28.
What is the first address in the block?
Solution
The binary representation of the given address is
11001101 00010000 00100101 00100111
If we set 32−28 rightmost bits to 0, we get
11001101 00010000 00100101 00100000
or
205.16.37.32.
This is actually the block shown in Figure 19.3.
19.22
Figure 19.6 A frame in a character-oriented protocol
19.23
Note
The last address in the block can be
found by setting the rightmost
32 − n bits to 1s.
19.24
Example 19.7
Find the last address for the block in Example 19.6.
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.
19.25
Note
The number of addresses in the block
can be found by using the formula
232−n.
19.26
Example 19.8
Find the number of addresses in Example 19.6.
Solution
The value of n is 28, which means that number
of addresses is 2 32−28 or 16.
19.27
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.
19.28
Example
Find all subnets for the following
204.17.5.0/27
Solution
204.17.5.0 - 11001100.00010001.00000101.00000000
255.255.255.224 - 11111111. 11111111. 11111111. 11100000
32-27 = 5
2^5 =32
204.17.5.0 255.255.255.224 host address range 1 to 30
204.17.5.32 255.255.255.224 host address range 33 to 62
204.17.5.64 255.255.255.224 host address range 65 to 94
204.17.5.96 255.255.255.224 host address range 97 to 126
204.17.5.128 255.255.255.224 host address range 129 to 158
204.17.5.160 255.255.255.224 host address range 161 to 190
204.17.5.192 255.255.255.224 host address range 193 to 222
204.17.5.224 255.255.255.224 host address range 225 to 254
19.29
Example
Determine the subnet for the given IP address
156.67.154.75/28
Solution
255.255.255.240 - 11111111. 11111111. 11111111. 11110000
32-28=4
2^4 =16
156.67.154.0/28
156.67.154.16/28
156.67.154.32/28
156.67.154.48/28
156.67.154.64/28
156.67.154.80/28
19.30
Example
Determine whether the following two addresses are in the same subnet?
10.21.45.137/13 and 10.23.156.198/13
Solution
255.248.0.0 - 11111111. 11111000. 00000000. 0000000
156.67.154.75/28
16-13 = 3
2^3 =8
10.0.0.0/13
10.8.0.0/13
10.16.0.0/13
10.24.0.0/13
19.31
Example
Determine whether the following two addresses are in the same subnet?
10.21.45.137/13 and 10.23.156.198/13
Solution
00001010 00010101
156.67.154.75/28 00101101 10001001
11111111 11111000 00000000 00000000
00001010 00010111 10011100 11000110
11111111 11111000 00000000 00000000
19.32
Example
Determine the subnet for the given IP address
192.168.13.44/29
Solution
255.255.255.248 - 11111111. 11111111. 11111111. 11111000
32-29=3
2^3 =8
156.67.154.0/29
156.67.154.8/29
156.67.154.16/29
156.67.154.24/29
156.67.154.32/29
156.67.154.40/29
19.33
Example
192.168.13.0/24
netA 120 Hosts
netB 60 Hosts
Solution
Since you need one subnet bits, that leaves you with 7 bits for the host
portion of the address. How many hosts does this support?
2^7 = 128 (128-2=126 usable)
netA: 192.168.13.0/? 25
netB: 192.168.13.?/? 25
128
19.34
Example
204.15.5.0/24
19.35
Example
Solution
Since you need three subnet bits, that leaves you with five bits for the host
portion of the address. How many hosts does this support?
156.67.154.75/28
2^5 = 32
(30 usable)
netA: 204.15.5.0/27 host address range 1 to 30
netB: 204.15.5.32/27 host address range 33 to 62
netC: 204.15.5.64/27 host address range 65 to 94
netD: 204.15.5.96/27 host address range 97 to 126
netE: 204.15.5.128/27 host address range 129 to 158
19.36
VLSM Example
Solution
netA: must support 14 hosts
netB: must support 28 hosts
156.67.154.75/28
netC: must support 2 hosts
netD: must support 7 hosts
netE: must support 28 host
netA: requires a /28 (255.255.255.240) mask to support 14 hosts
netB: requires a /27 (255.255.255.224) mask to support 28
hosts netC: requires a /30 (255.255.255.252) mask to support 2 hosts
netD*: requires a /28 (255.255.255.240) mask to support 7 hosts
netE: requires a /27 (255.255.255.224) mask to support 28 hosts
19.37
VLSM Example
Solution
204.15.5.0
netB: must support 28 hosts
156.67.154.75/28
2^5=32 (30 usable)
Subenet mask: 11111111 11111111 11111111 11100000/27
netE: must support 28 host
2^5 = 32 (30 usable)
Subenet mask: 11111111 11111111 11111111 11100000/27
netA: must support 14 hosts
2^4= 16 (14 usable)
Subenet mask: 11111111 11111111 11111111 11110000/28
19.38
VLSM Example
Solution
204.15.5.0
156.67.154.75/28
netD: must support 7 hosts
2^3= 8 or 2^4 = 16 why?
