help with /17 and /18 subnetting

[Akbarayub]

Dear all,
I was wondering if someone could help me with class B /17 and 18 subnetting i don't understand how the octets increment i find subnetting easy on class C now but get confuse with increments with class B. . I would appreciate if you can provide me with examples for the whole range so that i can make my own conclusions. Thanks in advance.

[JohnTylerPearce]

A /17 is using 17 bits for the network prefix, so with a class B , you are reserving one bit for the network portion 15 for host bits. So you are going to have 2 subnets and 32766 hosts if I did my math right. 184.64.0.1 - 184.64.127.254 and 184.64.128.1 - 184.64.255.254.



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[Akbarayub]

Thank you for the prompt response. But i thought on a /17 it increments at 128 and your example is showing 64 i think.

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[Richard Burts]

Your math is not correct. John is correct that a /17 would have only two subnets in which the third octet would start at 0 and increment to 128 as his post shows.

HTH

Rick

[Akbarayub]

Thank you for the response how do you figure out the number of hosts?

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[Joseph W. Doherty]

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Posting

Actually what John showed was the addresses for hosts, actual address blocks would be:

x.x.0.0/16 (x.x.0.0 - x.x.255.255)

x.x.0.0/17 (x.x.0.0 - x.x.127.255)

x.x.0.0/18 (x.x.0.0 - x.x.63.255)

x.x.64.0/18 (x.x.64.0 - x.x.127.255)

x.x.128.0/17 (x.x.128.0 - x.x.255.255)

x.x.128.0/18 (x.x.128.0 - x.x.191.255)

x.x.192.0/18 (x.x.192.0 - x.x.255.255)

Hosts are 2 to the N (host bits) - 2.  So for a /16 you have 2**16-2=65,534; for /17 2**15-2=32,766; and for /18 2**14-2=16,382

[Akbarayub]

Thank guys still not sure how to do class B subnetting.

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[Richard Burts]

You have told us that you are comfortable with subnetting of class C where the subnets are configured in the fourth octet. Subnetting with a class B is quite similar where the subnets are configured in the third octet.

HTH

Rick

[Akbarayub]

Dear Rick,
Thanks for the prompt response. Yeah i can easily subnet class C using the magic number method. I get all confused with class B for some strange reason. I just wanted some clear examples to help me understand it.

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[Richard Burts]

This works for me. I hope that it will work for you also:

! class C subnetting

mask /30  number of host bits 2 number of hosts (2^n - 2)  = 2

mask /29  number of host bits 3 number of hosts (2^n - 2)  = 6

mask /28  number of host bits 4 number of hosts (2^n - 2)  = 14

mask /27  number of host bits 5 number of hosts (2^n - 2)  = 30

mask /26  number of host bits 6 number of hosts (2^n - 2)  = 62

mask /25  number of host bits 7 number of hosts (2^n - 2)  = 126

mask /24  number of host bits 8 number of hosts (2^n - 2)  = 254

! class B subnetting

mask /23  number of host bits 9 number of hosts (2^n - 2)  = 510

mask /22  number of host bits 10 number of hosts (2^n - 2)  = 1022

mask /21  number of host bits 11 number of hosts (2^n - 2)  = 2046

mask /20  number of host bits 12 number of hosts (2^n - 2)  = 4094

mask /19  number of host bits 13 number of hosts (2^n - 2)  = 8190

mask /18  number of host bits 14 number of hosts (2^n - 2)  = 16382

mask /17  number of host bits 15 number of hosts (2^n - 2)  = 32766

mask /16  number of host bits 16 number of hosts (2^n - 2)  = 65534

HTH

Rick

[Sandeep Sharma]

Hi Akbar

There is simple rule of subnetting forgot class A,B,C. Just you need to remember the number bits you have to borrow from host part to network.

There would be 2 questions :

1. You know the host count and you need to create subnet accordingly.

For example you have 172.16.0.0/16 network and you required to create subnets so that each subnet have 500 hosts.

Answer> so for that you use the formula:      2^n-2 => 502  so from here  you need to get the value of n (where n is number of host bits)

here in our case n=9

172.16. _ _ _ _ _ _ _/ _ . _ _ _ _ _ _ _ _ :::::: So here you can see the / is distingusing the subnet or network part with host. 

So the first subnetwork will be  172.16.0.0/25    

First ip : 172.16.0.1

last IP : 172.16.1.254

Broadcast IP : 172.16.1.255

Hope this will help you.