Host autoassignment ipv6 » Historial » Revisió 3
Revisió 2 (Victor Oncins, 22-11-2012 10:55) → Revisió 3/7 (Victor Oncins, 22-11-2012 11:01)
h1. Auto-assignment Autoassignment 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. Hi