VLSM: subnets of different sizes

Why equal subnets waste addresses

With equal subnets, every subnet has the size of the biggest one. If your largest group needs 100 hosts, every subnet becomes a /25 with 126 hosts. A link between two routers needs only 2 addresses, so a /25 on that link wastes 124. A /24 holds only two /25s, so you run out after two groups.

VLSM, Variable Length Subnet Masking, means using different prefix lengths inside the same block. Big groups get big subnets, small groups get small ones. All modern routing protocols (OSPF, EIGRP, BGP, RIP version 2) send the mask with each route, so VLSM works everywhere today.

The largest-first method

1. List every group and how many hosts it needs. Count the router's gateway address as a host, and add room to grow.
2. Sort the list from the largest to the smallest.
3. For each group, pick the smallest prefix where 2^h − 2 is at least the host count.
4. Give the first group the start of the block. Give each next group the address right after the previous broadcast address.
5. Write down network, mask, host range and broadcast for each, and keep the leftover space as one or more free blocks.

Why largest first? A subnet must start at a multiple of its own block size: a /26 starts at 0, 64, 128 or 192. If you go from largest to smallest, the next free address is always a multiple of the next block size, so the subnets line up with no gaps and no overlap.

A full worked plan: an engineering college in Pune

The college has 192.168.10.0/24 for its new academic block. It needs: computer labs 100 hosts, staff 50 hosts, administration 25 hosts, and two point-to-point links to other buildings, 2 hosts each.

1. Labs, 100 hosts: 2^7 − 2 = 126 is enough, so /25. Start at the beginning: 192.168.10.0/25, hosts .1 to .126, broadcast .127.
2. Staff, 50 hosts: 2^6 − 2 = 62, so /26. Next free address is .128: 192.168.10.128/26, hosts .129 to .190, broadcast .191.
3. Admin, 25 hosts: 2^5 − 2 = 30, so /27. Next free is .192: 192.168.10.192/27, hosts .193 to .222, broadcast .223.
4. Link 1, 2 hosts: /30. Next free is .224: 192.168.10.224/30, hosts .225 and .226, broadcast .227.
5. Link 2, 2 hosts: /30. Next free is .228: 192.168.10.228/30, hosts .229 and .230, broadcast .231.

Everything fits in one /24 with 24 addresses to spare. With equal /25 subnets, the same needs would not have fitted at all.

import ipaddress

plan = ["192.168.10.0/25", "192.168.10.128/26", "192.168.10.192/27", "192.168.10.224/30", "192.168.10.228/30"]
nets = [ipaddress.ip_network(p) for p in plan]
for a in nets:
    for b in nets:
        if a != b and a.overlaps(b):
            print("overlap:", a, b)
print("checked", len(nets), "subnets")

Common mistakes

• Starting a subnet on a wrong boundary, such as 192.168.10.200/26. A /26 must start at 0, 64, 128 or 192.
• Forgetting the gateway: a group of 30 PCs plus the router needs 31 addresses, so a /27 (30) is too small.
• No room to grow: the lab with 100 PCs may have 130 next year. Leave spare space where you can.
• Going smallest first, which leaves gaps and breaks the alignment of later, bigger subnets.