• # Classless IPv4 Part 3 Question

**you are a network admin for demo.itpro.tv and you have been tasked with dividing the network into subnets. You have the following information

Network 209.123.221.0/24

You need 5 networks with 28 hosts per network
You need to provide additional networks for the future

Which of the following would accomplish this task
255.255.255.224
255.255.255.255
255.255.255.248
255.255.255.0**

Wes immediately says 255.255.255.0 cannot be right because it comes up with the same thing that we already have. I am confused by this, because wouldn't that already support 5+ networks and 30+ hosts already? Why do we need to further divide the network? Am I to assume that the question is to add 5 additional networks? I am probably being a bit nit picky here but I notice a lot of the practice questions try to catch you on things like this. It seems as though the existing network already allows for what they require.

• Remember that there's a purpose to doing this. It's function is merely division but it's PURPOSE to to be "less wasteful" or "more efficient use" of the address space itself.

You're given 209.123.221.0/24 . this is ONE network with a 254 possible hosts. So using this for only about 25 hosts will waste the address space of about 220 addresses that will not be used. So subnetting it is about efficiency.

With the requirements given, you're not creating an additional 5 subnets but you're dividing the ONE into 5. The key though is do yo know if can be done... and if so how do you do so.

Cordially,
Ronnie Wong
Edutainer, ITProTV

*if the post above has answered the question, please mark the topic as solved.
**All "answers" and responses are offered "as is" and my opinion. There is no implied support or guarantee by the ITProTV team.

• @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:

• 1- That subnetting can be thought of as subdividing a large network into smaller networks

• 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
• 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.

Best Regards,
Wes Bryan

Knowledge is a road to be traveled upon, not a destination to be reached~~

• Thank you both for the excellent responses!

After reading them and reviewing it again I now see where I was getting confused. I was assuming that "dividing the network into subnets" meant that I had access to other networks i.e 209.123.221.0/24, 209.123.222.0/24,209.123.223.0/24 and I could use this to get 5 networks with 28 hosts per network while still allowing networks for the future and then technically 255.255.255.0 could be an option. I believe this would only be correct if the network given was 209.123.0.0/16 but even then it would be a massive waste of addressing. I understood the concept, but was getting hung up on what I had access too and assuming the questions are trying to trick me haha! Thank you clearing this up for me and taking so much time for provide a detailed response.

• @Mike-Tennyson I want to make a correction to a typing error in my explanation:

"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:"

That should say:

""If we apply the first subnet mask then we added three 1 bits from the host portion to the network portion, how many "subdivisions" does this make? Let's look at the possible combinations below:"

I apologize for the error.

Best Regards,
Wes Bryan

Knowledge is a road to be traveled upon, not a destination to be reached~~

Posts 5Views 93