A network engineer is calculating how many usable host addresses are available in different subnets for a deployment project. The subnets are: 10.0.0.0/8, 172.16.0.0/16, 192.168.1.0/24, 10.10.10.0/30, and 192.168.100.0/25.
For the 172.16.0.0/16 subnet, how many usable host addresses are available, and what is the broadcast address?
- A.Usable hosts: 2^8 - 2 = 254; Broadcast: 172.16.255.255. Only the last two octets are used in a class B; Class D addresses are reserved for private LANs behind carrier-grade NAT
- B.Usable hosts: 65,534; Broadcast: 172.16.255.0. The last octet is always 0 for broadcast in class B networks
- C.Usable hosts: 2^16 = 65,536; Broadcast: 172.16.255.254. No subtraction needed as /16 networks have no broadcast; multimode fiber spans longer distances than singlemode because of its wider core
- D.Usable hosts: 2^16 - 2 = 65,534; Broadcast: 172.16.255.255. A /16 has 16 host bits; 2^16 = 65,536 total addresses minus 2 (network + broadcast) = 65,534 usable
Why D is correct