On Sun, 2002-11-17 at 02:09, L.V.Gandhi wrote: > PC2 IP address 192.168.1.1 > PC1 IP address 192.168.1.2 > PC3 IP address 192.168.1.3 > PC4 IP address 192.168.1.4 > > network address is 192.168.1.0 > netmask 255.255.255.249
Hi there, I suggest you visit the archives (if you don't keep your mail) and re-read the following message (and replies) contained in a thread you started here earlier this month: http://www.mandrake.com/en/archives/newbie/2002-11/msg00724.php Now, if the netmask you quoted above is merely a typo, kindly disregard this post :) Otherwise, please understand that what you are trying to do with your netmask (which should be 248 in any case) is totally unnecessary in your situation and you should stick to a classful mask (255.255.255.0) if you don't have a thorough understanding of IP addressing. If you insist on subnetting your network, please understand that the tightest mask that you can use on a 'C' class network requiring 4 host addresses is 248. I'll leave it to you to explain to me, why that is the case. If you don't know, do a little reading on TCP/IP and then try to explain your error. You won't believe how much it will help cement your understanding of the subject :) Sorry if I sound like a school teacher, but you really need to be solid in the fundamentals to understand networking when it becomes a little more complex. Kind regards, John...
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