If an Ethernet port on a router were assigned an IP address of 172.16....
A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
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If an Ethernet port on a router were assigned an IP address of 172.16....
Explanation:
To determine the valid subnet address of the host with IP address 172.16.112.1/25, we need to find out the subnet mask that corresponds to the given prefix length of 25 bits.
The subnet mask is a binary number that has 1's in the bit positions that are part of the network portion of the address, and 0's in the bit positions that are part of the host portion of the address.
In this case, the prefix length of 25 bits means that the first 25 bits of the IP address are part of the network portion, and the remaining 7 bits are part of the host portion.
Therefore, the subnet mask can be written as follows:
11111111.11111111.11111111.10000000
or
255.255.255.128
This means that the network address of the subnet is obtained by setting all the host bits to 0, and the result is:
172.16.112.0
Therefore, the valid subnet address for the host with IP address 172.16.112.1/25 is 172.16.112.0.
Summary:
- The subnet mask for the given prefix length of 25 bits is 255.255.255.128
- The network address of the subnet is obtained by setting all the host bits to 0
- The valid subnet address for the host with IP address 172.16.112.1/25 is 172.16.112.0