Catching up with TCP/IP fundamentals: IP address classes (2024)

Catching up with TCP/IP fundamentals: IPaddress classes

Jim McIntyre

Beforetwo or more computers can communicate with each other, a set of rules has to bein place describing the procedures for each host or router on the network tofollow when sending data to—or receiving data from—other hosts or routers. Thisset of rules is referred to as a protocol, and TCP/IP is the most commonset of rules in use for computing today. To understand computer networking aswell as internetworking, you must understand the TCP/IP protocol suite. In thisarticle, I'll provide you with an introduction to the TCP/IP protocol suite.

• The netid identifies the network.
• The hostid identifies the host on that network.

Class A addresses
Class A networks are assigned to organizations with very large numbers ofcomputers—including servers and routers—attached to their networks. Agovernment department or a Fortune 500 corporation would be examples of thistype of network. The following characteristics apply to class A networks:

• Class A addresses range from 0.0.0.0 to 127.255.255.255.
• The leftmost bit in a class A network is always 0.
• The first 8 bits (first byte) define the netid.
• The remaining 24 bits define the hostid.

It wouldappear that because the first 8 bits define the netid,the maximum number of class A networks available wouldbe 2⁷, or 128 networks. However, a netidwith all bits set to 1 and a netid with all bits setto 0 are reserved for special use. This means that the actual number of class A networks available is 2⁷ minus 2, or 126networks.

As I mentioned earlier, class Aaddresses use 24 bits to define the hostid. Intheory, there is a maximum of 2²⁴, or 16,777,216, hostids available. Like netids, ahostid with all bits set to 1 and a hostid with all bits set to 0 are reserved for special use.This means that the actual maximum number of hostidaddresses on a class A network is 2²⁴ minus2, or 16,777,214.

This limitation applies to all IP address classes. I'lldiscuss the limitation in more detail later.

Class B addresses
Class B networks are also assigned to organizations with large networks. ClassB networks have the following characteristics:

• Class B network addresses range from 128.0.0.0 to 191.255.255.255.
• The first 2 bits in a class B address are always 10 (128).
• In a class B network, the netid is determined by the first 8 bits (first 2 bytes).
• The hostid is determined by the last 8 bits (last 2 bytes).
• There are 2¹⁶ (65,536) possible IP addresses for each class B network.
• The netids 172.16.0.0 to 173.31.0.0 are reserved for special use. This makes the actual maximum number of class B netids 2¹⁶ minus 2, or 65,534.

Class C addresses
Class C networks are assigned to organizations with small to medium networks.The class C network is the most common network in use today. Class C networkshave the following characteristics:

• Class C network addresses range from 192.0.0.0 to 223.255.255.255.
• Class C networks use the first 24 bits to determine the netid.
• The three leftmost bits in a class C network are always 110 (192).
• The next 21 bits are used to define network.
• There are 2¹⁶, or 2,097,152, possible class C networks.
• In class C networks, 8 bits are used to define the hostid. Because an address with all bits set to 1 or all bits set to 0 is not allowed, the maximum number of hostids on a class C network is 2⁸ minus 2, or 254 hostids.

Class D IP addresses
Class D IP addresses are reserved for multicasting. Class D networks have thefollowing characteristics:

• Class D network addresses range from 224.0.0.0 to 239.255.255.255.
• In class D addresses, the first 4 bits are always 1110 (224).
• The remaining 28 bits are used to define multicast addresses.
• There is no netid or hostid in a class D address. The whole address is used for multicasting.

Class E IP addresses
Class E addresses are reserved for special use on the Internet. There is no netid or hostid in a class Eaddress. Class E network addresses range from 240.0.0.0 to 255.225.225.225. Thefirst 4 bits in a class address are always 1111 (240).

Determining the addressclass
For any given IP address, the address class is easy to determine. If theaddress is written in binary format, the leftmost bits in the address willdefine the class.

• If the first bit is 0, the address is a class A address.
• If the first bit is 1 and the second bit is 0, the address is a class B address.
• If the first 2 bits are 1 and the third bit is 0, the address is a class C address.
• If the first 3 bits are 1 and the fourth is 0, the address is a class D address.
• If the first 4 bits are 1, the address is a class E address.
• If the address is written in decimal format, the first number determines the address.
• If the first number is between 0 and 127 inclusive, the address is class A.
• If the first number is between 128 and 191 inclusive, the address is class B.
• If the first number is between 192 and 223 inclusive, the address is class C.
• If the first number is between 224 and 239 inclusive, the address is class D.
• If the first number is between 240 and 255 inclusive, the address is class E.

Determining the netid and hostid
Determining the netid and hostidcontained within any given IP address is not difficult. Once you've determinedthe class of the IP address, use the following process:

• If the class is class A, the first byte is the netid and the remaining 3 bytes are the hostid.
• If the class is class B, the first 2 bytes are the netid and the remaining 2 bytes are the hostid.
• For a class C address, the first 3 bytes are the netid and the last byte is the hostid.
• For a class D address, there is no netid or hostid. All class D addresses are reserved for multicasting.
• For class E addresses, there is no netid or hostid. All class E addresses are reserved for special uses.

Thetable below lists the number of netids and hostids available for each IP address class.

We've listed the number of netids and hostids available foreach IP address class.

Special IP addresses
Classes A, B, and C each have addresses set aside for special purposes. Youshould become familiar with six special addresses, which I'll describe in thefollowing sections.

