Cisco CCENT Practice and Study Guide: Subnetting IP Networks

Date: Feb 25, 2014

This chapter reviews the process of subnetting IP networks. First, by reviewing a process for subnetting IPv4 networks, then, by practicing subnetting skills, including several scenarios.

Understanding the hierarchical structure of the IP address and how to modify that hierarchy in order to more efficiently meet routing requirements is an important part of planning an IP addressing scheme. This chapter reviews the process of subnetting IP networks. First, we review a process for subnetting IPv4 networks. Then, you practice subnetting skills, including several scenarios. Then, we briefly review subnetting IPv6 networks.

Subnetting an IPv4 Network

The process of segmenting a network, by dividing it into multiple smaller network spaces, is called subnetting. These subnetworks are called subnets. Although subnetting calculators are plentiful and freely accessible on the Internet, you must know how to subnet without using a calculator when you sit for the CCENT exam. Furthermore, subnetting skill will serve you well when troubleshooting common IP addressing issues.

Subnetting in Four Steps

Everyone has a preferred method of subnetting. Each teacher will use a slightly different strategy to help students master this crucial skill. The method I prefer can be broken down into four steps:

Step 1. Determine how many bits to borrow based on the network requirements.

Step 2. Determine the new subnet mask.

Step 3. Determine the subnet multiplier.

Step 4. List the subnets, including subnetwork address, host range, and broadcast address.

Subnetting Example

The best way to demonstrate the four steps of subnetting is to use an example. Let’s assume that you are given the network address 192.168.1.0/24, you need 30 hosts per network, and want to create as many subnets as possible.

Determine How Many Bits to Borrow

Because our requirement specifies 30 host addresses per subnet, we need to first determine the minimum number of host bits to leave. The remaining bits can be borrowed:

Host Bits = Bits Borrowed + Bits Left

To provide enough address space for 30 hosts, we need to leave 5 bits. Use the following formula:

2^BL – 2 = number of host addresses

where the exponent BL is bits left in the host portion.

Remember, the “minus 2” is to account for the network and broadcast addresses that cannot be assigned to hosts.

In this example, leaving 5 bits in the host portion will provide the right number of host address:

2⁵ – 2 = 30

Because we have 3 bits remaining in the original host portion, we borrow all these bits to satisfy the requirement to “create as many subnets as possible.” To determine how many subnets we can create, use the following formula:

2^BB = Number of subnets

where the exponent BB is bits borrowed from the host portion.

In this example, borrowing 3 bits from the host portion will create 8 subnets: 2³ = 8.

As shown in Table 9-1, the 3 bits are borrowed from the far-left bits in the host portion. The highlighted bits in the table show all possible combinations of manipulating the 8 bits borrowed to create the subnets.

Table 9-1 Binary and Decimal Value of the Subnetted Octet

Subnet Number	Last Octet Binary Value	Last Octet Decimal Value
0	00000000	.0
1	00100000	.32
2	01000000	.64
3	01100000	.96
4	10000000	.128
5	10100000	.160
6	11000000	.192
7	11100000	.224

Determine the New Subnet Mask

Notice in Table 9-1 that the network bits now include the 3 borrowed host bits in the last octet. Add these 3 bits to the 24 bits in the original subnet mask and you have a new subnet mask, /27. In decimal format, you turn on the 128, 64, and 32 bits in the last octet for a value of 224. So, the new subnet mask is 255.255.255.224.

Determine the Subnet Multiplier

Notice in Table 9-1 that the last octet decimal value increments by 32 with each subnet number. The number 32 is the subnet multiplier. You can quickly find the subnet multiplier using one of two methods:

Method 1: Subtract the last nonzero octet of the subnet mask from 256. In this example, the last nonzero octet is 224. So, the subnet multiplier is 256 – 224 = 32.
Method 2: The decimal value of the last bit borrowed is the subnet multiplier. In this example, we borrowed the 128 bit, the 64 bit, and the 32 bit. The 32 bit is the last bit we borrowed and is, therefore, the subnet multiplier.

