Write the boundary before choosing an answer
Subnetting becomes much more reliable when every problem uses the same sequence: convert the prefix to a mask if needed, identify the changing octet, calculate the block size, locate the network boundary, then derive broadcast and usable range. This method also transfers directly into troubleshooting because the same math determines whether a gateway or peer is local.
High-value prefixes to know cold
| Prefix | Mask | Total addresses | Traditional usable | Block size |
|---|---|---|---|---|
| /24 | 255.255.255.0 | 256 | 254 | 256 |
| /25 | 255.255.255.128 | 128 | 126 | 128 |
| /26 | 255.255.255.192 | 64 | 62 | 64 |
| /27 | 255.255.255.224 | 32 | 30 | 32 |
| /28 | 255.255.255.240 | 16 | 14 | 16 |
| /29 | 255.255.255.248 | 8 | 6 | 8 |
| /30 | 255.255.255.252 | 4 | 2 | 4 |
Swipe horizontally to see the CIDR table.
Use 256 minus the mask octet
For a /26, the last mask octet is 192, so the block size is 256 − 192 = 64. That produces boundaries at 0, 64, 128, and 192. For a /27, 256 − 224 = 32, giving boundaries at 0, 32, 64, 96, 128, 160, 192, and 224.
Find the network and broadcast for 192.168.60.142/26
Work it in four steps
Mask: 255.255.255.192
Block size: 64
Boundaries: .0, .64, .128, .192
Host .142: falls inside .128–.191
Are 10.10.5.62/27 and 10.10.5.65/27 in the same subnet?
A /27 has a block size of 32. Address .62 is inside .32–.63. Address .65 is inside .64–.95.
How many usable hosts does /28 provide?
/28 leaves 4 host bits. 24 = 16 total addresses. Under traditional subnetting, subtract network and broadcast.
What prefix supports at least 50 traditional hosts?
Five host bits give 25 − 2 = 30 usable addresses, which is too small. Six host bits give 26 − 2 = 62 usable addresses.
Find the valid host range for 172.16.75.200/20
/20 equals 255.255.240.0. The changing octet is the third octet, and 256 − 240 = 16. Third-octet boundaries are 0, 16, 32, 48, 64, 80, and so on. The value 75 sits inside 64–79.
Client 192.168.100.130/26 uses gateway 192.168.100.1
.130 belongs to the 192.168.100.128/26 block. The usable range is .129–.190. Gateway .1 belongs to the 192.168.100.0/26 block.
Find network, broadcast, and usable range for 192.168.1.221/27
/27 has block size 32. Boundaries around .221 are .192 and .224.
Find network and last usable host for 10.40.12.70/28
/28 has block size 16. .70 lies in the .64–.79 block.
Find network and broadcast for 172.20.201.8/22
/22 equals 255.255.252.0, so the third-octet block size is 4. Third-octet boundaries around 201 are 200 and 204.
Which prefix supports at least 100 traditional hosts?
Six host bits give 62 usable, not enough. Seven host bits give 126 usable.
Are 10.1.1.127/25 and 10.1.1.128/25 local to each other?
/25 splits the last octet into .0–.127 and .128–.255.
Is 192.168.8.95/27 a valid host?
192.168.8.95 is the last address in the 192.168.8.64/27 block.
Is 192.168.8.94/27 a valid host with gateway 192.168.8.65?
The subnet is 192.168.8.64/27, with usable addresses .65 through .94.
Subnetting is really a local-versus-remote decision
Hosts do not use visual similarity to decide whether another address is local. They apply the prefix. Two addresses that differ by only one may still sit on opposite sides of a subnet boundary. Two addresses that look far apart may still be local under a broader prefix.
Same subnet
Attempt direct local delivery.
Different subnet
Send toward the default gateway.
Wrong mask
Can make the host choose the wrong behavior.
An overly broad mask can make a remote host look local
With /16, the client treats 10.20.40.50 as local and may ARP for it instead of using the gateway. The route table upstream is not yet the first problem.
An overly narrow mask can make a local host look remote
The /27 configuration places those addresses in different local subnets even though the design intended one /24. The client can send traffic toward the gateway unnecessarily or fail depending on the topology.
Work backward from host requirement
| Need at least | Smallest traditional subnet | Usable hosts |
|---|---|---|
| 2 | /30 | 2 |
| 6 | /29 | 6 |
| 14 | /28 | 14 |
| 30 | /27 | 30 |
| 50 | /26 | 62 |
| 100 | /25 | 126 |
| 200 | /24 | 254 |
Swipe horizontally to see the host-capacity table.
Do not subtract 2 before finding total addresses
The clean sequence is: determine host bits, calculate 2host bits total addresses, then apply the traditional network/broadcast subtraction when the problem expects usable host count.
Do not confuse block size with usable host count
A /27 block size is 32 addresses, but traditional usable host count is 30. Those are different quantities.
Do not assume the first or last address is a host
In traditional IPv4 subnetting, the first address is the network address and the last is the broadcast address. That is why 10.1.1.127/25 and 10.1.1.128/25 sit exactly on opposite boundaries.
Do not blame routing before checking the client's own subnet math
A wrong prefix can change the traffic path before any router receives the packet. Always verify the host's local/remote decision first when gateway or peer behavior looks strange.
Five questions to solve without scrolling back
- 192.168.40.201/26: network and broadcast?
- 10.0.0.94/27: network and last usable?
- 172.16.33.10/20: network and broadcast?
- What prefix supports 28 traditional hosts?
- Are 192.168.1.126/25 and 192.168.1.129/25 local to each other?
Strong subnetting answers show the intermediate boundary
Instead of writing only “192.168.60.128,” show why: “/26 = 255.255.255.192, block size 64, boundaries .0/.64/.128/.192, so .142 lies in .128–.191.” That makes the reasoning auditable and prevents lucky guessing.
24 Subnetting Practice digital flashcards
Practice CIDR masks, block sizes, host counts, network and broadcast addresses, usable ranges, same-subnet decisions, gateway logic, and troubleshooting implications.