Which of the following is the broadcast address of the network to which host 172.16.176.135/18 belongs?

To determine the correct broadcast address for the host 172.16.176.135 with a subnet mask of /18, we first need to identify the network range. The /18 notation indicates that the first 18 bits of the address are used for the network portion, and the remaining bits (14 bits, as there are 32 bits in total) are used for host addresses.

First, we convert 172.16.176.135 to binary:

• 172: 10101100

• 16: 00010000

• 176: 10110000

• 135: 10000111

In binary, the entire address is:

10101100.00010000.10110000.10000111

The subnet mask of /18 translates to:

11111111.11111111.11000000.00000000

When we perform a bitwise AND operation between the IP address and the subnet mask, we get the network address.

The first two octets (172.16) will remain the same because the subnet mask covers all bits in those octets (255.255). In the third octet, we will consider only