To determine the network address for a host with an IP address of 192.168.87.125/16, you need to consider the subnet mask. In this case, the subnet mask is /16, which means the first 16 bits of the IP address represent the network portion.
The network address is obtained by setting all the host bits to 0 in the IP address. So, in this scenario, the network address would be:
192.168.87.125 -> 11000000.10101000.01010111.01111101
Since the first 16 bits represent the network portion, the remaining bits (17-32) are set to 0:
11000000.10101000.01010111.01111101 -> 11000000.10101000.00000000.00000000
Converting the binary representation back to decimal, the network address is:
192.168.0.0
Therefore, the correct answer is option D: 192.168.0.0.
D is correct.
Since this is a /16
Subnet range from 192.168.0.0(Network address) to 192.168.255.255 (Broadcast address)
Valid hosts range from 192.168.0.1 to 192.168.255.254.
Don't understand, 192.168.0.0 is the address of 192.168.87.125/16 subnet and both A and B could be hosts in this network?
upvoted 2 times
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network_ninja_mx
10 months, 4 weeks agoCradical
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1 year, 11 months agowolfobi
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