Networking - Subnetting - Discussion

How many network will divide by this subnet mask ?

How you know the number of host to each subnet will be from 2.1 to 3.254 ?

If the range is like
.
.
172.16.2.253
172.16.2.254
172.16.2.255
172.16.3.0
172.16.3.1
172.16.3.2
.
.
How come 172.16.3.0 is not a valid address ?

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.

the option C is broadcast address and the others is host address.
that given address is class B (from 1st octet 172).

So here is sub network addr:

172.16.0.0
172.16.2.0
...
172.16.254.0

But as we always say that first and last ip is for networkk and host id than here why we are not considering 172.16.2.255 and 172.16.3.0.

Please tell since this network is divided into two bock then how?

I think the valid host IDs are between 172.16.0.1 to 172.16.1.254.

And broadcast address is 172.16.1.255.

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.].

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.

2 is the block size.

So 0, 2, 4, 6 is N.ID.

0.1, 0.2, 0.3 to 1.254 are host.

Similarly.

2.1, 2.2, 2.3, 2.4 to 2.254, 2.255, 3.1, 3.2 to 3.254 are host.

But 3.255 are B.id and 4.0 n.id.

Don't know if it helps but 172.16.1.100 and 172.16.1.198 are kinda off from the default gateway 172.16.2.1/23 notice that to there?