🧮 VLSM Calculator

Enter a base network in CIDR notation and a comma-separated list of required host counts (largest first works best) to plan variable length subnets that fit each requirement with minimal waste.

Base Network (CIDR)

Required Hosts (comma-separated)

How each subnet gets sized and placed

For every host count you list, the page walks prefixes from /30 upward and stops at the first block big enough, then hands out blocks back to back from the start of your base network:

block size = 2^(32 - prefix)
usable     = block size - 2      (block size itself at /31 and /32)
pick the largest prefix where usable >= hosts needed
next network = previous network + its block size

With the defaults, 192.168.1.0/24 and 50, 20, 10, 2, the 50-host group needs a /26 because a /27 offers only 30 usable addresses. That takes 192.168.1.0 to .63. Twenty hosts fit a /27, .64 to .95; ten fit a /28, .96 to .111; two fit a /30, .112 to .115. That uses 116 of the 256 addresses.

Reading and steering the plan

Enter hosts you need, not addresses. A group of 30 machines needs a /27 (30 usable), not a /28.
Order does not matter. The tool sorts your list largest first before allocating, then puts the rows back in the order you typed.
The smallest block it will issue is a /30. Ask for 1 host and you still get four addresses, because a /31 or /32 is treated as a special case rather than a general-purpose subnet.

Allocation begins at exactly the address you typed, not at the network address of the block. Enter 192.168.1.10/24 and the first subnet starts at .10, off its natural boundary. Type the network address itself.

Frequently asked questions

What is VLSM in networking?

Variable Length Subnet Masking: splitting one block into subnets of different sizes instead of equal ones, so a 50-user LAN and a two-address router link each get a mask that fits.

Can 50, 20, 10 and 2 hosts fit inside a single /24?

Yes, comfortably. They need a /26, a /27, a /28 and a /30 - 116 addresses in total, leaving 140 of the 256 spare.

Why allocate the biggest subnet first?

Every block has to start on a multiple of its own size. Place a /28 at the top of a /24 and the space left will not hold an aligned /26, so the largest requirement is fitted first and the smaller ones slot into what remains.