In this post I will explain the notation
/#.
In the OP the notation
/22 means that the subnet mask has \(\displaystyle 22\) ones \(\displaystyle (1\text{s})\) starting from the left:
Subnet mask \(\displaystyle = 11111111.11111111.11111100.00000000 = 255.255.252.0\)
All the \(\displaystyle 1\)s together represent the network portion while all the \(\displaystyle 0\)s together represent the host portion.
If we look at the options in the OP, we know that the answer must be \(\displaystyle E\) (if it is not a network or broadcast address) because it is the only one which matches the correct subnet mask.
Anyone who wants to explore further why \(\displaystyle E\) may be the correct answer from a different perspective, read what I will write next.
The octet that has a mixture of \(\displaystyle 1\)s and \(\displaystyle 0\)s is called the interesting octet. In this case it is the third octet (\(\displaystyle 11111100\)). We let \(\displaystyle x\) be the number of \(\displaystyle 1\)s in that octet (third octet) and we let \(\displaystyle y\) be the number of all remaining \(\displaystyle 0\)s (in third and fourth octets).
Then, the number of subnets is: \(\displaystyle 2^x = 2^6 = 64\)
And the number of hosts per subnet is: \(\displaystyle 2^y - 2 = 2^{10} - 2 = 1024 - 2 = 1022\)
We let \(\displaystyle z\) be the number in the interesting octet, in our case \(\displaystyle z = 252\). And we say:
The block size \(\displaystyle = 256 - z = 256 - 252 = 4\) (But remember the subnets always start at zero.)
The block size tells us that our subnets are:
\(\displaystyle 0, 4, 8, 12,\)
16\(\displaystyle , 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, \cdots, 252\) (total of \(\displaystyle 64\))
If we focus on the fith subnet, the network address will be \(\displaystyle 172.16.16.0\) and the broadcast address will be \(\displaystyle 172.16.19.255\) and anything between them is a valid host address.
The number of IP addresses (host addresses) between the network address \(\displaystyle (172.16.16.0)\) and the broadcast address \(\displaystyle (172.16.19.255)\) is \(\displaystyle 1022\). This is also true for all of the remaining \(\displaystyle 63\) subnets.
We were given in the OP: 172.16.17.0/22. One may think that 172.16.17.0 is a network address, but according to our analysis even 172.16.17.0 is a valid host address in the fifth subnet.
Back to the IP address in \(\displaystyle E\): 172.16.18.255, it is indeed the only valid host address in a network (or subnet) that also contains the host address 172.16.17.0
I will make a table to plant the idea directly in your head.
Address\Subnet | 1 | . . . . | 5 | 6 | . . . . | Last Subnet |
|---|
Network Address | 172.16.0.0 | . . . . | 172.16.16.0 | 172.16.20.0 | . . . . | 172.16.252.0 |
First Host Address | 172.16.0.1 | . . . . | 172.16.16.1 | 172.16.20.1 | . . . . | 172.16.252.1 |
Last Host Address | 172.16.3.254 | . . . . | 172.16.19.254 | 172.16.23.254 | . . . . | 172.16.255.254 |
Broadcast Address | 172.16.3.255 | . . . . | 172.16.19.255 | 172.16.23.255 | . . . . | 172.16.255.255 |
These formulas are worth to remember in subnetting:
Number of subnets: \(\displaystyle 2^x\)
Number of hosts per subnet: \(\displaystyle 2^y - 2\)
Block size: \(\displaystyle 256 - z\) (the length between two network addresses of any two neighboring subnets)
