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?
Correct Answer: B
IP address: 192.168.168.188
Subnet mask: 255.255.255.192 # /26 # Block size = 64
Subnets:
* 192.168.168.0/26 # 192.168.168.0 - 63
* 192.168.168.64/26 # 192.168.168.64 - 127
* 192.168.168.128/26 # 192.168.168.128 - 191 # Contains 192.168.168.188
* 192.168.168.192/26 # 192.168.168.192 - 255
Valid host range = 192.168.168.129 - 190
(Broadcast = 191, Network address = 128)
MTCNA Course Material - Subnetting Practice:
"To find valid hosts, exclude the subnet and broadcast address. In /26, each block is 64 addresses." Rene Meneses MTCNA Study Guide - IP Addressing:
"For /26 subnetting, calculate block size as 2^(32-26) = 64. Subnet starts at multiples of 64."