Network address
In class A, B, and C addresses, any address with a hostidwith all bits set to 0 is used to define the network address. This address isnever assigned to a host. This simply means that your network itself is givenan IP address with all the hostid bits set to 0. Thisis not the same as the netid. The network address isalways an address with the host bits set to 0. For example, if a host on anetwork has an IP address of 142.23.120.24, the network address is:
142.23.0.0

This host on this network
If an IP address consists of all zeros, the address refers to this host on thisnetwork. A host uses this address when it is booted, but the host does not knowits IP address. This address is always a class Aaddress, regardless of the network configuration.

Specific host on this network
An IP address with all netid bits set to 0 refers toa specific host on this network. Setting the netidbits to 0 ensures that any IP information sent using this address is notrouted. The information will remain within the local network. For example, ifmy IP address is 216.120.46.100 and I know I am on the 216.120.46 localnetwork, I could send information to the host with the IP address 216.120.46.69by using the address 0.0.0.60.

Direct broadcast address
Any class A, B, or C address with the hostid set toall ones is known as a direct broadcast address. Routers sending information toall hosts on a specific network use this address. For example, if you sent IPinformation to 204.36.120.255, all hosts on the 204.36.120.0 network wouldreceive the information sent.

Limited broadcast address
Any class A, B, or C address with all bits set to 1 (255.255.255.255) is usedto send information to all other hosts on the same network as the host sendingthe information. In other words, this address sends IP information to all hostson your local network.

Private IP networks
There are three blocks of IP addresses that may be used by any organization forTCP/IP networking. These IP addresses are nonroutable,meaning they are limited to use within the local network only. The three groupsof nonroutable IP addresses are listed in the tablebelow.

The three groups of nonroutable IP addresses are shown according to class.

Any organization may use any address from the blocks listedabove without registering the IP addresses used on its network.

Summary
The TCP/IP protocol suite provides the foundation for internetworking. In thisarticle, I introduced you to the TCP/IP protocol suite.

Getting to know your IP addresses

Todd Lammle

TheInternet Protocol (IP) was developed in the 1960s to provide packetfragmentation and reassembly across a packet-switched network. Thispacket-switched network became what we now call the Internet.

IP addressing is used to uniquely identify hosts on an internetwork. An internetwork ismade of LANs and WANs that are connected with a router or routers. To send datafrom a host on network A to a host on network B, a logical network addressingprotocol must be used. IP is the most popular logical addressing protocol.However, IPX in the Novell stack and Datagram Delivery Protocol (DDP) in theAppleTalk stack can also be used. Although you can find many other types ofrouted protocols that provide the same functionality as IP, in larger networksthey aren't as efficient as IP.

The dreaded IP address
Before we conquer IP addresses, I want to define some of the terminology usedin this article:

• Bit: One digit; either one or zero.
• Byte: 8 bits. This term is interchangeable with octet.
• Octet: 8 bits. This term is interchangeable with byte.
• Network address: Address with all host bits turned off.
• Broadcast address: Address with all host bits turned on.

To fullyunderstand IP addressing, you must be familiar with binary-to-decimalconversion. To find the decimal equivalent of a binary number, you must add thebinary values. Binary numbers use eight bits to create a byte. Each bit in thebyte has a certain value, and if a bit is turned on (assigned a value of 1),then the byte takes on that decimal value. Each bit has a value that starts at1 and doubles in value from right to left. Table 1 shows an example ofconverting a byte to a decimal value.

This table shows an example of convertinga byte to a decimal.

In the above example, there are three bits that are turnedon (1s). Add each of these values to get the bytes decimal value (16+8+2=26).Let's take a look at another example in Table 2.

In the above example, there are threebytes that are turned on (64+32+8=104).

Classes of IP addresses
The original designers of the IP stack came up with a hierarchical addressingscheme with five ranges, called classes. These ranges are named:

• Class A: IP address range from 0-127 in the first byte designed for very large companies.
• Class B: IP address range from 128-191 in the first byte designed for medium-size companies.
• Class C: IP address range from 192-223 in the first byte designed for small companies.
• Class D: IP address range from 224-239 in the first byte reserved for multicast addressing. Not used in the public sector.
• Class E: IP address range from 240-255 in the first byte reserved for scientific studies. Not used in the public sector.

• 10.0.0.0 is the network address because all of the host bits are turned off (0). The last three octets are zeros because the first octet is a 10, which is in the Class A range of 1-126.
• To find the broadcast address of the IP address, turn on the host bits: 10.255.255.255.
• The valid hosts are the numbers in between the network address and the broadcast address: 10.0.0.1 through 10.255.255.254.

• 152.93.0.0 is the network address (all host bits off).
• 152.93.0.1 is the first valid host.
• 152.93.255.254 is the last valid host.
• 152.93.255.255 is the broadcast address (all host bits on).

• 200.10.10.0 is the network address because all host bits are off.
• 200.10.10.1 is the first valid host.
• 200.10.10.254 is the last valid host.
• 200.10.10.255 is the broadcast address because all host bits are on.

Conclusion
In today's computing age, IP addressing is critical in every network. To becomecertified in any network program, you must have a fundamental understanding ofIP addressing. If you want a networking job, it's imperative that you'refamiliar with IP addressing. You should be able to determine a valid IP addressjust by glancing at the IP address and mask. To do this, you must be able toquickly and efficiently determine the network and broadcast addresses.

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Using variable-length subnettingwith TCP/IP networks

Jim McIntyre

Theprocess of subnetting large networks is oftenrestricted by the network mask (netmask) employed.Quite often, we are required to create several subnetworks,each with a different number of host computers (workstations). Problems likethis are often solved using variable-length subnetting.With variable-length subnetting, the administratoruses two or more subnet masks and each mask is a different length. In thisarticle, we will look at how variable subnetting maybe used to solve some typical subnetting problems.

