Chapter 2 - Internet Protocol (IP) Addresses

Chapter 2: Internet Protocol (IP) Addresses

	Overview

A complete understanding of unicast and multicast IP addressing is required in order to design and implement robust IP networks. Concepts such as subnetting and Variable Length Subnet Masks (VLSM) should be mastered so that IP addressing plans make efficient use of your assigned address space. The concept, operation, and configuration of IP unicast routing protocols, RIP, IGRP, EIGRP, and OSPF, also need to be mastered because most multicast routing protocols rely on the underlying unicast routing configuration.

IP Address Format

	An IP address is a 32-bit number that can be represented in many formats. Routers and computers are designed to operate efficiently on binary numbers, so a binary representation is a natural way for them to store and manipulate IP addresses. A typical 32-bit IP address to a router would look something like this:

	10011100000110100001111000111100

	This may be a fine representation for routers, but for us it is not the most appealing method. So let s take a look at the binary representation and see if we can find a way to represent these numbers using a method that may be a bit more palatable. One way is to simply represent the IP address as a decimal number. The binary number used in the example above has a decimal value of

	2,618,957,372

	This may be easier to read, but the size of the number makes it cumbersome to work with. Another representation scheme is to break up the binary number into pieces and represent each piece as a decimal number. A natural size for binary pieces is 8 bits, which is the familiar byte or not-as-familiar octet (octet is the telecommunication term, but the two words can be used interchangeably). So let s take our binary number, write it using groups of 8 bits (four octets) and then represent each group as a decimal number.

	10011100

	00011010

	00011110

	00111100

156

26

30

60

	Table 2-1: Range of IP Addresses

Low

High

	Binary

0000000000000000000 0000000000000

11111111111111111111111111111111

	Decimal

0

4,294,967,295

	Dotted Decimal

0.0.0.0

255.255.255.255

	We don t need all that space between the numbers, so let s use a period, or dot, as a separator. Now our IP address has the form

	156.26.30.60

	which is referred to as dotted decimal notation. How many IP addresses are there? The range of IP addresses in all our representation schemes is shown in Table 2-1.

	Theoretically, there are 4,294,967,296 possible IP addresses, although we will discover in this chapter that the actual usable number of IP addresses is much smaller.

	Classful IP Addressing

	For a protocol to be routable, its address structure must be hierarchical, meaning that the address must contain at least two parts. For IP addresses, these parts are the network portion and the host portion. A host is an end station such as a computer workstation, router interface, or printer, while a network consists of one or more hosts. Figure 2-1 is a simple network consisting of two networks connected by a two-port router. The address of each host on this network, including the router interfaces, is given by its network and host numbers.

	Figure 2-1: Hierarchical addressing

	When the IP address scheme was designed, the decision was made to create five classes of IP addresses simply named Class A, B, C, D, and E. The logic behind the first three network classes was that the IP addressing scheme would be used for a few networks with a large number of hosts (Class A), a moderate number of networks with a moderate number of hosts (Class B), and a large number of networks with a small number of hosts (Class C). Class D addresses would be used for multicasting and Class E addresses would be reserved for experimental use.

	Having three classes of IP addresses to handle different size networks requires that the network part and the host part for each address class have unequal sizes. The breakdown for the allocation of bits for the network and host portion for the first four IP address classes is shown in Figure 2-2.

	Figure 2-2: Classful IP address structure

	Class A addresses use 8 bits to identify the network and 24 bits to identify the host with the most significant bit of the first octet set to zero. Class B addresses use 16 bits to identify the network and 16 bits to identify the host with the first two bits of the first octet set to 1 0. Class C addresses use 24 bits to identify the network and 8 bits to identify the host with the first 3 bits of the first octet set to 1 1 0.

	Class D or multicast addresses differ from unicast addresses in their interpretation.

	A Class A, B, or C address is used to identify a network and a host on that network. A Class D multicast address is used to identify a group of receivers and senders of multicast traffic. Additionally, multicast senders and receivers can be present on any network.

