Networking - Subnetting - Discussion

B) is the correct answer.

The correct explanation is;

First, understand the question, Describe the IP 10.16.3.65/23
let's start from the beginning 11111111.11111111.11111110.00000000/23 which mean we are in the third octet.

How many zeroes we have in the third octet? One so 2^1=2 each network.

Now the answer will be:
First =10.16.0.0 (to) 10.16.1.255
10.16.0.0 this is N/W
10.16.0.1 first IP
10.16.1.254 last IP
10.16.1.255 Broadcast

Second= 10.16.2.0 (to) 10.16.3.255
10.16.2.0 N/W
10.16.2.1 First IP
10.16.3.254 last IP
10.16.3.255 Broadcast

And the question asking IP of 10.16.3.65 which is between the second N/W the first or lowest host in second N/W is 10.16.2.1 and last 10.16.3.254 the 255 is broadcast.

Note: 10.16.3.65/23 lock to the 23 which mean 8+8+7+0. (11111111+11111111+11111110+00000000) How many Zeros we have in the third octet? One So you need to use this formula to get the IP each N/W 2^(number of zeros) in the octet where we are.

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.

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.

IP address converted to binary (128 64 32 16 8 4 2 1):
10.16.3.65 is 00001010-00010000-00000011-01000001

Subnet /23 is 11111111-11111111-11111110-00000000.
Network id: 00001010-00001000-00000010-00000000 (10.16.2.0).
Broadcastid: 00001010-00001000-00000011-11111111 (10.16.3.255).
Therefore host address range: 10.16.2.1 to 10.16.3.254.

I think first address should be 10.16.2.0 instead of 10.16.2.1.

because 2^(32-23) = 2^9,
3 : 0 0 0 0 0 0 1 1,
65 : 0 1 0 0 0 0 0 1.

9 means 1 bit from 3rd octet and rest of them from 4th octet,

0 0 0 0 0 0 1 [0. 0 0 0 0 0 0 0 0] = 2.0,
0 0 0 0 0 0 1[1. 1 1 1 1 1 1 1 1 ] = 3.255.

Why option 3 is wrong?

(The last valid host address in the subnet is 10.16.2.254, 255.255.254.0).

I think option 3 is wrong because 10.16.3.254 is the last. Since in 3rd octet its 3 not 2. Therefore, The last valid host address in the subnet is 10.16.3.254.

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.

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.

Why is option 1 not correct, is it true to say that a subnet address is the same as a subnet mask?


According to me, it is option 2 and 4.

Anyone, please explain in detail.