# Subnetting a Class B Network Using Binary

#### Andrey Volkov

System, network administration + DBA. And a little programmer!)) See Author profile.

You have an address of 172.16.0.0 /16. You need nine subnets. What is the IP plan of network numbers, broadcast numbers, and valid host numbers? What is the subnet mask needed for this plan?

You cannot use N bits, only H bits. Therefore, ignore 172.16. These numbers cannot change.

Step 1. Determine how many H bits you need to borrow to create nine valid subnets.

2N ≥ 9

N = 4, so you need to borrow 4 H bits and turn them into N bits.

 Start with 16 H bits HHHHHHHHHHHHHHHH (Remove the decimal point for now) Borrow 4 bits NNNNHHHHHHHHHHHH

Step 2. Determine the first valid subnet in binary (without using decimal points).

 0000HHHHHHHHHHHH 0000000000000000 Subnet number 0000000000000001 First valid host … 0000111111111110 Last valid host 0000111111111111 Broadcast number

Step 3. Convert binary to decimal (replacing the decimal point in the binary numbers).

 00000000.00000000 = 0.0 Subnetwork number 00000000.00000001 = 0.1 First valid host number … 00001111.11111110 = 15.254 Last valid host number 00001111.11111111 = 15.255 Broadcast number

Step 4. Determine the second subnet in binary (without using decimal points).

 0001HHHHHHHHHHHH 0001000000000000 Subnet number 0001000000000001 First valid host … 0001111111111110 Last valid host 0001111111111111 Broadcast number

Step 5. Convert binary to decimal (returning the decimal point in the binary numbers).

 00010000.00000000 = 16.0 Subnetwork number 00010000.00000001 = 16.1 First valid host number … 00011111.11111110 = 31.254 Last valid host number 00011111.11111111 = 31.255 Broadcast number

Step 6. Create an IP plan table.

 Subnet Network Number Range of Valid Hosts Broadcast Number 1 0.0 0.1–15.254 15.255 2 16.0 16.1–31.254 31.255 3 32.0 32.1–47.254 47.255

Notice a pattern? Counting by 16.

Step 7. Verify the pattern in binary. (The third subnet in binary is used here.)

 0010HHHHHHHHHHHH Third valid subnet 00100000.00000000 = 32.0 Subnetwork number 00100000.00000001 = 32.1 First valid host number … 00101111.11111110 = 47.254 Last valid host number 00101111.11111111 = 47.255 Broadcast number

Step 8. Finish the IP plan table.

 Subnet Network Address (0000) Range of Valid Hosts (0001–1110) Broadcast Address (1111) 1 (0000) 172.16.0.0 172.16.0.1–172.16.15.254 172.16.15.255 2 (0001) 172.16.16.0 172.16.16.1–172.16.31.254 172.16.31.255 3 (0010) 172.16.32.0 172.16.32.1–172.16.47.254 172.16.47.255 4 (0011) 172.16.48.0 172.16.48.1–172.16.63.254 172.16.63.255 5 (0100) 172.16.64.0 172.16.64.1–172.16.79.254 172.16.79.255 6 (0101) 172.16.80.0 172.16.80.1–172.16.95.254 172.16.95.255 7 (0110) 172.16.96.0 172.16.96.1–172.16.111.254 172.16.111.255 8 (0111) 172.16.112.0 172.16.112.1–172.16.127.254 172.16.127.255 9 (1000) 172.16.128.0 172.16.128.1–172.16.143.254 172.16.143.255 10 (1001) 172.16.144.0 172.16.144.1–172.16.159.254 172.16.159.255 11 (1010) 172.16.160.0 172.16.160.1–172.16.175.254 172.16.175.255 12 (1011) 172.16.176.0 172.16.176.1–172.16.191.254 172.16.191.255 13 (1100) 172.16.192.0 172.16.192.1–172.16.207.254 172.16.207.255 14 (1101) 172.16.208.0 172.16.208.1–172.16.223.254 172.16.223.255 15 (1110) 172.16.224.0 172.16.224.1–172.16.239.254 172.16.239.255 16 (1111) 172.16.240.0 172.16.240.1–172.16.255.254 172.16.255.255 Quick Check Always in form even #.0 First valid host is always even #.1Last valid host is always odd #.254 Always odd #.255

Use any nine subnets—the rest are for future growth.

Step 9. Calculate the subnet mask. The default subnet mask for a Class B network is as follows:

 Decimal Binary 255.255.0.0 11111111.11111111.00000000.00000000

1 = Network or subnetwork bit

0 = Host bit

You borrowed 4 bits; therefore, the new subnet mask is the following:

 11111111.11111111.11110000.00000000 255.255.240.0

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