Subenet mask: 11111111 11111111 11111111 11100000/28
netC: must support 2 hosts
2^2 = 4 (4-2 usable)
Subenet mask: 11111111 11111111 11111111 11111100/30
19.39
VLSM Example
Solution
net: 204.15.5.0/27
net: 204.15.5.32/27
net: 204.15.5.64/27
net: 204.15.5.96/27
net: 204.15.5.128/27
netB: 204.15.5.0/27 host address range 1 to 30
netE: 204.15.5.32/27 host address range 33 to 62
19.40
VLSM Example
Solution
net: 204.15.5.0/27 netB
net: 204.15.5.32/27 netE
net: 204.15.5.64/27 subnet
net: 204.15.5.96/27
net: 204.15.5.128/27
netA& netD : must support 14 hosts
2^4= 16 (14 usable)
Subenet mask: 11111111 11111111 11111111 11110000/28
netA: 204.15.5.64/28 host address range 65 to 78
netD: 204.15.5.80/28 host address range 81 to 94
19.41
VLSM Example
Solution
net: 204.15.5.0/27 netB
net: 204.15.5.32/27 netE
net: 204.15.5.64/28 netA
net: 204.15.5.80/28 netD
net: 204.15.5.96/27 Subnet
net: 204.15.5.128/27 Future
netC: must support 2 hosts
2^2 = 4 (4-2 usable)
Subenet mask: 11111111 11111111 11111111 11111100/30
netC: 204.15.5.96/30 host address range 97 to 98
19.42
VLSM Example
Solution
netB: 204.15.5.0/27 host address range 1 to 30
netE: 204.15.5.32/27 host address range 33 to 62
156.67.154.75/28
netA: 204.15.5.64/28 host address range 65 to 78
netD: 204.15.5.80/28 host address range 81 to 94
netC: 204.15.5.96/30 host address range 97 to 98
19.43
Example
192.168.1.0/24
19.44
VLSM Example
Solution
192.168.1.0
156.67.154.75/28
netA: must support 125 hosts
2^7= 128
32-7 = 25
Subenet mask: 11111111 11111111 11111111 10000000/25
netB: must support 60 hosts
2^6=64
32-6= 26
Subenet mask: 11111111 11111111 11111111 11000000/26
19.45
VLSM Example
Solution
192.168.1.0
156.67.154.75/28
netC: must support 28 hosts
2^5= 32
32-5 = 27
Subenet mask: 11111111 11111111 11111111 11100000/27
netD: must support 12 hosts
2^4=16
32-4= 28
Subenet mask: 11111111 11111111 11111111 11110000/28
19.46
VLSM Example
Solution
192.168.1.0
156.67.154.75/28
WAN network: must support 2 hosts
2^2= 4
32-2 = 30
Subenet mask: 11111111 11111111 11111111 11111100/30
19.47
Note
Each address in the block can be
considered as a two-level
hierarchical structure:
the leftmost n bits (prefix) define
the network;
the rightmost 32 − n bits define
the host.
19.48
Figure 19.7 Configuration and addresses in a subnetted network
19.49
Figure 19.8 Three-level hierarchy in an IPv4 address
19.50
Example 19.10
An ISP is granted a block of addresses starting with
190.100.0.0/16 (65,536 addresses). The ISP needs to
distribute these addresses to three groups of customers as
follows:
a. The first group has 64 customers; each needs 250
addresses.
b. The second group has 128 customers; each needs 124
addresses.
c. The third group has 128 customers; each needs 60
addresses.
Design the subblocks and find out how many addresses
are still available after these allocations.
19.51
Example 19.10 (continued)
Solution
Figure 19.9 shows the situation.
Group 1
For this group, each customer needs 250 addresses. This
means that 8 (log2 256) bits are needed to define each
host. The prefix length is then 32 − 8 = 24. The addresses
are
19.52
Example 19.10 (continued)
Group 2
For this group, each customer needs 124 addresses. This
means that 7 (log2 128) bits are needed to define each
host. The prefix length is then 32 − 7 = 25. The addresses
are
19.53
Example 19.10 (continued)
Group 3
For this group, each customer needs 60 addresses. This
means that 6 (log264) bits are needed to each host. The
prefix length is then 32 − 6 = 26. The addresses are
Number of granted addresses to the ISP: 65,536
Number of allocated addresses by the ISP: 40,960
Number of available addresses: 24,576
19.54
Figure 19.9 An example of address allocation and distribution by an ISP
19.55
Figure 19.10 A NAT implementation
19.56
Figure 19.11 Addresses in a NAT
19.57
Figure 19.12 NAT address translation
19.58
Table 19.4 Five-column translation table
19.59
Figure 19.13 An ISP and NAT
19.60
19-2 IPv6 ADDRESSES
Despite all short-term solutions, address depletion is
still a long-term problem for the Internet. This and
other problems in the IP protocol itself have been the
motivation for IPv6.
Topics discussed in this section:
Structure
Address Space
19.61
Note
An IPv6 address is 128 bits long.
19.62
Figure 19.14 IPv6 address in binary and hexadecimal colon notation
19.63
Figure 19.15 Abbreviated IPv6 addresses
19.64
Example 19.11
Expand the address 0:15::1:12:1213 to its original.
Solution
We first need to align the left side of the double colon to
the left of the original pattern and the right side of the
double colon to the right of the original pattern to find
how many 0s we need to replace the double colon.
This means that the original address is.
19.65
Table 19.5 Type prefixes for IPv6 addresses
19.66
Table 19.5 Type prefixes for IPv6 addresses (continued)
19.67
Figure 19.16 Prefixes for provider-based unicast address
19.68
Figure 19.17 Multicast address in IPv6
19.69
Figure 19.18 Reserved addresses in IPv6
19.70
Figure 19.19 Local addresses in IPv6
19.71
