Computer Repair Student Part.5 [subnetting] Finnish

Subnetting

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Appendix E–Subnetting

OBJECTIVES After completing this chapter you will • Understand the difference between major classes of IP addresses. • When given an IP address, be able to identify its IP class. • Determine the appropriate mask to use with each IP class. • Understand and be able to subnet IP addresses.

KEY TERMS broadcast ISP subnet useable host numbers useable subnets

Subnetting Overview

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SUBNETTING OVERVIEW Any company that needs IP addresses can lease them from an ISP (Internet Service Provider. ISPs are organizations that provide individuals and businesses access to the Internet. If your company decides to connect to the Internet, a network administrator would contact an ISP and make arrangements for a connection to the Internet. Your connection to the Internet would be through the ISPs own network. You would have to provide the ISP with the number of IP addresses the company needs. ISPs have a limited number of IP addresses available to give out to their customers. The ISP will ask how many computers your company presently has and how many computers are planned for the near future. Public IP addresses are in short supply due to the overwhelming popularity and success of the Internet. ISPs are reluctant to lease more IP addresses than a customer needs. The ISP may have 100 Class C IP addresses available to lease, but that does not mean that the ISP will lease an entire Class C just because it was requested. What if your company doesn’t have enough computers to warrant a full Class C? For example, the Smiley Company has 30 computers and needs Internet access. The company predicts a growth of 20 additional computers over the next two years. If the Smiley Company contacts an ISP and asks for a full Class C block of addresses, the request would probably be denied. Remember that a full Class C contains 256 host addresses, and the Smiley Company needs only 50 addresses. The ISP would probably lease a portion of the Class C to Smiley. As another example, the WebBook Company has more than 450 computers on its network with expected growth of 50 new computers this year. The ISP has 100 Class Cs available for its customers and will have to dedicate at least two Class Cs for the WebBook Company. The WebBook Company is not centralized in one building; it has several offices throughout the city. Each office has 30–60 computers that need Internet access. The ISP gives the company two full Class Cs to organize as the network administrator sees fit. The network administrator can subdivide the Class C addresses to enable all external sites to have access to the Internet. The ISP does not care if the Class C is broken into smaller segments to fit a company’s needs. When an IP addresses range is subdivided like this, it is called subnetting. Although the above examples are very simplified, this gives you a basic understanding of the process involved in acquiring and using IP addresses. A subnet is a method used that divides the IP address into three parts rather than two. A normal IP address consists of two parts—a network number and a host number. Remember from the Network chapter that the number of bits used for the two parts depends on the class of IP address. Refer back to Network Figure #14 to refresh your memory. When subnetting is used, the IP address has three parts—a network number, a subnetwork number, and a host number. Subnetting involves borrowing bits from the host portion of the IP address and creating the third part of the IP address—the subnetwork number, which is commonly called the subnet. Take a Class C IP address of 192.168.10.4. Because it is a Class C

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Appendix E–Subnetting address, the first three octets are the network number—192.168.10. The last octet is the host or network device located on network 192.168.10. In this example, the host number is 4. Without subnetting, this IP address has one network number and 256 different host numbers. If this IP address is subnetted, the network 192.168.10 can be divided into more than just one network. It can have a varying number of subnetworks. Subnetting has three primary functions: (1) efficient use of one or more IP addresses, (2) reduces the money spent to lease IP addresses, and (3) divides the network into easier to manage portions. The effects of these three functions will be seen in the sections on how to subnet and why to subnet.

HOW TO SUBNET When subnetting, bits are borrowed from the host portion of the IP address. Borrowing bits from the host creates a new number called a subnet.

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When subnetting, always borrow bits from the left-most host bits. Subnetting reduces the number of hosts, but allows more networks using a single IP address. Take the example of a standard Class C IP address, 207.193.204.0. The number 207.193.204 is the network number, and the last octet is used for host numbers. Subnet Figure #1 shows the bit positions for the last octet (Octet 4) for a standard Class C IP address.

Subnet Figure #1 bit positions

Class s C Network No. Octet 1 Octet 2 Octet 3 128 207.

193.

64

32

Hosts Octet 4 16 8

4

1䊴

2

204.

In order to subnet, bits are borrowed from the host bits to create a subnet field. The borrowed bits make up the subnet field. Subnet Figure #2 shows two bits borrowed from the hosts field for subnetting.

Subnet Figure #2 Class C Network No. Octet 1

Octet 2

Subnet

Octet 3

Octet 4 128

207.

193.

Hosts

204.

64

32

16

8

4

2

1

Subnetwork Numbers

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Remember that bits are always borrowed from the left-most bits in the octet. Since two bits are being borrowed, bit positions 128 and 64 now represent subnet numbers. In Subnet Figure #2, you can see that the IP address still contains a network number and host numbers. The new addition to the IP address is called the subnet and is formed by taking left-most bits away from the host field. The standard Class C has eight bits that represent hosts, but now only six host bits remain because two bits were borrowed to create subnets.

SUBNETWORK NUMBERS The subnet portion of an IP address can have varying combinations of 1s and 0s. For example, if two bits are borrowed for subnetting, the combinations of 1s and 0s are four different numbers—00, 01, 10, and 11. Subnet Figure #3 shows this concept.

Subnet Figure #3 Class C Network No. Octet 1

Octet 2

Subnet

Hosts

Octet 3

Octet 4 128

64

207.

193.

204.

0

0

207.

193.

204.

0

1

207.

193.

204.

1

0

207.

193.

204.

1

1

32

16

8

4

2

1

A mathematical formula can be used to determine how many subnets are formed when borrowing bits.

0

The number of subnets can be found by taking 2X where x is the number of bits borrowed. For example, if two bits are borrowed, 22 = 4 or four subnetworks. If three bits are borrowed, 23 = 8 or eight subnetworks. Look back at Subnet Figure #3. The 00 combination in the subnet field represents Subnetwork 0. The 01 combination in the subnetwork column designates it as Subnetwork 64. The 64 is obtained by a 1 being set in the 64 bit position. What subnetwork number is designated by a bit combination of 10 in the subnetwork column? The answer is 128 because there is a 1 set in the 128 bit position. Subnet Figure #4 shows how the various combinations of 1s and 0s create different subnetwork numbers.

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Appendix E–Subnetting

Subnet Figure #4 Class C Network No. Octet 1

Octet 2

Subnet

Hosts

Octet 3

Subnet No.

Octet 4 128

64

32

16

8

4

2

1

207.

193.

204.

0

0

0

207.

193.

204.

0

1

64

207.

193.

204.

1

0

128

207.

193.

204.

1

1

192

Now that subnetwork numbers are understood, let’s see how this applies to networks. Subnet Figure #5 shows two networks connected by a router. The networks are subnetted.

Subnet Figure #5

192.168.10.64

Router

192.168.10.128

In Subnet Figure #5, one network is 192.168.10.64 and the other network is 192.168.10.128. Even though a company purchased one Class C IP address, two networks can be created because of subnetting.

NUMBER OF HOSTS An IP addressing rule is that every device on the network must have a unique IP address. How does this rule affect subnetting? Each subnet can have a varying number of hosts. The number of hosts on each subnet depends on how many host bits have been borrowed to subnet. The more bits borrowed for subnetting, the fewer host bits remain for network devices. Subnet Figure #6 shows how this works in a Class C IP address with two bits borrowed for subnets.

Number of Hosts

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Subnet Figure #6 Class C Network No.

Octet 1

Octet 2

Subnet

Hosts

Octet 3

Subnet No.

Octet 4 128

64

32

16

8

4

2

1

207.

193.

204.

0

0

0

0

0

0

0

0

207.

193.

204.

0

0

0

0

0

0

0

1

207.

193.

204.

0

0

0

0

0

0

1

0

207.

193.

204.

0

0

0

0

0

0

1

1

207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207. 207.

193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193. 193.

204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204. 204.

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

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

0

0

0

1

0

0

1 0 0 0

1 0 0 0

1 0 0 0

1 0 0 0

1 0 0 1

1 0 1 0

1 0 0 0

1 0 0 0

1 0 0 0

1 0 0 0

1 0 0 1

1 0 1 0

1 0 0 0

1 0 0 0

1 0 0 0

1 0 0 0

1 0 0 1

1 0 1 0

1

1

1

1

1

1

0

64

128

192

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Appendix E–Subnetting The easiest way to determine the total number of hosts is to count the number of bits that are left for hosts and raise the number 2 to that power. For example, in Subnet Figure #6, there are six hosts bits remaining. Take the number 2 and raise it to the number of host bits—26 = 64. This means that there are 64 different combinations of 1s and 0s in the host field for each subnetwork. One of the most important rules about subnetting concerns the first and last subnet and the first and last host addresses.

