Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 1)
Your router has the following IP address on Ethernet0: Which of the following can be valid host IDs on the LAN interface attached to the router?
1 only
2 and 3 only
3 and 4 only
None of the above
Answer: Option
The router's IP address on the E0 interface is, which is 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.
35 comments Page 1 of 4.

Tonychrys said:   2 months ago
Well done for the detailed explanation @Waseem.

From h= 32- 23 = 9.

Here, How did you get that term 32? Please explain to me.

Rahul Gawale said:   3 years ago
The valid host id is from to [excluding this is itself routers id)].

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

Malede said:   3 years ago
It is very helpful. Thanks everyone for explaining.

Babu said:   4 years ago

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 to then next change to 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:   4 years ago
Your router has the address 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:   6 years ago
Thanks for the explanation @Vikash_Jaiswal.

Waseem said:   7 years ago
(Q).Given ip

(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
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)
from formula 2^h
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
As we got hosts from hosts formula we should divide our N/W with 512 hosts
so the range is as bellow
(1st subnet). to
(2nd subnet). to
(3rd subnet). to

From given

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

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

Jaanu said:   7 years ago
First host -
Last host -
Valid range - -
Broad cast -

Is this calculate in correct? Can anyone tell me?

Chandni said:   7 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:   7 years ago
I'm not getting this. Please anyone help it to me.

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