	If we examine the first octet of each class, we can see that the range of values for the four classes is

	00000001 (1) 01111110 (126) for Class A

10000000 (128) 10111111 (191) for Class B

11000000 (192) 11011111 (223) for Class C

11100000 (224) 11101111 (239) for Class D

	Looking at the first octet of the IP address can easily identify the network class. For example, the address used previously, 156.26.30.60, is a Class B address because the first octet is between 128 and 191. Another (and more tedious) way to identify the class is to represent the first octet of the address in binary and see what the first couple of bits are set to. For example, 156 equals 10011100 in binary. The first two bits are 1 0, so according to Figure 2-1, this is a Class B address.

	How many Class A, B, and C networks are there? Class A networks use 7 bits for the network ID, so 128 Class A networks are possible. Class B addresses use 6 bits from the first octet and all 8 bits of the second octet, so there are 16,384 networks (64 x 256), 64 from the first octet and 256 from the second octet. Class C addresses use 5 bits from the first octet, 8 bits from the second octet, and 8 bits from the third octet, so there are 2,097,152 possible Class C networks (32 x 256 x 256). Class D addresses are not associated with networks but with multicast groups.

	Class A, B, and C addresses are unicast addresses. Each IP address in the first three classes is used to identify a particular and unique Internet host, while a Class D address is used to identify a group of hosts belonging to a particular IP multicast group. The multicast addresses are in the range 224.0.0.0 through 239.255.255.255 (currently assigned multicast addresses are listed in Appendix B). The range of addresses between 224.0.0.0 and 224.0.0.255, inclusive, is reserved for the use of routing protocols and other low-level topology discovery or maintenance protocols, such as gateway discovery and group membership reporting. Multicast routers should not for ward any multicast datagram with destination addresses in this range, regardless of the TTL.

	How many hosts can each network have? Class A networks have 24 bits to identify a host; this equals 1,677,216 possible hosts per network! Class B networks have 16 bits to identify a host, which equals 65,536 hosts, and Class C networks have 8 bits to identify a host, which equals 256 possible hosts. Table 2-2 lists the capabilities for Class A, B, and C addresses.

	Table 2-2: IP Classful Address Capabilities

	Class

Networks

Hosts

	A

0,000,126

16,777,214

	B

0,016,384

00,065,534

	C

2,097,152

00,000,254

	You may have noticed that the number of hosts listed in Table 2-1 is always two less than the number calculated. The reason for this discrepancy is that two special addresses can t be assigned to a host. A host address of all ones is the broadcast address for a particular network, and a host address of all zeros is used by a host to temporarily identify itself ( this host ) until it has been assigned an IP address. Only 126 Class A networks exist because network 0 cannot be used, and network 127 is reserved for the loopback address that is used for testing interprocess communication. When a host sends a packet to 127.0.0.1, the data is not sent on the network but is returned immediately to the sending host.

	The IP address blocks listed below have been reserved for private Internets.

	11110.0.0.0 10.255.255.2550

1172.16.0.0 172.31.255.2550

192.168.0.0 192.168.255.255

	These private IP addresses should never be advertised on the Internet because they can be used by any private Internet. If these addresses are used, then a technique such as network address translation would need to be used in the private Internet to be connected to the public Internet.

	Classful IP address assignments can be extremely inefficient as the following design problem demonstrates. Assume we are designing a network for a campus that has approximately 1500 nodes or end-stations. Also assume that the predicted future growth of the network over the next five years will be no more than 5000 nodes. At first glance, it would seem that a Class B network would suffice for the current network requirements and also leave plenty of room for future growth. Having 1500-plus nodes (5000-plus in the future) would be a very large ethernet collision domain. If we want to limit the number of nodes on an ethernet segment to no more than 100, then we need 50 networks to accomplish our design. Regardless of which class of IP network addresses we decide to use (assuming we could choose any addresses we want), there will be an enormous waste of IP addresses as shown in Table 2-3.

	Table 2-3: IP Address Design Inefficiencies

	Network Class

Addresses Required

Addresses Available

Addresses Wasted

	A

100

16,777,214

16,777,114

	B

100

00,065,534

00,065,434

	C

100

00,000,254

00,000,154

Now multiply each entry in Table 2-3 by the 50 networks that are required and you can easily see that regardless of which address class we choose, an enormous number of IP addresses will be wasted. Also, if we are to have connectivity to the Internet, then the network will have to advertise 50 networks to the Internet routers. Multiply that by the number of campuses in the world and you have a situation where the size of the Internet routing tables becomes unmanageable. How do we overcome these problems? In a word, subnetting.

