Question

You have an IP address of 172.16.13.5 with a 255.255.255.128 subnet mask. What is your class of address, subnet address, and broadcast address?

(a) Class A, Subnet 172.16.13.0, Broadcast address 172.16.13.127

(b) Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.127

(c) Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.255

(d) Class B, Subnet 172.16.0.0, Broadcast address 172.16.255.255

I have been asked this question in an online interview.

The doubt is from Analyzing Subnet Masks topic in section TCP/IP Protocol Suite of Computer Network

Answer 1

Right choice is (b) Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.127

Explanation: We know that the prefix 172 lies in class B (128 to 191) of IPv4 addresses. From the subnet mask, we get that the class is divided into 2 subnets: 172.16.13.0 to 172.16.13.127 and 172.16.13.128 to 172.16.13.255. The IP 172.16.13.5 lies in the first subnet. So the starting address 172.16.13.0 is the subnet address and last address 172.16.13.127 is the broadcast address.