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An IPv4 address is 32 bits long.

Note

The IPv4 addresses are uniqueand universal.

Note


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The address space of IPv4 is232or 4,294,967,296.

Note

Numbers in base 2, 16, and 256 arediscussed in Appendix B.

Note


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Binary to Decimal Conversion

Bit 7 6 5 4 3 2 1 0

27= 26= 25= 24= 23= 22= 21= 20=

Deci 128 64 32 16 8 4 2 1

Binary 1 0 1 0 1 1 1 0

Deci 128 0 32 0 8 4 2 0

Binary (10101110)2 = (128+32+8+4+2)10

Decimal = 174

Example: Convert the binary number (10101110)2 into decimalequivalent


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Figure 5.1 Dotted-decimal notation

IPv4Least Address= 00000000. 00000000. 00000000. 00000000 = 0.0.0.0

Max Addr = 11111111 .11111111. 11111111 .11111111 = 255.255.255.255

Number of address = 232 = 4,294,967,296

Or Address Space


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Find the number of addresses in a range if the first address is146.102.29.0 and the last address is 146.102.32.255.Solution

Last address 146 . 102 . 32 . 255 (-)First Address 146 . 102 . 29 . 0

--------------------------

0 . 0 . 3 . 255

--------------------------

Example 5.5


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The first address in a range of addresses is 14.11.45.96. If thenumber of addresses in the range is 32, what is the lastaddress?SolutionWe convert the number of addresses minus 1 to base 256,which is 0.0.0.31. We then add it to the first address to get thelast address. Addition is in base 256.

Example 5.6


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Figure 5.2 Bitwise NOT operation


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Example 5.7


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Figure 5.3 Bitwise AND operation


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Example 5.8

Shortcut#1 for AND operation

When at least one of the number is

0 or 255selects the smaller byte

or one of them if equal

Shortcut#2 for AND operation

When none of the two bytes iseither 0 or 255,

write each byte as the sum of

eight terms, where each term is a

power of 2.

select the smaller term in each (or

one of them if equal)

and add them to get the result.


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Figure 5.4 Bitwise OR operation


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Example 5.9

Shortcut#1 for OR operation

When at least one of the number is

0 or 255selects the Largest byte

or one of them if equal

Shortcut#2 for OR operation

When none of the two bytes is

either 0 or 255,

write each byte as the sum ofeight terms, where each term is a

power of 2.

select the larger term in each (or

one of them if equal)

and add them to get the result.


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5-2 CLASSFUL ADDRESSING

IP addresses, when started a few decades ago,used the concept of classes. This architecture iscalled classful addressing. In the mid-1990s, a newarchitecture, called classless addressing, wasintroduced that supersedes the original architecture.In this section, we introduce classful addressingbecause it paves the way for understandingclassless addressing and justifies the rationale formoving to the new architecture. Classlessaddressing is discussed in the next section.


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Figure 5.5 Occupation of address space


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Figure 5.6 Finding the class of address


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Figure 5.7 Finding the class of an address using continuous checking

1

Class: A

0Start

1

0

Class: B

1

0

Class: C

1

0

Class: D Class: E


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Find the class of each address:a. 00000001 00001011 00001011 11101111b. 11000001 10000011 00011011 11111111c. 10100111 11011011 10001011 01101111d. 11110011 10011011 11111011 00001111SolutionSee the procedure in Figure 5.7.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 Caddress.c. The first bit is 1; the second bit is 0. This is a class Baddress.d. The first 4 bits are 1s. This is a class E address.

Example 5.10


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Find the class of each address:a. 227.12.14.87b. 193.14.56.22c. 14.23.120.8d. 252.5.15.111Solutiona. 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.

Example 5.11


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Figure 5.8 Netid and hostid


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Figure 5.9 Blocks in Class A

27 =

0


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Figure 5.10 Blocks in Class B

214=

10


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Many class B addresses are wasted.

