Discussion
Which of the following is the valid host range for the subnet on which the IP address 192.168.168.188 255.255.255.192 resides?
Answer: A
Correct answer: 192.168.168.129-190
The mask 255.255.255.192 is /26, leaving 6 host bits, so the block size is
\[ 256 - 192 = 64 \]and the subnets begin at 0, 64, 128 and 192. The address 188 falls in the 128 subnet, whose broadcast is the address before the next subnet, 191. The usable hosts are therefore 129 to 190: 128 is the network address and 191 the broadcast, and neither can be assigned to a host.
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