Networking - Subnetting - Discussion

D is the correct answer.

As I solved this question :

Nework id :- 192.168.19.24.
First vaild host :- 192.168.19.25.
Last vaild host - 192.168.19.30.
Broadcost IP address :- 192.168.19.31.
Subnet mask :- 255.255.255.248.

Please check it.

The first host address should be :192.168.19.25.

Because up to 29 bits it is fixed part and the remaining part is 3 bits.
192.168.19.00011000 -> subnet block id.
.00011001-> this will be the first host for this subnet.

The first host address should be:

Address: 192.168.19.25.
Subnet: 255.255.255.248.

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

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 got this as the answer 192.168.19.23.

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 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?