[aclug-L] Re: IP address question

[<Todd_Lundstedt@xxxxxxxxxxxxxxx>]

One thing that I noticed.. there are 156^4 (not 155^4) possible numbers with
three digits... 100-255 is 156 numbers (Gotta count that zero, yes?)


|--------+----------------------->
|        |          jonhall@futur|
|        |          eks.net      |
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|        |          02/29/00     |
|        |          10:11 AM     |
|        |          Please       |
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|        |          discussion   |
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  |       To:     aclug-l@xxxxxxxxxxxx                                         |
  |       cc:     (bcc: Todd Lundstedt/VCHS)                                   |
  |       Subject:     [aclug-L] IP address question                           |
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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?  :-)


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