IP Subnetting

megan1989

I am having some problems grasping this concept and am hoping someone can guide me in the right direction for a question like this:
An engineer is performing the IP subnet design for an organization and needs to find the best mask that should be used. The organization has been assigned the 172.52.239.0/24 network. They must have the addresses split up so they can be equally organized into 10 subnets of at least 32 addresses. Which mask would best fit this requirement?
A. 255.255.255.240
B. 255.255.255.224
C. 255.255.255.248
D. None

Any help would be appreciated, thank you

fuz1on

I think it's best to look at the question in 2 parts. First, they've been assigned 172.52.239.0/24 so you know the restraints. Second, you need to make 10 subnets and at least 32 addresses so just look for the correct increment. GL!

megan1989

Okay I think I am headed in the right direction, thank you. What do you mean by the increment, magic number, interesting octet? So far I am trying to use network, subnet and host bits to try to figure this out. It needs 10 subnets so, 4 bits for the subnet 2^4=16, and for the host bits I multiplied the subnet 10 and the addresses together so 320 so 9 bits for that 2^9 = 512. 16 network bits as it is class B, does any of this make sense, or I am even headed in the right direction? Right now the subnet would be missing 4 bits I assume just fill that in with more subnet or network bits. Thank you

Jon_Cisco

Your math looks ok but you need to apply it to the question. Also I'm not sure if this is an attempt to trick you but for 32 hosts you will need a network size of at least 64. Since 32-2 only provides 30 hosts.

According to your question you are given 8 bits to work with.

clarson

i guess if you like doing it the hard way that is how you would start.
but, with a /24 network that leaves 8 host bits. which mean you can easily place an upper limit on the number of hosts at 256. there will be addresses that can not be used (network and broadcast addresses). So, you aren't even going to be able to have 256 hosts. 10 subnets of 32 hosts requires at least 320 host addresses. 256 is less than 320 so it can not be done. The answer is none.

clarson

another way is you need 32 hosts per subnet. as jon pointed out, the 32 increment/magic number only gives you 30 host addresses. Therefore, you will need to use 64 as your increment. that requires 6 host bits. you need 10 subnets. 10 is greater than 8 but less than 16 so you require 4 bits for subnets. 6+4 equals 10. you require 10 bits for subnet and hosts. your network at most can be a /22 network. once again can't be done with a /24 network.

megan1989

Thank you for explaining how this process works, with your help I think I now understand this process!

binarysoul

When I first looked at it, I thought it's not possible hence answer None.

Wording is a bit unclear; I'm assuming it says 32 hosts/subnet, and not 32 hosts in total. In that case, B would be the answer (doing math in my head)! But this is highly unlikely.

satishtech

B may not be the answer as 2^5 gives 32-2 = 30 hosts ?
and 32 hosts are required...
correct me if i am wrong
what is the answer ?

clarson

but that does make the question ambiguous. it does just say 32 addresses and not 32 host addresses. So 30 host address, a network address and a broadcast address adds up to 32 addresses. But, still that would mean 5 hosts bits, 4 bits for subnet. Meaning that you would require a /23 network. still cant be done with a /24 network.

clarson

b could not be the answer as you require 4 subnet bits. with a /24 network, b only gives you 3 subnet bits. if you are only required a total of 32 host addresses on all 10 subnets then A gives you 16 subnets with 14 host addresses so that works. but so does C with 32 subnets with 6 host address.

satishtech

yes 3 subnet bits with a /24 network
2^3 = 8 Subnets and 10 are required hmmmm krikey !!
maybe D none is the answer