The administrator must first consider which subnet maskswill work in this situation. The best approach here is to look at thelimitations of Class C subnet masks. The limitations include:

• If the standard Class C netmask of 255.255.255.0 is used, only one network is possible, with a maximum of 256 hosts. Using this mask eliminates any subnets.
• If the mask 255.255.255.128 is used, only one subnet with a maximum of 128 hosts is possible.
• Using two bits in the subnetid section of the IP address, 255.255.255.192, allows four subnets, each with a maximum of 62 hosts (256 / 4 – 2 = 62). Sixty-two hosts are more than we will have on any one subnet, but this allows for only four subnets. The situation requires five subnets.
• The next subnet mask, 255.255.255.224, allows for eight subnets to be created, but each subnet is limited to 30 hosts (256 / 8 =32, and 32 –2 = 30).

Thesolution is to use variable-length subnetting. Whenvariable-length subnetting is employed, the router issupplied with two subnet masks. These masks are applied one after the other. Inthis situation, the administrator first creates four subnets using the mask255.255.255.192. The original subnet mask is 255.255.255.0, written in binaryas:
11111111.11111111.11111111.00000000= 255.255.255.0

and the original network IP address was 192.168.3.0, writtenin binary as:

1. The first step is to apply the mask to the original 192.168.3.0 network address. This new mask creates four new subnetworks (subnets). The subnetwork identifiers (subnetid) available to the administrator are:
   192.168.3.0
   192.168.3.64
   192.168.3.128
   192.168.3.192
   Three of the new subnetworks will each contain 60 hosts. When these first three subnetworks are created, our original Class C network, 192.168.3.0, looks like the example in Figure A.
2. To create two more subnets, a second subnet mask of 255.255.255.224 is applied. This divides the remaining subnet into two separate subnets. With five bits remaining to be used as host identifiers (hostids), each of these new subnets may hold up to 32 (2⁵) hosts. Once this step is completed, the network now takes on the configuration shown in Figure B.

With that step completed, we have now solved our originalproblem of how to create five subnetworks with 50hosts on three of the subnets and 20 hosts on the two subnetworkscreated using variable-length subnet masking.

Conclusion
The rapid growth of the Internet has led to many situations where Class Cnetwork addresses are the only ones available to administrators. The ability tocreate subnetworks using the procedures outlined inthis article will allow administrators more flexibility in how solutions areapplied to some common networking problems.

Subnetting a Class C network address

Todd Lammle

Let'sface it, some day you are going to have to subnet a network. Although IPaddressing isn't a network administrator's favorite task, it's a critical skillthat you must have. In this article, I will explore the process of taking alarge IP address range and dividing it into smaller, more manageable pieces.Since this is a complicated, involved process, I will only discuss the processof breaking up Class C networks. First, however, I need to discuss why you'dwant to subnet a network and the advantages of doing so.

Subnetting a Class C network
So you understand why you want to subnet, but how do you do it? Your goal is tolook at an IP address and subnet mask of a host and then determine three thingsquickly:

1. The subnet the host is located in
2. The broadcast address of the subnet
3. The valid host range of the subnet used to configure hosts

Once the subnet is determined, the broadcast address mustbe found. Why? Because these are not valid host addresses and cannot beassigned to host configurations. Also, by determining the subnet and broadcastaddresses, you can easily determine the host addresses because the valid hostrange is always the numbers between the subnet address and the broadcastaddress.

If you use the default subnet mask with a Class C networkaddress, then you already know that three bytes are used to define the networkand only one byte is used to define the hosts on each network.

The default Class C mask is: 255.255.255.0. To makesmaller networks, called subnetworks, you will borrowbits from the host portion of the mask. Since the Class C mask only uses thelast octet for host addressing, you only have 8 bits at your disposal.Therefore, only the following masks can be used with Class C networks (TableA).

Class C masks

You can see in Table A that the bits that are turned on(1s) are used for subnetting, while the bits that areturned off (0s) are used for addressing of hosts. You can use some easy math todetermine the number of subnets and hosts per subnet for each different mask.

To determine the number of subnets, use the 2^x-2,where the x exponent is the number of subnet bits in the mask.

To determine the number of hosts, use the 2^x-2,where the x exponent is the number of host bits in the mask.

To determine the mask you need for your network, you mustfirst determine your business requirements. Count the number of networks andthe number of hosts per network that you need. Then determine the mask by usingthe equations shown above—and don't forget to factor for growth.

For example, if you have eight networks and each requires10 hosts, you would use the Class C mask of 255.255.255.240. Why? Because 240in binary is 11110000, which means you have four subnet bits and four hostbits. Using this math, you'd get the following:
2⁴-2=14 subnets
2⁴-2=14 hosts

Many people find it easy to memorize the Class Cinformation because Class C networks have few bits to manipulate. However,there is an easier way to subnet.
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Easy subnetting
Instead of memorizing the entire table (Table A), it's possible to glance at ahost address and quickly determine the necessary information if you'vememorized key parts of the table. First, you need to know yourbinary-to-decimal conversion. Memorize the number of bits used with each maskthat are shown in Table A. Second, you need to remember the following:
256-192=64
256-224=32
256-240=16
256-248=8
256-252=4

Once you have the two steps memorized, you can begin subnetting.Our first example will use the Class C mask of 255.255.255.192. Ask five simplequestions to gather all the facts:

1. How many subnet bits are used in this mask?
2. How many host bits are available per subnet?
3. What are the subnet addresses?
4. What is the broadcast address of each subnet?
5. What is the valid host range of each subnet?