IP Subnets

	The solution to our design problem is to divide whatever class of IP address we are assigned into a number of smaller networks with fewer hosts per network. This is accomplished by borrowing bits from the host portion of our IP address and using them in the network portion. How do we and, more importantly, how does a router know how many bits to use for the network and how many to use for the host? The answer is by using a subnet mask.

	A subnet mask is a 32-bit binary number that identifies which bits in the address are used for the host and which bits are used for the network. A one in the mask identifies the corresponding bit in the IP address as a network bit, and a zero in the mask identifies the corresponding bit in the IP address as a host bit. A router accomplishes this operation by performing a bitwise AND operation with the IP address and the subnet mask.

	0 AND 1 5 0

1 AND 0 5 01 AND 1 5 1

	As an example, consider the IP address/subnet mask pair

	156.26.30.60/255.255.240.0

	which has the binary representations

	Address

	10111100

	00011010

	00011110

	00111100

Mask

11111111

11111111

11110000

00000000

	Performing the AND operation yields

	000110100001000000000000

	Converting the result to dotted decimal notation yields the network portion of the IP address

	156.26.16.0

	One subnet mask restriction is that the 1 bits in the mask must be contiguous. Because of this, an alternative representation for the mask is just to indicate how many 1 bits are in the mask. For example, the IP address/ subnet mask pair in the previous example can be written as 156.26.30.60/20. The subnet masks for non-subnetted networks are shown in Figure 2-3.

	Figure 2-3: Standard IP subnet masks

	Subnet masks never have fewer ones than the masks listed in Figure 2-3. A Class C address, for example, cannot have a subnet mask of 255.255.0.0. Request for Comment (RFC) 950 first defined the subnetting of IP addresses and does not allow the use of the all-zeros and all-ones subnet, so we will initially look at subnetting examples that obey these restrictions. In later examples, we will see how we can remove these restrictions with the use of an appropriate routing protocol, such as OSPF. The number of subnet bits cannot be one because of the restriction in RFC 950 (see Tables 2-4, 2-5, and 2-6). A 1-bit subnet mask would have a value of either zero (all zeros) or one (all ones) and this is not allowed.