0

When subnetting, the first and last subnetwork numbers and the first and last host numbers within each subnet cannot be used. As previously discussed in the Network chapter, a network has an IP address that cannot be used by any network device. An example of a Class C network address is 192.107.10.0. A Class B network address is 152.124.0.0 and a Class A example is 11.0.0.0. When all of the host bits are 0s, that is considered the network or subnetwork number and that number cannot be assigned to a network device as a host number. Look back to Subnet Figure #7. The first subnetwork shown is 0. When all host bits are 0, that is considered the subnetwork number. Some people call it the wire. Each combination of 1s and 0s after that point is a host number on that subnetwork until the subnetwork number changes. The only exception to this is when all host bits are set to 1. When all of the host bits are a binary 1, this designates a broadcast for that network or subnetwork. Using the same network numbers given above as examples, the broadcast addresses would be 192.107.10.255 for the Class C network, 152.124.255.255 for the Class B network, and 11.255.255.255 for the Class A network. The broadcast address cannot be assigned to a network device. The broadcast address is used to communicate with all network devices simultaneously. When borrowing two host bits from a Class C IP address, subnetworks 0, 64, 128, and 192 were created. Based on the rule stated above, subnetworks 0 and 192 cannot be used, so all that are left are subnetworks 64 and 128. These subnetworks that can be used are known as useable subnets. The first and the last subnet are considered unuseable because they contain all 0s and all 1s in the host bits. Subnet 0 contains all 0s in the host bits. Subnet 192 can contain all 1s in the host bits. On subnetwork 64, the host numbers that are possible are 64 through 127. On subnetwork 64, the host numbers that are useable are 65 through 126. The host numbers that can be assigned to network devices are known as useable host numbers. For subnetwork 128, the possible host numbers are 128 through 191, but only 129 through 190 are used. This is found by using varying combinations of 1s and 0s through the host bits. Subnet Figure #7 illustrates this concept.

Number of Hosts

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Subnet Figure #7 Class C Network No.

Octet 1

Octet 2

Subnet

Octet 3

Hosts Subnet No.

Octet 4 128

64 32 16 8

4

2 1

Host No.

207.

193.

204.

0

1

0

0

0

0

0 0

64

207.

193.

204.

0

1

0

0

0

0

0 1

65

207.

193.

204.

0

1

0

0

0

0

1 0

66

207.

193.

204.

0

1

0

0

0

0

1 1

67

207.

193.

204.

0

1

0

0

0

1

0 0

68

207.

193.

204.

0

1

0

0

0

1

0 1

69

207.

193.

204.

0

1

0

0

0

1

1 0

70

207.

193.

204.

0

1

207.

193.

204.

0

1

207.

193.

204.

0

1

1

1

1

1

1 0

126

207.

193.

204.

1

1

1

1

1

1

1 1

Broadcast

207.

193.

204.

1

0

0

0

0

0

0 0

207.

193.

204.

1

0

0

0

0

0

0 1

129

207.

193.

204.

1

0

0

0

0

0

1 0

130

207.

193.

204.

1

0

0

0

0

0

1 1

131

207.

193.

204.

1

0

0

0

0

1

0 0

132

207.

193.

204.

1

0

0

0

0

1

0 1

133

207.

193.

204.

1

0

207.

193.

204.

1

0

207.

193.

204.

1

0

1

1

1

1

1 0

190

207.

193.

204.

1

0

1

1

1

1

1 1

191

128

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Appendix E–Subnetting One of the hardest concepts for students to grasp is that, when determining what the decimal number is for an octet, you must use all eight octet bits in the number. Ignore the line drawn for the subnet bits. For example, in Subnet Figure #7, for subnetwork 64, the first useable host number is 65. Eight bits designate the number 65—01000001. Since there is a 1 in bit position 64 and a 1 in bit position 1, 64 + 1 = 65. The subnetwork number is 64 because the subnetwork columns have a 01 combination. The 1 is set in the 64 column. The host number is 67 because the decimal IP number represents all eight bits in an octet. Subnet Figure #8 illustrates how a subnetted network has host numbers assigned to each network device. Notice that each device has numbers that relate to the range valid for each subnetwork.

Subnet Figure #8 Host Addresses on a Subnetted Network 192.168.10.64

.67

.68

.69

.65

.70

Router

.129

.130

192.168.10.128

.131

.132

.133

Notice in Subnet Figure #8 how the router received two host numbers. This is because, inside the router, there are two Ethernet ports. Each of these ports receives a host number just as if it were a network card inside a computer. The host number assigned to the router’s Ethernet port corresponds to a host number on the subnetwork the Ethernet port attaches to. Since the left side of the router is connected to Subnetwork 64, the router’s port host address is .65, the first available host number on the subnetwork. The router’s port does not have to receive the first host number in the subnetwork, but it is done this way in the figure to illustrate how a host number is assigned.

0

When subnetting, each device still has a unique IP address. The only difference that subnetting makes is that each subnetwork has a range of valid host numbers for that individual subnetwork.

MASK REVIEW In order to subnet, the subnet mask is used and is very important to understand. In the Network chapter you learned that a Class A IP address has a standard mask of 255.0.0.0. A Class B IP address has a standard mask of 255.255.0.0, and a Class C IP address has a mask of 255.255.255.0. For example, consider a computer that has an IP address of 150.150.150.150. The IP address is a Class B address. If the computer uses a standard mask, the mask entered would be 255.255.0.0.

Solving IP Subnetting Problems

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Keep in mind that the mask distinguishes the network number from the host number. Using the same example used above of 150.150.150.150 and a mask of 255.255.0.0 yields the network number of 150.150.0.0 because of the mask used. An IP address of 193.206.52.4 with a mask of 255.255.255.0 has a network address of 193.206.52.

THE MASK WHEN SUBNETTING The way that any network device knows that a subnet is being used is through the mask. This is why the mask is sometimes known as the subnet mask. The mask number stays the same up to the point that bits are borrowed. Then, in the octet where bits are borrowed, the new mask number is found by adding the bit positions together that are borrowed to create the subnet number. Look back to Subnet Figure #3. Since this is a Class C address, the normal mask is 255.255.255.0. However, since subnetting is implemented, the mask changes to 255.255.255.192. The 192 is found by adding the value of the bit positions being borrowed—bit position 128 and bit position 64 (the two bit positions borrowed to create the subnet number). 128 + 64 = 192. So, the new mask is 255.255.255.192. If three bits are borrowed in a Class C IP address, the last octet mask would be 224 (128 + 64 + 32). If four bits are borrowed in a Class C IP address, the last octet mask would be 240 (128 + 64 + 32 + 16). Examples of the new mask with each IP class are given later in the chapter.

SOLVING IP SUBNETTING PROBLEMS When asked a subnetting problem, you can be presented with several pieces of information that describe the situation. Given that information, it will be up to you to figure out the remaining pieces of information to solve the problem. The types of information that you must be able to identify are as follows:

Information Class of IP address Network number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

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Appendix E–Subnetting

Class C Problem 1 Let’s do one example without subnetting to make it simple and to explain how the chart works. Given the XYZ Company’s Class C address of 201.15.6.0 and a mask of 255.255.255.0, you should be able to fill in the following information:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 201.15.6.0 255.255.255.0 1 256 N/A because there are no subnets 201.15.6.255 N/A because there are no subnets 254 N/A because there are no subnets

Now that the purpose of the chart is clear, let’s try an example with subnetting. Suppose the XYZ Company has four different networks throughout its factory, but only one full Class C address. One option the company can do is to divide the Class C address into four different subnetworks. The Class C IP address is 201.15.6.0. The network administrator decides that the new subnet mask is 255.255.255.224. The following information is what we know so far:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 201.15.6.0 255.255.255.224 ?? ?? ?? ?? ?? ?? ??

