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Host autoassignment ipv6 » Historial » Revisió 3

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Victor Oncins, 22-11-2012 11:01


Auto-assignment of IPv6 host-oriented prefixes and collision estimation

One of the desired features is qmp nodes are able to offer native IPv6 addressing to final hosts. IPv6 based networks usually assign /64 prefix to each host-oriented interface, subnetted from a bigger /48. This implies the nodes have 2^16 possible /64 prefixes to auto-assign.

If we want to use MAC address numbering as a mapping value, we'll need a mapping function from 48 bits of MAC to 48 bits of IPv6 prefix.

According to this auto-configure addressing philosophy, we deduce that is impossible to avoid
the election of the same prefix /64 (collision) by one or more network interfaces.

Thus the collision probability is greater than 0 if more than one network interface is auto-addressed. Anyway, we
could calculate this probability and then estimate the maximum number of network interfaces that can be auto-addresses with
a probability collision less than certain value.

Let l as the last l-LSB of a MAC address, where 1<=l<=24. We left the first 24 bits from OUI. Thus we have a k=2^l possible endings. If
we have in the network N different OUI we'll have p=N*2^(24-l) possible MAC addresses for each possible ending. Obviously n=k*p is the
total space of possible MAC present in a network. If the network randomly includes two or more MAC with the same l-bit ending from
different OUI we'll have a collision, i.e. the same IPv6 /64 prefix will be auto-assigned on different network interfaces.

If we select randomly m < k interfaces (MAC addresses) from N OUI, we obtain

n!/m!n-m!

different combinations without replacement. This set of m MAC addresses will contain all its MAC addresses with different l-bits ending. These
combinations will generate /64 prefixes without collision. The number of combinations with all different ending is

k!/m!k-m! * p^m

thus the non-collision probability is

Pĉ(m) = k!/n!*n-m!/k-m! * p^m

If only one interface is present in the network Pĉ(1) = kp/n = 1, i.e. the collision is impossible.
We can now calculate the maximum number M of interfaces such that the probability of collision was less than certain value. The following table
shows this number for a maximum collision probability of 4%.

Prefix l M for N=2 M for N=20 M for N=200
----------------------------------------------------
/51 13 27 27 27
/50 14 38 38 38
/49 15 53 53 53
/48 16 74 74 74
/47 17 105 104 104
/46 18 148 147 147
/45 19 210 208 208
/44 20 298 294 294
/43 21 428 416 415
/42 22 627 520 587
/41 23 957 839 830
/40 24 1656 1202 1174

We note that the variation of M with the number of OUI is very small. Thus the maximum number of interfaces in a /48 network with a collision
probability 4% is 74.

Actualitzat per Victor Oncins fa més de 11 anys · 3 revisions