Urgent Subnetting help needed

Unanswered Question
Aug 3rd, 2007
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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

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Edison Ortiz Fri, 08/03/2007 - 12:03
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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 - 12:06
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Your explantion is better :)

Edison Ortiz Fri, 08/03/2007 - 12:23
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Both equal in value to the community :)

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