2009/9/16 Lick A Prize <lick.prize(a)gmail.com>
In the ROBIN address allocation described here
http://robin-mesh.wik.is/Documentation/Router_Configuring/Advanced_Configur…
curious how collisions are avoided.
If you generate IP 5.dd:ee:ff from aa:bb:cc:dd:ee:ff won't you get a
collision if you other devices with MACs ending in dd:ee:ff like
aa:aa:cc:dd:ee:ff, aa:aa:aa:dd:ee:ff, and so on.
What is the occurrence probability of "r" MAC-Adress twon agree on their
last two bytes?
This is similar to the birthday paradox: At a meeting of r> 2 people which
is the number of people needed for the probability that two men have same
birthday sould be 99%? The solution is in many books and that can view at:
http://en.wikipedia.org/wiki/Birthday_problem
Then, At a meeting of 60 people there is the 99% probability of the least
two men have a birthday the same day. So,, Similarly to the case of
MAC-Adress is:
P(r)=1-65536!/(65536^r * (65536-r)!)
where *! is factorial operator, and "r" is the Mac-Adress.
Well if the calculations are not mistake, we obtain r = 13,107 MAC-Adress
minimun in the same site then 99% probability two matches in his last two
bytes.
Now a new MAC entering, which is the probability that given "n"
MAC-Adress,,, all with two last byte differents this new MAC matches with
any MAC-Adress in its last two bytes? Well this is:
P(n)=1-(65535/65536)^n then when n=13107 is P(13107)=0.18 or 18%. That is
MAC-paradox...
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