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Complete the following table; define each of the subnets, the range of host addresses on the subnet, and the directed broadcast address on the subnet

Explanation of this table: In part b, we indicated that there were eight subnets. The best way to fill out the table is to identify the subnet addresses for all subnets.

 The very first subnet’s IP address is when all the bits in the last byte are 0, giving us the following decimal value: 192.168.1.0. Recall from part f that the incremental value is 32, which means that the second subnet’s IP address will have the third placeholder equal to 1, giving us the following address: 192.168.1.32.

To find the third subnet’s IP address, we need to multiply the increment value (32) by 2, resulting in 192.168.1.64. You would continue until the eighth subnet, in which all the first 3 bits in the last byte equal 1, giving us 192.168.1.224. The direct broadcast address’s value is one less than the next subnet’s IP address. Also, this address will have all the host bits in the last byte equal to 1. For simplicity, I will only convert the last byte of several broadcast addresses to binary to illustrate this: Broadcast Address Last Byte Converted to Binary (network bits | host bits) 192.168.1.31 0 0 0 | 1 1 1 1 1 192.168.1.63 0 0 1 | 1 1 1 1 1 192.168.1.95 0 1 0 | 1 1 1 1 1 192.168.1.127 0 1 1 | 1 1 1 1 1 The addresses between the subnet address and the broadcast address can be assigned to any hosts on the network