Solving IP Subnetting Problems

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If the mask is given, you can solve the rest of the unknowns and fill in the chart. The mask in this problem is 255.255.255.224. Since this is a Class C network and the default mask is 255.255.255.0, we know that some bits are being borrowed in the last octet because the number in the last octet has changed to 255.255.255.224. The first step is to break the last octet into binary to see how many bits are being borrowed for the subnetting. Subnet Figure #9 shows the last octet broken down into bits.

Subnet Figure #9 224 broken into bits

Class C Network No. Octet 1

Octet 2

255.

255.

Subnet

Hosts

Octet 3

Octet 4

255.

128

64

32

16

8

4

2

1

1

1

1

0

0

0

0

0䊴

Notice in Subnet Figure #9 how the first three bits are set to 1. Since there is a 1 in the 128 column, a 1 in the 64 column, and a 1 in the 32 column, 128 + 64 + 32 = 224. This is just the process of converting decimal to binary as shown in earlier chapters. The next step is to draw a vertical line between the 1s and the 0s. In the case of the 224 mask, a vertical line is drawn between the 32 and the 16 column. See Subnet Figure #10 to see where the vertical line is placed.

Subnet Figure #10 Class C Network No. Octet 1

255.

Octet 2

255.

Subnet

Hosts

Octet 3

255.

Octet 4 128

64

32

16

8

4

2

1

1

1

1

0

0

0

0

0

Once you have figured out where this line goes, you can answer many questions. Now you know that the number of bits borrowed is three because there are three 1s set when you translate 224 into binary. Let’s update the chart.

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Appendix E–Subnetting

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 201.15.6.0 255.255.255.224 ?? ?? ?? ?? 3 ?? ??

Once you know the number of bits borrowed, you can determine the number of subnetworks using the formula 2x = number of subnetworks where x is the number of bits borrowed. So 23 = 8 meaning that there are eight subnetworks. Again, update the chart with the new information.

Information

Value

Class of IP address

C

Network number

201.15.6.0

Mask

255.255.255.224

Total no. of networks/subnetworks

8

Total no. of hosts per network/subnetwork

??

Subnet numbers

??

Broadcast addresses

??

No. of bits borrowed

3

Useable hosts per network/subnetwork

??

Useable subnets

??

Also, since you know the number of bits borrowed, you can determine how many bits are left for hosts (network devices). Since three of the eight bits in the octet are borrowed for subnetworks, there are five bits left for hosts. The number of hosts per subnetwork can be found by using the formula 2x = number of hosts where x is the number of host bits remaining after bits have been borrowed for subnetting. In our example, three bits were borrowed for subnetting. Five bits were left for hosts. 25 = 32 hosts per subnetwork. Updating the chart shows the following:

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

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Value C 201.12.6.0 255.255.255.224 8 32 ?? ?? 3 ?? ??

Once you have determined the total number of subnets and the total number of hosts per subnet, you can determine the number of useable subnets and useable hosts per subnet. The number of useable subnets is the number of subnets minus 2. Since there are eight possible subnetworks, 8 – 2 = 6 useable networks. The number of useable hosts is the number of hosts per subnetwork minus 2. Since there are 32 possible hosts per subnetwork, 32 – 2 = 30.

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 201.15.6.0 255.255.255.224 8 32 ?? ?? 3 30 6

We have determined that there are eight subnetworks available using the subnet mask of 255.255.255.224. Subnet Figure #11 shows the eight different subnetworks converted into decimal values.

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Appendix E–Subnetting

Subnet Figure #11 Class C Network No. Octet 1

Octet 2

Subnet

Hosts

Octet 3

Subnet No.

Octet 4 128

64

32

16

8

4

2

1

201.

15.

6.

0

0

0

0

0

0

0

0

0

201.

15.

6.

0

0

1

0

0

0

0

0

32

201.

15.

6.

0

1

0

0

0

0

0

0

64

201.

15.

6.

0

1

1

0

0

0

0

0

96

201.

15.

6.

1

0

0

0

0

0

0

0

128

201.

15.

6.

1

0

1

0

0

0

0

0

160

201.

15.

6.

1

1

0

0

0

0

0

0

192

201.

15.

6.

1

1

1

0

0

0

0

0

224

Notice in Subnet Figure #11 that the subnetworks are grouped into multiples of 32. Also notice that the first column to the left of the line (the darker gray area) has a value of 32. The total subnetworks are as follows: 0, 32, 64, 96, 128, 160, 192, and 224. These are all multiples of 32. Now, we can update our list and put in the subnet numbers as shown below:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 201.15.6.0 255.255.255.224 8 32 201.15.6.0, 201.15.6.32, 201.15.6.64, 201.15.6.96, 201.15.6.128, 201.15.6.160, 201.15.6.192, 201.15.6.224 ?? 3 30 6

Solving IP Subnetting Problems

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Instead of running 1s and 0s in the subnet columns, you can determine that the subnetworks are in groups of 32, which is the value in the column left of the subnet line.

0

Now let’s determine the broadcast addresses for each subnetwork. The broadcast address can be found by placing all 1s in the host bits for each subnetwork. Subnet Figure #12 shows the broadcast address calculation.

Subnet Figure #12 4th Octet in Binary Subnet 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8

IP Address Range (201.15.6.0) (201.15.6.31) (201.15.6.32) (201.15.6.63) (201.15.6.64) (201.15.6.95) (201.15.6.96) (201.15.6.127) (201.15.6.128) (201.15.6.159) (201.15.6.160) (201.15.6.191) (201.15.6.192) (201.15.6.223) (201.15.6.224) (201.15.6.255)

128 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

64 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

32 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

16 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

8 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

4 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

2 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

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

Notice in Subnet Figure #12 how each subnetwork is shown with the subnetwork number and the broadcast address for that subnetwork. Now for the final chart update.

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Appendix E–Subnetting

Information

Value

Class of IP address

C

Network Number

201.15.6.0

Mask

255.255.255.224

Total no. of networks/subnetworks

8

Total no. of hosts per network/subnetwork

No. of bits borrowed

32 201.15.6.0, 201.15.6.32, 201.15.6.64, 201.15.6.96, 201.15.6.128, 201.15.6.160, 201.15.6.192, 201.15.6.224 201.15.6.31, 201.15.6.63, 201.15.6.95, 201.15.6.127, 201.15.6.159, 201.15.6.191, 201.15.6.223, 201.15.6.255 3

Useable hosts per network/subnetwork

30

Useable subnets

6

Subnet numbers

Broadcast addresses

Class C Problem 2 The Hi-IQ Company has leased one Class C address—200.200.200.0. The company has ten networks with 12 computers on each network. With this information, the chart appears as follows:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 200.200.200.0 ?? (10 needed) (12 needed) ?? ?? ?? ?? ??

Solving IP Subnetting Problems

E-19

To solve this problem, the first step is to determine the mask and, from there, the rest is the same as the first example. To determine what mask is needed, you must discover how many bits to borrow. Remember to use the formula 2x = total number of subnets – 2 = the number of useable subnets, where x is the number of bits borrowed. If two host bits are borrowed (the minimum number for a Class C network), only two subnets are created (22 = 4 – 2 = 2). That number of subnets is not enough for the Hi-IQ Company. If three host bits are borrowed, six subnets are useable (23 = 8 – 2 = 6). Again, there are not enough subnets. If four host bits are borrowed, 14 subnets are useable (24 = 16 – 2 = 14). This is the correct number of bits to borrow for the Hi-IQ Company; however, the mask must still be determined. When borrowing four host bits to create the subnets, the mask is found by adding the bit values for the highest most bits. See Subnet Figure #13.

Subnet Figure #13 Class C Network No. Octet 1

Octet 2

Subnet

Octet 3

Octet 4 128

255.

255.

255.

Hosts

1

64 32 16

8

4

2

1

1

0

0

0

0

1

1

The normal Class C mask is 255.255.255.0, but we are borrowing bits from the last octet, so we know the mask is going to be different. By adding the bit values for the bits being borrowed, the mask is found—128 + 64 + 32 + 16 = 240. The mask for this subnetted Class C address is 255.255.255.240. Updating the chart shows the following:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 200.200.200.0 255.255.255.240 16 (12 needed) ?? ?? 4 ?? 14

E-20

Appendix E–Subnetting Since the number of bits borrowed has been determined, it is easy to see how many host bits are left to determine the total number of hosts per subnetwork. Look back at Subnet Figure #13. At a quick glance, it is apparent that four bits remain for hosts. Using the formula 2x = total number of hosts – 2 = the number of useable hosts where x is the number of host bits remaining, if four host bits are remaining, 14 host addresses are useable (24 = 16 – 2 = 14). This works well for the Hi-IQ Company since they have 12 computers on each network. Updating the chart with this information provides the following:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 200.200.200.0 255.255.255.240 16 16 ?? ?? 4 14 14

The only thing left to do is to figure out the subnetwork numbers and the broadcasts. Subnet Figure #14 shows only the subnetwork numbers by placing 0s in each of the host bits.