Note


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Figure 5.11 Blocks in Class C

110

221=


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Not so many organizations are so smallto have a class C block.

Note

Fi 5 12 Th i l bl k i Cl D


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Figure 5.12 The single block in Class D


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Class D addresses are made of oneblock, used for multicasting.

Note

Fi 5 13 Th i l bl k i Cl E


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Figure 5.13 The single block in Class E


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The range of addresses allocated to anorganization in classful addressing

was a block of addresses in

Class A, B, or C.

Note

Fig re 5 14 T l l add ssi g i lassf l add ssi g


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Figure 5.14 Two-level addressing in classful addressing

Figure 5 15 Information extraction in classful addressing


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Figure 5.15 Information extraction in classful addressing

netid

First address

000 ... 0

Example 5 13


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An address in a block is given as 73.22.17.25. Find the number of

addresses in the block, the first address, and the last address.

Solution

73.22.17.15 belongs class B with n=8

1. The number of addresses in this block is

N = 232n = 232

8 =224 = 16,777,216.2. To find the First address,

we keep the leftmost 8 bits and

set the rightmost 24 bits all to 0s.

The first address is 3.0.0.0/8, in which 8 is the value ofn.

3. To find the last address, we keep the leftmost 8 bits and

set the rightmost 24 bits all to 1s.

The last address is 73.255.255.255.

Example 5.13


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Figure 5 17 Solution to Example 5 14


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Figure 5.17 Solution to Example 5.14

Example 5 15


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An address in a block is given as 200.11.8.45. Find the numberof addresses in the block, the first address, and the lastaddress.SolutionFigure 5.17 shows a possible configuration of the network thatuses this block.1. The number of addresses in this block is N = 232n = 256.2. To find the first address, we keep the leftmost 24 bits and setthe rightmost 8 bits all to 0s. The first address is200.11.8.0/24, in which 24 is the value of n.3. To find the last address, we keep the leftmost 24 bits and setthe rightmost 8 bits all to 1s. The last address is200.11.8.255/24.

Example 5.15


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Figure 5 19 Sample Internet


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Figure 5.19 Sample Internet


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The network address is the identifier of anetwork.

Note

Figure 5.20 Network addresses


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Figure 5.20 Network addresses

Figure 5.21 Network mask


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Figure 5.21 Network mask


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Given To find Method

First

Address

and last

address

Number of addresses (Range) Subtract first and last address

Convert the result into

decimal equivalent plus 1.

IP address

(classful

addressing)

Number of addresses = 232-n

First Address = AND operation with

network address

Last Address = we keep the leftmost

mask bits and set the rightmost

remaining all bits to 1s

Identify the class of IP address

Identify the netid of the IP

address

Identify the network

address/mask by placing 1s for

netid and 0s for hostid.

Figure 5.22 Finding a network address using the default mask


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g g g f

Example 5 16


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A router receives a packet with the destination address201.24.67.32. Show how the router finds the network address ofthe packet.SolutionSince the class of the address is B, we assume that the routerapplies the default mask for class B, 255.255.0.0 to find thenetwork address.

Example 5.16

Example 5.18


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Figure 5.23 shows a network using class B addresses beforesubnetting. We have just one network with almost 216 hosts.The whole network is connected, through one singleconnection, to one of the routers in the Internet. Note that wehave shown /16 to show the length of the netid (class B).

Example 5.18


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NetworkSubnetwork

Computer lab # 1 Computer lab # 2

Computer lab # 4Computer lab # 3

Example 5.19


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Figure 5.24 shows the same network in Figure 5.23 aftersubnetting. The whole network is still connected to the Internetthrough the same router. However, the network has used aprivate router to divide the network into four subnetworks. Therest of the Internet still sees only one network; internally thenetwork is made of four subnetworks. Each subnetwork cannow have almost 214 hosts. The network can belong to auniversity campus with four different schools (buildings). Aftersubnetting, each school has its own subnetworks, but still thewhole campus is one network for the rest of the Internet. Notethat /16 and /18 show the length of the netid and subnetids.