	Table 2-4: Class A Subnet Masks

	Number of Subnet bits

Subnet Mask

Number of Subnetworks

Number of Hosts/Subnet

Total Number of Hosts

	1

	2

255.192.0.0

0000002

4194302

08388604

	3

255.224.0.0

0000006

2097150

12582900

	4

255.240.0.0

0000014

1048574

14680036

	5

255.248.0.0

0000030

0524286

15728580

	6

255.252.0.0

0000062

0262142

16252804

	7

255.254.0.0

0000126

0131070

16514820

	8

255.255.0.0

0000254

0065534

16645636

	9

255.255.128.0

0000510

0032766

16710660

	10

255.255.192.0

0001022

0016382

16742404

	11

255.255.224.0

0002046

0008190

16756740

	12

255.255.240.0

0004094

0004094

16760836

	13

255.255.248.0

0008190

0002046

16756740

	14

255.255.252.0

0016382

0001022

16742404

	15

255.255.254.0

0032766

0000510

16710660

	16

255.255.255.0

0065534

0000254

16645636

	17

255.255.255.128

0131070

0000126

16514820

	18

255.255.255.192

0262142

0000062

16252804

	19

255.255.255.224

0524286

0000030

15728580

	20

255.255.255.240

1048574

0000014

14680036

	21

255.255.255.248

2097150

0000006

12582900

	22

255.255.255.252

4194302

0000002

08388604

	23

	24

	Table 2-5: Class B Subnet Masks

	Number of Subnet Bits

Subnet Mask

Number of Subnetworks

Number of Hosts/Subnet

Total Number of Hosts

	1

	2

00255.255.192.0

00002

16382

32764

	3

00255.255.224.0

00006

08190

49140

	4

00255.255.240.0

00014

04094

57316

	5

00255.255.248.0

00030

02046

61380

	6

00255.255.252.0

00062

01022

63364

	7

00255.255.254.0

00126

00510

64260

	8

00255.255.255.0

00254

00254

64516

	9

255.255.255.128

00510

00126

64260

	10

255.255.255.192

01022

00062

63364

	11

255.255.255.224

02046

00030

61380

	12

255.255.255.240

04094

00014

57316

	13

255.255.255.248

08190

00006

49140

	14

255.255.255.252

16382

00002

32764

	15

	16

	Table 2-6: Class C Subnet Masks

	Number of Subnet Bits

Subnet Mask

Number of Subnetworks

Number of Hosts/Subnet

Total Number of Hosts

	1

	2

255.255.255.192

02

62

124

	3

255.255.255.224

06

30

180

	4

255.255.255.240

14

14

196

	5

255.255.255.248

30

06

170

	6

255.255.255.252

62

02

124

	7

	8

	A 15-bit subnet mask for Class B and a 7-bit subnet mask for Class C is also illegal because it would leave only 1-bit for the host, which we have seen cannot be all zeros or all ones. A 16-bit subnet mask for Class B or an 8-bit subnet mask for Class C makes no sense because this would leave zero host bits.

	Subnet Examples

	In the following examples, determine if the address/subnet pair is legal. If it is legal, determine the network number and the range of host addresses for that network. Also determine for the mask, the number of available networks and available hosts per network.

	1.		IP address = 193.144.233.130

		Subnet mask 5 255.255.255.192

			For a Class C address, we only need to look at the last octet of the address and the mask.

		130 = 1000 0010

		192 = 1100 0000

			This is a legal pair because neither the subnet nor the host is all zeros or all ones.

	Network equals 193.144.233.128 because the mask selects the upper two bits of the address (130) and the rest of the bits are set to zero to identify the network.

	Range of hosts = 193.144.233.129 193.144.233.190.

	The host portion (last six bits) can have values ranging from 000001 to 111110 (remember they can t be all zeros or all ones). Add in the subnet portion, which is the upper two bits of the address (in this case, 1 0), and you have 10 000001 to 10 111110 for the host addresses.

		From Table 2-6, the number of available networks is 2 and the number of hosts is 62.

	2.		IP address = 156.26.30.60

		Subnet Mask = 255.255.255.0

	This is relatively easy because the entire third octet is used for the subnet and the entire fourth octet is used for the host. This is a legal pair because neither the subnet nor the host is all zeros or all ones.

		Network = 156.26.30.0

		Range of hosts = 156.26.30.1 156.26.30.254

		From Table 2-5, the number of networks is 254 and the number of hosts is 254.

	3.		IP address = 199.200.201.50

		Mask = 255.255.255.128

	This is illegal because the subnet mask only borrows 1 bit from the host and that bit has to be either zero or one.

	4.		IP address = 191.200.201.50

		Mask = 255.255.255.128

			This is a legal pair because the address is Class B and we are borrowing 9 bits from the host portion.

		Network = 191.200.201.0

		Range of hosts = 191.200.201.1 191.200.201.126

			From Table 2-5, the number of networks is 510 and the number of hosts is 126.

	Subnetting can be viewed as creating a three-part hierarchical address. The network portion of the address can be found by applying the standard subnet mask to the IP address (refer to Figure 2-3). The subnet is determined from the bits borrowed from the host portion and the host number is simply those bits that are left over. For an example, we will examine the Class B address/mask pair

	144.223.0.0/255.255.255.0

	and determine the network number, the subnetwork numbers, and the range of host numbers. The network number is found by applying the standard Class B 16-bit subnet mask, which yields the network

	144.223.0.0

	The subnet is the entire third octet, so the 254 subnets are

	144.223.1.0

144.223.2.0

.

.

.

