Let me reconstruct my message.
After having read through some more RFC's, I now believe that the following
is the truth. If you do not agree, please reply.
CLASS A : 1.h.h.h - 126.h.h.h = 126 networks
CLASS B : 128.1.h.h - 191.254.h.h = 16382 networks
CLASS C : 192.0.1.h - 223.255.254.h = 2097150 networks
and
CLASS A : n.0.0.1 - n.255.255.254 = 16777214 hosts
CLASS B : n.n.0.1 - n.n.255.254 = 65534 hosts
CLASS C : n.n.n.1 - n.n.n.254 = 254 hosts
Thanks,
Ole
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ole Drews Jensen
Systems Network Manager
CCNA, MCSE, MCP+I
RWR Enterprises, Inc.
[EMAIL PROTECTED]
http://www.oledrews.com/ccnp
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
NEED A JOB ???
http://www.oledrews.com/job
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-----Original Message-----
From: Ole Drews Jensen [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 18, 2000 8:21 AM
To: '[EMAIL PROTECTED]'
Subject: IP addressses - Are networks 2^x or (2^x)-2 ???
I thought I knew it by heard now, but when I had to prove a point to
someone, I grapped some of my books, and they all have different
explanations. I therefore jumped to IETF's homepage and started surfing
their RFC's. BUT...
On RFC 943 (Assigned Numbers) it says on page 1:
Class A has 7 bit network number which allows 128 (2^7) networks.
Class B has 14 bit network number which allows 16384 (2^14) networks
Class C has 21 bit network number which allows 2097152 (2^21) networks
This looks right, because the MSB is 0 in a Class A, thus the 7 bits. It is
only 127 networks though, since 127.x.x.x is reserved. That is also
explained later in the RFC. The MSB in Class B is 10, thus the 14 bits. And
finally, the MSB in Class C is 110, thus the 21 bits.
On page page 10 however, it says that the maximum allowed networks are:
Class A : (2^7 - 2) 126
Class B : (2^14 - 2) 16382
Class C : (2^21 - 2) 2097150
It now looks like it follows the same rule as host calculations, where you
cannot use all 0's or all 1's.
I have looked through all my books (I have a lot) and they all have slightly
different opinions about this.
What is the right answer?
Class A is from 0.x.x.x to 126.x.x.x OR 1.x.x.x to 126.x.x.x ?
Class B is from 128.0.x.x to 191.255.x.x OR 128.1.x.x to 191.254.x.x ?
Class C is from 192.0.0.x to 223.255.255.x OR 192.0.1.x to 223.255.254.x ?
Regarding SUBNET's, they all have the same explanation, but here it is
anyway. If you use 192.168.1.0/28, you have 4 bits for the subnet and 4 bits
for the hosts, which give you 2^4 - 2 = 14 subnets with 14 hosts each. I am
pretty sure that that is the right explanation.
Thanks for any comments on this,
Ole
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ole Drews Jensen
Systems Network Manager
CCNA, MCSE, MCP+I
RWR Enterprises, Inc.
[EMAIL PROTECTED]
http://www.oledrews.com/ccnp
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
NEED A JOB ???
http://www.oledrews.com/job
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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