Solving IP Subnetting Problems

E-21

Subnet Figure #14 Class C Network No. Octet 1

Octet 2

Subnet

Octet 3

Hosts Subnet No.

Octet 4 128

64

32

16

8

4

2

1

201.

15.

6.

0

0

0

0

0

0

0

0

0

201.

15.

6.

0

0

0

1

0

0

0

0

16

201.

15.

6.

0

0

1

0

0

0

0

0

32

201.

15.

6.

0

0

1

1

0

0

0

0

48

201.

15.

6.

0

1

0

0

0

0

0

0

64

201.

15.

6.

0

1

0

1

0

0

0

0

80

201.

15.

6.

0

1

1

0

0

0

0

0

96

201.

15.

6.

0

1

1

1

0

0

0

0

112

201.

15.

6.

1

0

0

0

0

0

0

0

128

201.

15.

6.

1

0

0

1

0

0

0

0

144

201.

15.

6.

1

0

1

0

0

0

0

0

160

201.

15.

6.

1

0

1

1

0

0

0

0

176

201.

15.

6.

1

1

0

0

0

0

0

0

192

201.

15.

6.

1

1

0

1

0

0

0

0

208

201.

15.

6.

1

1

1

0

0

0

0

0

224

201.

15.

6.

1

1

1

1

0

0

0

0

240

Notice that the first subnet has all 0s in the last octet. This is why none of the first subnets cannot be used as a useable subnet. Now let’s get the broadcast numbers. Subnet Figure #15 shows only the broadcasts by placing 1s in each of the host bits.

E-22

Appendix E–Subnetting

Subnet Figure #15 Class C Network No. Octet 1

Octet 2

Subnet

Octet 3

Hosts Broadcast Address

Octet 4 128

64

32

16

8

4

2

1

201.

15.

6.

0

0

0

0

1

1

1

1

15

201.

15.

6.

0

0

0

1

1

1

1

1

31

201.

15.

6.

0

0

1

0

1

1

1

1

47

201.

15.

6.

0

0

1

1

1

1

1

1

63

201.

15.

6.

0

1

0

0

1

1

1

1

79

201.

15.

6.

0

1

0

1

1

1

1

1

95

201.

15.

6.

0

1

1

0

1

1

1

1

111

201.

15.

6.

0

1

1

1

1

1

1

1

127

201.

15.

6.

1

0

0

0

1

1

1

1

143

201.

15.

6.

1

0

0

1

1

1

1

1

159

201.

15.

6.

1

0

1

0

1

1

1

1

175

201.

15.

6.

1

0

1

1

1

1

1

1

191

201.

15.

6.

1

1

0

0

1

1

1

1

207

201.

15.

6.

1

1

0

1

1

1

1

1

223

201.

15.

6.

1

1

1

0

1

1

1

1

239

201.

15.

6.

1

1

1

1

1

1

1

1

255

Notice how the last subnet has all 1s in the last octet. This is why the last subnet cannot be used as a useable subnet. Now let’s update the chart with the subnetwork numbers and their associated broadcast addresses.

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork

Subnet numbers

Broadcast addresses

No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

E-23

Value C 200.200.200.0 255.255.255.240 16 16 200.200.200.0, 200.200.200.16, 200.200.200.32, 200.200.200.48, 200.200.200.64, 200.200.200.80, 200.200.200.96, 200.200.200.112, 200.200.200.128, 200.200.200.144, 200.200.200.160, 200.200.200.176, 200.200.200.192, 200.200.200.208, 200.200.200.224, 200.200.200.240 200.200.200.15, 200.200.200.31, 200.200.200.47, 200.200.200.63, 200.200.200.79, 200.200.200.95, 200.200.200.111, 200.200.200.127, 200.200.200.143, 200.200.200.159, 200.200.200.175, 200.200.200.191, 200.200.200.207, 200.200.200.223, 200.200.200.239, 200.200.200.255 4 14 14

All important pieces of information needed to set up the network are now provided. Class C Problem 3 A network administrator for the Total Cool Company is working on a computer. The computer’s IP address is 204.210.179.142 with a mask of 255.255.255.192. The network administrator needs to know on which subnet the computer belongs. The information found by looking at the computer is the Class of IP address, the network portion of the IP address, and the mask. The following chart shows this information.

E-24

Appendix E–Subnetting

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 204.210.179.0 255.255.255.192 ?? ?? ?? ?? ?? ?? ??

The first step in solving this problem is to discover how many bits are borrowed. A normal Class C mask is 255.255.255.0, but the one on this computer is 255.255.255.192. Finding out how many bits are borrowed requires you to put 1s in Octet 4’s bits until the bit positions add up to 192. Look at Subnet Figure #16 to see how this is done.

Subnet Figure #16 Class C Network No. Octet 1

Octet 2

Subnet

Octet 3

Octet 4 128

255.

255.

255.

Hosts

1

64 32 16

8

4

2

1

1

Bit position 128 plus bit position 64 added together gives you 192. So, two bits are borrowed. Updating the chart with the number of bits borrowed shows the following:

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

E-25

Value C 204.210.179.0 255.255.255.192 ?? ?? ?? ?? 2 ?? ??

Now that the number of bits borrowed is solved, you can determine the number of subnetworks using the formula 2x = number of subnetworks where x is the number of bits borrowed and subtracting 2 because you cannot use the first or the last subnetworks. So 22 = 4 minus 2 equals 2, meaning that there are two subnetworks. You can also determine the number of hosts on each network using the same method. 2x = number of hosts where x is the number of host bits remaining after bits have been borrowed for subnetting. Then subtract 2 for the number of useable hosts. In our example, two bits were borrowed for subnetting. Six bits are left for hosts. 26 = 64 hosts per subnetwork minus 2 for the subnetwork number and broadcast leaves 62 hosts per subnetwork. Updating the chart shows the following:

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value C 204.210.179.0 255.255.255.192 4 64 ?? ?? 2 62 2

E-26

Appendix E–Subnetting Now, the only thing left to do is to determine the subnet numbers. Look back to Subnet Figure #16. You can see that the line is drawn between the 32 and 64 bit positions. A shortcut is to look at the number to the left of the line and you can tell that the subnetwork numbers will be incremented in steps of 64, but filling in the chart with 0s in the host bits proves this shortcut. Subnet Figure #17 shows the subnetwork numbers.

Subnet Figure #17 Class C Network No. Octet 1

Octet 2

Subnet

Hosts

Octet 3

Octet 4

Subnet No.

128

64

32

16

8

4

2

1

204.

210.

179.

0

0

0

0

0

0

0

0

0

204.

210.

179.

0

1

0

0

0

0

0

0

64

204.

210.

179.

1

0

0

0

0

0

0

0

128

204.

210.

179.

1

1

0

0

0

0

0

0

192

A shortcut for finding the broadcast address is to subtract 1 from the subnetwork below the one you are working on because the broadcast address is always 1 less than the subnetwork number. Also remember that the last subnet will always have a broadcast address of 255. However, filling in the chart shows the full version. Subnet Figure #18 shows broadcasts for each of the subnetworks by putting 1s in the host positions.

Subnet Figure #18 Class C Network No. Octet 1

Octet 2

Subnet

Hosts

Octet 3

Broadcast Address

Octet 4 128

64

32

16

8

4

2

1

204.

210.

179.

0

0

1

1

1

1

1

1

63

204.

210.

179.

0

1

1

1

1

1

1

1

127

204.

210.

179.

1

0

1

1

1

1

1

1

191

204.

210.

179.

