Let’s add some of this information to our table:
Host IP address|172.25.114.250|
Major network mask|255.255.0.0 (/16)|
Major (base) network address|172.25.255.255|
Total Number of Host BitsNumber of hosts|16 bits or 216 or 65,536 total hosts|
Subnet mask|255.255.255.192 (/26)|
Number of subnet bitsNumber of subnets|261024 (210)|
Number of host bits per subnetNumber of usable hosts per subnet|664 – 2 = 62 (from 26 – 2)|
IP for first host on this subnet|172.25.114.192|
IP address of the last host on this subnet|172.25.114.254|
Broadcast address for this subnet|172.25.114.255|
Create a subnetting worksheet to show and record all work for each problem.
Problem 1:
Host IP address|172.30.1.33|
Subnet mask|255.255.255.0|
Number of subnet bits|8|
Number of subnets|256 – 2 = 254|
Number of host bits per subnet|8|
Number of usable hosts per subnet|256 – 2 = 254|
Subnet address for this IP address|172.30.1.0|
IP address of first host on this subnet|172.30.1.1|
IP address of last host on this subnet|172.30.1.254|
Broadcast address for this subnet|172.30.1.255|
Problem 2:
Host IP address|172.30.1.33|
Subnet mask|255.255.255.252|
Number of subnet bits|14|
Number of subnets|16,384 – 2 = 16,382|
Number of host bits per subnet|4 – 2 = 2|
Number of usable hosts per subnet|2|
Subnet address for this IP address|172.30.1.32|
IP address of first host on this subnet|172.30.1.33|
IP address of last host on this subnet|172.30.1.34|
Broadcast address for this subnet|172.30.1.35|
Problem 3:
Host IP address|192.192.10.234|
Subnet mask|255.255.255.0|
Number of subnet bits|24|
Number of subnets|16 777 216 (224)|
Number of host bits per subnet|8|
Number of usable hosts per subnet|256 – 2 = 254|
Subnet address for this IP address|192.192.10.0|
IP address of first host on this subnet|192.192.10.1|
IP address of last host on this subnet|192.192.10.254|
Broadcast address for this