You already know how to answer questions one and two. Toanswer question three, use the formula 256-subnetmask to get the firstsubnet and your variable. Keep adding this number to itselfuntil you get to the subnet mask value to determine the valid subnets. Once youverify all of the subnets, you can determine the broadcast address by lookingat the next subnet's value. The broadcast address is the number just before thenext subnet number. Once you have the subnet number and broadcast address, thevalid hosts are the numbers in between.

Here are the answers using 255.255.255.192:

1. How many subnet bits are used in this mask?
   Answer: 2
   2²-2=2 subnets
2. How many host bits are available per subnet?
   Answer: 6
   2⁶-2=62 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-192=64 (the first subnet)
   64+64=128 (the second subnet)
   64+128=192. However, although 192 is the subnet mask value, it's not a valid subnet. The valid subnets are 64 and 128.
4. What is the broadcast address of each subnet?
   Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.

Let's do a second example using the Class C mask of255.255.255.224. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 3 bits or 2³-2=6 subnets
2. How many host bits are available per subnet?
   Answer: 5 bits or 2⁵-2=30 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-224 =32, 64, 96, 128, 160 and 192 (Six subnets found by continuing to add 32 to itself.)
4. What is the broadcast address of each subnet?
   Answer: The broadcast address for the 32 subnet is 63. The broadcast address for the 64 subnet is 95. The broadcast address for the 96 subnet is 127. The broadcast address for the 160 subnet is 191. The broadcast address for the 192 subnet is 223 (since 224 is the mask).
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-62.

Let's do a third example using the Class C mask of255.255.255.240. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 4 bits or 2⁴-2=14 subnets
2. How many host bits are available per subnet?
   Answer: 4 bits or 2⁴-2=14 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-240 =16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, and 224 (14 subnets found by continuing to add 16 to itself).
4. What is the broadcast address of each subnet?
   Answer: Here are some examples of the broadcast address: The broadcast address for the 16 subnet is 31. The broadcast address for the 32 subnet is 47. The broadcast address for the 64 subnet is 79. The broadcast address for the 96 subnet is 111. The broadcast address for the 160 subnet is 175. The broadcast address for the 192 subnet is 207.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. The 32 subnet valid hosts are 33-46.

Let's do a fourth example using the Class C mask of255.255.255.248. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 5 bits or 2⁵-2=30 subnets
2. How many host bits are available per subnet?
   Answer: 3 bits or 2³-2=6 hosts per subnet
3. What are the subnet addresses?
   Answer 256-248 =8, 16, 24, 32, 40, 48, and so forth. The last subnet is 240 (30 subnets found by continuing to add 8 to itself).
4. What is the broadcast address of each subnet?
   Answer: The broadcast address for the 8 subnet is 15. The broadcast address for the 16 subnet is 23. The broadcast address for the 48 subnet is 55.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-38.

Let's do a fifth example using the Class C mask of255.255.255.252. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 6 bits or 2⁶-2=62 subnets
2. How many host bits are available per subnet?
   Answer: 2 bits or 2²-2=2 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-252 =4, 8, 12, 16, 20, and so forth. The last subnet is 248 (62 subnets found by continuing to add 4 to itself).
4. What is the broadcast address of each subnet?
   Answer: The broadcast address for the 4 subnet is 7. The broadcast address for the 8 subnet is 11. The broadcast address for the 12 subnet is 15. The broadcast address for the 20 subnet is 23.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 16 subnet valid hosts are 17 and 18.

Conclusion
It is important to be able to subnet quickly and efficiently. After studyingthe examples presented in this article, you should be familiar with thisprocess with Class C addresses. Practice your subnetting as much as possible, and the process will geteasier and easier.

All possible Class B subnets

There are quite a few more masks we can use with a Class Bnetwork address than we can with a Class C network address. Remember that thisis not harder than subnetting with Class C, but itcan get confusing if you don't pay attention to where the subnet bits and hostbits are in a mask. This takes practice!

In this article, we will use the same techniques we usedin "Subnetting a Class C network address" to subnet anetwork. We'll start with the Class B subnet mask of 255.255.192.0 and figureout the subnets, broadcast address, and valid host range. You will need toanswer the same five questions that you answered for the Class C subnet masks:

1. How many subnets does this mask provide?
2. How many hosts per subnet does this mask provide?
3. What are the valid subnets?
4. What is the broadcast address for each subnet?
5. What is the host range of each subnet?

Before answering these questions, there is one differenceyou need to be aware of when subnetting a Class Bnetwork address. When subnetting in the third octet,you need to add the fourth octet. For example, on the 255.255.192.0 mask, the subnetting will be done in the third octet. To create avalid subnet, you must add the fourth octet of all 0s and all 1s for thenetwork and broadcast address (0 for all 0s and 255 for all 1s).
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Example 1: Answers for the 255.255.192.0 mask

1. 2²-2=2 subnets
2. 2¹⁴-2=16,382 hosts per subnet
3. 256-192=64.0, 128.0
4. Broadcast for the 64.0 subnet is 127.255. Broadcast for the 128.0 subnet is 191.255.
5. The valid hosts are:

Notice that the numbers in the third octet are the same numbersyou used in the fourth octet when subnetting the 192mask. The only difference is that you add 0 and 255 in the fourth octet.

For the 64.0 subnet, all the hosts between 64.1 and127.254 are in the 64 subnet. In the 128.0 subnet, the hosts are 128.1 through191.254.

Work through a few more with me, and it should start tobecome clearer.

