Network+ N10-005 - Subnetting Example #7

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In Subnetting Example#7 (which can be seen below) I understand that there are 8 possible subnets available within the last octet of the subnet mask for the solution, but what about all other possible subnets availble from the first 24 bits (first 3 octects of the ip addresses). Why is the answer not 134,217,728 (2^27 for the /27 subnet)? Is it the way the question is phrased or am I just confused? Why does it only pertain to the possible subnets in the last octet?

------------------------------------Subnetting Example#7---------------------------------
Problem:

Your company uses the 172.16.20.0/24 network for its LAN. After a security breach, it has been decided that each department should be placed on its own subnet. Each department currently has 20 computers. What is the maximum number of subnets you can create by dividing the 172.16.20.0/24 network? Each subnet must be able to contain at least 20 hosts.

Solution:

Each subnet must be able to have 20 hosts per subnet. You currently have a single subnet with 254 host. 172.16.20.0/24 (1 network, 254 hosts) 172.16.20.00000000/24 (1 network, 254 hosts) 1st questions how many bits must remain as HOST bits to have 20 hosts per network? 172.16.20.00000000

1 bit = 2 host (not useable) 172.16.20.00000000
2 bits = 4 hosts (not useable) 172.16.20.00000000
3 bits = 8 hosts (not useable) 172.16.20.00000000
4 bits = 16 hosts (not useable) 172.16.20.00000000
5 bits = 32 hosts (useable) 172.16.20.00000000
the original single network: 172.16.20.00000000
the required network solution: 172.16.20.00000000 show that we would be able to create 2^3 power (8 networks)
Each of those 8 would support 30 hosts per network. 172.16.20.0/27, 172.16.20.32/27, 172.16.20.64/27, 172.16.20.96/27, 172.16.20.128/27, 172.16.20.160/27, 172.16.20.192/27, 172.16.20.224/27

@Yousef-Hegazi - Maybe it will be more understandable from this perspective:

The original 172.16.20.0/24 network has a Network address of 172.16.20.0, and a Broadcast address of 172.16.20.255 (these are the delimiters for the /24 subnet), with 254 possible hosts (usable IP addresses) between them. Let's split that /24 network into two /25 subnets:

Network 1: Defined as 172.16.20.0/25, Network address of 172.16.20.0, Broadcast address of 172.16.20.127, with 126 possible hosts
Network 2: Defined as 172.16.20.128/25, Network address of 172.16.20.128, Broadcast address of 172.16.20.255, with 126 possible hosts

Further subdivided into four /26 subnets:
#1: 172.16.20.0 Network address, 172.16.20.63 Broadcast address, 62 possible hosts
#2: 172.16.20.64 Network address, 172.16.20.127 Broadcast address, 62 possible hosts
#3: 172.16.20.128 Network address, 172.16.20.191 Broadcast address, 62 possible hosts
#4: 172.16.20.192 Network address, 172.16.20.255 Broadcast address, 62 possible hosts

The provided answer is further subnetting into eight /27 networks, using the same conventions. A table or spreadsheet would be an easier way to demonstrate separating a /24 network into smaller blocks of IP addresses. Does my example help? Remember you are "losing" addresses to the Network and Broadcast delimiters.

Yousef-Hagazi,

David Beem is correct in his explanation. The problem begins with a PROVIDED network. Think of the fact that other networks are not or may not be under your administrative control, therefore you cannot encroach and use them. Your recourse is to take that given network and subdivide that network into equal number of networks you have for your department or building, etc.

David also indicates that the use of a spreadsheet would have been more helpful, I agree. The only thing though that keeps me from doing that is the fact that you will not have access to a spreadsheet program when it comes to any exam dealing with subnetting. So I show it to you in the simplified way that I would work it out if I were writing the exam on exam day. You may wonder why I then use notepad because that won't be on the exam either. This is more for you eyesight, since my penmanship is so bad, you may wonder if I was writing Klingon or Vulcan rather than English.

Cordially,
Ronnie Wong
Host, ITProTV

I can understand that there is no substitute for the math, this example of a subnetting word problem was very "real-world" in how it is applied (I am responsible for our IPv4 space, and had to subnet from /24 to /32 based on what I estimated we would need). If you commonly encounter situations of breaking up a /24 you start to remember the structure. Also remember to read a word problem thoroughly, and what it is trying to describe.

Ok I understand now. I just need to pay closer attention to the boundaries set and think more real world. I was just worried that the exam would turn something like that into a trick question where they would want to know the theoretical limit as opposed to just the limit within the given range for the real world case given by the question.

Should I expect trick questions like that on the exams?

@Yousef-Hegazi - I know nothing of the exams, but would hope it would be real-world examples (usually subnetting a larger block into smaller parts). Practice with the examples, and be aware of what the question is. Good luck to you.

Thank you