Question

How many hexadecimal digits are needed to define the netid?

Answer 1

Answer:

Class B has 2 bytes (16 bits) netid and from the binary notation we see that Class B address starts with 10, so there are total 14 bits that can be changed out of 16.

Therefore total number of blocks in Class B = 214 = 16,384

There are 2 bytes (16 bits) for hostid in Class B so total number of host in each block = 216 = 65,536

So total number of addresses in Class B = No. of Blocks in Class B x No. of Hosts in each block of Class B

= 16,384 x 65,536

= 1,073,741,824

This is 25% of the total addresses in IPv4.

Explanation: