Refer to the exhibit. The routers are running RIPv2. Which addressing scheme would satisfy the
needs of this network yet waste the fewest addresses?

A.
Network 1: 192.168.10.0/26Network 2: 192.168.10.64/27Network 3: 192.168.10.96/28Serial link
1: 192.168.10.112/30Serial link 2: 192.168.10.116/30

B.
Network 1: 192.168.10.0/27Network 2: 192.168.10.64/28Network 3: 192.168.10.96/29Serial link
1: 192.168.10.112/30Serial link 2: 192.168.10.116/30

C.
Network 1: 192.168.10.0/26Network 2: 192.168.10.64/28Network 3: 192.168.10.80/29Serial link
1: 192.168.10.88/30Serial link 2: 192.168.10.96/30

D.
Network 1: 192.168.10.0/26Network 2: 192.168.10.64/26Network 3: 192.168.10.128/26Serial
link 1: 192.168.20.0/24Serial link 2: 192.168.30.0/24

Explanation:
Network 1
Required Number of hosts :50
When We use the 26 bits for Network : 11111111.11111111.11111111.11000000 so 62 usable
host can be in one network. 50 host for now and remaining hosts address for further growth.
Network 2
Required Number of Hosts: 20
When we use the 27 bits for Network: 11111111.11111111.1111111.11100000 so 30 usable hosts
can be in one network.
Network 3
Required Number of Hosts: 10
When we use the 28 bits for Network: 11111111.11111111.11111111.11110000 so 14 usable
hosts can be in one network.
Connection between R1, R2 and R3 is WAN so when you use 30 bits network IP, you will not lose
any IP addresses from network since this subnet allows for only 2 host addresses.