Networking - Subnetting - Discussion

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.

How the block size is 8 in the fourth octet?

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?

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

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.

I think it the range coming in 192.168.19.24 to 192.168.19.32.

So the 1st host coming is 192.168.19.25.

Please check it.

I got this as the answer 192.168.19.23.

First host address 192.168.19.25 255.255.255.248 is assigned to the router as said in the question, after that next available address is (as given in options) 192.168.19.26 255.255.255.248 which would assign to the server.

I think here first usable host is 192.168.19.25. Because here 192.168.19.24 is the network address.

The first host address should be:

Address: 192.168.19.25.
Subnet: 255.255.255.248.