Example 5.19

Figure 5.24 Example 5.19


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Computer lab # 1 Computer lab # 2

Computer lab # 3Computer lab # 4

Figure 5.25 Network mask and subnetwork mask


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Formula to find the subnet id : nsub = n + log2s

Where, nsub: length of each subnetidn : length of netid

s : number of subnet (must power of 2)

Example 5.20


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Divided a class B network into four subnetworks.SolutionFormula to find the subnet id : nsub = n + log2s

Where, nsub: length of each subnetid

n : length of netid

s : number of subnet (must power of 2)

The value of n = 16

The value of n1 = n2 = n3 = n4 = 16 + log24 = 18.

This means that the subnet mask has eighteen 1s and fourteen 0s.

In other words, the subnet mask is 255.255.192.0

Different from the network mask for class B (255.255.0.0).

Example 5.20

Example 5.21


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Since one of the addresses in subnet 2 is 141.14.120.77, wecan find the subnet address as:

Example 5.21

The values of the first, second, and fourth bytes are calculatedusing the first short cut for AND operation. The value of the thirdbyte is calculated using the second short cut for the ANDoperation.

Figure 5.26 Comparison of subnet, default, and supernet mask


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nsub=n+log2s

nsuper=n-log2c


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Type To find Method

IP address

(classful

addressing)

Number of addresses = 232-n

First Address = AND operation with

network address

Last Address = we keep the

leftmost mask bits and set the

rightmost remaining all bits to 1s



address

Identify the network

address/mask by placing 1s for

netid and 0s for hostid.

Subnetting

Classful

Addressing

Subnet id Network : n+log2S

S No. of subnets required (mustbe in power of 2)



address (n)

-Sbunetting

-Classless

Addressing

Prefix of network: n+log2(N/Ni)

N: total number of address granted

Ni : number of address per subnet


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5-3 CLASSLESS ADDRESSING

Subnetting and supernetting in classful addressing didnot really solve the address depletion problem. Withthe growth of the Internet, it was clear that a largeraddress space was needed as a long-term solution.Although the long-range solution has already beendevised and is called IPv6, a short-term solution wasalso devised to use the same address space but tochange the distribution of addresses to provide a fairshare to each organization. The short-term solutionstill uses IPv4 addresses, but it is called classlessaddressing.

Topics Discussed in the Section


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Variable Length Blocks

Two-Level Addressing

Block Allocation

Subnetting

Figure 5.27 Variable-length blocks in classless addressing


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In classless addressing, the prefixdefines the network and the suffix

defines the host.

Note

Figure 5.28 Prefix and suffix


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The prefix length in classlessaddressing can be 1 to 32.

Note

Example 5.22


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What is the prefix length and suffix length if the whole Internet isconsidered as one single block with 4,294,967,296 addresses?SolutionIn this case, the prefix length is 0 and the suffix length is 32. All32 bits vary to define 232 = 4,294,967,296 hosts in this singleblock.

p

Example 5.23


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What is the prefix length and suffix length if the Internet isdivided into 4,294,967,296 blocks and each block has onesingle address?SolutionIn this case, the prefix length for each block is 32 and the suffixlength is 0. All 32 bits are needed to define 232 = 4,294,967,296blocks. The only address in each block is defined by the blockitself.

p

Example 5.24


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The number of addresses in a block is inversely related to thevalue of the prefix length, n. A small n means a larger block; alarge n means a small block.

Figure 5.29 Slash notation


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In classless addressing, we need toknow one of the addresses in the blockand the prefix length to define the block.

