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Beginner

Class B Subnetting

ok i been going through subnetting and when it comes to class B subnetting i get confused. I am stuck at 2 problems and here they are. If can help me out,

Problem 1:-

172.16.0.0 = Network address

255.255.255.128 = Subnet mask

in other words 172.16.0.0 / 25

Problem 2:-

172.16.0.0 = network address

255.255.255.192 = Subnet mask

so what will be the valid subents, host range and broadcast range.

Thank you

1 ACCEPTED SOLUTION

Accepted Solutions
Highlighted

Ahmed

Alain's answer is correct. The confusion comes because -

172.16.0.0 255.255.255.128 does indeed mean -

172.16.0.1 - 172.16.0.126 broadcast 172.16.0.127

and you can see the same answer from the first column in Lammle's example.

But then you can also have -

172.16.0.128/25  -> hosts 172.16.0.129 -> 254 broadcast 172.16.0.255

172.16.1.0/25  -> 172.16.1.1 -> 127 broadcast 172.16.1.127

etc..

this is what Lammle's example is showing. You can with a class B address subnetted with a 255.255.255.128 subnet mask have 512 subnets each with 126 hosts per subnet.

So Alain was referring specifically to the first subnet only 172.16.0.0/25. He wasn't saying you can't have other subnets from the same class B address with the same subnet mask (apologies if i am misrepresenting you Alain).

Jon

View solution in original post

6 REPLIES 6
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Advisor

Hi,

subnetting a class B is not different from subnetting another class.

ok first problem:

magic number= 256-128=128 is the subnet increment so subnets are:

172.16.0.0-172.16.0.128 and in the first subnet 172.16.0.1 is first IP, 172.16.0.127 is broadcast so last ip is 172.16.0.126

now on to 2nd problem:

magic= 256-192=64 so subnets are 172.16.0.0-172.16.0.64-172.16.0.128-172.16.0.192

and in first subnet: 172.16.0.1 is first ip, 172.16.0.63 is broadcast so 172.16.0.62 is last ip.

Regards.

Alain.

Don't forget to rate helpful posts.
Highlighted

thank you for replying to my post.

Todd lammale explained it in a bit different way so heres what i read.

For problem 1,

according to him. the subnets will be 0 and 128 but since we cant forget the one subnet bit used in the fourth octet so the subnets will be like this

Subnet - 0.0  0.128 - 1.0 1.128 -        2.0    2.128 -     3.0     3.128 - 4.0 4.128 - ..........  255.0 255.128

1st H   - 0.1 0.129 -  1.1 1.129 -        2.1     2.129 -    3.1     3.129 - 4.1 4.129 -.............255.1 255.129

lst H    - 0.126 0.254 - 1.126 - 1.254 - 2.126 2.254 -    3.126 3.254 -..............           255.126 255.254

Brdcst - 0.127 0.255 - 1.127 1.255 -  2.127 2.255 -    3.127 3.255 -..........               255.127 255.255

this is what i have read for the first problem the second problem he solved like this i will post that to but first we have to sort out this first problem

Thank you

Highlighted

Ahmed

Alain's answer is correct. The confusion comes because -

172.16.0.0 255.255.255.128 does indeed mean -

172.16.0.1 - 172.16.0.126 broadcast 172.16.0.127

and you can see the same answer from the first column in Lammle's example.

But then you can also have -

172.16.0.128/25  -> hosts 172.16.0.129 -> 254 broadcast 172.16.0.255

172.16.1.0/25  -> 172.16.1.1 -> 127 broadcast 172.16.1.127

etc..

this is what Lammle's example is showing. You can with a class B address subnetted with a 255.255.255.128 subnet mask have 512 subnets each with 126 hosts per subnet.

So Alain was referring specifically to the first subnet only 172.16.0.0/25. He wasn't saying you can't have other subnets from the same class B address with the same subnet mask (apologies if i am misrepresenting you Alain).

Jon

View solution in original post

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Hi Jon,

perfect understanding of my reply and thanks for clarifying for OP.

Regards.

Alain.

Don't forget to rate helpful posts.
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I also studied in the same way in a textbook, but I couldn't understand how!! could you explain it to me? thanks in advance

 

Highlighted

Agree. The confusing part in Lammle's explanation is why he subtracted 255 (not 128 as in all other examples, where from 256 are subtracted forth octet values) in 255.255.255.128 subnet.

 

Anyone can help please?
Thank you

 

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