Question

given the ip address, 121.211.216.245 and a subnet mask of 255.255.255.248 provide the broadcast address
○ 121.211.216.246
○ 121.211.216.247
○ 121.211.216.248
○ 121.211.216.240

Explanation:

Step1: Analyze subnet mask

The subnet mask is 255.255.255.248. In binary, 248 is $11111000$. So, the number of host bits is $3$ (since $8 - 5 = 3$, where $5$ is the number of 1s in 248's binary). The block size is $2^3 = 8$? Wait, no, the block size for the last octet is calculated as $256 - 248 = 8$. Wait, no, the block size is determined by the subnet mask. For the last octet, the subnet mask is 248, so the block size is $8$? Wait, no, let's find the network address first.

The IP address is 121.211.216.245. Let's convert the last octet (245) and the subnet mask's last octet (248) to binary. 245 in binary is $11110101$, 248 is $11111000$. To find the network address, we perform a bitwise AND on the IP address and the subnet mask.

For the last octet: $245 \& 248 = 240$ (since $11110101 \& 11111000 = 11110000$ which is 240). So the network address is 121.211.216.240.

Now, the broadcast address is the network address plus the block size minus 1. The block size is $256 - 248 = 8$. So the network address is 240, block size 8, so the next network would be 240 + 8 = 248. Therefore, the broadcast address is 248 - 1 = 247. So the broadcast address is 121.211.216.247.

Step2: Verify

Let's check the options. The network address is 121.211.216.240, block size 8, so the hosts in this subnet are from 241 to 246 (usable), and the broadcast address is 247. So the correct option is 121.211.216.247.

Answer:

B. 121.211.216.247