Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 1)
1.
Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?
  1. 172.16.1.100
  2. 172.16.1.198
  3. 172.16.2.255
  4. 172.16.3.0
1 only
2 and 3 only
3 and 4 only
None of the above
Answer: Option
Explanation:
The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.
Discussion:
34 comments Page 1 of 4.

Rahul Gawale said:   1 year ago
The valid host id is from 172.16.2.0 to 172.16.3.254 [excluding 172.16.2.1(as this is itself routers id)].

And the broadcast add is 172.16.3.255 [here we are making all the last nine bits as 1.].

Malede said:   1 year ago
It is very helpful. Thanks everyone for explaining.

Babu said:   2 years ago
@All.

We have to understand host range is 1 to 254,0 and 255 is network & broadcast ID REMEMBER Only first and last bits Network & Broadcast ID.

So 172.16.2.1 to 172.16.2.255 then next change 172.16.3.0 to 172.16.3.255 Because we have 2^9-2=510 hosts.

That's why 3.254 is valid one So 2.55 is not Broadcast ID.

I hope you will understand.

Mohamad Fazil M said:   3 years ago
Your router has the address 172.16.2.1/23 on the ethernet interface which IP address can be assigned to a host connected to the ethernet interface of the router and what will be that host default gateway.

Sunil said:   5 years ago
Thanks for the explanation @Vikash_Jaiswal.

Waseem said:   5 years ago
(Q).Given ip 172.16.2.1/23

(A). by default the given IP is class B(N.N.H.H), slash notation for Class B is /16 but from given it is /23, so the converted network bits are 7 (23-16).
formula for subnets is 2^n (note:- it indicates how many parts we need to divide a N/w)

Here n represents no. of converted N/w bits
n=7
from formula 2^7=128
Formula for Hosts is 2^h-2 (note:- it indicates how many IP's consists in a N/w)

Here h represents no. of host bits
-2 represents (one is N/W IP and another one is Broadcast IP, we can't use
these 2 IP's these are default)
h=32-23
=9
from formula 2^h
2^9=512
The given IP is defaultly class B(N.N.H.H) we should vary Host bits only,N/W bits are constant in every N/W.

So the range starts from 172.16.0.0.
As we got hosts from hosts formula we should divide our N/W with 512 hosts
so the range is as bellow
(1st subnet). 172.16.0.0 to 172.16.1.255
(2nd subnet). 172.16.2.0 to 172.16.3.255
(3rd subnet). 172.16.4.0 to 172.16.5.255

From given

The given ip 172.16.2.1 belongs to 2nd subnet range, we should neglect 1st and last IP's in this range those are 172.16.2.0 and 172.16.3.255(N/W ID and Broad cast ID).

Finally the valid IP's are 172.16.2.1 to 172.16.3.254
From given option's only option C belongs to above range.. So answer is 'C'.

Jaanu said:   5 years ago
First host - 172.16.2.1
Last host - 172.16.255.254
Valid range - 172.16.2.1 - 172.16.255.254
Broad cast - 172.16.255.255.

Is this calculate in correct? Can anyone tell me?

Chandni said:   5 years ago
Thanks for sharing this Subnetting problem, I was looking for idea for solving these kind of problems. Finally I got it.

Basir said:   6 years ago
I'm not getting this. Please anyone help it to me.

Ansar said:   6 years ago
According to me, the answer will be 172.16.2.255.


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