Simeon M. answered • 07/30/22
B.S. In Cyber Security and Information Assurance
Ok so, with each multiple of 8 you know which octets you should be working in
With a CIDR of 28, we should know that we are working inside of the 4th octet.
So the first three will stay the same.
28 is higher than 24. We know there are 4 on bits in the last octet from this.
We take the following to do our binary conversion.
128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0
So we have 16 IPs in each set, 14 that are usable in each set
Sense the range of IPs is greater than a 2x multiple of 16, and not 3x, we know that it's the 3rd range of IPs.
Remember we start with 0 in the first range. So x.x.x.0 - x.x.x.15 is the first range. x.x.x.16-x.x.x.31 is our second range.
This leaves us with our third range, x.x.x.32-x.x.x.47
With this information, you should be able to to deduce the following.
The first address in the block is 205.16.37.32
The last address is in block is 205.16.37.47
And there are 16 addresses in the block.
However, the questions can be confusing.
If they use the word "usable" then the answers change a bit.
Remember that the network and broadcast addresses are always predefined based on the IP plus subnetmask/CIDR. So you always lose two usable addressees.
The 205.16.37.32 is the network address and 205.16.37.47 is the broadcast for this range.
That in mind we know that 205.16.37.33 - 205.16.37.46 is our usable range and we know we have only 14 IP addresses available for our hosts.
If you want a more detailed explanation of how this works, I highly recommend the following video.
Subnetting is best taught with visual aides. https://www.youtube.com/watch?v=ZxAwQB8TZsM
