[ale] Telnet problems - continued

[Gary Maltzen]

(Part 1 of 2 part answer)

Your subnet is 10.0.2.0/255.255.255.0 (or "10.0.2")
Your "mask" is 255.255.255.0

  0000-1010 0000-0000 0000-0010 0000-1111 = 10.0.2.15
& 1111-1111 1111-1111 1111-1111 0000-0000 = 255.255.255.0
  =======================================
  0000-1010 0000-0000 0000-0010 0000-0000 = 10.0.2.0

Since 255.255.255.0 mask contains 24 bits, the subnet is also called

 10.0.2/24

"10.0.2.0/255.255.255.0" is the old (pre-IPv6) notation.
"10.0.2/24" is the new notation.

A subnet with a 24-bit mask is commonly referred-to as a "Class C" subnet.

-----Original Message-----
 From: Brubakken, Eric <eric.brubakken at SouthernEnergy.Com>


Yesterday I posted a problem using Telnet and being forced to use the modem
to connect to my LAN at home. From numerous responses here I have solved
that delema and thanks to all that responded to my question. But now for
the next round.

I have 2 WinXX boxes on my LAN and I can open a MS-DOS command window on
each box and ping my Linux box without a problem. But on the new Win98 box
when I now use Telnet every thing 'appears' to be working but, I do not get
any prompt for a login when trying to connect the the Linux box. I can type
in characters, but nothing else happens. When I select disconnect from the
menu I get this message "You still have open remote connections - do you
want to disconnect?". Each PC connected to the LAN has a subnet address of
255.255.255.0 is this correct or should each subnet be unique? 

Thanks

Eric