Final answer:
The CIDR block from 232.45.160.0 to 232.45.175.255 has a net ID of /21, a host ID of 11 bits, a mask of 11111111.11111111.11111000.00000000, a maximum of 2046 valid IP addresses, and with 127 subnets, the subnet mask becomes /28.
Step-by-step explanation:
a. The network ID is the portion of the IP address that identifies the network. In this case, the CIDR block is from 232.45.160.0 to 232.45.175.255, so the net ID is the first 21 bits: 232.45.160.0/21.
b. The host ID is the portion of the IP address that identifies the specific device on the network. Since the net ID uses 21 bits, the remaining 11 bits are used for the host ID.
c. The mask in slash notation is /21, and in binary notation, it is 11111111.11111111.11111000.00000000.
d. To calculate the maximum number of valid IP addresses, we subtract 2 for the network and broadcast addresses. In this case, there are 2^(32-21) - 2 = 2^11 - 2 = 2046 valid IP addresses.
e. If you want 127 subnets, you would need to increase the number of bits used for the network ID. Since 127 can be represented as 2^7, you would need 7 additional bits. The new subnet mask would be /28, and the first two subnets would be 232.45.160.0/28 and 232.45.160.16/28.
f. The maximum number of valid IP addresses available per subnet is calculated in the same way as in part d. With a /28 subnet mask, there are 2^(32-28) - 2 = 2^4 - 2 = 14 valid IP addresses per subnet.