6 IPv4 Subnetting Practice

IPv4 subnetting and prefixes

Subnetting divides an IPv4 address block into smaller networks. It helps organize devices, limit broadcast traffic to each subnet, and allocate address space to fit network needs.

An IPv4 address is 32 bits long. CIDR notation states how many leading bits identify the network, and the remaining bits identify addresses within it. CIDR uses explicit prefix lengths rather than relying on address classes.

Subnet masks and address counts

A subnet mask has 32 bits: network bits are 1 and host bits are 0. The prefix length is the number of leading 1 bits. Dotted-decimal masks show those bits in four octets.

Common prefix sizes have these address counts and ordinary usable host counts:

• /24: mask 255.255.255.0; 256 total addresses and 254 usable hosts.

• /25: mask 255.255.255.128; 128 total addresses and 126 usable hosts.

• /26: mask 255.255.255.192; 64 total addresses and 62 usable hosts.

• /27: mask 255.255.255.224; 32 total addresses and 30 usable hosts.

• /28: mask 255.255.255.240; 16 total addresses and 14 usable hosts.

• /29: mask 255.255.255.248; 8 total addresses and 6 usable hosts.

• /30: mask 255.255.255.252; 4 total addresses and 2 usable hosts.

For ordinary IPv4 subnets, usable host addresses exclude the network address and broadcast address. /31 is a special case commonly used on point-to-point links, where both addresses can be assigned to the two endpoints. /32 represents one address.

A prefix can fall within any octet. For example, /20 has mask 255.255.240.0: eight network bits in each of the first two octets and four more in the third, for 20 network bits total.

A method for calculating subnets

For a prefix /p, the host-bit count is $32-p$. The total address count is $2^{32-p}$. For ordinary subnets, the usable host count is usually the total address count minus $2$.

To calculate a subnet:

1. Find the host bits: $32-p$.

2. Find the total addresses: $2^{\text{host bits}}$.

3. Find the usable hosts: usually the total addresses minus $2$.

4. Find the network address by keeping the network bits and setting all host bits to 0. Equivalently, bitwise-AND the IP address with its subnet mask.

5. Find the broadcast address by keeping the network bits and setting all host bits to 1.

6. Find the usable host range: one address after the network address through one before the broadcast address.

When the mask changes within an octet, calculate the block size as $256-\text{mask value}$ in that octet. Subnet network numbers advance by that block size.

Worked subnet calculation

For 192.168.10.77/26, the mask is 255.255.255.192. The last-octet block size is $256-192=64$, so subnet boundaries are .0, .64, .128, and .192. The address .77 falls between .64 and .127, so it belongs to that subnet.

• Network address: 192.168.10.64

• Broadcast address: 192.168.10.127

• Usable host range: 192.168.10.65–192.168.10.126

• Total addresses: $2^6=64$

• Usable hosts: $64-2=62$

The IP address identifies a device within the block; the prefix determines which block it belongs to.

Planning subnet sizes

Choose a prefix that provides enough host addresses for the devices on a subnet, allowing room for growth. For example, 25 devices need at least 27 addresses under the ordinary network-and-broadcast rule. A /27 provides 32 total addresses and 30 usable host addresses, so it fits.

A /24 can be divided into eight /27 subnets because /27 has three more network bits than /24. The number of equal subnets is $2^3=8$. For 10.20.30.0/24, the subnet network addresses are:

• 10.20.30.0/27

• 10.20.30.32/27

• 10.20.30.64/27

• 10.20.30.96/27

• 10.20.30.128/27

• 10.20.30.160/27

• 10.20.30.192/27

• 10.20.30.224/27

Private IPv4 ranges often used for internal addressing are 10.0.0.0/8, 172.16.0.0/12, and 192.168.0.0/16.

Practice and answers

For each exercise, find the network address, broadcast address, usable host range, and usable host count. Assume ordinary subnets unless noted.

1. 192.168.5.130/25

2. 172.16.4.200/22

3. Divide 192.168.8.0/24 into equal subnets that each support at least 50 usable hosts. Find the number of subnets and the first subnet’s range.

Answers

1. /25 has 128 addresses, and boundaries in the last octet advance by 128. The network address is 192.168.5.128; the broadcast address is 192.168.5.255; the usable host range is 192.168.5.129–192.168.5.254; and there are 126 usable hosts.

2. /22 has mask 255.255.252.0, so the third-octet block size is $256-252=4$. The third-octet boundary containing 4 is 4. The network address is 172.16.4.0; the broadcast address is 172.16.7.255; the usable host range is 172.16.4.1–172.16.7.254; and there are 1,022 usable hosts.

3. A /26 provides 62 usable hosts, so use /26. A /24 yields four /26 subnets. The first is 192.168.8.0/26: network address 192.168.8.0, broadcast address 192.168.8.63, and usable host range 192.168.8.1–192.168.8.62.