Note

Example 5.25


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In classless addressing, an address cannot as such define theblock the address belongs to. For example, the address230.8.24.56 can belong to many blocks some of them areshown below with the value of the prefix associated with thatblock:

Example 5.26


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The following addresses are defined using slash notations.a. In the address 12.23.24.78/8, the network mask is 255.0.0.0.The mask has eight 1s and twenty-four 0s. The prefix lengthis 8; the suffix length is 24.b. In the address 130.11.232.156/16, the network mask is255.255.0.0. The mask has sixteen 1s and sixteen 0s.Theprefix length is 16; the suffix length is 16.c. In the address 167.199.170.82/27, the network mask is

255.255.255.224. The mask has twenty-seven 1s and five0s. The prefix length is 27; the suffix length is 5.

Example 5.27


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One of the addresses in a block is 167.199.170.82/27. Find thenumber of addresses in the network, the first address, and thelast address.SolutionThe value of n is 27. The network mask has twenty-seven 1sand five 0s. It is 255.255.255.240.a. The number of addresses in the network is 232 n = 32.b. We use the AND operation to find the first address (networkaddress). The first address is 167.199.170.64/27.

167.199.170.64/27

255.255.255.240

167.199.170.82

Example 5.27 Continued


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c. To find the last address, we first find the complement of thenetwork mask and then OR it with the given address: The lastaddress is 167.199.170.95/27.


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c. To find the last address, we use the complement of thenetwork mask and the first short cut method wediscussed before. The last address is 17.63.110.255/24.

Example 5.29


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One of the addresses in a block is 110.23.120.14/20. Find thenumber of addresses, the first address, and the last address inthe block.SolutionThe network mask is 255.255.240.0.a. The number of addresses in the network is 232 20 = 4096.b. To find the first address, we apply the first short cut tobytes 1, 2, and 4 and the second short cut to byte 3. Thefirst address is 110.23.112.0/20.



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c. To find the last address, we apply the first short cut tobytes 1, 2, and 4 and the second short cut to byte 3. TheOR operation is applied to the complement of the mask.The last address is 110.23.127.255/20.

Example 5.30


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An ISP has requested a block of 1000 addresses. The followingblock is granted.a. Since 1000 is not a power of 2, 1024 addresses aregranted (1024 = 210).b. The prefix length for the block is calculated as n = 32

log21024 = 22.c. The beginning address is chosen as 18.14.12.0 (which isdivisible by 1024).The granted block is 18.14.12.0/22. The first address is18.14.12.0/22 and the last address is 18.14.15.255/22.

First address 18.14.12.0/22 18.14.00001100.00000000

Last Address 18.14.15.255/22 18.14.00001111.11111111


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Example 5.31


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Assume an organization has given a class A block as 73.0.0.0in the past. If the block is not revoked by the authority, theclassless architecture assumes that the organization has ablock 73.0.0.0/8 in classless addressing.


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The restrictions applied in allocatingaddresses for a subnetwork are

parallel to the ones used to allocateaddresses for a network.

Note

Example 5.32


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An organization is granted the block 130.34.12.64/26. Theorganization needs four subnetworks, each with an equalnumber of hosts. Design the subnetworks and find theinformation about each network.SolutionThe number of addresses for the whole network can be foundas N = 232 26 = 64.The first address in the network is 130.34.12.64/26 and the lastaddress is 130.34.12.127/26. We now design the subnetworks:1. We grant 16 addresses for each subnetwork to meet thefirst requirement (64/16 is a power of 2).2. The subnetwork mask for each subnetwork is:

N: total number of address granted

Ni : number of address per subnet



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3. We grant 16 addresses to each subnet starting from thefirst available address. Figure 5.30 shows the subblock foreach subnet. Note that the starting address in eachsubnetwork is divisible by the number of addresses inthat subnetwork.



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Example 5.33


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An organization is granted a block of addresses with thebeginning address 14.24.74.0/24. The organization needs tohave 3 subblocks of addresses to use in its three subnets asshown below: One subblock of 120 addresses. One subblock of 60 addresses. One subblock of 10 addresses.SolutionThere are 232 24 = 256 addresses in this block. The firstaddress is 14.24.74.0/24; the last address is 14.24.74.255/24.a. The number of addresses in the first subblock is not a

power of 2. We allocate 128 addresses. The subnetmask is 25. The first address is 14.24.74.0/25; the lastaddress is 14.24.74.127/25.