1

1

1

1

1

1

1

1

255

Now, update the chart to include the subnetwork numbers and broadcast addresses:

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of networks/subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

E-27

Value C 204.210.179.0 255.255.255.192 4 64 204.210.179.0, 204.210.179.64, 204.210.179.128, 204.210.179.192 204.210.179.63, 204.210.179.127, 204.210.179.191, 204.210.179.255 2 62 2

Now that the chart is complete, look at the sequence of network numbers or back at Subnet Figure #17 to determine in what subnetwork IP address 204.210.179.142 falls. Subnetwork 0 is IP addresses 204.210.179.0 through 204.210.179.63. This does not include .142, so look at the next subnetwork. Subnetwork 1 is IP addresses 204.210.179.64 through 204.210.179.127. Again, .142 is not in this range. Subnetwork 2 is IP addresses 204.210.179.128 through 204.210.179.191. This range of addresses does include .142, so IP address 204.210.179.142 is on subnetwork 204.210.179.128. Another way of solving for the subnetwork number is to “and” the IP address with the mask. Remember when “anding” that two 1s together make a 1. All other combinations of 1s and 0s make a 0. Subnet Figure #19 shows the “anding” of the IP address 204.210.179.142 with the mask—255.255.255.192 with the result being the subnetwork number.

Subnet Figure #19 1 1 1 1 2 6 3 1 2 6 3 1 2 6 3 1 2 6 3 1 8 4 2 6 8 4 2 1 8 4 2 6 8 4 2 1 8 4 2 6 8 4 2 1 8 4 2 6 8 4 2 1 IP in binary Mask in binary Subnet in binary

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

E-28

Appendix E–Subnetting

CLASS B SUBNETTING Class B IP addresses are handled the same as Class Cs with the exception of how many host bits can be borrowed for subnetting. With Class B IP addresses, the first two octets (16 bits) represent the network number and the last two octets (16 bits) represent host bits. Subnet Figure #20 shows this concept.

Subnet Figure #20 Class B Network No. Octet 1

Hosts

Octet 2

Octet 3

Octet 4

When subnetting, host bits are borrowed to create subnets, and they are always borrowed from the left-most host bits. Subnet Figure #21 shows a Class B IP address with three bits borrowed.

Subnet Figure #21 Class B Network No. Octet Octet 1 2

Subnets

Hosts Octet 3

1 2 8

64 32 16

8

Octet 4 4

2

1

1 2 8

64 32 16

8

4

2

1

The number of subnets is still found using the formula 2x = total number of subnets where x is the number of bits borrowed. The useable subnets is found by subtracting 2 from the result just like it is done with Class C addresses. Look back to Subnet Figure #21. When 3 bits are borrowed, there are eight subnets and six useable subnets possible because 23 = 8 and 8 – 2 = 6. To determine the total number of hosts, the same formula used when working with Class C addresses is applied: 2x = number of hosts where x is the number of host bits remaining after bits have been borrowed for subnetting. Subtract the number 2 from this result to obtain the useable number of hosts per subnetwork. Look back to Subnet Figure #21. With three bits borrowed to perform subnetting, 13 bits remain. 213 = 8,192, which is the total number of hosts. To find the useable number of hosts per subnetwork, subtract 2 from 8,192. 8,192 – 2 = 8,190. So, 8,190 hosts are allowed on each subnet when borrowing three bits from a Class B address.

Solving IP Subnetting Problems

E-29

The normal subnet mask used with Class B IP addresses is 255.255.0.0. When implementing subnets, the third (and possible fourth) octet number changes. The mask is found by looking at the octet(s) where host bits are borrowed and adding the bit values together. Subnet Figure #22 shows how the mask is obtained when borrowing three bits from a Class B address.

Subnet Figure #22 Mask

Class B Network No. Octet Octet 1 2

䊴 255

255

Subnets

Hosts Octet 3

1 2 8 1

6 4

3 2

1

1

1 6

8

Octet 4 4

2

1

1 2 8

6 4

3 2

1 6

8

4

2

1

When three bits are borrowed, the mask has 1s set in the first three bits of the third octet. These bit positions are 128, 64, and 32. Add these bit values together to get 224 (128 + 64 + 32 = 224). The mask for a Class B network with three bits borrowed is 255.255.224.0. Now, for some more practice. The best way to learn Class B IP address subnetting is to practice, practice, practice. Class B Problem 1 The PDQ, Inc. company has ten different networks located throughout the country, but has leased only one Class B address, 180.10.0.0. One option for the company is to divide the Class B address into subnetworks. The network administrator must determine how many bits to borrow by looking at how many subnets the company needs. Since the company has ten networks, a minimum of four bits must be borrowed. 24 = 16 and 16 – 2 = 14 useable subnet numbers. (If three bits were borrowed, there would not be enough subnets because 23 = 8 and 8 – 2 = 6.) The network administrator has heard that a merger is imminent, so, playing it safe, she decides to borrow five bits for subnetting. Borrowing five bits allows for 30 subnetworks because 25 = 32 and 32 – 2 = 30. Subnet Figure #23 shows how the mask is determined.

E-30

Appendix E–Subnetting

Subnet Figure #23 Mask

Class B Network No. Octet Octet 1 2

䊴 255

255

Subnets

Hosts

Octet 3 1 2 8 1

6 4

3 2

1 6

8

1

1

1

1

Octet 4 4

2

1

1 2 8

6 4

3 2

1 6

8

4

2

1

1s are placed in all subnetwork bit positions and those bit values are added together: 128 + 64 + 32 + 16 + 8 = 248. The mask used in the PDQ, Inc. network is 255.255.248.0. The chart below shows the information gathered thus far:

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value B 180.10.0.0 255.255.248.0 32 ?? ?? ?? 5 ?? 30

Solving IP Subnetting Problems

E-31

The number of hosts can be determined by looking at how many host bits remain. Look back to Subnet Figure #13 and see that the number of remaining host bits is 11. 211 = 2,048 total number of hosts for each subnet. Subtract 2 to obtain the number of useable hosts on each subnet: 2,048 – 2 = 2,046 useable hosts on each subnet. The updated chart shows the following:

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per subnetwork Useable subnets

Value B 180.10.0.0 255.255.248.0 32 2,048 ?? ?? 5 2,046 30

Now for the real work—determining subnet numbers and broadcast addresses. Subnet Figure #24 shows the 1 and 0 patterns for determining subnetwork numbers. Due to the lack of space, not all subnetworks are shown, but enough subnets are shown to illustrate the pattern.

E-32

Appendix E–Subnetting

Subnet Figure #24 Class B Network No. Octet Octet 1 2

180 180 180 180 180 180 180 180 180 180 180 180 180 180 180 180 180 180 180 180

10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

Subnets

Hosts

Octet 3 1 2 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

1

6 4

3 2

1 6

8

4

2

1

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

0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 • • 1

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

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

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 2 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1

1

0

0

0

0

1

Subnet No.

Octet 4 6 4

3 2

1 6

8

4

2

1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0

0

0

0

0

0

0

0.0 8.0 16.0 24.0 32.0 40.0 48.0 56.0 64.0 72.0 80.0 88.0 96.0 104.0 112.0 120.0 128.0 • • 248.0

Notice how the subnetwork numbers are in increments of eight. When doing Class A and B subnetting, there is normally not enough time to write every combination of 1s and 0s. When first learning subnetting, you should definitely write out a few, but once you see the pattern emerge, you should do the first couple of subnets and the last subnet. Updating the chart with the subnetwork numbers yields the following:

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork

Subnet numbers

Broadcast addresses No. of bits borrowed Useable hosts per subnetwork Useable subnets

E-33

Value B 180.10.0.0 255.255.248.0 32 2,048 180.10.0.0, 180.10.8.0, 180.10.16.0, 180.10.24.0, 180.10.32.0, 180.10.40.0, 180.10.48.0, 180.10.56.0, 180.10.64,0, 180.10.72.0, 180.10.80.0, 180.10.88.0, 180.10.96.0, 180.10.104.0, 180.10.112.0, 180.10.120.0, 180.10.128.0, 180.10.136.0, 180.10.144.0, 180.10.152.0, 180.10.160.0, 180.10.168.0, 180.10.176.0, 180.10.184.0, 180.10.192.0, 180.10.200.0, 180.10.208.0, 180.10.216.0, 180.10.224.0, 180.10.232.0, 180.10.240.0, 180.10.248.0 ?? 5 2,046 30

All that is left to find in the chart is the broadcast address for each subnetwork. This is found by placing 1s in all of the host bits after the subnetworks are found. This, too, will show an emerging pattern as Subnet Figure #25 illustrates:

