What is the network address for a host with an IP address of 192.168.87.125/16?
What is the network address for a host with an IP address of 192.168.87.125/16?
To determine the network address for a host with an IP address of 192.168.87.125/16, you must apply the subnet mask, which in this case is /16. This means the first 16 bits (or the first two octets) of the IP address are used for the network address, while the remaining bits are for host addresses. Converting 192.168.87.125 into binary, we get 11000000.10101000.01010111.01111101, where the first 16 bits represent the network portion. By setting all the host bits (the last 16 bits) to 0, we get 11000000.10101000.00000000.00000000, which translates back to 192.168.0.0 in decimal form. Therefore, the network address is 192.168.0.0.
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.
Subnet calculator: Host: 192.168.87.125/16 Network ID: 192.168.0.0 Usable hosts range: 192.168.0.1 - 192.168.255.254 Total numbers of hosts: 65,536 More info: https://www.calculator.net/ip-subnet-calculator.html?cclass=any&csubnet=16&cip=192.168.87.125&ctype=ipv4&printit=0&x=65&y=28
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?
The first 16 bits are the network mask, so the network address is 192.168.0.0/16 192 = 8 bits + 168 = 8 bits