2009/9/16 Lick A Prize
In the ROBIN address allocation described here http://robin-mesh.wik.is/Documentation/Router_Configuring/Advanced_Configuration/Robin_IP_Space I'm curious how collisions are avoided.
Ā
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" ĀMAC-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 Āof the least two men have a Ābirthday 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 Āmatches with any MAC-Adress in its last two bytes? Well this is:

P(n)=1-(65535/65536)^n Āthen when n=13107 is ĀP(13107)=0.18 or 18%. That is MAC-paradox...

Ā

_______________________________________________
B.A.T.M.A.N mailing list
B.A.T.M.A.N@lists.open-mesh.net
https://lists.open-mesh.net/mm/listinfo/b.a.t.m.a.n