Trancenders Question of the day ?

Irish Man

My Network IP Address is 204.99.99.0/24

I require 3 Subnets
SubNet 1 Requires 10 hosts
SubNet 2 Requires 25 hosts
Subnet 3 Requires 60 hosts

An additional Subnet will be added in the future. I must devide the ip addresses to reserve the max number of ips for the future Subnet.

Seems simple enough.
I break down in to subnetting as follows.
SN1 - 204.99.99.240/29
SN2 - 204.99.99.224/28
SN3 - 204.99.99.192/27

But they tell me the answer is incorrect, am I wrong here ?

Thanks for the help.
Colin

royal

Subnet 1 requires 10 hosts:
2^1-2=0 which is not enough
2^2-2=2 which is not enough
2^3-2=6 which is not enough
2^4-2=14 which gives enough

This means for Subnet 1, we require 4 bits in the client portion of the ip address. Since we are using /24, we have 8 bits available for client ip addresses. Since this is too much, to be efficient, we will change this to a /28 to provide for the 4 bits needed for clients.

Subnet 1 will be 204.99.99.240 /28

The reason it is .240 is because since we are using 4 bits for the client, we will be using the left-most 4 bits for the network. Binary works as 128 64 32 16 8 4 2 1. The 4 left most bits added up = 240. 128+64+32+16=240.


Subnet 2 requires 25 hosts:
2^1-2=0 which is not enough
2^2-2=2 which is not enough
2^3-2=6 which is not enough
2^4-2=14 which is not enough
2^5-2=30 which is enough

This means for Subnet 2, we require 5 bits in the client portion of the ip address. Since we are using /24, we will have 8 bits available for the client ip addresses. Since this is too much, to be efficient, we will change this to a /27 to provide the 5 bits needed for clients.

Subnet 2 will be 204.99.99.224 /27

The reason it is .224 is because since we are using 5 bits for the client, we will be using the left-most 3 bits for the network. Binary works as 128 64 32 16 8 4 2 1. The 3 left most bits added up = 224. 128+64+32=224.


Subnet 3 requires 60 hosts:
2^1-2=0 which is not enough
2^2-2=2 which is not enough
2^3-2=6 which is not enough
2^4-2=14 which is not enough
2^5-2=30 which is not enough
2^6-2=62 which is enough

This means for Subnet 2, we require 6 bits in the client portion of the ip address. Since we are using /24, we will have 8 bits available for the client ip addresses. Since this is too much, to be efficient, we will change this to a /26 to provide the 6 bits needed for clients.

Subnet 3 will be 204.99.99.192 /26

The reason it is .192 is because since we are using 6 bits for the client, we will be using the left-most 2 bits for the network. Binary works as 128 64 32 16 8 4 2 1. The 2 left most bits added up = 192. 128+64=192.


So to sum it all up

Subnet 1 = 204.99.99.240 /28
Subnet 2 = 204.99.99.224 /27
Subnet 3 = 204.99.99.192 /26

Hopefully this helps

Irish Man

Brilliant Explaination

Cheers
Colin

georgemc

Irish Man wrote:

My Network IP Address is 204.99.99.0/24

I require 3 Subnets
SubNet 1 Requires 10 hosts
SubNet 2 Requires 25 hosts
Subnet 3 Requires 60 hosts

An additional Subnet will be added in the future. I must devide the ip addresses to reserve the max number of ips for the future Subnet.

Seems simple enough.
I break down in to subnetting as follows.
SN1 - 204.99.99.240/29
SN2 - 204.99.99.224/28
SN3 - 204.99.99.192/27

But they tell me the answer is incorrect, am I wrong here ?

Thanks for the help.
Colin

I would break it down as follows:

Subnet 1 mask = 255.255.255.240 or /28 = 14 hosts
Subnet 2 mask = 255.255.255.224 or /27 = 30 hosts
Subnet 3 mask = 255.255.255.192 or /26 = 62 hosts

I would then place subnet 3 at the end of the available space

Subnet 3 = 204.99.99.192/26 (204.99.99.192 - 204.99.99.255)

I would leave subnet 204.99.99.128/26 (204.99.99.128 - 204.99.99.191) free for future use as another 62 host subnet

I would assign Subnet 2 as 204.99.99.96/27 (204.99.99.96 - 204.99.99.127)

I would leave subnet 204.99.99.64/27 (204.99.99.64 - 204.99.99.95) free for future use as another 30 host subnet

I would assign Subnet 1 as 204.99.99.16/28 (204.99.99.16 - 204.99.99.31)

I would leave subnet 204.99.99.0/28 (204.99.99.0 - 204.99.99.15) free for future use as another 14 host subnet

This also leaves subnet 204.99.99.32/27 (204.99.99.32 - 204.99.99.63) free for future use as another 30 host subnet or two 14 host subnets.

You should place you similarly masked subnets continuously within you available range. I prefer to place the larger subnets near the end and smaller ones near the beginning.

To summarize:

204.99.99.0/28 = 14 host subnet free for future use
204.99.99.16/28 = Subnet 1 (10 hosts required, 14 total available)
204.99.99.32/27 = 30 host subnet free for future use
204.99.99.64/27 = 30 host subnet free for future use
204.99.99.96/27 = Subnet 2 (25 hosts required, 30 total available)
204.99.99.128/26 = 62 host subnet free for future use
204.99.99.192/26 = Subnet 3 (60 hosts required, 62 total available)

I hope this helps.

Georgemc

georgemc

I was kinda looking for a response...

Raven Ramon

I will give you a response bro.........excellent job !!!

I was looking for exactly what your response was. "How-To" & "Why" Assign the IP addressing the way you did makes perfect sense for organizing and designing a network. Thanks alot bro !

Raven Ramon

mattd75

Hey george, since you have already answered the op's question, I have one for you. which was harder: CCNA or MCSA?

Thanks!

knownhero

Royal thanks for that help man. You're a dude! Helped me alot!