Example 2: 255.255.240.0

1. 2⁴-2=14 subnets
2. 2¹²-2=4,094 hosts per subnet
3. 256-240=16.0, 32.0, 48.0, 64.0, etc.
4. Broadcast for the 16.0 subnet is 31.255. Broadcast for the 32.0 subnet is 47.255, etc.
5. The valid hosts are:

Example 3: 255.255.248.0

1. 2⁵-2=30 subnets
2. 2¹¹-2=2,046 hosts per subnet
3. 256-248=8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, etc.
4. Broadcast for the 8.0 subnet is 15.255. Broadcast for the 16.0 subnet is 23.255, etc.
5. The valid hosts are:

Example 4: 255.255.252.0

1. 2⁶-2=62 subnets
2. 2¹⁰-2=1,022 hosts per subnet
3. 256-252=4.0, 8.0, 12.0, 16.0, 20.0, 24.0, 28.0, 32.0, etc.
4. Broadcast for the 4.0 subnet is 7.255. Broadcast for the 8.0 subnet is 11.255, etc.
5. The valid hosts are:

Example 5: 255.255.255.0

1. 2⁸-2=254 subnets
2. 2⁸-2=254 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, etc.
4. Broadcast for the 1.0 subnet is 1.255. Broadcast for the 2.0 subnet is 2.255, etc.
5. The valid hosts are:

That last example was pretty simple. You should notice apattern now. All the numbers were basically the same except we added the fourthoctet into the address.

The more difficult process of subnettinga Class B network address is when you start using bits in the fourth octet for subnetting. For example, what happens when you use thismask with a Class B network address: 255.255.255.128? Is that valid?Absolutely! There are nine bits for subnetting andseven bits for hosts. That is 510 subnets, each with 126 hosts. However, it isthe most difficult mask to figure out the valid hosts for.

Example 6: The Class B 255.255.255.128 subnet mask:

1. 2⁹-2=510 subnets
2. 2⁷-2=126 hosts per subnet
3. For the third octet, the mask would be 256-255=1, 2, 3, 4, 5, 6, etc.
4. For the fourth octet, the mask would be 256-128=128, which is one subnet if it is used. However, if you turn the subnet bit off, the value is 0. This means that for every subnet in the third octet, the fourth octet has two subnets: 0 and 128, for example 1.0 and 1.128.
5. Broadcast for the 0.128 subnet is 128.255; the broadcast for the 1.0 subnet is 1.127. Broadcast for the 1.128 subnet is 1.255, etc.
6. The valid hosts are:

The thing to remember is that for every subnet in thethird octet, there are two in the fourth octet: 0 and 128. For the 0 subnet,the broadcast address is always 127. For the 128 subnet, the broadcast addressis always 255.

Let's continue with more subnettinginto the fourth octet. This is exactly like subnettinga Class C network address, but the third octet is part of the subnet address.
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Example 7: Class B network 255.255.255.192

1. 2¹⁰-2=1,022 subnets
2. 2⁶-2=62 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-192=64, 128, 192 for the fourth octet. For every valid subnet in the third octet, we get four subnets in the fourth octet: 0, 64, 128, and 192.
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are as follows:

On this one, the 0 and 192 subnets are valid, since we areusing the third octet as well. The subnet range is 0.64 through 255.128. 0.0 isnot valid since no subnet bits are on. 255.192 is notvalid because then all subnet bits would be on.

Example 8: Class B network 255.255.255.224

1. 2¹¹-2=2,046 subnets
2. 2⁵-2=30 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-224=32, 64, 96, 128, 160, 192 for the subnet value. (For every value in the third octet, we get eight subnets in the fourth octet: 0, 32, 64, 96, 128, 160, 192, 224.)
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are:

For this subnet mask, the 0 and 224 subnets are valid aslong as not all subnet bits in the third octet are off or all subnet bits inthe fourth octet are on.

When would you use this valuable information? All thetime! For example, if you have a host configuration of 172.16.10.33255.255.255.224, what subnet, broadcast address, and valid host range is thishost a member of? (We would solve this question with the information presentedabove.)

• FTP (File Transfer Protocol)
• HTTP (Hypertext Transfer Protocol)
• SMTP (Simple Mail Transfer Protocol)
• Telnet

Transportlayer

• TCP (Transmission Control Protocol)
• UDP (User Datagram Protocol)

Internetlayer

• ARP (Address Resolution Protocol)
• RARP (Reverse Address Resolution Protocol)
• IP (Internet Protocol)
• ICMP (Internet Control Message Protocol)
• IGMP (Internet Group Membership Protocol)
• RIP (Routing Information Protocol)

NetworkAccess layer

• Physical Transmission Layer (Cat 5, etc.)
• Framing Protocols (Ethernet, etc.)

The Application layer
Let's start at the top. The Application layer runs its services via the layerimmediately below it—the Transport layer. In essence, it exploits TCP and UDPto deliver its goods. The Application layer is no slouch, however, as itfunctions to infiltrate and interact. DNS (Domain Name System) and FTP performat this level, as does HTTP, Telnet, SMTP, SNMP (Simple Network ManagementProtocol), and a myriad of other applications. Windows Sockets operate here inthe Microsoft scheme.

The Network Access layer
The Network Access layer is the equivalent of a loading dock, where the dataframes are put on the 10Base-T (or media of your choice) by token ring (orEthernet, etc.), and are also taken off.

Diagnostic utilities
Every good mechanic has a toolbox; if you're going to fix a problem, you needto know at what root it lies in order to troubleshoot it. TCP/IP is noexception.

Here are some common PING switches.

