I am having a major brain-cloud and it has been many years since I got down and dirty with IP routing, but anyway. If I have the following netmask: 255.255.255.248 Would my corresponding network be, for example, 192.168.1.0/29 ? Which would mean 8 adresses, thus 5 usable? If my memoery serves me, /29 means 29 bits for the network segment, leaving 3 for the host address. ie: 2^3 = 8 addresses? Thx for any help, sorry if kinda of off-topic. -CC ---------- This message contains confidential information and is intended only for the individual named. If you are not the named addressee you should not disseminate, distribute or copy this e-mail. Please notify the sender immediately by e-mail if you have received this e-mail by mistake and delete this e-mail from your system. E-mail transmission cannot be guaranteed to be secure or error-free as information could be intercepted, corrupted, lost, destroyed, arrive late or incomplete, or contain viruses. The sender therefore does not accept liability for any errors or omissions in the contents of this message, which arise as a result of e-mail transmission. If verification is required please request a hard-copy version.
On Sat, Feb 22, 2003 at 03:58:54PM -0800, Carson, Chuck wrote:
I am having a major brain-cloud and it has been many years since I got down and dirty with IP routing, but anyway.
If I have the following netmask: 255.255.255.248 Would my corresponding network be, for example, 192.168.1.0/29 ?
Which would mean 8 adresses, thus 5 usable? If my memoery serves me, /29 means 29 bits for the network segment, leaving 3 for the host address. ie: 2^3 = 8 addresses?
Yes. Except it would be 6 usable, not 5. -- Brad Shelton On Line Exchange http://ole.net Phone: 313-526-1111 Fax: 313-526-3333
participants (2)
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Brad Shelton
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Carson, Chuck