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Posted by on April 28, 2007, 12:26 pm
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> On Sat, 28 Apr 2007 07:12:06 -0400
>
>
>
>
> > > Hi all -I saw this question and cant figure out how to work it out-all
> > > help gratefully appreciated-
> > > The mask used is 255.255.254.0
> > > It asks which of the following 3 addresses can be used-
> > > 113.10.4.0
> > > 186.54.3.0
> > > 175.33.3.255
> > > 26.35.2.255
> > > 152.135.7.0
> > > 17.35.36.0
>
> > > Am I right in thinking the mask allows 510 usable hosts per subnet in
> > > a class C and only 1 host in a Class B?
> > > Thanks alot-I know the answers but dont understand why.
>
> > 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
>
> 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??
Yep, i was banging my head as well. I concur with your selection.
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