E-34

Appendix E–Subnetting

Subnet Figure #25 Class B Network No. Octet Octet 1 2

Subnet

Hosts

Octet 3

Octet 4

Broadcast

180

10

1 1 6 3 1 6 3 1 2 8 4 2 1 2 8 4 2 1 4 2 6 4 2 6 8 8 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1

180

10

0

0

0

0

1

1

1

1

1

1

1

1

1

1

1 1

15.255

180

10

0

0

0

1

0

1

1

1

1

1

1

1

1

1

1 1

23.255

180

10

0

0

0

1

1

1

1

1

1

1

1

1

1

1

1 1

31.255

180

10

0

0

1

0

0

1

1

1

1

1

1

1

1

1

1 1

39.255

180

10

0

0

1

0

1

1

1

1

1

1

1

1

1

1

1 1

47.255

180

10

0

0

1

1

0

1

1

1

1

1

1

1

1

1

1 1

55.255

180

10

0

0

1

1

1

1

1

1

1

1

1

1

1

1

1 1

63.255

180

10

0

1

0

0

0

1

1

1

1

1

1

1

1

1

1 1

71.255

180

10

0

1

0

0

1

1

1

1

1

1

1

1

1

1

1 1

79.255

180

10

0

1

0

1

0

1

1

1

1

1

1

1

1

1

1 1

87.255

180

10

0

1

0

1

1

1

1

1

1

1

1

1

1

1

1 1

95.255

180

10

0

1

1

0

0

1

1

1

1

1

1

1

1

1

1 1

103.255

180

10

0

1

1

0

1

1

1

1

1

1

1

1

1

1

1 1

111.255

180

10

0

1

1

1

0

1

1

1

1

1

1

1

1

1

1 1

119.255

180

10

0

1

1

1

1

1

1

1

1

1

1

1

1

1

1 1

127.255

180

10

1

0

0

0

0

1

1

1

1

1

1

1

1

1

1 1

135.255

180

10

•

1

1

1

1

1

1

1

1

1

1 1

•

180

10

•

1

1

1

1

1

1

1

1

1

1 1

•

180

10

1

1

1

1

1

1

1

1

1

1 1

255.255

0

1

1

1

1

1

7.255

Do not forget to look at the entire octet when determining the decimal value for the octet.

Solving IP Subnetting Problems

E-35

The pattern that emerges is that the third octet increments by eight each time and the fourth octet is always 255. Also notice that the broadcast number is one less than the subnetwork number that follows. The completed chart lists below:

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork

Subnet numbers

Broadcast addresses

No. of bits borrowed Useable hosts per subnetwork Useable subnets

Value B 180.10.0.0 255.255.248.0 32 2,048 180.10.0.0, 180.10.8.0, 180.10.16.0, 180.10.24.0, 180.10.32.0, 180.10.40.0, 180.10.48.0, 180.10.56.0, 180.10.64,0, 180.10.72.0, 180.10.80.0, 180.10.88.0, 180.10.96.0, 180.10.104.0, 180.10.112.0, 180.10.120.0, 180.10.128.0, 180.10.136.0, 180.10.144.0, 180.10.152.0, 180.10.160.0, 180.10.168.0, 180.10.176.0, 180.10.184.0, 180.10.192.0, 180.10.200.0, 180.10.208.0, 180.10.216.0, 180.10.224.0, 180.10.232.0, 180.10.240.0, 180.10.248.0 180.10.7.255, 180.10.15.255, 180.10.31.255, 180.10.39.255, 180.10.47.255, 180.10.55.255, 180.10.63.255, 180.10.71.255, 180.10.79.255, 180.10.87.255, 180.10.95.255, 180.10.103.255, 180.10.111.255, 180.10.119.255, 180.10.127.255, 180.10.135.255, 180.10.143.255, 180.10.151.255, 180.10.167.255, 180.10.175.255, 180.10.183.255, 180.10.191.255, 180.10.199.255, 180.10.207.255, 180.10.215.255, 180.10.223.255, 180.10.231.255, 180.10.239.255, 180.10.255.255 5 2,046 30

E-36

Appendix E–Subnetting

Class B Problem 2 One of the hardest concepts for students to grasp is when borrowed host bits are in more than one octet. In the next scenario, the Top Hats Co. has 2,000 locations throughout the world. Each location has a network with approximately 20 computers. The Top Hats Co. has leased the Class B IP address of 189.208.0.0. The first step in solving this problem is determining how many bits to borrow. Subnet Table #1 helps with this decision:

Subnet Table #1 Borrowed Bits 5 6 7 8 9 10 11 12 13

No. of Host Bits 11 10 9 8 7 6 5 4 3

Total Subnets 32 34 128 256 512 1024 2048 4096 8192

Useable Subnets 30 32 126 254 510 1022 2046 4094 8190

Total Hosts 2048 1024 512 256 128 64 32 16 8

Useable Hosts 2046 1022 510 254 126 62 30 14 6

Looking at Subnet Table #1, you can see that borrowing 11 bits allows for the 2,000 Top Hats Co.’s locations throughout the world. By borrowing 11 bits, there are also enough remaining host bits to accommodate the computers at each site. The following chart summarizes the information gathered so far:

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per subnetwork Useable subnets

Value B 189.208.0.0 ?? 2,048 32 ?? ?? 11 30 2,046

Solving IP Subnetting Problems

E-37

To determine what mask is needed throughout the Top Hats Co.’s network, place 1s in the subnetwork field and add the bit value positions together for each octet. Subnet Figure #26 shows this concept:

Subnet Figure #26 Mask

Class B Network No. Octet Octet 1 2

䊴 255

255

Subnets

Hosts

Octet 3

Octet 4

1 2 8

64 32 16

8

4

2

1

1 2 8

64 32 16

1

1

1

1

1

1

1

1

1

1

8

4

2

1

1

The first two octets are the standard 255.255 numbers. The third octet is filled with 1’s, so the third octet mask is 255—(128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255). The fourth octet has three bits set, so the mask is 224—(128 + 64 + 32 = 224). The final mask for this problem is 255.255.255.244, and the information can be inserted into the chart:

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per subnetwork Useable subnets

Value B 189.208.0.0 255.255.255.224 2,048 32 ?? ?? 11 30 2,046

Now, the subnetworks must be determined. Subnet Figure #27 shows the breakdown of the subnets with 1s and 0s.

0

Keep in mind when putting 1s and 0s in multiple subnetwork octets that you must treat them as one big group.

E-38

Appendix E–Subnetting

Subnet Figure #27 Class B Network No. Octet Octet 1 2

189 189 189 189 189 189 189 189 189 189 189 189 189 189 189 189 189 189 189 189 189

208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208

Subnets

Hosts

Octet 3 1 2 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1

Octet 4

6 4

3 2

1 6

8

4

2

1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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

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

1

1

1

1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 • • 1

1 2 8 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0

1

1

1

Subnets

6 4

3 2

1 6

8

4

2

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

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

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1

1

0

0

0

0

0.0 0.32 0.64 0.96 0.128 0.160 0.192 0.224 1.0 1.32 1.64 1.96 1.128 1.160 1.192 1.224 2.0 2.32 • • 255.224

Updating the chart could take several pages with these subnetworks, but entering in a few of them shows the following:

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork

Subnet numbers

Broadcast addresses No. of bits borrowed Useable hosts per subnetwork Useable subnets

E-39

Value B 189.208.0.0 255.255.255.224 2,048 32 189.208.0.0, 189.208.0.32, 189.208.0.64, 189.208.0.96, 189.208.0.128, 189.208.0.160, 189.208.0.192, 189.208.0.224, 189.208.1.0, 189.208.1.32, …189.208.1.224, 189.208.2.0, 189.208.2.32, 189.208.2.64, 189.208.2.96, 189.208.2.128, . . . 189.208.255.224 ?? 11 30 2,046

The last bit of information left to find is the broadcast numbers. Simply put 1s in the host bits for each subnetwork. Subnet Figure #28 shows this process.

