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This Question is Answered 2 Helpful Answers available (2 pts)
1517 Views 2 Replies Latest reply: Sep 27, 2009 11:29 PM by Conwyn

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## Subnetting Q?

### Sep 27, 2009 10:55 PM

58 posts since
Sep 23, 2009

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

• subnet mask
• broadcast address
• 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

Please help me with the broadcast address, host bits, subnet bits, and if you could explain how you get that...

I would appreciate similar answered questions.

ab

• 7,913 posts since
Sep 10, 2008
Currently Being Moderated
1. Sep 27, 2009 11:29 PM (in response to abkaniki)
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

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