144.223.254.0

	and the range of hosts for each subnet is 1 to 254. Now let s try a bit more complicated example. Consider the address/mask pair

	144.223.0.0/255.255.255.224

	The network number is still 144.223.0.0. The subnet mask borrows 11 bits from the host portion of the address. The first 8 bits borrowed include the entire third octet, which has a value of 0 to 255. The 3 bits borrowed from the third octet have the values

	00000000=0

00100000=32

01000000=64

01100000=96

10000000=128

10100000=160

11000000=192

11100000=224

	Why are the values 0 (all zeros) and 255 (all ones) for the third octet, and 0 (all zeros) and 224 (all ones) from the fourth octet included? The third octet can be 0 if the 3 bits in the fourth octet are not zero. The third octet can also be all ones if the 3 bits in the fourth octet are not all ones. The 3 bits in the fourth octet can be all zeros if the third octet is not all zeros, and the 3 bits from the fourth octet can be all ones if the third octet is not all ones. In other words, the 11 subnet bits cannot be all zeros or all ones. Therefore, the range of subnet numbers is

	144.223.0.32

144.223.0.64

.

.

.

144.223.0.224

144.223.1.0

144.223.1.32

.

.

.

144.223.255.0

.

.

.

144.223.255.192

	Determining the range of host addresses for each subnet requires more effort. The bit pattern for the fourth octet of network 144.223.0.32 is

	001 hhhhh

	where hhhhh represents the host number, which cannot be all zeros or all ones. Therefore, the first legal host number is 00001, making the fourth octet

	00100001 = 33

	so the first host address is

	144.223.0.33

	and the last legal host bit pattern for the fourth octet is

	00111110 = 62

	which gives the range of hosts addresses for the first subnet as

	144.223.0.33 144.223.0.62

	The broadcast address for each subnet is found by setting all the bits in the host portion to 1. The broadcast address for subnet 144.223.0.32 is determined by setting the last 5 bits of the fourth octet to 1 yielding

	00111111 = 63

	Putting it all together gives us the broadcast address

	144.223.0.63

	5.		Determine all the subnet numbers for the address/mask pair 193.128.55.0/255.255.255.240. Also determine the range of host addresses and the broadcast address for the fourth subnet.

	Network

Hosts

	193.128.55.0

1 14 (If IP subnet-zero is used)

	193.128.55.16

17 30

	193.128.55.32

33 46

	193.128.55.48

49 62, Broadcast address = 193.128.55.63

	193.128.55.64

65 78

	193.128.55.80

81 94

	193.128.55.96

97 110

	193.128.55.112

113 126

	193.128.55.128

129 142

	193.128.55.144

145 158

	193.128.55.160

161 174

	193.128.55.176

177 190

	193.128.55.192

193 206

	193.128.55.208

209 222

	193.128.55.224

225 238

	193.128.55.240

241 254

	IP Address Design Example 1

	Assume your company has been assigned the Class C address 198.28.61.0 and you have determined that you require four networks with a maximum of 25 hosts per network. From Table 2-6, you will need three subnet bits, resulting in a subnet mask of 255.255.255.224. The subnet numbers for this design are any four of the following, as shown in Figure 2-4.

	Figure 2-4: IP address design example 1

	198.28.61.32

198.28.61.64

198.28.61.96

198.28.61.128

198.28.61.160

198.28.61.192

	Although subnets solve some of the problems associated with the inefficient use of IP address space, situations occur when simple subnetting does not suffice. Consider the network in Figure 2-5 in which two routers are connected by a serial link. This serial link is a point-to-point connection, so there are only two hosts on the link, the two router interfaces. Each network must also be on a separate subnet, so no matter which subnet mask we choose, we will be wasting IP addresses. If we are using a Class B address with a 24-bit subnet mask, then the subnet assigned to the serial link will only use two out of a possible 254 host addresses.

	Figure 2-5: Limitations of simple subnetting

	If we could use different subnet masks for different subnetworks, then the limitations of Figure 2-5 could be solved. A subnet mask of 255.255.255.252 (or /30) can accommodate only two hosts, which is perfect for a point-to-point serial link. Unfortunately, this mask, if used throughout the network, would limit all subnets to two hosts. The ideal solution would be to vary the length of the subnet mask and adjust it according to the needs of each individual network.

	Variable Length Subnet Masks

	RFC 1009, 1987, specifies the procedures for using multiple subnet masks. This technique is referred to as variable length subnet masks (VLSM). The term VLSM can be confusing because the subnet mask for a specific network does not vary but is fixed. VLSM means that the subnet masks for different subnets can have unequal lengths. As an example, it would allow a subnet mask of 255.255.255.252 to be assigned to a serial link and 255.255.255.0 to an ethernet network. Once the masks are assigned, however, they do not change, at least by themselves.

