Date: Thu, 15 Feb 1996 13:48:01 -0700 From: Nate Williams <nate@sri.MT.net> To: "Garrett A. Wollman" <wollman@lcs.mit.edu> Cc: Nate Williams <nate@sri.MT.net>, hackers@FreeBSD.org, wollman@FreeBSD.org Subject: Re: Broadcast, Netmask, and other such information Message-ID: <199602152048.NAA01295@rocky.sri.MT.net> In-Reply-To: <9602152037.AA13188@halloran-eldar.lcs.mit.edu> References: <199602152029.NAA01202@rocky.sri.MT.net> <9602152037.AA13188@halloran-eldar.lcs.mit.edu>
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> > hosts: hosts: > > 10.5.5.1 - 10.5.5.31 10.5.5.32 - 10.5.5.63 > > netmask: netmask: > > 255.255.255.224 255.255.255.192 > > broadcast: broadcast: > > 10.5.5.31 10.5.5.63 > > 10.5.5.31 is not a valid host address. (Neither is 10.5.5.0, which > you correctly avoided using; it means ``this host''.) Ok, so neither is 10.5.5.63 then either? This should be (using from below that all netmasks should be the same): hosts: hosts: 10.5.5.1 - 10.5.5.30 10.5.5.32 - 10.5.5.62 netmask: netmask: 255.255.255.224 255.255.255.224 broadcast: broadcast: 10.5.5.31 10.5.5.63 > The netmask must be the same for all subnets of the same size. In > this example, it should be 255.255.255.224. The network you describe > would then be referred to as 10.5.5.0/227, 10.5.5.32/227, etc. Where does /227 come from? I see 224, but it looks like you've pulled 227 out of thin air. > > Am I really confused, or can someone else tell me how to calculate > > these? > > You are really confused. Hopefully this will unconfuse you. So to determine the netmask, you need to know how many addresses are in the subnet, and subtract that from 255.255.255.255, right? And, to determine the broadcast, you add one to the highest host # assigned to you (which should be one-less than a 2^n boundary based on the # of hosts you are using), right? Nate
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