# IP Subnetting

• Ops

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

### Quick Method

Figure out the subnets:

1. network and host split in third octect
2. subtract: 256 – 240 = 16, it means that network are incrementing in values of 16: 0, 16, 32, 48…
3. 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

First host = Subnet + 1

Last host = Broadcast – 1

### 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

1. “stealing” or “taking away” bits from the host portion of an address, and
2. 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

1. draw the line with /18 to split network and host
2. 2^5 > 30, need 5 subnet bit, draw the line to split subnet and host
3. 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