How do you subnet more than 2 networks at once with VLSM?

Question

if I have a class C address of 192.168.10.0/24 and I need 4 subnetworks, one requiring 60 hosts, one requiring 28 hosts, and 2 requiring 12 hosts each; how many bits must I borrow and what is the network address for each subnet?

Answer 1

One simple solution here is that you need 4 subnets, so just use a /26 mask and you have your 192.168.10.0 broken into four subnets that support up to 62 hosts each and you're done. Breakdown is like this:

192.168.10.0 subnet 1 address
192.168.10.1 first host
192.168.10.62 last host
...
192.168.10.63 broadcast address
192.168.10.64 subnet 2 address
192.168.10.65 first host
...
192.168.10.126 last host
192.168.10.127 broadcast address
192.168.10.128 subnet 3 address
192.168.10.129 first host
...
192.168.10.190 last host
192.168.10.191 broadcast address
192.168.10.192 subnet 4 address
192.168.10.193 first host
...
192.168.10.254 last host
192.168.10.255 broadcast address

Alternatively, what you might be after is that you have 192.168.10.0/24 and you require 4 subnets, one with 60 hosts, one with 28 hosts and two with 12 hosts. This can be accommodated by making your first subnet cover the range from 192.168.10.0 to 10.63 for a maximum of 62 hosts, the 2nd from 192.168.10.64 to 10.95 for up to 30 hosts, the 3rd subnet from 192.168.10.96 to 10.111 for up to 14 hosts and the 4th subnet from 192.168.10.112 to 10.127 for up to 14 hosts.

192.168.10.0 /26 subnet 1 address (borrow 2 bits)
192.168.10.1 first host
192.168.10.62 last host
...
192.168.10.63 broadcast address
192.168.10.64 /27 subnet 2 address (borrow 3 bits)
192.168.10.65 first host
...
192.168.10.94 last host
192.168.10.95 broadcast address
192.168.10.96 /28 subnet 3 address (borrow 4 bits)
192.168.10.97 first host
...
192.168.10.110 last host
192.168.10.111 broadcast address
192.168.10.112 /28 subnet 4 address
192.168.10.113 first host
...
192.168.10.126 last host
192.168.10.127 broadcast address