The mask of an IP subnet is 28 bits, then the number of hosts in the subnet is?

Question

Options

• A30
• B14
• C16
• D32

• A
  3% (1)
• B
  74% (28)
• C
  16% (6)
• D
  8% (3)

Explanation

With a /28 mask, 28 bits are reserved for the network, leaving 4 bits for hosts (32 - 28 = 4). Those 4 bits yield 2⁴ = 16 total addresses, but you must subtract 2 - one for the network address and one for the broadcast address - leaving 14 usable host addresses, making B correct.

Why the distractors are wrong:

• C (16) is the total address count (2⁴), but it ignores the unusable network and broadcast addresses.
• A (30) = 2⁵ − 2, which applies to a /27 subnet (5 host bits), not /28.
• D (32) = 2⁵, the total address space of a /27 - off by an entire subnet size.

Memory tip: Use the formula 2ⁿ − 2, where n = 32 − prefix length. For /28: n = 4, so 16 − 2 = 14. The "− 2" never changes - always subtract network + broadcast.

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