So does it mean on the first subent - the host range is 192.168.50.1 -
192.168.50.126, and the second subnet host range is 192.168.50.129 -
192.168.50.254?
Best Regards,
Hunt Lee
G Z wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
The 25 bit can be 0 or 1. It looks like the
, February 24, 2002 4:34 PM
To: [EMAIL PROTECTED]
Subject: Re: VLSM Question [7:35827]
So does it mean on the first subent - the host range is 192.168.50.1 -
192.168.50.126, and the second subnet host range is 192.168.50.129 -
192.168.50.254?
Best Regards,
Hunt Lee
G Z wrote in message
[EMAIL
As for VLSM, I found an example in Jeff Doyle (TCP/IP Vol 1) on
p290 that I don't understand.
192.168.50.0 /25, and it states that the reason it has /25 is
because it needs to have 100 hosts = so 2^7-2=126 hosts (as 2^6 would be
too small), so it makes sense.
What confuses me is that since
Don't think of the Class C at allif you have a /25, that means that, of
the 32 bits in the address, the most significant 25 represent the network
address. Dotted decimal notation is for human convenience, nothing more. The
address is a string of binary digits coming over the wire, and it is
The 25 bit can be 0 or 1. It looks like the book chose 0. The CCNA books use
2^n - 2 for both hosts and subnets. But you also can use the
zero and one subnet so that would add 2. Then you would have 2^n for
subnets. So, 192.168.50.0 is one, and 192.168.50.128 is another subnet.
Using the first
Sorry about the oversight:
192.168.50.0001 first host
192.168.50.0110 last host
should be:
192.168.50.0111 broadcast
Message Posted at:
http://www.groupstudy.com/form/read.php?f=7i=35832t=35827
--
FAQ, list archives, and
6 matches
Mail list logo
