Fasmy98,

The /20 represents the subnet bits that are locked to "1", or in simple english, it means 20 subnet bits.

I'll analyse for you the 172.16.0.0/19 to help you understand whats happening.

172.16.0.0 with 19 subnet bits

19 subnet bits= 1111 1111 . 1111 1111 . 1110 0000 . 0000 0000

In decimal, that means 255.255.224.0

So, 19 bits= 255.255.224.0

Now, we need to find how many bits are used for the subnets.

Since we have a Class B network (172.X.X.X), the default subnet mask for it will be 255.255.0.0.

Default mask= 255.255.0.0

Our mask is = 255.255.224.0

Lets look at it in binary to give us a better idea:

Default Mask= 1111 1111 . 1111 1111 . 0000 0000 . 0000 0000

Our Mask is_= 1111 1111 . 1111 1111 . 1110 0000 . 0000 0000

Now you can clearly see the 3 bits used --^

These 3 bits give us 2^3-2=6 useable subnets

Let's take a look at them:

172.16.0.0 -Subnet Zero - not used according to Cisco

172.16.32.0 - First Valid Subnet

172.16.64.0

172.16.96.0

172.16.128.0

172.16.160.0

172.16.192.0 - Last Valid Subnet

172.16.224.0 - Subnet broadcast - not used according to Cisco

Now take notice that according to Cisco, the first and last subnets, that is, subnet zero and the broadcast subnet, are NOT valid.

In practice though, you can use them without a problem!

Let's now take a look at the available Hosts per Subnet. We will use the first one as our example:

Subnet : 172.16.32.0

Bits assigned to Network+Subnet=19

Remaining bits for hosts= 32-19= 13

Valid number of hosts= 2^13-2= 8190 hosts PER subnet.

First valid IP for this subnet= 172.16.32.1

Last valid IP for this subnet= 172.16.63.254

Broadcast for this subnet= 172.16.63.255

Now what you have here is a full analysis of the 172.16.0.0/19 network and its first useable subnet.

I hope that answers your questions. If anyone spots a mistake, please correct me!

Cheers,