Questions of Classful IP Addressing

The question is, if the given IP address = 201.20.31.65, then find out the following parts.

Part 1: Find the Class of the given IP.

Solution: The first Octet of the given IP address exists in between the range of Class C (192 – 223). So, the given IP belongs to class C.

Part 2: Find the Subnet Mask of the given IP.

Solution: After checking the class of the given IP. A subnet mask can be found by putting all octet-bits of network ID to “255” and Host bits to “0”. So, the Class C subnet mask is 255.255.255.0.

Part 3: Find the Network ID.

Solution: To find the network ID, Perform AND operation of the given IP with Subnet mask in binary. It will provide the network ID where that particular IP exists. So, the network ID will be 201.20.31.0.

Part 4: Find the First and Last IP address of the given Network IP.

Solution: We know that there are two IPs for each network, which are reserved. As the Network ID is 201.20.31.0. So, the first IP address will be 201.20.31.0, which is reserved for network identification. The last IP address is 201.20.31.255, which is reserved for direct broadcasting.

Part 5: Find the first and Last Host ID of the given Network IP.

Solution: Due to the reservation of two Hosts, the Network ID is 201.20.31.0. So, the first host IP address will be 201.20.31.1, and the Last host IP address will be 201.20.31.254.

Part 6: Find the Limited broadcast IP address.

Solution: Replacing all octets of IP address to 255 (decimal) is called Limited broadcast IP addressing. So, Limited broadcast address = 255.255.255.255.

Part 7: Find the Direct broadcast IP addressing.

Solution: Replacing the Host-bits-octets to 255 of a given IP is called direct broadcast IP addressing. So, direct broadcast IP addressing = 201.20.31.255.