	Figure 2-6: VLSM example 1

	The VLSM technique is very useful for allocating IP addresses more efficiently (less waste) and for reducing the size of routing tables. However, VLSM can also cause a number of massive network headaches if not used properly.

	VLSM Example 1

	Let s apply VLSM to the network in Figure 2-5. Assume we have been assigned the Class B network 156.26.0.0. The ethernet networks are assigned addresses using a /24 subnet mask; we will use the first two networks with this mask, 156.26.1.0 and 156.26.2.0. The third network, 156.26.3.0, will be sub-subnetted using a /30 subnet mask, which will give us a possible 62 sub-subnets we can use for serial connections. Notice that we are subnetting an already subnetted network, 156.26.3.0. Figure 2-6 illustrates this technique.

	Figure 2-6 visually represents the technique that should be used when using VLSM. Start with the standard subnet mask (/8, /16, or /24 for Class A, B, or C). Determine the network with the required maximum number of hosts, in this case 254. Then subnet using a mask that will give you networks that can handle the largest number of hosts you need. For smaller networks, sub-subnet the large networks and keep going until you have satisfied your requirements.

	VLSM Example 2

	The best way to master a technique is practice, practice, practice, so here we go. Given the IP network 202.128.236.0, design a network with the following requirements:

		Four networks with a maximum of 26 hosts

		Three networks with a maximum of 10 hosts

		Four point-to-point serial links

	Starting with the greatest number of hosts per network, we can use a /27 subnet mask to satisfy the first requirement. From Table 2-6, this gives us six networks of 30 hosts each with two networks left over to sub-subnet. To satisfy the next requirement, we can sub-subnet the two leftover /27 networks using a /28 subnet mask to give us four networks with 14 hosts each. Finally, take one of the four sub-subnetted networks and sub-sub-subnet using a /30 subnet mask.

	How did I arrive at the diagram in Figure 2-7? Let s take a closer look as to where these network numbers came from; then we ll look at another VLSM design problem to ensure that you have mastered the technique.

	Figure 2-7: VLSM example 2

	1.		Determine the mask for the networks containing the greatest number of hosts.

	The first requirement is for four networks with a maximum of 26 hosts. Using Table 2-6, we need three subnet bits or a /27 subnet mask. The fourth octet of our IP network would be segmented as

	S S S H H H H H

	where S S S indicates the subnet bits and H H H H H indicates the host bits. The subnets then are

	0 0 1 0 0 0 0 0=32

0 1 0 0 0 0 0 0=64

0 1 1 0 0 0 0 0=96

1 0 0 0 0 0 0 0=128

1 0 1 0 0 0 0 0=160

1 1 0 0 0 0 0 0=192

	and we are using subnets 96 through 192 for the networks containing 26 hosts because these subnets can handle a maximum of 30 hosts.

	2.		Sub-subnet the subnetted networks as needed.

	The second requirement calls for three networks with a maximum of 10 hosts each. Again, we consult Table 2-6 and see that we need four subnet bits or a /28 subnet mask. We will sub-subnet network 202.128.236.32 and 202.128.236.64. The first three subnet bits are fixed with the values 001 (subnet 32) and 010 (subnet 64), so now we have

	0 0 1 S H H H H

0 1 0 S H H H H

	Network 32 S can be 0 or 1, giving us

	0 0 1 0 H H H H

0 0 1 1 H H H H

	Setting the host bits to 0, the sub-subnets are

	0 0 1 0 0 0 0 0 = 32

0 0 1 1 0 0 0 0 = 48

	Applying the same procedure to subnet 64, we get

	0 1 0 0 0 0 0 0 = 64

0 1 0 1 0 0 0 0 = 80

	3.		To satisfy the last requirement of four point-to-point serial links, we will sub-sub-subnet sub-subnet 32, which now is equal to

	0 0 1 0 S S H H

	S S can be either 0 0, 0 1, 1 0 , or 1 1 yielding

	0 0 1 0 0 0 0 0 = 32

0 0 1 0 0 1 0 0 = 36

0 0 1 0 1 0 0 0 = 40

0 0 1 0 1 1 0 0 = 44

	As a final task for this exercise, determine the range of hosts and the broadcast addresses for networks 202.128.236.192, 202.128.236.80, and 202.128.236.40.