By using the subnet multiplier, you no longer have to convert binary subnet bits to decimal.

List the Subnets, Host Ranges, and Broadcast Addresses

Listing the subnets, host ranges, and broadcast addresses helps you see the flow of addresses within one address space. Table 9-2 documents our subnet addressing scheme for the 192.168.1.0/24 address space. Fill in any missing information.

Table 9-2 Subnet Addressing Scheme for 192.168.1.0/24: 30 Hosts Per Subnet

Subnet Number	Subnet Address	Host Range	Broadcast Address
0	192.168.1.0	192.168.1.1–192.168.1.30	192.168.1.31
1	192.168.1.32	192.168.1.33–192.168.1.62
2		192.168.1.65–192.168.1.94	192.168.1.95
3	192.168.1.96
4	192.168.1.128	192.168.1.129–192.168.1.158	192.168.1.159
5		192.168.1.161–192.168.1.190	192.168.1.191
6	192.168.1.192
7

Use the four subnetting steps to complete the following scenarios.

Subnetting Scenario 1

Subnet the address space 10.10.0.0/16 to provide at least 100 host addresses per subnet while creating as many subnets as possible.

How many bits should your borrow?
What is the new subnet mask in dotted-decimal and prefix notation?
What is the subnet multiplier?

In Table 9-3, list the first three subnets, host ranges, and broadcast addresses.

Table 9-3 Subnet Addressing Scheme for Scenario 1

Subnet Number	Subnet Address	Host Range	Broadcast Address
0
1
2

Subnetting Scenario 2

Subnet the address space 10.10.0.0/16 to provide at least 500 subnet addresses.

How many bits should your borrow?
What is the new subnet mask in dotted-decimal and prefix notation?
What is the subnet multiplier?

In Table 9-4, list the first three subnets, host ranges, and broadcast addresses.

Table 9-4 Subnet Addressing Scheme for Scenario 2

Subnet Number	Subnet Address	Host Range	Broadcast Address
0
1
2

Subnetting Scenario 3

Subnet the address space 10.10.10.0/23 to provide at least 60 host addresses per subnet while creating as many subnets as possible.

How many bits should your borrow?
What is the new subnet mask in dotted-decimal and prefix notation?
What is the subnet multiplier?

In Table 9-5, list the first three subnets, host ranges, and broadcast addresses.

Table 9-5 Subnet Addressing Scheme for Example 3

Subnet Number	Subnet Address	Host Range	Broadcast Address
0
1
2

VLSM Addressing Schemes

Variable-length subnet masking (VLSM) subnetting is similar to traditional subnetting in that bits are borrowed to create subnets. The formulas to calculate the number of hosts per subnet, and the number of subnets created still apply. The difference is that subnetting is not a single-pass activity.

VLSM Review

You probably noticed that the starting address space in Subnetting Scenario 3 is not an entire classful address. In fact, it is subnet 5 from Subnetting Scenario 2. So in Subnetting Scenario 3, you “subnetted a subnet.” That is what VLSM is in a nutshell: subnetting a subnet.

Let’s use a small example. Given the address space 172.30.4.0/22 and the network requirements shown in Figure 9-1, apply an addressing scheme that conserves the most amount of addresses for future growth.

Figure 9-1 VLSM Example Topology

We need five subnets: four LAN subnets and one WAN subnet. Starting with the largest host requirement on LAN 3, begin subnetting the address space.

To satisfy the 250 hosts requirement, we leave 8 hosts bits (2⁸ – 2 = 254 hosts per subnet). Because we have 10 host bits total, we borrow 2 bits to create the first round of subnets (2² = 4 subnets). The starting subnet mask is /22 or 255.255.252.0. We turn on the next two bits in the subnet mask to get /24 or 255.255.255.0. The multiplier is 1. The four subnets are as follows:

Subnet 0: 172.30.4.0/24
Subnet 1: 172.30.5.0/24
Subnet 2: 172.30.6.0/24
Subnet 3: 172.30.7.0/24

Assigning Subnet 0 to LAN 3, we are left with three /24 subnets. Continuing on to the next largest host requirement on LAN 4, we take Subnet 1, 172.30.5.0/24, and subnet it further.

