2 Replies Latest reply: Sep 27, 2009 11:29 PM by Conwyn

# Subnetting Q?

Could someone calculate and explain this for me. I am able to start but I always miss calculate and make mistakes. I need help getting the following:

172.17.250.105/27

• host bit range

Please explain. I can go this far for now

172.17.250.105          10110010.00010001.11111010.01101001

255.255.255.224        11111111.11111111.11111111.11100000

subnet ID                    172.17.250.96

I would appreciate similar answered questions.

ab

• ###### 1. Re: Subnetting Q?

Hi  Abkaniki

Let us try another way.

/32-/27 = /5   2 to the power of 5 is 32  so the sub net mask is 255.255.255.256-32 = 255.255.255.224 which you identified.

So the host block is 32 so lets retrurn to 172.17.250.105

If you know your 16 times table then you know 6 * 16 = 96 so 3 * 32 =96 so this subnet is 172.17.250.96

So the network is 172.17.250.96 the 1st host is .97 and the broadcast is 96+32-1 = .127 and the last  host.126

Regards Conwyn

• ###### 2. Re: Subnetting Q?

conwyn.flavell wrote:

Hi  Abkaniki

Let us try another way.

/32-/27 = /5   2 to the power of 5 is 32  so the sub net mask is 255.255.255.256-32 = 255.255.255.224 which you identified.

So the host block is 32 so lets retrurn to 172.17.250.105

If you know your 16 times table then you know 6 * 16 = 96 so 3 * 32 =96 so this subnet is 172.17.250.96

So the network is 172.17.250.96 the 1st host is .97 and the broadcast is 96+32-1 = .127 and the last  host.126

Regards Conwyn

Thanks for that Conwyn, you explained it really easy for me. I will try out a few questions, similar to this and see how I go. I guess the trick will be for me to chart down a few columns in my 10minutes of reading time before the final exam.

Thanks again, and hope you dont mind me calling on you when Im stuck on one of such Q?