CCNA Subnetting Practice: 8 IPv4 Problems with Worked Answers

A /26 IPv4 subnet contains 64 addresses. On a conventional LAN, 62 are usable host addresses. For example, 192.168.10.77/26 belongs to the subnet that starts at 192.168.10.64 and ends at 192.168.10.127. Here is how to find those boundaries, followed by eight problems you can solve on paper.

The three-step subnetting method

1. Subtract the prefix length from 32 to find the number of host bits. A /26 leaves 6 host bits, so the subnet has 2⁶ = 64 addresses.
2. Find the subnet mask and the increment in the octet where the prefix ends. For /26, the mask is 255.255.255.192. In the fourth octet, 256 − 192 = 64, so the blocks start at 0, 64, 128 and 192.
3. Locate the block that contains the IP address. The network address is the first address in the block and the broadcast address is the last. On a conventional LAN, the usable range runs from the address after the network to the address before the broadcast.

For 192.168.10.77/26, the fourth octet, 77, falls in the 64–127 block. The usable range is 192.168.10.65–192.168.10.126. Write down the block endpoints before counting usable hosts.

Eight practice problems

For each IP address, find the subnet mask, network address, broadcast address, usable host range and usable host count. Try the questions before opening the answer key below.

1. 192.168.10.77/26
2. 192.168.10.126/26
3. 192.168.10.130/26
4. 10.4.8.73/27
5. 10.4.8.158/28
6. 172.16.9.41/29
7. 172.16.9.254/30
8. 172.16.19.22/22

Worked answer key

How to spot the boundary

• Problems 1–3: /26 blocks are 64 addresses wide. Both 77 and 126 fall in 64–127. The address ending in 130 falls in the next block, 128–191.
• Problem 4: A /27 has an increment of 32. The fourth octet, 73, sits in 64–95.
• Problem 5: A /28 has an increment of 16. The fourth octet, 158, sits in 144–159. It is the last usable host in that block.
• Problem 6: A /29 has an increment of 8. The fourth octet, 41, sits in 40–47. It is the first usable host.
• Problem 7: A /30 has an increment of 4. The last block starts at 252 and ends at 255. It has two usable host addresses, 253 and 254.
• Problem 8: A /22 uses a mask of 255.255.252.0, so the increment of 4 is in the third octet. The third octet, 19, is in the 16–19 block. The network starts at 172.16.16.0 and the broadcast is 172.16.19.255.

Common mistakes to check

• Use the octet where the prefix ends. In a /22, the third octet determines the subnet boundary.
• Calculate the broadcast as the address immediately before the next network. For a /26 starting at .64, the broadcast is .127.
• Keep the address count and the usable host count separate. The two-address deduction here applies to the conventional subnets in these exercises. /31 point-to-point links and /32 host routes are special cases.

Quick check

Are 192.168.10.126/26 and 192.168.10.130/26 on the same subnet? Compare their network addresses in the answer key. One belongs to 192.168.10.64/26 and the other to 192.168.10.128/26, so they are on different subnets.

Further reading Cisco’s IP addressing and subnetting guide