To satisfy the 100 hosts requirement, we leave 7 bits (2⁷ – 2 = 128 hosts per subnet). Because we have 8 host bits total, we can borrow only 1 bit to create the subnets (2¹ = 2 subnets). The starting subnet mask is /24 or 255.255.255.0. We turn on the next bit in the subnet mask to get /25 or 255.255.255.128. The multiplier is 128. The two subnets are as follows:

Subnet 0: 172.30.5.0/25
Subnet 1: 172.30.5.128/25

Assigning Subnet 0 to LAN 4, we are left with one /25 subnet and two /24 subnets. Continuing on to the next largest host requirement on LAN 1, we take Subnet 1, 172.30.5.128/25, and subnet it further.

To satisfy the 60 hosts requirement, we leave 6 bits (2⁶ – 2 = 62 hosts per subnet). Because we have 7 host bits total, we borrow 1 bit to create the subnets (2¹ = 2 subnets). The starting subnet mask is /25 or 255.255.255.128. We turn on the next bit in the subnet mask to get /26 or 255.255.255.192. The multiplier is 64. The two subnets are as follows:

Subnet 0: 172.30.5.128/26
Subnet 1: 172.30.5.192/26

Assigning Subnet 0 to LAN 1, we are left with one /26 subnet and two /24 subnets. Finishing our LAN subnetting with LAN 2, we take Subnet 1, 172.30.5.192/26, and subnet it further.

To satisfy the 10 hosts requirement, we leave 4 bits (2⁴ – 2 = 14 hosts per subnet). Because we have 6 host bits total, we borrow 2 bits to create the subnets (2² = 4 subnets). The starting subnet mask is /26 or 255.255.255.192. We turn on the next two bits in the subnet mask to get /28 or 255.255.255.240. The multiplier is 16. The four subnets are as follows:

Subnet 0: 172.30.5.192/28
Subnet 1: 172.30.5.208/28
Subnet 2: 172.30.5.224/28
Subnet 3: 172.30.5.240/28

Assigning Subnet 0 to LAN 2, we are left with three /28 subnets and two /24 subnets. To finalize our addressing scheme, we need to create a subnet only for the WAN link, which needs only two host addresses. We take Subnet 1, 172.30.5.208/28, and subnet it further.

To satisfy the two hosts requirement, we leave 2 bits (2² – 2 = 2 hosts per subnet). Because we have 4 host bits total, we borrow 2 bits to create the subnets (2² = 4 subnets). The starting subnet mask is /28 or 255.255.255.240. We turn on the next 2 bits in the subnet mask to get /30 or 255.255.255.252. The multiplier is 4. The four subnets are as follows:

Subnet 0: 172.30.5.208/30
Subnet 1: 172.30.5.212/30
Subnet 2: 172.30.5.216/30
Subnet 3: 172.30.5.220/30

We assign Subnet 0 to the WAN link. We are left with three /30 subnets, two /28 subnets, and two /24 subnets.

VLSM Addressing Design Exercises

In the following VLSM addressing design exercises, you apply your VLSM addressing skills to a three router topology. Each exercise is progressively more difficult than the last. There may be more than one correct answer in some situations. However, you should always practice good addressing design by assigning your subnets contiguously.

Exercise 1

Assume that 4 bits were borrowed from the host portion of 192.168.1.0/24. You are not using VLSM. Starting with Subnet 0, label Figure 9-2 contiguously with subnets. Start with the LAN on RTA and proceed clockwise.

Figure 9-2 Addressing Design Exercise 1 Topology: Subnets

How many total valid host addresses will be wasted on the WAN links?

Now come up with a better addressing scheme using VLSM. Start with the same 4 bits borrowed from the host portion of 192.168.1.0/24. Label each of the LANs with a subnet. Then subnet the next available subnet to provide WAN subnets without wasting any host addresses. Label Figure 9-3 with the subnets.

Figure 9-3 Addressing Design Exercise 1 Topology: VLSM Subnets