An engineer warns to map a multicast IP address to a multicast MAC. How many bits are used to make the conversion?

Question

Options

• Ahigh-order 24 bits
• Bhigher-order 23 bits
• Clow order 23 bits
• Dlower-order 24 bits

• A
  3% (1)
• C
  93% (27)
• D
  3% (1)

Explanation

Option C is correct because when mapping a multicast IPv4 address to a multicast MAC address, only the low-order 23 bits of the IP address are copied into the low-order 23 bits of the MAC address (using the IANA-reserved prefix 01:00:5E). The MAC multicast block only provides 23 usable bits because the 24th bit of the MAC's address space is reserved as 0, meaning the mapping is intentionally lossy (32 IP multicast addresses share a single MAC address).

Why the distractors fail:

• A & B (high/higher-order bits) - These are the network/group identifier bits of the IP address, which are discarded, not mapped. You always take from the low (rightmost) end.
• D (lower-order 24 bits) - The bit count is off by one. Even though the MAC has a 24-bit variable portion, only 23 bits are available for mapping because the most significant of those 24 bits is fixed at 0 by the MAC address block structure.

Memory tip: Think "23 = lower" - the lower 23 bits drop into the MAC. If you remember the number 23, rule out anything that says 24 or high-order, and you'll nail it.

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