Subnetting question?

Yes there are 510 usable hosts in a /23. Classes are really meaningless once you break the /8, /16 or /24 boundry. Example 10.0.0.0/23 is technically a class "A" address, it still contains 510 hosts. 172.16.0.0/23 is technically a class "B" address, it still contains 510 hosts. 192.168.0.0/23 is technically a class "C" address, it still contains 510 hosts.

When using a /23 mask there is a very quick shortcut you can use. If the 3rd octet is an even number it immeadiately tells you it contains the network address and the rest are valid hosts. If the 3rd octet is an odd number it immeadiately tells you it contains all hosts and the broadcast address.

113.10.4.0 is a network address 186.54.3.0 is a valid host, it belongs to the 186.54.2.0/23 network. 175.33.3.255 is the broadcast address for the 175.33.2.0/23 network 26.35.2.255 is a valid host on the 26.35.2.0/23 network 152.135.7.0 is a network address 17.35.36.0 is a valid host on the 17.35.35.0/23 network

Thanks alot for the explanation-will definitely remember the /23 shortcut you explained-I think what threw me was looking at the

26.35.2.255 I thought it was a b/c because of the .255 -however I am assuming the b/c would be 26.35.255.255 ?

This sort of stuff gives me brain-ache :-( Isn't it:

152.135.7.0 is the valid address and 17.35.36.0 the network address??

We've added new IP addressing tutorial links to our webpage. I suggest the 3Com IP Addressing Document. It has a workbook at the end...quite useful.

Yep, i was banging my head as well. I concur with your selection.

You guys are absolutely correct! Not enough coffee!

try this. it's good practice on subnetting

and also subnetmask calculator and explaination

http://subnetmask.>> Hi all -I saw this question and cant figure out how to work it out-all