Networking - Subnetting - Discussion

I think answer is (D).

Solution:

Given CIDR is 192.168.19.24/29 which means the NID is 29 bits and HID is 3 bits.

So the subnet mask becomes 255.255.255.248.

Given that the router has the first available host address (i.e, SubnetID), which is given by:

(192.168.19.24) AND (255.255.255.248) = 192.168.19.24, which is the IP address of the router.

We know that 192.168.19.24 = 192.168.19.00011000.

Here the NID part is 192.168.19.00011 (29 bits) and host id part is remaining 3 bits.

In order to get the broadcast address (address of the server), set all host id bits to 1. Then it becomes 192.168.19.00011111.

i.e, 192.168.19.31 which is the IP address of the server.

Guys if I'm correct you count up to 256 but I see some where people counted to 255 maybe?

I think D is the right answer.

192.168.19.24/29 --> 11000000.10110000.00010011.00011000 with the first 5 bits of the 4th octet defining the sub net 24 and last 3 bits defining the host number. So if the next sub net is 32--> 11000000.10110000.00010011.00100000 then the broadcast address for sub net 24 would be 11000000.10110000.00010011.00011111.

Am I missing something?

How the block size is 8 in the fourth octet?

Subnet id is 192.168.19.24/29 or mask is 255.255.255.248.

First host of the subnet is 11000000.10110000.00010011.00011001 that is 192.168.19.25/29.