The main idea of this exercise is to fill the following table:
Network Address | 172.16.0.0 | 172.16.0.32 | . . . . . . | 172.16.0.224 |
First Host Address | 172.16.0.1 | 172.16.0.33 | . . . . . . | 172.16.0.225 |
Last Host Address | 172.16.0.30 | 172.16.0.62 | . . . . . . | 172.16.0.254 |
Broadcast Address | 172.16.0.31 | 172.16.0.63 | . . . . . . | 172.16.0.255 |
I have shown only \(\displaystyle 3\) subnets out of \(\displaystyle 8\). I have shown the first, second, and last subnet in this network. In later posts, you will understand why this network has \(\displaystyle 8\) subnets and why they are divided in this range.
For now this information is sufficient to show you what kind of ip address is \(\displaystyle 172.16.3.159\)
I started with the network address \(\displaystyle 172.16.0.0\) and I got the table above. If I start with the network address \(\displaystyle 172.16.3.0\), I will get the same table above, but instead of having \(\displaystyle 0\) in the third octet, I will have \(\displaystyle 3\).
The length between the first and second network address is \(\displaystyle 32\), this means that each time I add \(\displaystyle 32\) I get the next network address. Let us count.
First Network Address: \(\displaystyle 172.16.0.0\)
Second Network Address: \(\displaystyle 172.16.0.32\)
Third Network Address: \(\displaystyle 172.16.0.64\)
Fourth Network Address: \(\displaystyle 172.16.0.96\)
\(\displaystyle 5^{\text{th}}\) Network Address: \(\displaystyle 172.16.0.128\)
\(\displaystyle 6^{\text{th}}\) Network Address: \(\displaystyle 172.16.0.160\)
\(\displaystyle 7^{\text{th}}\) Network Address: \(\displaystyle 172.16.0.192\)
\(\displaystyle 8^{\text{th}}\) Network Address: \(\displaystyle 172.16.0.224\)
This information tell us that since \(\displaystyle 172.16.0.160\) is a network address, the ip that comes before it \(\displaystyle 172.16.0.159\) must be a broadcast address.
Then
\(\displaystyle 172.16.3.159\) is a broadcast address.
