@Mike-Tennyson great question, like Ronnie stated the process of subnetting divides an existing network into smaller networks. With a subnet mask of 24 bits (/24) this is the equivalent of 255.255.255.0, so in this example, if you chose 255.255.255.0 then the network still has the same amount of hosts (like Ronnie mentioned, 254 hosts), our larger network has not be "subdivided."This is why I say that we cannot use 255.255.255.0, because that value is exactly the same as /24. While this will put 28 hosts on the network, it will not create additional smaller networks (subnets/subdivisions) which is also a condition of the question.
Remember a couple of concepts:
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1- That subnetting can be thought of as subdividing a large network into smaller networks
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2- Subnetting is done to more efficiently utilize the IP addresses that you have for a given network
- In this case we started with 254 IP addresses in the block
- To further divide this block we need to move some of the bits that are in the "host" portion of the address into the network portion
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3- /24 is the equivalent to 255.255.255.0 in dotted decimal notation.
- It is important to be able to determine the dotted decimal equivalents of slash notation
- This will help you to identify the right value given a scenario above
Take 255.255.255.255 for instance, this is a /32 slash notation which only allows for a single IP address (all binary 1 bits)
This leaves us will only two answers:
255.255.255.224 and 255.255.255.248
And the last part is to determine how many bits in the subnet mask would it take to create 5 networks. So let's look at the last octet
255.255.255.224 or 255.255.255.(11100000)
-------------------------------------.224
255.255.255.248 or 255.255.255. (11111000)
-------------------------------------.248
If we apply the first subnet mask the we added two 1 bits to the host portion. How many "subdivisions" does this make? Let's look at the possible combinations below:
00 000000
01 000000
10 000000
11 000000
There are 4 possibilities with two binary 1 bits added to the network portion of the subnet mask (also called borrowing host bits)
Each of these possibilities or combinations are 4 subnetworks/subdivisions of the larger network we started with (as mentioned by Ronnie). So this does not accomplish the "first half" of the condition as we needed 5 networks, we could stop there and save exam time as the only answer left is 255.255.255.248. but let's see if it does accomplish the second half of that condition? We need "28 hosts per network," to determine the amount of hosts you can use the formula 2^hostbits or 2^6=64 hosts.
Remember the first address in a network is the network ID so all nodes use this, and is denoted by all binary 0 bits in the host portion (00 000000, 01 000000, 10 000000) or the last address as this is the broadcast address (denoted by all binary 1 bits in the host portion...(00 111111, 01 11111,1 10 1111111). So we have 62 addresses available for hosts. This means it will accomplish the 28 hosts. But the answer is wrong
Let's look at the last possible answer and break it down piece by piece:
255.255.255.248 or 255.255.255. (11111000)
-------------------------------------.248
What possibilities do we have with 5 bits added to the network portion?
000 00000
001 00000
010 00000
011 00000
100 00000
101 00000
110 00000
111 00000
We can see above that there are 8 possibilities that we can make with 3 bits added to the network portion.
Another way to look at it is 2^3=8
These 8 possibilities are 8 subnetted (subdivided) smaller networks from our starting point of 1 network with 254 hosts. We have accomplished the first condition 5 networks (subdivided networks) with an addition 3 networks that we can use as the we add hosts to the network.
Now how about the hosts? Well let's count how many bits are remaining as those bits are for hosts:
2^5= 32 possibilities.
Remember the first address in a network is the network ID so all nodes use this (denoted by all binary 0 bits in the host portion) or the last address as this is the broadcast address (denoted by all binary 1 bits in the host portion....ie....(000 11111, 001 11111, 010 111111). This leaves us with 30 hosts if we subtract the two possibilities we cannot use. This will accomplish the first condition, allowing for 28 hosts on each of the 8 networks, which also meets the second condition of providing for additional networks (subnets/subdivisions) for the futures.
So building on what Ronnie said "The key though is do you know if can be done... and if so how do you do so.." I hope this walkthrough shows you that it can be done, also the "how" and "why" it is done via a step by step walkthrough.