E-40

Appendix E–Subnetting

Subnet Figure #28 Class B Network No. Octet Octet 1 2

189 189 189 189 189 189 189 189 189 189 189 189

208 208 208 208 208 208 208 208 208 208 208 208

Subnets

Hosts

Octet 3 1 2 8 0 0 0 0

Octet 4

6 4

3 2

1 6

8

4

2

1

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 1 1

1

1

1

1

1

1

1

0 0 0 0 • • 1 1 0 0 • 1

1 2 8 0 0 0 0

Subnets

6 4

3 2

1 6

8

4

2

0 0 1 1

0 1 0 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 0 0

1 1 0 0

0 1 0 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1

1

1

1

1

1

1

0.31 0.63 0.95 0.127 • • 1.223 1.255 2.31 2.63 • 255.255

Of course updating the list with the broadcasts is quite lengthy too, but some have been inserted into the chart to illustrate the point.

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork

Subnet numbers

Broadcast addresses

No. of bits borrowed Useable hosts per subnetwork Useable subnets

E-41

Value B 189.208.0.0 255.255.255.224 2,048 32 189.208.0.0, 189.208.0.32, 189.208.0.64, 189.208.0.96, 189.208.0.128, 189.208.0.160, 189.208.0.192, 189.208.0.224, 189.208.1.0, 189.208.1.32, …189.208.1.224, 189.208.2.0, 189.208.2.32, 189.208.2.64, 189.208.2.96, 189.208.2.128, . . . 189.208.255.224 189.208.0.31, 189.208.0.63, 189.208.0.95, 189.208.0.127, …189.208.1.223, 189.208.1.255, 189.208.2.31, 189.208.2.63, 189.208.2.95, 189.208.2.127 …189.208.254.63, 189.208.254.95, 189.208.254.127, 189.208.254.159, 189.208.254.191, 189.208. 254.255, 189.208.255.63, 189.208.255.95, 189.208.255.127, 189.208.255.159, 189.208.255.191, 189.208.255.255 11 30 2,046

Of course, each of the broadcast addresses is in groups of 32 just like the subnetworks are. Do not forget to treat each octet as a group of eight bits when determining subnetwork numbers and broadcast addresses!

E-42

Appendix E–Subnetting

Class B Problem 3 The network administrator is working on a computer with the IP address of 157.208.190.144. The mask shows as 255.255.255.192. On what subnet is the computer attached? The information known so far is summarized in the following table:

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

Value B 157.208.0.0 255.255.255.192 ?? ?? ?? ?? ?? ?? ??

The first step is to determine how many host bits have been borrowed for subnetting. Subnet Figure #29 shows the mask (the borrowed bits).

Subnet Figure #29 Mask

Class B Network No. Octet Octet 1 2

䊴 255

255

Subnets

Hosts

Octet 3 1 2 8 1

Octet 4

6 4

3 2

1 6

8

4

2

1

1

1

1

1

1

1

1

1 2 8 1

6 4

3 2

1 6

8

4

2

1

1

As seen in Subnet Figure #29, there are ten borrowed host bits. Knowing this information, the total number of subnets, useable subnets, total number of hosts, and useable hosts can be determined by using the formula 2x (where x is either the number of bits borrowed or the remaining bits) and then entered into the chart as shown below.

Solving IP Subnetting Problems

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per network/subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per network/subnetwork Useable subnets

E-43

Value B 157.208.0.0 255.255.255.192 1,024 64 ?? ?? 10 62 1,022

The next thing is to determine the subnetwork numbers. Subnet Figure #30 shows a partial illustration of subnetwork numbers for this problem.

E-44

Appendix E–Subnetting

Subnet Figure #30 Class B Network No. Octet Octet 1 2

157 157 157 157 157 157 157 157 157 157 157 157 157 157 157 157 157 157 157 157

208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208 208

Subnets

Hosts

Octet 3 1 2 8 0 0 0 0 0 0 0

Octet 4

6 4

3 2

1 6

8

4

2

1

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

0 0 0 0 0 0 1 1

1

1

1

1

1

0 0 0 0 0 0 0 • • 1 1 1 1 1 1 1 1 • • 1

1 2 8 0 0 0 1 1 1 1

1

1

Subnets

6 4

3 2

1 6

8

4

2

0 1 1 0 0 1 1

1 0 1 0 1 0 1

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 1 1 1 1 0 0

1 1 0 0 1 1 0 0

0 1 0 1 0 1 0 1

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

1

1

1

0

0

0

0

0.32 0.64 0.96 0.128 0.160 0.192 0.224 • • 190.64 190.96 190.128 190.160 190.192 190.224 191.0 191.32 • • 255.224

The answer to the problem is actually in Subnet Figure #30. The subnetwork numbers shown are the beginning address (the number for “the wire”) for the subnet. For example, subnetwork 157.208.190.0 has addresses that extend from 157.208.190.0 through 157.208.190.31. Subnetwork 157.208.190.32 has addresses that extend from 157.208.190.32 through 157.208.190.63. The computer in this example has an IP address of 157.208.190.144. The solution is found by looking at a subnetwork number that is the smallest number below 190.144. The answer is subnetwork 157.208.190.128. Since the solution is solved, there is no need to finish the chart or determine broadcast addresses.

Solving IP Subnetting Problems

E-45

However, to determine the broadcast address for each subnetwork, the same method is used as shown before—put 1s in all of the host addresses and determine the decimal value for the octet. A shortcut for solving a problem that gives an IP address and a mask and asks for the subnetwork number is to put the IP address and mask in binary and “and” the two numbers together. Then convert the result to dotted decimal notation. Remember when “anding,” the only way to get a 1 is by “anding” two 1s together. Subnet Figure #31 shows this process.

0

Subnet Figure #31 1 1 1 1 2 6 3 1 2 6 3 1 2 6 3 1 2 6 3 1 8 4 2 6 8 4 2 1 8 4 2 6 8 4 2 1 8 4 2 6 8 4 2 1 8 4 2 6 8 4 2 1 IP in binary Mask in binary Subnet in binary

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

Subnet in dotted decimal notation: 157.208.190.128 CLASS A SUBNETTING Class A subnetting is handled the same as Class Bs and Cs with the exception of how many host bits can be borrowed for subnetting. With Class A IP addresses, the first octet (eight bits) represents the network number and the last three octets (24 bits) represent host bits. Subnet Figure #32 shows this concept.

Subnet Figure #32 Class A Network No. Octet 1

Hosts Octet 2

Octet 3

Octet 4

When subnetting Class A IP addresses, bits are borrowed from the left-most host bits and can extend across octets 2, 3, and 4 because these are the Class A host bits. Subnet Figure #33 shows a Class A IP address with eleven bits borrowed from the first and second octets.

E-46

Appendix C–Subnetting

Subnet Figure #33 Class A Network No. Octet 1

Hosts Octet 2

Octet 3

Octet 4

1 1 1 6 3 1 6 3 1 6 3 1 2 8 4 2 1 2 8 4 2 1 2 8 4 2 1 4 2 6 4 2 6 4 2 6 8 8 8

With Class A subnets, the same formula, 2x = total number of subnets (where x is the number of bits borrowed) is still used. The useable subnets is found by subtracting 2 from the result, just like it was done with Class B and Class C subnets. In Subnet Figure #33, 11 bits are borrowed from octets 2 and 3. 211 = 2,048 total subnets and subtracting 2 yields the useable subnets—2,048 – 2 = 2,046. The same formula is also used for determining total number of hosts. In Subnet Figure #33, 13 host bits remain. 213 = 8,192 total host addresses. Subtracting 2 yields the useable host addresses—8,192 – 2 = 8,190. Keep consistent in how you solve IP subnetting problems and no exam can trip you up. The normal subnet mask used with Class A IP addresses is 255.0.0.0. When implementing subnets, the second, third, and fourth octets can be used and therefore the mask changes for these octets. Subnet Figure #34 shows how the mask is obtained when borrowing 11 bits from a Class A address.

Subnet Figure #34 Mask

Class A Network No. Octet 1

䊴 255

Hosts Octet 2 1 6 3 1 2 8 4 2 1 4 2 6 8 1 1 1 1 1 1 1 1

Octet 3

Octet 4

1 1 6 3 1 6 3 1 2 8 4 2 1 2 8 4 2 1 4 2 6 4 2 6 8 8 1 1 1

Solving IP Subnetting Problems

E-47

When borrowing 11 bits, the mask has 1s set in the second octet and the first three bits of the third octet. Octet 2 is all 1s, so the mask for octet 2 is 255. Octet 3 has 1s set in the first three bit positions. Add these bit values together to get 224. (128 + 64 + 32 = 224). The mask for a Class A network with 11 bits borrowed is 255.255.224.0. The best way to learn Class A addresses (as it has been for the other classes) is to practice. They are done the exact same way as the other addresses except there are more host bits from which to borrow. Class A Problem 1 The Super Duper Company has 5,000 locations worldwide. In each location, there are more than 1000 computers. The Super Duper Company has leased one Class A IP address, 19.0.0.0. One option for the company is to divide the class A address into subnetworks. The first task for the network administrator is to determine how many bits to borrow for subnetting. Subnet Table #2 summarizes some borrowed bits with corresponding number of subnets.