The most under-utilized aspect of PINGis its ability to diagnose the local machine. To do this, type either ping127.0.0.1 or ping localhost at the commandprompt. This will send a packet down the loopbackaddress and back up without sending it out on the network. A successfulresponse will verify that TCP/IP is installed on your local machine.

Here are some common IPCONFIGswitches.

On Windows 9x boxes, WINIPCFG will perform these functionsin a neat little GUI package.

ROUTE
ROUTE tells you everything you want to know about routes and routing at thelocal level. Not only does it provide you with data to view, it also allowsroute modification. Some of the most common switches are shown in Table C.

Here are some common ROUTE switches.

TRACERT
TRACERT is my personal favorite. As the name implies, it discovers, or TRACE ROUTEs the path from your local host to your destinationhost. It helps designate failed or slow links, as well as provides informationabout where all your packets travel on their way to a particular destination.Common TRACERT commands are shown in Table D.

Here are some common TRACERTswitches.

ARP
The Address Resolution Protocol will resolve IP addresses to MAC addresses.It's useful in discovering network configurations on the fly. Common ARPswitches are shown in Table E.

Here are some common ARP switches.

HOSTNAME
HOSTNAME provides your local host's name, which is useful to know if you'regoing to PING from it.

NETSTAT
NETSTAT provides Network (protocol) statistics and their current state. Thiscan encapsulate details for the following protocols: TCP, IP, ICMP, and UDP.Several commands you can use for NETSTAT are shown in Table F.

Here are some common NETSTATswitches.

NBTSTAT
As with NETSTAT, NBTSTAT provides network protocol statistics; however, it willalso provide NetBIOS over TCP/IP statistics. It's also useful for updating theLMHOSTS cache. Common NBTSTAT switches are shown in Table G.

Here are some common NBTSTATswitches.

NSLOOKUP
NSLOOKUP (Name Server Lookup) roughly looks up entries from DNS databases. TableH shows a limited list of common NSLOOKUP switches.

Here are some common NSLOOKUPswitches.

Data transfer tools
This is what networking is all about—the sharing of data. Of course, you haveto move data from point A to point B and back again. Throwing a floppy diskacross the office is not acceptable. Across a TCP/IP connection, FTP is the wayto go. FTP allows for the transfer of information when you either download itfrom or upload it to a remote host. The data transfer commands are shown in TableI.

Here are some common data transfercommands.

TFTP (Trivial File Transfer Protocol) is similar to FTP;however, while FTP demands authentication from the user, TFTP does not. TFTPsimply transfers data.

Remote execution tools
In order to control or merely interact with a remote host, you'll need to workfrom an interface. Telnet is perhaps the best-known and most widely usedprotocol. Its flexibility can provide access across server ports. RSH (RemoteShell) provides access to run commands on UNIX hosts. REXEC allows remoteexecution on remote hosts.

Printing utilities
The commands shown in Table J are used primarily to interact with lineprinters.

Troubleshooting TCP/IP

Talainia Posey

TheTCP/IP protocol is the backbone of the Internet. It's also heavily used in widearea networks. Unfortunately, due to the complexity of large networks, it canbe especially difficult to obtain an accurate diagnosis when problems occur. Inthis article, I'll explain several techniques that you can use to troubleshootTCP/IP when things go wrong.

Why troubleshootTCP/IP?
So, why is troubleshooting TCP/IP such a big deal? TCP/IP isn't one of thosecomponents that you can say is either working or not working. Instead, TCP/IPmay be partially functional, thus giving the illusion that it's working. Thissituation is possible because TCP/IP consists of many subcomponents.

Where do I begin?
Now that you have a basic understanding of some of the major portions ofTCP/IP, you may be puzzled as to where to begin the troubleshooting process.When troubleshooting TCP/IP, I recommend starting close to home and workingoutward. Basically, that means making sure that your own machine is functional,testing your network in general, and then checking your Internet connection.

The local address
The first step in the troubleshooting process is to verify that your PC has avalid IP address with which to work. As I mentioned earlier, most of the PCs ona network can use either a static or dynamic IP address. To determine whichcategory your PC falls into, open Control Panel and double-click the Networkicon. When you see the Network properties sheet, select the copy of TCP/IPthat's bound to your network card and click the Properties button. When you do,you'll see the TCP/IP properties sheet, as shown in Figure A.

If the Specify An IP Addressradio button is selected, you should make sure that valid entries have beenmade in the IP Address and Subnet Mask boxes. If, on the other hand, the ObtainAn IP Address Automatically radio button has beenselected, you must verify that you're receiving an IP address from a DHCPserver.

DHCP
DHCP stands for Dynamic Host Configuration Protocol, and it's a service thatcan run on one or more Windows NT Servers within your organization. When theObtain An IP Address Automatically radio button isselected, Windows will scan your network automatically for a DHCP server andobtain an IP address. There are a couple of ways to find out if you're gettingan address from the DHCP server. The easiest method involves opening an MS-DOSprompt window. If you're running Windows 9x, type WINIPCFG and press[Enter]. When you do, you'll see a summary of your TCP/IP configuration asWindows sees it. As you can see in Figure B, the IP Configuration dialogbox contains a drop-down list of your system's various adapters. Make sure thatyou've selected the correct network adapter from the list before you jump toany conclusions. It's easy to select a nonexistent dial-up adapter by mistake.

If you're using Windows NT, the command is slightlydifferent. Windows NT users should type IPCONFIG. Doing so will displaya summary of the configuration information for all network devices, as shown inFigure C.

