Networking - Subnetting - Discussion

Convert the given IP subnet into binary operations & then perform the operations to get the network ID.
Then combine the IP and subnet to get the broadcast address.

10101100.00010000.00000010.00000001
11111111.11111111.11111111.11111111

172.16.2.0 network id /23

11111111.11111111.11111111.11111111

Network ID IP: 172.16.2.0
So, Ip address range 172.16.2.1 to 172.16.3.254
broadcast IP:172.16.3.255.

Well done for the detailed explanation @Waseem.

From h= 32- 23 = 9.

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

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.].

It is very helpful. Thanks everyone for explaining.

@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.

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.

Thanks for the explanation @Vikash_Jaiswal.

(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'.

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?

Thanks for sharing this Subnetting problem, I was looking for idea for solving these kind of problems. Finally I got it.