Subnetting drills for exam speed

Why speed matters

In the CCNA exam, many questions hide a subnetting step inside them: "which host can ping this server?", "which route will the router use?". If each one takes you five minutes, you run out of time. At work, you will subnet while someone waits on a call. The aim is a correct answer in under a minute, without a calculator.

The cheat table

Write this table from memory on a blank sheet at the start of every practice session (and on the whiteboard you get in an exam centre, if allowed). After a week you will not need it.

The six-step routine

1. Write the mask for the prefix.
2. Find the interesting octet (not 255, not 0).
3. Block size = 256 minus that mask octet.
4. Network: the largest multiple of the block size that is not more than the address octet. Octets to the right are 0.
5. Broadcast: next multiple minus 1. Octets to the right are 255.
6. First host = network + 1, last host = broadcast − 1, hosts from the table.

Ten worked drills

1. 192.168.5.130/26: block 64, 130 is in 128 to 191. Network .128, broadcast .191, hosts .129 to .190, 62 hosts.
2. 10.10.10.10/29: block 8, 10 is in 8 to 15. Network 10.10.10.8, broadcast .15, hosts .9 to .14, 6 hosts.
3. 172.16.100.50/22: block 4 in the third octet, 100 is in 100 to 103. Network 172.16.100.0, broadcast 172.16.103.255, hosts 172.16.100.1 to 172.16.103.254, 1,022 hosts.
4. 192.168.20.250/28: block 16, 250 is in 240 to 255. Network .240, broadcast .255, hosts .241 to .254, 14 hosts.
5. 10.200.37.5/12: mask 255.240.0.0, block 16 in the second octet, 200 is in 192 to 207. Network 10.192.0.0, broadcast 10.207.255.255, 1,048,574 hosts.
6. 172.31.200.1/19: block 32 in the third octet, 200 is in 192 to 223. Network 172.31.192.0, broadcast 172.31.223.255, 8,190 hosts.
7. 192.168.100.63/27: block 32, 63 is in 32 to 63. Network .32, broadcast .63. Trap: 192.168.100.63 is the broadcast address, so it cannot be given to a host.
8. How many /28 subnets fit in a /24, and how many hosts each? 4 borrowed bits: 16 subnets of 14 hosts.
9. Smallest subnet for 500 hosts? /23, with 510 usable. A /24 (254) is too small.
10. Is 10.1.0.255/23 a usable host? The /23 block of 2 in the third octet is 0 to 1, so the subnet is 10.1.0.0 to 10.1.1.255. 10.1.0.255 is in the middle: yes, usable.

Route summarisation

Summarisation (also called supernetting or route aggregation) is the reverse of subnetting. A router that knows 172.16.0.0/24, 172.16.1.0/24, 172.16.2.0/24 and 172.16.3.0/24 can tell its neighbours about one route instead of four. Smaller routing tables mean less memory, faster lookups, and fewer updates when one small link goes up and down.

Method: write the octet that changes in binary, and count how many bits from the left are the same in all the networks. Third octets 0, 1, 2 and 3 are 00000000, 00000001, 00000010 and 00000011. The first 6 bits match. 16 bits from the first two octets plus 6 gives /22. The summary is 172.16.0.0/22.

Second example: 192.168.8.0/24 to 192.168.15.0/24. 8 is 00001000 and 15 is 00001111: the first 5 bits match, so the summary is 192.168.8.0/21.

import ipaddress

nets = [ipaddress.ip_network(f"172.16.{i}.0/24") for i in range(4)]
print(list(ipaddress.collapse_addresses(nets)))   # [IPv4Network('172.16.0.0/22')]

Common mistakes

• Doing the drill steps in a different order each time. Use the same routine until it is automatic.
• Missing the broadcast trap: always check whether the "host" in the question is actually the network or broadcast address.
• Summarising 10.1.3.0/24 and 10.1.4.0/24 as a /23. 3 is 00000011 and 4 is 00000100: only the first 5 bits match, so the smallest summary is 10.1.0.0/21.