Networking - Subnetting - Discussion

Discussion :: Subnetting - Subnetting (Q.No.2)

2. 

Which two statements describe the IP address 10.16.3.65/23?

  1. The subnet address is 10.16.3.0 255.255.254.0.
  2. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
  3. The last valid host address in the subnet is 10.16.2.254 255.255.254.0.
  4. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.

[A]. 1 and 3
[B]. 2 and 4
[C]. 1, 2 and 4
[D]. 2, 3 and 4

Answer: Option B

Explanation:

The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.

Ikechukwu said: (Oct 26, 2013)  
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.

Mothebe said: (Mar 10, 2014)  
Why is it class A address? I thought that's a class B address.

Vaishali said: (Jul 13, 2014)  
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?

Santhosh said: (Jul 27, 2014)  
The valid host id is from 10.16.2.1 to 10.16.3.254.

And the broadcast add is 10.16.3.255.

Sagar Madgundi said: (Sep 23, 2014)  
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

Atif said: (Jan 28, 2015)  
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.

Davit said: (Jun 5, 2015)  
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.

Vishwajeet Mishra said: (Aug 18, 2015)  
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.

Vishwajeet said: (Aug 18, 2015)  
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.

Rohit said: (Feb 3, 2016)  
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?

Mainodin said: (Mar 8, 2016)  
Why option 3 is wrong ?

Priyanka said: (Mar 29, 2016)  
Why it is started from 10.16.2.0? Why not from 10.16.1.0 or from 10.16.3.0 please tell.

Priyanka said: (Mar 29, 2016)  
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.

Abdulrazig Mohammed said: (Apr 3, 2016)  
What is the broadcast for 10.1.1.1/8?

Milhem said: (Apr 20, 2016)  
Why option 4 is wrong? Explain me.

Sia said: (May 30, 2016)  
Can anyone know the answer for Abdulrazig Mohammed's question?

Ali said: (Jun 30, 2016)  
After calculating how did we come to know that "the possible networks will be 10.16.2.1 to 10.16.3.254"?

Teshie said: (Jul 11, 2016)  
I don't understand how this will be the right answer.

Davies said: (Jul 24, 2016)  
Please, what is the broadcast for 10.1.1.1.8?

Aryan said: (Sep 2, 2016)  
255.255.254.0 is the subnet mask for the network right?

Then why not option C is correct?

Arunkumar said: (Dec 26, 2016)  
Option C is correct.

The subnet is 255.255.254.0 for10.16.3.0 because the CIDER value is 23.

Arunkumar said: (Dec 26, 2016)  
Hi @Davies.

The answer :10.255.255.255.

John said: (Jan 31, 2017)  
Good and enough explanation thanks to all.

Nana said: (Apr 25, 2017)  
Buy I thought the up address is class A. Am I right?

Colly said: (May 19, 2017)  
What is the broadcast for 10.1.1.1.8? Please, explain me.

Nia said: (Jul 14, 2017)  
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.

Shailendra Kumar Yadav said: (Nov 22, 2017)  
The lowest host address in the subnet is 10.16.2.1 255.255.254.0.

It is incorrect because the lowest host address in the subnet is 10.16.3.254 255.255.254.0.

Monty said: (Jan 20, 2018)  
Why not option A? Please explain.

Sourav Chettr said: (Jan 25, 2018)  
Why option A is not correct?

Usama said: (Apr 11, 2018)  
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.

Venkatesh said: (Jul 26, 2018)  
The subnet mask for the mentioned IP address is 255.255.254.0 then option 1 is also correct right.

Thishmi said: (Aug 21, 2018)  
Why subnet mask can't take as 255.255.254.0 in option A?

Gexun said: (Feb 23, 2019)  
Thank you for explaining the answer.

Christian Luora said: (Jul 14, 2020)  
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.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.