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b. The number of addresses in the second subblock is not apower of 2 either. We allocate 64 addresses. The subnetmask is 26. The first address in this block is14.24.74.128/26; the last address is 14.24.74.191/26.c. The number of addresses in the third subblock is not apower of 2 either. We allocate 16 addresses. The subnetmask is 28. The first address in this block is14.24.74.192/28; the last address is 14.24.74.207/28.d. If we add all addresses in the previous subblocks, theresult is 208 addresses, which means 48 addresses are leftin reserve. The first address in this range is 14.24.74.209.The last address is 14.24.74.255.e. Figure 5.31 shows the configuration of blocks. We haveshown the first address in each block.



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Example 5.34


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Assume a company has three offices: Central, East, and West.The Central office is connected to the East and West offices viaprivate, WAN lines. The company is granted a block of 64addresses with the beginning address 70.12.100.128/26. Themanagement has decided to allocate 32 addresses for theCentral office and divides the rest of addresses between the twoother offices.1. The number of addresses are assigned as follows:

2. We can find the prefix length for each subnetwork:



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3. Figure 5.32 shows the configuration designed by themanagement. The Central office uses addresses70.12.100.128/27 to 70.12.100.159/27. The company has usedthree of these addresses for the routers and has reserved thelast address in the subblock. The East officeuses the addresses 70.12.100.160/28 to 70.12.100.175/28. Oneof these addresses is used for the router and the company hasreserved the last address in the subblock. The West office usesthe addresses 70.12.100.160/28 to 70.12.100.175/28. One ofthese addresses is used for the router and the company hasreserved the last address in the subblock. The company usesno address for the point-to-point connections in WANs.

Figure 5.32 Example 5.34


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SolutionLet us solve the problem in two steps. In the first step, weallocate a subblock of addresses to each group. The totalnumber of addresses allocated to each group and the prefixlength for each subblock can found as

Figure 5.33 shows the design for the first hierarchical level.Figure 5.34 shows the second level of the hierarchy. Note thatwe have used the first address for each customer as the subnetaddress and have reserved the last address as a specialaddress.

Figure 5.33 Solution to Example 5.35: first step


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Figure 5.34 Solution to Example 5.35: second step


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5-4 SPECIAL ADDRESSES


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5 4 SPECIAL ADDRESSES

In classful addressing some addresses werereserved for special purposes. The classlessaddressing scheme inherits some of these specialaddresses from classful addressing.


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Figure 5.35 Example of using the all-zero address


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Source: 0.0.0.0Destination: 255.255.255.255

Packet

Figure 5.36 Example of limited broadcast address


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221.45.71.20/24 221.45.71.178/24

221.45.71.64/24 221.45.71.126/24

Network

Figure 5.37 Example of loopback address


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Transport layer

Application layer

Network layer

Process 1 Process 2

Destination address:127.x.y.z

Packet


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Figure 5.38 Example of a directed broadcast address


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221.45.71.0/24

221.45.71.20/24 221.45.71.178/24

221.45.71.64/24 221.45.71.126/24

Network:

Packet

5-5 NAT


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5 5 NAT

The distribution of addresses through ISPs hascreated a new problem. If the business grows or thehousehold needs a larger range, the ISP may not beable to grant the demand because the addressesbefore and after the range may have already beenallocated to other networks. In most situations,however, only a portion of computers in a smallnetwork need access to the Internet simultaneously.A technology that can help in this cases is networkaddress translation (NAT).



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Address Translation

Translation Table

Figure 5.39 NAT


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Figure 5.40 Address resolution


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Internet

Site using private addresses

172.18.3.1

172.18.3.2

172.18.3.20

Source: 172.18.3.1 Source: 200.24.5.8

Destination: 200.24.5.8Destination: 172.18.3.1

Figure 5.41 Translation


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Chap 05 Network Layer

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