|Subnet Mask||Last octet (in bits)||CIDR||Networks||Hosts per network|
Question: How many nodes can be allocated on 10.10.3.3/29?
10.10.3.3 is an IP (v4) address. Let's expand the address into binary octets:
/29 of CIDR notation tells us a lot about this address. It says that 29 bits belong to the network (subnet) portion of the address. From this, we can derive both the subnet mask and the network address.
Let's separate the binary address after the 29th bit.
00001010.00001010.00001010.00000 | 011
The subnet mask is given by replacing all bits to the left with 1, and all to the right with 0:
11111111.11111111.11111111.11111000 = 255.255.255.248
Looking at our binary IP address, the network address is given by replacing all bits to the right of the 29th bit with 0. This is essentially a bitwise AND operation of the IP address and the subnet mask.
00001010.00001010.00001010.00000000 = 10.10.3.0
The broadcast address is given by replacing all bits to the right with 1.
00001010.00001010.00001010.00000111 = 10.10.3.7
So, we see that IP addresses between
10.10.3.7 are available for hosts. In this case, we have
.6, total of 6 possible hosts.
A quick way to calculate the number of available hosts using the CIDR notation:
2^(32 - <CIDR SUBNET MASK>) - 2 = Number of hosts
So, we have:
2^(32 - 29) - 2 =
2^3 - 2 = 6 hosts
Updated May 22, 2020.