Urgent Subnetting help needed

Unanswered Question
Aug 3rd, 2007

Hi All,

Could anyone be kind to help with the first six subnets of the following address WITHOUT inclusion of subnet zero and please explain to help my understanding.



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Overall Rating: 4 (2 ratings)
Edison Ortiz Fri, 08/03/2007 - 12:03

27 bits are on


5 bits are variable

2^5 = 32

You need a subnet ID and a broadcast address on each subnet, so starting with the first subnet. [subnet ID] [IP for Host] [broadcast for that subnet] [subnet ID] [IP for Host] [broadcast for that subnet] [subnet ID] [IP for Host] [broadcast for that subnet]

And I will let you do the last 3. The logic is adding 32 to the subnet ID and broadcast address, whatever you have in between those 2 values, they are your IPs for the hosts.


Jon Marshall Fri, 08/03/2007 - 12:05


You can use binary to write this out but a quick way to work it out

/27 =

256 - 224 = 32

So you subnets go up in increments of 32.



newnetman Fri, 08/03/2007 - 12:49

Hi All,

I am most delighted for all your magnificent responses.

I am so proud to call you all my invisible mentors.

May you all be bountifully blessed.



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