Subnet Table #2 Borrowed Bits 5 6 7 8 9 10 11 12 13 14 15 16 17

No. of Host Bits 19 18 17 16 15 14 13 12 11 10 9 8 7

Total Subnets 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072

Useable Subnets 30 62 126 254 510 1022 2046 4094 8190 16382 32766 65534 131070

Total Hosts 524288 262144 131072 65536 32768 16384 8192 4096 2048 1024 512 256 128

Useable Hosts 524286 262142 131070 65534 32766 16382 8190 4094 2046 1022 510 254 126

Looking at Subnet Table #2, one can see that, to assign subnetwork numbers to 5,000 locations, the Super Duper Company must borrow 13 host bits. This also allows for 2,046 host addresses per location. Subnet Figure #35 shows how the subnet mask is determined using the 13 host bits for subnetworks.

E-48

Appendix E–Subnetting

Subnet Figure #35 Mask

Class A Network No. Octet 1

䊴255

Hosts Octet 2 1 6 3 1 2 8 4 2 1 4 2 6 8 1 1 1 1 1 1 1 1

Octet 3

Octet 4

1 1 6 3 1 6 3 1 2 8 4 2 1 2 8 4 2 1 4 2 6 4 2 6 8 8 1 1 1 1 1

With 13 bits borrowed, the mask has the standard 1s in the first octet (255), 1s in the second octet (255), and 1s in five bits of the third octet (248—128 + 64 + 32 + 16 + 8). So, the mask for the Super Duper Company is 255.255.248.0. The chart below shows the information determined so far.

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork Subnet numbers Broadcast addresses No. of bits borrowed Useable hosts per subnetwork Useable subnets

Value A 19.0.0.0 255.255.248.0 8,192 2,048 ?? ?? 11 2,046 8,190

Determining subnetwork numbers is done exactly the same way as when Class B and Class C addresses are subnetted. Subnet Figure #36 shows a partial view of the Class A subnetworks. Octet 1 is not divided into bit positions because it is always 19. Octet 4 is not subdivided because it always contains 0s for the subnetwork number. For presentation, only Octets 2 and 3 are shown where the subnetting occurs; Octet 4 is not shown.

Solving IP Subnetting Problems

E-49

Subnet Figure #36 Mask

Class A Network No. Octet 1

䊴

19 19 19 19 19 19 19 19 19 19 19 19

Hosts Octet 2 1 2 8 0 0 0 0

0 0 0 0 1

1 6 3 1 6 8 4 2 1 2 4 2 6 4 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 • • 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 • 1 1 1 1 1 1 1 1 1

Octet 3

Subnets

3 1 8 4 2 1 2 6 0 0 0 0

0 0 1 1

0 1 0 1

1 0 0 0

1 0 0 1

1 0 1 0

1 1 1

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

19.0.0.0 19.0.8.0 19.0.16.0 19.0.24.0 • • 19.0.248.0 19.1.0.0 19.1.8.0 19.1.16.0 • 19.255.248.0

As you can see in Subnet Figure #36, there are 13 borrowed host bits. Subnetwork numbers increment in groups of eight (19.0.0.0, 19.0.8.0, 19.0.16.0, and so on up to 19.255.248.0). A few of the subnetwork numbers are filled into the chart so you can see the trend. To determine the broadcast address for each subnetwork, the same method used with Class C and Class B addresses is used—put all 1s in the host address and determine the decimal value (or take the shortcut and subtract 1 from the next subnetwork number). The final chart is as follows:

E-50

Appendix E–Subnetting

Information Class of IP address Network Number Mask Total no. of subnetworks Total no. of hosts per subnetwork Subnet numbers

Broadcast addresses

No. of bits borrowed Useable hosts per subnetwork Useable subnets

Value A 19.0.0.0 255.255.248.0 8,192 2,048 19.0.8.0, 19.0.16.0, 19.0.24.0, 19.0.32.0 through 19.0.248.0, 19.1.0.0, 19.1.8.0, 19.1.16.0 through 19.1.248, 19.2.0.0 through 19.255.248.0 19.0.15.255, 19.0.23.255, 19.0.31.255, 19.0.39.255 etc. through 19.0.255.255, 19.1.7.255, 19.1.15.255, 19.1.31.255 etc. through 19.1.255.255, 19.2.7.255 through 19.255.255.255 11 2,046 8,190

Writing all of the subnet numbers and broadcast addresses would take up page after page of this text, so enough numbers are inserted for you to get the idea of the pattern. Once you do a couple of numbers and see the patterns, you can determine all of the subnet numbers.

Subnetting Review Questions

E-51

Name _______________________________

SUBNETTING REVIEW QUESTIONS 1. A company has received a Class C IP address for their four networks. How many bits need to be borrowed? 2. A company uses a Class C mask of 255.255.255.224. What is the maximum number of hosts per subnetwork? 3. How many bits are borrowed in a Class C address if the mask is 255.255.255.240? 4. Given the IP address 199.14.180.4, what class IP address is this? 5. Given the IP address of 201.60.250.91 and a mask of 255.255.255.248, what is the subnetwork number? 6. Given the IP address 210.199.184.66 and the fact that a company borrows three bits to subnet, what mask is used? 7. What is the standard subnet mask for a Class C address? 8. What is the maximum number of bits that can be borrowed when using a Class C address? 9. What is the minimum number of bits that can be borrowed when using a Class C address? 10. Given the IP address 204.16.8.0 and a mask of 255.255.255.240. What is the first useable subnetwork number? 11. Given the IP address 197.56.2.141 and a mask of 255.255.255.192, what is the broadcast address for this subnetwork?

E-52

Appendix E–Subnetting 12. Given the broadcast address of 202.202.159.159 and a mask of 255.255.255.248, what is the subnetwork number? 13. A company has a policy of only 25 hosts per subnet. They have 20 networks. How many Class C addresses does the company need? 14. How many bits are set in a standard Class C mask? 15. What is the maximum number of hosts on a Class C network? 16. Given the mask of 255.255.255.224 and an IP address of 200.200.200.200, on what subnetwork is the device? 17. Given an IP address of 193.15.10.105 and a mask of 255.255.255.252, what is the subnetwork number? 18. Given the mask of 255.255.255.248 and the fact that a Class C address is being used, how many hosts are on each subnet? 19. Given the IP address of 206.19.1.186 and a mask of 255.255.255.192, what are the two unuseable subnets? 20. Given the IP address of 199.199.144.43 and a mask of 255.255.255.224, what is the last useable subnetwork number? 21. Given the IP address 130.14.207.39 and a mask of 255.255.240.0, how many total subnets are available? 22. Given the IP address 130.14.207.39 and a mask of 255.255.240.0, what is the subnetwork number associated with this IP address? 23. Given the IP address 188.188.188.188 and a mask of 255.255.255.128, what is the subnetwork number associated with this IP address? 24. Given the IP address 191.10.59.63 and with six bits borrowed, what is the mask?

Subnetting Review Questions

E-53

25. A company has a Class B IP address. What is the maximum number of bits that can be borrowed and still have 100 hosts per subnetwork? 26. A company is leasing a Class A IP address and has 3000 networks. How many bits do they need to borrow? 27. Given the IP address 15.200.166.41 and a mask of 255.252.0.0, what is the subnetwork number? 28. What is the mask when 15 bits are borrowed and a Class A network address is being used? 29. Given the IP address 14.168.29.180 and a mask of 255.255.192.0, how many bits are borrowed for subnetting? 30. Given the IP address 14.168.29.180 and a mask of 255.255.192.0, what is the broadcast for this subnetwork? 31. How many bits are set with a Class A subnet mask of 255.255.240.0? 32. Given the IP address 120.150.150.150 and a mask of 255.255.240.0, what is the subnetwork number and broadcast address?

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Appendix E–Subnetting

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