Question

Which of the following is the broadcast address for a Class B network ID using the default subnetmask? (a) 172.16.10.255 (b) 255.255.255.255 (c) 172.16.255.255 (d) 172.255.255.255

Comments

in case of Braodcast address  all host bit is 1 means 172.16.11111111.11111111=172.16.255.255

Answer 1

172.16.255.255

$For$ $Class$ $B$:  255.255.[00000000].[0000000]

                                HOST BITs (represented by 0’s)

 

In this case, the class B network ID is 172.16.0.0,

In Broadcast, all host bits will be replaced by ‘1’

$Broadcast$ address $will$ $be$ $172.16.[11111111].[11111111]$

i.e.  $172.16.225.225$

Answer 2

The default subnet mask for a Class B network is 255.255.0.0. The network ID for a Class B network takes up the first two octets (16 bits) and the remaining two octets (16 bits) are used for host addressing.

The broadcast address is the address at which all hosts on the network can receive a message simultaneously. To calculate the broadcast address, we set all host bits to 1s. For a Class B network, the host bits are in the last two octets (16 bits).

Therefore, the broadcast address for a Class B network with a default subnet mask would be 172.16.255.255. So, the correct option is (c) 172.16.255.255.

Comments

can D not be answer as well. 172.255.0.0 also falls under class B. The broadcast IP of this address is 172.255.255.255. even though this coincides with the class A network 172.0.0.0’s broadcast IP, since we always specify the length in the classless notation we will have to present it as 172.255.255.255/16. The question also didn't specify the network has to fall within the private IP address space.