Match each statement with the example subnet and wildcard that it describes. (Not all options are used.)
Place the options in the following order:
192.168.15.65 255.255.255.240 ==> the first valid host address in a subnet
192.168.15.144 0.0.0.15 ==> subnetwork address of a subnet with 14 valid host addreses
host 192.168.15.2 ==> all IP address bits must match exactly
192.168.5.0 0.0.3.255 ==> hosts in a subnet with SM 255.255.252.0
192.168.3.64 0.0.0.7 ==> address with a subnet 255.255.255.248
Explanation: Converting the wildcard mask 0.0.3.255 to binary and subtracting it from 255.255.255.255 yields a subnet mask of 255.255.252.0.
Using the host parameter in a wildcard mask requires that all bits match the given address.
192.168.15.65 is the first valid host address in a subnetwork beginning with the subnetwork address 192.168.15.64. The subnet mask contains 4 host bits, yielding subnets with 16 addresses.
192.168.15.144 is a valid subnetwork address in a similar subnetwork. Change the wildcard mask 0.0.0.15 to binary and subtract it from 255.255.255.255, and the resulting subnet mask is 255.255.255.240.
192.168.3.64 is a subnetwork address in a subnet with 8 addresses. Convert 0.0.0.7 to binary and subtract it from 255.255.255.255, and the resulting subnet mask is 255.255.255.248. That mask contains 3 host bits, and yields 8 addresses.
Please login or Register to submit your answer