Networking - Subnetting - Discussion

I do not understand how option B is the correct answer. I would have thought that option 1 is the only correct description for the IP address.

Why is it class A address? I thought that's a class B address.

I thought it is a class B address. How can it be class A address? and how the subnet bits can be 15 in this?

The valid host id is from 10.16.2.1 to 10.16.3.254.

And the broadcast add is 10.16.3.255.

Why is it class A address? Why is it not class B address?

Answer:

IP address consists of network and host portion. For example, /24 notation means that first 24 bits are dedicated to the network portion and they remain the same for all hosts in the network, whereas, the last 8 bits uniquely identify each host in the network.

Classful addressing divides IP addresses into 5 different classes, each with its own predefined address range and subnet mask. On the contrary, classless addressing can set the network boundary practically anywhere, thus breaking the classful limitations.

Classful ----> When a IPaddress remains in its default given Subnet mask..
eg. 10.1.1.1/8 or 172.168.2.10/16 or 192.168.100.75/24

ClassLess----> When a IPaddress is out of its default given Subnet mask..
eg. 10.1.1.1/24 or 172.168.2.10/24 or 192.168.100.75/28

Option 2 and 4 is correct.

Explanation:

10.16.3.65/23.
11111111.11111111.11111110.00000000.

Magic Number is 2. So Range is:

1)10.16.0.0-------------10.16.1.255.
2)10.16.2.0-------------10.16.3.255.
3)10.16.4.0------------- 10.16.5.255.

Sub net is No.2:

The range between second is 10.16.2.1 to 10.16.3.254.

Therefore; 10.16.2.1 255.255.254.0 is the lowest host address in sub net and Broadcast address 10.16.3.255 255.255.254.0.

00000000.00000000.0000000|0.00000000 its subnet is 255.255.254.0.

Number 2. IP starts from 10.16.2.1 and range of IP is from 2.1 to 3.254. So broadcast address will be 10.16.3.255.

Yes, The answer (2 & 4) is a correct option, because now we are working in CIDR Value of (/23), which means we will work on only those bits which have the both combination's of 0's and 1's. So here this happens on third octat of subnet mask.

CIDR Value = /23. So the subnet in binary is 255.255.254.0 => 11111111.11111111.11111110.00000000.

So if we see the third octat then we will see that here is the combination of 0's and 1's. Now, let's find out the value of 1's in third octat. 2^7= 128.

And the total no. Of hosts in a subnetwork= 2^1=2. So here the subnetwork will be 0, 2, 4, 6, 8 and so on because the block size = 2.

So every subnet will made after the gap of 2. How to find the block size = we have to subtract the octat third from 256.

So here we subtract 256-254= 2. It defines the interval of the subnetwork in a subnetted network.

Then the possible networks will be 10.16.2.1 to 10.16.3.254, because this is the one subnet. So if we write 10.16.4.0 then this the another network. That's why we take the IP of 10.16.2.254 and 10.16.3.1. For more explanation see my previous post.

How come 10.16.2.254 255.255.254.0 is not a valid address keeping in mind that valid address for hosts range from 10.16.2.1 to 10.16.3.254?