Networking - Subnetting - Discussion

(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'.

Guys What I think is.

Block size is 2(256-254:FROM LAST NON-ZERO OCTET IN SUBNET MASK).
So the subnets would be divided by a factor of 2.

Starting from
172.16.0.0-172.16.1.255(including a network address and broadcast address).

As we are considering E0 starting from _._.2.1.

The subnet it comes under is as 127.16.2.0(being n/w id) and the flow is as follows:
172.16.2.0 NID.
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.0 *VALID ADDRESS(as it is not a n/w id).
172.16.3.1.
172.16.3.254 LAST VALID HOST.
172.16.3.255. BROAD CAST.

According to which it should be none of these networks.
Hope it will help you.

As given, router's IP address as 172.16.2.1/23 ;

it means last 9 bits (32-23) of binary form of 172.16.2.1 are responsible for deciding host ID.

When you make those last 9 bits 0 then u will get address as
172.16.2.0 and when you make those last 9 bits as 1 then you will get address as 172.16.3.255 ;

But 172.16.2.0 is used for network-ID (as 0 in last) and
172.16.3.255 is used for host-ID (as 255 in last).

So, rest of IPs will be vallid IPs for hosts.

23 is a prefix length. So IPV4 has 32 bit of address.

32 - 23 = 9. Now IP address of ethernet 0 is 172.62.2.1/23.

If we want to find out first address than we have to set 9 (32 - 23 = 9) rightmost bit as zero.

10101100.00111110.00000010.00000000 = 172.62.2.1.

So, the first address is 172.62.2.1.

last address -> put 9 rightmost bit as 1.

10101100.00111110.00000011.11111111 = 172.62.3.255.

So the range = 172.62.2.1 to 172.62.3.255.

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.

@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.

Answer C is correct because;

Given IP- 172.16.2.1/23.
First IP used as Network Id: 172.16.00000010.00000000(172.16.2.0).
First Valid Host IP: 172.16.00000011.00000000(172.16.3.0).
Broadcast IP: 172.16.3.255 & NOT 172.16.2.255.

So, 172.16.2.255 is a valid Host IP.

The valid host range will be (172.16,2.1-172.16.3.254).

The network address : 172.16.2.0.

The broadcast address: 172,16.3.255.

The network 172.16.2.0 will ends 172.16.2.55 then it will become 172.16.3.0 which is still on the range of the network.

172.16.2.0 NID.
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.1.
172.16.3.255.
172.16.4.1.
172.16.4.254 LAST VALID HOST.
172.16.4.255 BROAD CAST.

172.16.3.0 it's not a valid address.

Between in these option valid address is 172.16.2.255

00000000.00000000.0000000|0.00000000.
76543210.76543210.7654321|0.76543210.
1.255.

IP starts from 2.1 and ends with 3.254. So we can't use these numbers. It means we can use 172.16.2.1 - 172.16.3.254. I think so:x.