/25 Subnet: Mask 255.255.255.128

A /25 is half a /24: 128 addresses, 126 usable. Every /24 holds exactly two of them: .0 to .127 and .128 to .255.

/25 at a glance

Subnet mask
255.255.255.128
Wildcard mask
0.0.0.127
Usable hosts
126
Total addresses
128 (27)
Block size (fourth octet)
128
Network / host bits
25 / 7
Mask in hex
0xFFFFFF80

Mask in binary: 11111111.11111111.11111111.10000000

Network bits Host bits

Try it: which /25 is an address in?

Network address
10.10.5.128/25
First usable host
10.10.5.129
Last usable host
10.10.5.254
Broadcast address
10.10.5.255
Next /25 starts at
10.10.6.0

How many /25 networks fit

How /25 relates to /8, /16 and /24 blocks
Parent block Relationship
/8 131,072 × /25 fit inside one /8
/16 512 × /25 fit inside one /16
/24 2 × /25 fit inside one /24

2 × /25 = one /24. A /25 holds 2 × /26 or 4 × /27.

Worked example: 10.10.5.200/25

A /24 was split in two: wired desks on the lower half, Wi-Fi on the upper half. A phone has 10.10.5.200. Which half is it in, and what should its gateway be?

  1. The mask is 255.255.255.128. The fourth octet (128) is the "interesting" one — the first that is not 255.
  2. Block size = 256 − 128 = 128. Networks start at multiples of 128 in the fourth octet: 0, 128
  3. The fourth octet of 10.10.5.200 is 200. 200 ÷ 128 = 1 remainder 72, so the block starts at 1 × 128 = 128.
  4. Network = 10.10.5.128. Broadcast = 128 + 128 − 1 = 255, giving 10.10.5.255.
  5. Usable hosts: 10.10.5.129 to 10.10.5.254, which is 27 − 2 = 126 addresses.

Same thing in binary (network bits | host bits):

Address  00001010.00001010.00000101.1|1001000
Mask     11111111.11111111.11111111.1|0000000
Network  00001010.00001010.00000101.1|0000000

Where /25 is used in real networks

  • Splitting one /24 between two VLANs

    If you only have one /24 left, two /25s give you two VLANs of 126 hosts each — for example staff and guests, or wired and wireless.

  • Public IP blocks

    Hosting providers sometimes hand out a /25 of public space to a customer rack: one gateway, 125 addresses for servers.

Common /25 mistakes

  • Using .1 as the gateway for the upper /25. The upper half starts at .128, so its first usable address is .129. A gateway of .1 is in the other subnet and hosts can't reach it.
  • Forgetting that .127 and .128 are now reserved. In a /24 they were normal hosts; after the split, .127 is the broadcast of the lower half and .128 is the network address of the upper half.

/25 FAQ

How many hosts are in a /25?

126: 2^7 = 128 addresses minus network and broadcast.

What are the two /25 subnets of 192.168.1.0/24?

192.168.1.0/25 (hosts .1 to .126, broadcast .127) and 192.168.1.128/25 (hosts .129 to .254, broadcast .255).

What is the wildcard mask for /25?

0.0.0.127. In an ACL, 192.168.1.128 0.0.0.127 matches the upper half of the /24: .128 to .255.

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