A. 172.16.3.0/1/24B. 10.10.10.0/8C. 192.168.2.0/24D. 172.10.17.0/24E. 60.7.8.0/24F. 15.17.8.14/24G. 12.12.17.0/16H. 13.9.8.0/8QUESTION 1Given one of the IP address above as a choice, use VLSM concept to create 4 subnetworks consistingof the following number of host addresses. In each case provide the network, usable host addressesand broadcast addresses respectively. Show your workings and justification for each case.(10MKS)

Let's use the IP address 192.168.2.0/24 for this example.

1. First, we need to determine the number of bits needed for each subnet. We have 4 subnets and the number of hosts needed for each subnet is not specified, so we'll assume that we need an equal number of hosts for each subnet.

   The /24 in the IP address means that the first 24 bits are used for the network address, leaving 8 bits for host addresses. Since we need 4 subnets, we'll need to borrow 2 bits from the host portion (since 2^2 = 4). This leaves us with 6 bits for host addresses.

2. The new subnet mask is /26 (24 original network bits + 2 borrowed bits).

3. Now we can calculate the network addresses for each subnet:

   • Subnet 1: 192.168.2.0/26. The usable host addresses are from 192.168.2.1 to 192.168.2.62 and the broadcast address is 192.168.2.63.

   • Subnet 2: 192.168.2.64/26. The usable host addresses are from 192.168.2.65 to 192.168.2.126 and the broadcast address is 192.168.2.127.

   • Subnet 3: 192.168.2.128/26. The usable host addresses are from 192.168.2.129 to 192.168.2.190 and the broadcast address is 192.168.2.191.

   • Subnet 4: 192.168.2.192/26. The usable host addresses are from 192.168.2.193 to 192.168.2.254 and the broadcast address is 192.168.2.255.

This is a simplified example and the actual number of hosts needed for each subnet may require a different configuration.