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 172.23.0.0/27 WITHOUT inclusion of subnet zero and please explain to help my understanding.

Regards

Newnetman

I have this problem too.
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Edison Ortiz Fri, 08/03/2007 - 12:03

27 bits are on

11111111.11111111.11111111.11100000

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.

172.23.0.0 [subnet ID]

172.23.0.1-30 [IP for Host]

172.23.0.31 [broadcast for that subnet]

172.23.0.32 [subnet ID]

172.23.0.33-62 [IP for Host]

172.23.0.63 [broadcast for that subnet]

172.23.0.64 [subnet ID]

172.23.0.65-94 [IP for Host]

172.23.0.95 [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.

HTH,

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

Hi

172.23.0.32

172.23.0.64

172.23.0.96

172.23.0.128

172.23.0.160

172.23.0.192

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

/27 = 255.255.255.224

256 - 224 = 32

So you subnets go up in increments of 32.

HTH

Jon

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.

regards

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