[aclug-L] Re: IP address question

[Jonathan Hall <jonhall@xxxxxxxxxxxx>]

Taking this into account, and the observation that I should have used 156
instead of 155, my new "official"(?) answer is:

463162752

:-)

156^4 - 156^3 - 156^2 - (255-223)*(156^3)



On Tue, Feb 29, 2000 at 10:31:21AM -0600, Steven Saner wrote:
> Hmmm... Well I suppose this depends on your definition of
> routable. RFC1918 (private) addresses are perfectly routable, you just
> shouldn't route them outside of your network. You should probably
> exclude class D and E networks. Class D addresses are for multicast
> and they begin with 224 to 239. Class E addresses are for
> experimentation purposes. They begin with 240 to 255.
> 
> 
> On Tue, Feb 29, 2000 at 10:11:14AM -0600, Jonathan Hall wrote:
> > Well, I got a few replies in responce to my question, "How many _routeable_
> > IP addresses have exactly 12 digits?"  I think one was correct (I can't find
> > it right now).  That person doesn't live in Kansas any more, though, so I
> > guess that relieves us from furnishing a prize?  :-)
> > 
> > Here's the answer as I've figured it... someone tell me if I'm wrong or
> > overlooking something:
> > 
> > 573452724
> > 
> > This is how I figured it:
> > 
> > Normal 32-bit IP space contains 4294967296 IP addresses,
> > 0.0.0.0 - 255.255.255.255.  This number can be found by taking 256^4; 256
> > possible numbers for each of 4 locations.
> > 
> > To get 12 digits in an IP address, each of the 4 numbers must have 3 
> > digits. 
> > This means that there are only 155 possibilities for each number (100
> > through 255).  So, we can take 155^4 (155 possibilities in each of 4
> > locations) and come up with 577200625 possible IP addresses--that's only
> > 13.4% of the original IP space.
> > 
> > Not all of these 570 million IP addresses are routeable, though.
> > 
> > There are three subnets that are deemed 'non-routable' and for use in
> > private networks only:
> > 
> > 10.0.0.0/255.0.0.0
> > 192.168.0.0/255.255.0.0
> > 172.16.0.0/255.255.240.0
> > 
> > So, we need to remove these three subnets from our calculations.  Only one
> > of these (192.168.0.0) even needs to be considered, though, b/c theother two
> > have numbers with less than 3 digits.  In the 192.168.0.0 subnet, we have
> > 155^2 or 24025 hosts with 12 digits, so we subtract that from 577200625 for
> > 577176600.
> > 
> > Also, 127.0.0.0/255.0.0.0 is private and non-routable.  Here we have 155^3
> > or 3723875 12-digit IPs we must discount leaving us with 573452725.
> > 
> > Then, for my final answer I subtracted one more for 255.255.255.255, and
> > get... 573452724
> > 
> > So the answer is: 155^4 - 155^3 - 155^3 - 1 = 573452724
> > 
> > Did I miss anything?  :-)
> > 
> > 
> > --
> > "Yes, the president should resign. He has lied to the American people, time
> > and time again, and betrayed their trust. He is no longer an effective
> > leader.  Since he has admitted guilt, there is no reason to put the American
> > people through an impeachment. He will serve absolutely no purpose in
> > finishing out his term, the only possible solution is for the president to
> > save some dignity and resign."
> >       -- William Jefferson Clinton, 1974 on President Nixon.
> > --
> > - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
> >   Jonathan Hall  *  jonhall@xxxxxxxxxxxx  *  PGP public key available
> >  Systems Admin, Future Internet Services; Goessel, KS * (316) 367-2487
> >          http://www.futureks.net  *  PGP Key ID: FE 00 FD 51
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