	The fourth octet of network 202.128.236.192 is

	1 1 H H H H H H

	and the host bits can range from

	0 0 0 0 0 1 1 1 1 1 1 0

	which gives us a range of

	1 1 0 0 0 0 0 1 (193) 1 1 1 1 1 1 1 0 (254)

	The broadcast address is determined by setting the host bits to 1, which is

	1 1 1 1 1 1 1 1 5 255

	so the broadcast address is 202.128.236.255. For network

	Figure 2-8: Realization of VLSM example 2

	202.128.236.80, the fourth octet contains

	0 1 0 1 H H H H

	so the range of host addresses is

	0 1 0 1 0 0 0 1 (81) 0 1 0 1 1 1 1 0 (94)

	and the broadcast address is

	0 1 0 1 1 1 1 1 (95)

	For network 202.128.236.40, the fourth octet contains

	0 0 1 0 1 0 H H

	Because H H cannot be 0 0 or 1 1, the host addresses for this network are 202.128.236.41 and 202.128.236.42 with a broadcast address of 202.128.236.243. The realization of this network design is shown in Figure 2-8.

	For the final VLSM example, design a network using the Class C address 200.100.50.0 that satisfies the following requirements:

		_Nine serial point-to-point links

		_Four networks with a maximum of 30 hosts

		_Three networks with a maximum of five hosts

	Determine the address host ranges and the broadcast address for each subnet.

	From Table 2-6, a 3-bit subnet mask will give us six networks of 30 hosts each.

	Subnet mask = 255.255.255.224

	Networks

Hosts

Broadcast Address

	200.100.50.0

1 30

200.100.50.31 (if we use IP subnet-zero)

	200.100.50.32

33 62

200.100.50.63

	200.100.50.64

65 94

200.100.50.95

	200.100.50.96

97 126

200.100.50.127

	200.100.50.128

129 158

200.100.50.159

	200.100.50.160

161 190

200.100.50.191

	200.100.50.192

193 222

200.100.50.223

	One solution is to use the first four networks to satisfy the requirement of four networks with 30 hosts each. For the requirement of three networks with five hosts each, we can sub-subnet network 200.100.50.160 using the 5-bit subnet mask 200.100.50.160/255.255.255.248, which gives us the networks listed below.

	Network

Hosts

Broadcast Address

	200.100.50.160

161 166

200.100.50.167

	200.100.50.168

169 174

200.100.50.175

	200.100.50.176

177 182

200.100.50.183

	200.100.50.184

185 190

200.100.50.191

	We can use any three of the four networks to satisfy the requirement of three networks with five hosts.

	Finally we can sub-subnet the 200.100.50.192 network using a 30-bit subnet mask that gives us the networks listed below.

	Network

Hosts

Broadcast Address

	200.100.50.192

193 194

200.100.50.195

	200.100.50.196

197 198

200.100.50.199

	200.100.50.200

201 202

200.100.50.203

	200.100.50.204

205 206

200.100.50.207

	200.100.50.208

209 210

200.100.50.211

	200.100.50.212

213 214

200.100.50.215

	200.100.50.216

217 218

200.100.50.219

	200.100.50.220

221 222

200.100.50.223

	200.100.50.224

225 226

200.100.50.227

	200.100.50.228

229 230

200.100.50.231

	200.100.50.232

233 234

200.100.50.235

	200.100.50.236

237 238

200.100.50.239

	200.100.50.240

241 242

200.100.50.243

	200.100.50.244

245 246

200.100.50.247

	200.100.50.248

249 250

200.100.50.251

	Choose any nine of the networks for the serial links.

	Subnet masks can also be used with Class D multicast addresses. As an example, assume we have the following Class D address/mask pair.

	225.250.250.0/255.255.255.0

	This address mask/pair would then represent all the multicast groups from 225.250.250.0 225.250.250.255. The multicast address/mask pair can be used to summarize the range of groups that a router will allow or that a multicast entity will service. We will learn more about the use of a mask with a multicast address later in the book.