doubt in multicast address translation

cisco.com says "For IP networks, Class D addresses have been set aside for multicast addressing. A Class D address consists of 1110 as the higher order bits in the first octet followed by an unstructured 28 bit group address. For mapping IP multicast addresses to Ethernet addresses, the lower 23 bits of the Class D address are mapped into a block of Ethernet addresses that have been reserved for multicast. With this mapping scheme, each Ethernet multicast address corresponds to 32 IP multicast addresses."

Can someone explain how does each ethernet multicast address here corresponds to 32 multicast IP addresses?

thanks in advance

avinase

Hi Avinase,

Both Ethernet and FDDI frames have a 48 bit destination address field. In order to avoid a kind of multicast ARP to map multicast IP addresses to ethernet/FDDI ones, the IANA reserved a range of addresses for multicast: every ethernet/FDDI frame with its destination in the range 01-00-5e-00-00-00 to 01-00-5e-ff-ff-ff (hex) contains data for a multicast group. The prefix 01-00-5e identifies the frame as multicast, the next bit is always 0 and so only 23 bits are left to the multicast address. As IP multicast groups are 28 bits long, the mapping can not be one-to-one. Only the 23 least significant bits of the IP multicast group are placed in the frame. The remaining 5 high-order bits are ignored, resulting in 32 different multicast groups being mapped to the same ethernet/FDDI address. This means that the ethernet layer acts as an imperfect filter, and the IP layer will have to decide whether to accept the datagrams the data-link layer passed to it. The IP layer acts as a definitive perfect filter

Hope it will help u.

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Dove