Something you need to know first: Binary Odometer
10.1.1.254 + 1 = 10.1.1.255
10.1.1.255 + 1 = 10.1.2.0
10.1.2.0 + 1 = 10.1.2.1
in reverse:
10.1.2.0 – 1 = 10.1.1.255
Example 1
172.16.35.123/20 or 172.16.35.123 with the mask 255.255.240.0
Binary Method
Quick Method
Figure out the subnets:
- network and host split in third octect
- subtract: 256 – 240 = 16, it means that network are incrementing in values of 16: 0, 16, 32, 48…
- 35 in the range of 32 and 48, so 172.16.35.123 is on subnet 172.16.32.0; next subnet is 172.16.48.0
First subnet = 172.16.32.0
Next subnet = 172.16.48.0
Broadcast address = next subnet – 1
1 2 |
172.16.48.0 - 1 = 172.16.47.255 |
First host = Subnet + 1
1 2 |
172.16.32.0 + 1 = 172.16.32.1 |
Last host = Broadcast – 1
1 2 |
172.16.47.255 - 1 = 172.16.47.254 |
Subnetting
- Class A subnetting (255.0.0.0) support 1677214 (2^24) host per network, that way too much
- Class B subnetting (255.255.0.0) support 16382 (2^16) host per network, that way too much
- Class C subnetting (255.255.255.0) support 254 (2^8) host, more likely we subnet down to at least 254 hosts or even further
If you subnetting a network only has 2 hosts, you can subnet with (255.255.255.254) or CIDR as /31
Network, host number
- Networks: 2^(network bits)
- one allocate for the subnet
- one allocate for the broadcast
- Hosts: 2^(host bits) – 2
Subnetting to be short
- “stealing” or “taking away” bits from the host portion of an address, and
- allocating those bits to network portion
Example 2
Origin network 10.128.192.0/18 need at least 30 subnets as many hosts as possible
- draw the line with /18 to split network and host
- 2^5 > 30, need 5 subnet bit, draw the line to split subnet and host
- network/subnet portion is 8+8+7=23 bits, host portion is 32-23=9 bits
- First subnet: 10.128.192.0/23
- Second subnet: 10.128.194.0/23
- Last subnet: 10.128.254.0/23