Final answer:
The minimum number of usable IP addresses in network E is 1022. The network address 192.168.208.0/22 has a total of 1024 networks. The IP addresses for each network can be calculated by setting the host bits to zero.
Step-by-step explanation:
The network address 192.168.208.0/22 has a subnet mask of 255.255.252.0. With a /22 subnet, there are 1024 IP addresses available. However, the first and last IP addresses in a subnet are reserved for network address and broadcast address, respectively.
Therefore, the usable IP addresses for network E would be 1024 - 2 = 1022.
To calculate the number of networks in the given network address, we need to determine the subnet mask. A /22 subnet mask provides 4 blocks of 256 addresses each, resulting in a total of 4 x 256 = 1024 networks.
The IP addresses for each network can be calculated by considering the subnet mask. Each network's IP address will be the network address, which is obtained by setting all the host bits (those not covered by the subnet mask) to zero.
So for the given network 192.168.208.0/22, each network's IP address can be determined by setting the host bits to zero: 192.168.0.0/22, 192.168.4.0/22, 192.168.8.0/22, ..., 192.168.248.0/22.
As for network B, to find the first broadcast IP address, we subtract 1 from the network address.
To find the last broadcast IP address, we add the number of usable IP addresses to the network address and then subtract 1. So for network B (192.168.4.0/22), the first broadcast IP address would be 192.168.4.0 - 1 = 192.168.3.255, and the last broadcast IP address would be 192.168.4.0 + 1022 - 1 = 192.168.5.221.