19.1
Chapter 19
Network Layer:
Logical Addressing
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19.2
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.
Address Space
Notations
Classful Addressing
Classless Addressing
Network Address Translation (NAT)
Topics discussed in this section:
19.3
The address space of IPv4 is
232 or 4,294,967,296.
Note
19.4
Figure 19.1 Dotted-decimal notation and binary notation for an IPv4 address
19.5
Change the following IPv4 addresses from binary
notation to dotted-decimal notation.
Example 19.1
Solution
We replace each group of 8 bits with its equivalent
decimal number and add dots for separation.
19.6
Change the following IPv4 addresses from dotted-decimal
notation to binary notation.
Example 19.2
Solution
We replace each decimal number with its binary
equivalent .
19.7
Find the error, if any, in the following IPv4 addresses.
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.
19.8
In classful addressing, the address
space is divided into five classes:
A, B, C, D, and E.
Note
19.9
Figure 19.2 Finding the classes in binary and dotted-decimal notation
19.10
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
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.
19.11
Table 19.1 Number of blocks and block size in classful IPv4 addressing
19.12
In classful addressing, a large part of the
available addresses were wasted.
Note
19.13
Table 19.2 Default masks for classful addressing
19.14
Classful addressing, which is almost
obsolete, is replaced with classless
addressing.
Note
19.15
Figure 19.3 shows a block of addresses, in both binary
and dotted-decimal notation, granted to a small business
that needs 16 addresses.
We can see that the restrictions are applied to this block.
The addresses are contiguous. The number of addresses
is a power of 2 (16 = 24), and the first address is divisible
by 16. The first address, when converted to a decimal
number, is 3,440,387,360, which when divided by 16
results in 215,024,210.
Example 19.5
19.16
Figure 19.3 A block of 16 addresses granted to a small organization
Restrictions on classless
addressing
 The addresses in a block must be contiguous, one after
another.
 The number of addresses in a block must be a power of
2 (i.e. I, 2, 4, 8, ... ).
 The first address must be evenly divisible by the number
of addresses.
19.17
19.18
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.
Note
19.19
The first address in the block can be
found by setting the rightmost
32 − n bits to 0s.
The last address in the block can be
found by setting the rightmost
32 − n bits to 1s.
The number of addresses in the block
can be found by using the formula
232−n
19.20
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 & last 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 0010000
or
205.16.37.32.
Example 19.6
19.21
Last address
The binary representation of the given address is
11001101 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 1, we get last address
11001101 00010000 00100101 00101111
or
205.16.37.47
Numbers of addresses
The value of n is 28, which means that number
of addresses is 2 32−28 or 16.
Cont..
19.22
Another way to find the first address, the last address, and
the number of addresses is to represent the mask as a 32-
bit binary (or 8-digit hexadecimal) number. This is
particularly useful when we are writing a program to find
these pieces of information. For example the /28 can be
represented as
11111111 11111111 11111111 11110000
(twenty-eight 1s and four 0s).
Find
a. The first address
b. The last address
c. The number of addresses.
19.23
Solution
a. The first address can be found by ANDing bit by bit the
given addresses with the mask. (205.16.37.39/28)
Ccntinued
b. The last address can be found by ORing the given
addresses bit by bit with the complement of the mask.
Continued
The number of addresses can be found by
complementing the mask, interpreting it as a decimal
number, and adding 1 to it.
19.25
A network configuration for the block 205.16.37.32/28
19.26
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
Two levels ofhierarchy in an IPv4 address
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.
19.28
Configuration and addresses in a subnetted network
19.29
Figure 19.8 Three-level hierarchy in an IPv4 address
19.30
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 256
addresses.
b. The second group has 128 customers; each needs 128
addresses.
c. The third group has 128 customers; each needs 64
addresses.
Design the subblocks and find out how many addresses
are still available after these allocations.
Example 19.10
19.31
Solution
Figure 19.9 shows the situation.