If you're trying to use a DHCP server to configure TCP/IPbut aren't having any luck, there are some things that you should check. First,try reloading TCP/IP on your computer. It's possible that a file has becomecorrupted. If that doesn't solve the problem, verify that the DHCP server isrunning and that it's part of the same domain and subnet as your workstation.Next, check to see whether your co-workers' PCs are functional. If they arefunctional and rely on the DHCP server, it's a good bet that either the DHCPserver has run out of addresses or that there's a physical problem with yourmachine, such as a bad network card or a disconnected network cable.

The rest of the network
Once you've established with reasonable certainty that no IP address conflictsexist, that there isn't a problem with Windows, and that your networkconnection isn't unplugged, it's time to see if your computer can talk to therest of the LAN. One of the easiest ways of doing so is to log on. If thecomputer logs on, that's a step in the right direction. Next, check NetworkNeighborhood to see if any other computers show up. If they do, you're probablyokay on the local level. However, you should double-check to make sure that youdon't have any other protocols loaded. That way, you can be certain that it'sactually TCP/IP that's allowing you to see the other computers on your network.

IP routing in 40 short steps

Todd Lammle

• IP addressing and subnetting
• The difference between a router and a switch
• How collision and broadcast domains work within an internetwork

IPaddressing, subnetting, and IP routing are veryimportant fundamentals; once you have a good grasp on them you can move on toexploring more advanced, really exciting subjects like supernettingand Variable Length Subnet Masking (VLSM)!

Move that data!
Okay, you know routers are hardware devices that employ software to perform thetask of routing packets throughout an internetwork,and you know that routing is the term used to describe the process of taking apacket of data from a device on one network and switching it through thatrouter over to another device on a different network—packet distribution anddelivery. So if your network has no routers, you're not routing.

Figure A highlights the routing raison d'etre: to make it possible to connect multiple networks,thereby creating an internetwork so that all hostswithin that internetwork can communicate with eachother by sending and receiving data.

Now take a closer look at Figure A so we can gostep-by-step through the IP routing process. Let's begin by pretending that HostA wants to send a ping request (packet internet groper)to HostB.

By looking at the IP networks and addresses in the figure,we can see that HostA is on the 192.168.10.32 networkand that the /27 is a subnet block of 32. What this tells us is that our validhosts are 33-62. Also, notice that HostA's IP addressis 192.168.10.34 with a configured default gateway of 192.168.10.33.

HostB is on the 192.168.10.96 subnet and it has an IP address of192.168.10.98 with a configured default gateway of 192.168.10.97. It's veryimportant to make sure that a host's default gateway is configured correctlybecause it's used to tell the host how to send packets out of the localnetwork.

Okay, that said, let's type a ping request to192.168.10.98 from HostA at the command prompt andfollow it through from beginning to end. Here's what happens, starting at HostA (these are our 40 steps to IP routing):

