Can someone tell me if the the answer to this question is wrong. The explanation says that bit 16 would be 16,384 but isn't that 65,536 or 2 to the power of 16?
You’re designing a network scheme from a Class A network address.You want to be able to
have about 16,000 hosts on each subnet. Based on this, what is the maximum number of
host address bits you can take to still allow up to 16,000 hosts per subnet?
A. 8
B. 16
C. 24
D. 17
B.There are a number of ways to calculate this answer. One way is to start with your
knowledge that bit 8 (left-most) of the first octet is equal to 128. As we move to the left,
each bit is twice the one to its right.Thus, the string becomes (bit 9 =256), (bit 10 =
512)…(bit 16 = 16,384), (bit 17 = 32,76
.Therefore, we need no fewer than 16 bits for
our host address space to allow for up to 16,382 addresses per subnet.
Answer A is incorrect, because eight bits would give us only 254 address spaces (128 +
64 + 32 + 16 + 8 + 4 + 2 + 1 = 256 – 2 – 254). Answer C is incorrect, because 24 bits
would allow us far too many host address spaces. 224 = 16,277,214 useable addresses.
Answer D is incorrect, because 17 bits would give us just about double the number we
need, 32,766. It’s one more bit than we need. It’s more common to take one more network
bit than you think you need versus taking one more host bit, as it’s typically better
to have fewer hosts on more subnets for faster, more efficient networks.