Example 19.10 (continued)
Group 1
For this group, each customer needs 256 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
19.32
Example 19.10 (continued)
Group 2
For this group, each customer needs 128 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
19.33
Example 19.10 (continued)
Group 3
For this group, each customer needs 64 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
19.34
Figure 19.9 An example of address allocation and distribution by an ISP
19.35
Table 19.3 Addresses for private networks
No router will forward a packet that has one of these addresses
as the destination address.
19.36
NAT enables a user to have a large set of addresses internally and one
address, or a small set of addresses, externally.
Figure 19.10 A NAT implementation
19.37
Figure 19.11 Addresses in a NAT
All the outgoing packets go through the NAT router, which replaces the
source address in the packet with the global NAT address.
All incoming packets also pass through the NAT router, which replaces the
destination address in the packet (the NAT router global address) with the
appropriate private address.
19.38
Figure 19.12 NAT address translation
19.39
Table 19.4 Five-column translation table
19.40
19-2 IPv6 ADDRESSES
Despite all short-term solutions, address depletion is
still a long-term problem for the Internet.
lack of accommodation for real-time audio and video
transmission, and encryption and authentication of
data for some applications, have been the motivation
for IPv6.
19.41
An IPv6 address is 128 bits long.
Note
19.42
Figure 19.14 IPv6 address in binary and hexadecimal colon notation
19.43
Figure 19.15 Abbreviated IPv6 addresses
19.44
Expand the address 0:15::1:12:1213 to its original.
Example 19.11
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.
19.45
Table 19.5 Type prefixes for IPv6 addresses
19.46
Table 19.5 Type prefixes for IPv6 addresses (continued)
19.47
Figure 19.16 Prefixes for provider-based unicast address
19.48
Figure 19.17 Multicast address in IPv6
19.49
Figure 19.18 Reserved addresses in IPv6
19.50
Figure 19.19 Local addresses in IPv6

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ch19new.ppt

  • 1. 19.1 Chapter 19 Network Layer: Logical Addressing Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
  • 2. 19.2 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. Address Space Notations Classful Addressing Classless Addressing Network Address Translation (NAT) Topics discussed in this section:
  • 3. 19.3 The address space of IPv4 is 232 or 4,294,967,296. Note
  • 4. 19.4 Figure 19.1 Dotted-decimal notation and binary notation for an IPv4 address
  • 5. 19.5 Change the following IPv4 addresses from binary notation to dotted-decimal notation. Example 19.1 Solution We replace each group of 8 bits with its equivalent decimal number and add dots for separation.
  • 6. 19.6 Change the following IPv4 addresses from dotted-decimal notation to binary notation. Example 19.2 Solution We replace each decimal number with its binary equivalent .
  • 7. 19.7 Find the error, if any, in the following IPv4 addresses. 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.
  • 8. 19.8 In classful addressing, the address space is divided into five classes: A, B, C, D, and E. Note
  • 9. 19.9 Figure 19.2 Finding the classes in binary and dotted-decimal notation
  • 10. 19.10 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 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.
  • 11. 19.11 Table 19.1 Number of blocks and block size in classful IPv4 addressing
  • 12. 19.12 In classful addressing, a large part of the available addresses were wasted. Note
  • 13. 19.13 Table 19.2 Default masks for classful addressing
  • 14. 19.14 Classful addressing, which is almost obsolete, is replaced with classless addressing. Note
  • 15. 19.15 Figure 19.3 shows a block of addresses, in both binary and dotted-decimal notation, granted to a small business that needs 16 addresses. We can see that the restrictions are applied to this block. The addresses are contiguous. The number of addresses is a power of 2 (16 = 24), and the first address is divisible by 16. The first address, when converted to a decimal number, is 3,440,387,360, which when divided by 16 results in 215,024,210. Example 19.5
  • 16. 19.16 Figure 19.3 A block of 16 addresses granted to a small organization
  • 17. Restrictions on classless addressing  The addresses in a block must be contiguous, one after another.  The number of addresses in a block must be a power of 2 (i.e. I, 2, 4, 8, ... ).  The first address must be evenly divisible by the number of addresses. 19.17
  • 18. 19.18 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. Note
  • 19. 19.19 The first address in the block can be found by setting the rightmost 32 − n bits to 0s. The last address in the block can be found by setting the rightmost 32 − n bits to 1s. The number of addresses in the block can be found by using the formula 232−n
  • 20. 19.20 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 & last 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 0010000 or 205.16.37.32. Example 19.6
  • 21. 19.21 Last address The binary representation of the given address is 11001101 00010000 00100101 00100111 If we set 32 − 28 rightmost bits to 1, we get last address 11001101 00010000 00100101 00101111 or 205.16.37.47 Numbers of addresses The value of n is 28, which means that number of addresses is 2 32−28 or 16. Cont..