1. Internet Control Message Protocol (ICMP) creates an echo request payload (which is just the alphabet forward and backwards).
2. ICMP hands that payload to IP, which then creates a packet. At a minimum, this packet contains an IP source address, an IP destination address, and a protocol field with 01h. All of that tells the receiving host whom to hand the payload to—in this example, ICMP.
3. Once the packet is created, IP works with the Address Resolution Protocol (ARP) to determine whether the destination IP address is on the local network or on a remote one.
4. Since this is a remote request, the packet needs to be sent to the default gateway. The registry in Windows is parsed to find the configured default gateway.
5. The default gateway of host 192.168.10.34 is 192.168.10.33. To be able to send this packet to the default gateway, the hardware address of the router's interface Ethernet1 (configured with the IP address of 192.168.10.33) must be known. Why? So the packet can be handed down to the Data Link Layer, framed, and sent to the router's interface for the 192.168.10.32 network.
6. Now, the ARP cache is checked to see if the IP address of the default gateway has been already resolved to a hardware address. If it has, the packet is then free to be handed to the Data Link Layer for framing. (The hardware destination address is also handed down with that packet.)
7. If the hardware address isn't already in the ARP cache of the host, an ARP broadcast will be sent out onto the local network to search for the hardware address of 192.168.10.33. The router will respond to the request and provide the hardware address of Ethernet1, and the host will cache this address. The 2500A router will also cache the hardware address of HostA in the ARP cache.
8. Once the packet and destination hardware address are handed to the Data Link Layer, the LAN driver is used to provide media access via the type of LAN being used. (In this example, it's Ethernet.) A frame is then generated, encapsulating the packet with control information. Within that frame are the hardware destination and source address, plus an Ethernet-Type field that describes the Network Layer protocol that handed the packet to the Data Link Layer (in this case, IP). At the end of the frame is the Frame Check Sequence (FCS) field, which houses the answers to the Cyclic Redundancy Check (CRC).
9. Once the frame is completed, the frame is handed down to the Physical Layer to be put on the physical medium (in this example, twisted-pair wire), one bit at a time.
10. Every device in the collision domain receives these bits and builds the frame. They each run a CRC and check the answer in the FCS field. If the answers don't match, the frame is discarded. If the CRC matches (in this example, it is the router's interface Ethernet1), then the hardware destination address is checked to see if it matches too. If it's a match, then the Ethernet-Type field is checked to find the protocol used at the Network Layer.
11. The packet is pulled from the frame and the frame is discarded. The packet is handed to the protocol listed in the Ethernet-Type field, meaning it's given to IP.
12. IP receives the packet and checks the IP destination address. Since the packet's destination address doesn't match any of the addresses configured on the receiving router itself, the router will look up the destination IP network address in its routing table.
13. The routing table must have an entry for the network 192.168.10.96 or the packet will be discarded immediately and an ICMP message will be sent back to the originating device with a Destination Network Unavailable message.
14. If the router does find an entry for the destination network in its table, the packet will be switched to the exit interface—in this example, interface serial 0.
15. Serial interfaces don't use Ethernet framing techniques, but by default, Cisco routers use something called High-Level Data Link Control (HDLC) to frame the packet. When serial interfaces are used, though, it's not called framing; it's known as "encapsulating." And because this is a point-to-point circuit, hardware addresses aren't used or needed. Instead, HDLC encapsulation is sent out one bit at a time to the next router.
16. Serial 0/0 on 2600A then extracts the packet from the HDLC encapsulation and checks the IP destination address. Since the IP addresses of the different router interfaces do not match, the router will check its routing to determine whether it knows how to get the packet over to the destination network.
17. The routing table will show that to get to network 192.168.10.96, it must use exit interface FastEthernet 0/0 so the packet is switched to interface FastEthernet 0/0.
18. The FastEthernet interface has the packet in the buffers and needs to send the packet to HostB, or IP address 192.168.10.98, but the hardware address must be known in order to deliver the packet to the correct host. The router checks the ARP cache, and if no match is found, an ARP broadcast is sent out interface FastEthernet 0/0.
19. HostB will respond with its hardware address, and the packet and destination hardware address are both sent to the Data Link Layer for framing.
20. The Data Link Layer will create a frame with the destination and source hardware address, Ethernet-Type field, and a FCS field at the end of the frame. The frame is handed to the Physical Layer to be sent out on the physical medium one bit at a time.
21. HostB receives the frame and immediately runs a CRC. If the answer matches what's in the FCS field, the hardware destination address is then checked. If it finds a match, the Ethernet-Type field is then checked to determine where the packet should be received on the Network Layer.
22. At the Network Layer, IP gets the packet and checks the IP destination address. Since there's finally a match made, the protocol field is checked to find out to whom the payload should be given.
23. The payload is handed to ICMP, which understands that this is an echo request. ICMP responds to this by immediately discarding the packet and generating a new payload as an echo reply.
24. A packet is then created including the source and destination address, protocol field, and payload.
25. ARP then checks to see if the destination IP address is a local device on the local LAN or if it's a device on a remote network. Since the destination device is on a remote network, the packet needs to be sent to the default gateway.
26. The default gateway address is found in the registry of the Windows device, and the ARP cache is checked to see if the hardware address has already been resolved from an IP address.
27. Once the hardware address of the default gateway is found, the packet and destination hardware address are handed down to the Data Link Layer for framing.
28. The Data Link Layer frames the packet of information and includes the following in the header: the destination and source hardware address, Ethernet-Type field with IP in it, and the FCS field with the CRC answer in tow.
29. The frame is now handed down to the Physical Layer (of the OSI model) to be sent out over the network medium one bit at a time.
30. The router's FastEthernet 0/0 interface will receive the bits and build a frame. The CRC is run, and the FCS field is checked to make sure the answers match.
31. Once the CRC is found to be okay, the hardware destination address is checked. Since the router's interface is a match, the packet is pulled from the frame and the Ethernet-Type field is checked to see what protocol at the Network Layer the packet should be delivered to.
32. That's determined to be IP, so it gets the packet. IP runs a CRC check on the IP header first and then checks the destination IP address. (Note: IP does not run a complete CRC like the Data Link Layer—it only checks the header for errors.) Since the IP destination address doesn't match any of the router's interfaces, the routing table is checked to see if it has a route to 192.168.10.32. If it doesn't have a route over to the 32 network, the packet will be discarded immediately. (This is the source point of confusion for a lot of administrators. When a ping fails, most people think the packet never reached the destination host. But as we see here, that's not always the case! All it takes is for just one of the remote routers to be lacking a route back to the originating host's network and POOF! The packet is dropped on the return trip, not en route to the host.)
33. But Router 2600A does know how to get to network 192.168.10.32—the exit interface is serial 0/0—so the packet is switched to serial interface 0/0.
34. The serial interface builds an HDLC encapsulation and sends the packet inside an HDLC encapsulation method out interface 0/0, one bit at a time, to the 2500A router. The 2500A router receives the HDLC encapsulation and hands over the packet to IP.
35. IP checks the destination address, and since no interface on the router matches, the routing table is checked for a path to network 192.168.10.32.
36. The routing table tells IP that the path to network 192.168.10.32 is out interface Ethernet1, so over it goes—the packet is switched to interface Ethernet1.
37. The hardware address must first be found in order to send the packet to the destination IP address of 192.168.10.34, and the ARP cache is checked for that first. Since the address is in the ARP cache, the packet and the destination hardware address are sent to the Data Link Layer to be framed.
38. The frame adds the hardware source and destination address and the Ethernet-Type field, and puts the CRC answer in the FCS field.
39. The frame is then given to the Physical Layer to be sent out onto the local network, one bit at a time.
40. The destination host receives the frame, runs a CRC, checks the destination hardware address, and then looks in the Ethernet-Type field to find who to hand the packet to. IP, at the Network Layer, is the designated receiver, and after the packet is handed to IP at the Network Layer, it checks the protocol field for further direction. IP finds instructions to give the payload to ICMP, and ICMP determines the packet to be an ICMP echo reply. It acknowledges that it has received the reply by sending an exclamation point (!) to the user interface. ICMP will then attempt to send four more echo requests to the destination host.

• Some things never change, and this includes packets. They never change in any way; they are only encapsulated in control information to enable them to be sent from one router to another.
• Routers are impersonal; they do not keep host addresses in their routing tables. Routers only care about networks and the best path to each one.
• Each and every hardware address is unique. These are what devices use to find a unique host on a local network.