  • 22. 19.22 Another way to find the first address, the last address, and the number of addresses is to represent the mask as a 32- bit binary (or 8-digit hexadecimal) number. This is particularly useful when we are writing a program to find these pieces of information. For example the /28 can be represented as 11111111 11111111 11111111 11110000 (twenty-eight 1s and four 0s). Find a. The first address b. The last address c. The number of addresses.
  • 23. 19.23 Solution a. The first address can be found by ANDing bit by bit the given addresses with the mask. (205.16.37.39/28) Ccntinued
  • 24. b. The last address can be found by ORing the given addresses bit by bit with the complement of the mask. Continued The number of addresses can be found by complementing the mask, interpreting it as a decimal number, and adding 1 to it.
  • 25. 19.25 A network configuration for the block 205.16.37.32/28
  • 26. 19.26 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
  • 27. Two levels ofhierarchy in an IPv4 address 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.
  • 28. 19.28 Configuration and addresses in a subnetted network
  • 29. 19.29 Figure 19.8 Three-level hierarchy in an IPv4 address
  • 30. 19.30 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 256 addresses. b. The second group has 128 customers; each needs 128 addresses. c. The third group has 128 customers; each needs 64 addresses. Design the subblocks and find out how many addresses are still available after these allocations. Example 19.10
  • 31. 19.31 Solution Figure 19.9 shows the situation. Example 19.10 (continued) Group 1 For this group, each customer needs 256 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
  • 32. 19.32 Example 19.10 (continued) Group 2 For this group, each customer needs 128 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
  • 33. 19.33 Example 19.10 (continued) Group 3 For this group, each customer needs 64 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
  • 34. 19.34 Figure 19.9 An example of address allocation and distribution by an ISP
  • 35. 19.35 Table 19.3 Addresses for private networks No router will forward a packet that has one of these addresses as the destination address.
  • 36. 19.36 NAT enables a user to have a large set of addresses internally and one address, or a small set of addresses, externally. Figure 19.10 A NAT implementation
  • 37. 19.37 Figure 19.11 Addresses in a NAT All the outgoing packets go through the NAT router, which replaces the source address in the packet with the global NAT address. All incoming packets also pass through the NAT router, which replaces the destination address in the packet (the NAT router global address) with the appropriate private address.
  • 38. 19.38 Figure 19.12 NAT address translation
  • 39. 19.39 Table 19.4 Five-column translation table
  • 40. 19.40 19-2 IPv6 ADDRESSES Despite all short-term solutions, address depletion is still a long-term problem for the Internet. lack of accommodation for real-time audio and video transmission, and encryption and authentication of data for some applications, have been the motivation for IPv6.
  • 41. 19.41 An IPv6 address is 128 bits long. Note
  • 42. 19.42 Figure 19.14 IPv6 address in binary and hexadecimal colon notation
  • 44. 19.44 Expand the address 0:15::1:12:1213 to its original. Example 19.11 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.
  • 45. 19.45 Table 19.5 Type prefixes for IPv6 addresses
  • 46. 19.46 Table 19.5 Type prefixes for IPv6 addresses (continued)
  • 47. 19.47 Figure 19.16 Prefixes for provider-based unicast address
  • 48. 19.48 Figure 19.17 Multicast address in IPv6
  • 49. 19.49 Figure 19.18 Reserved addresses in IPv6
  • 50. 19.50 Figure 19.19 Local addresses in IPv6