CyberCertPrepNetwork FundamentalsIP Addressing & SubnettingEASYFree question
Scenario
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: 65,534; Broadcast: 172.16.255.0. The last octet is always 0 for broadcast in class B networks
- B.Usable hosts: 2^16 = 65,536; Broadcast: 172.16.255.254. No subtraction needed as /16 networks have no broadcast
- C.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
- D.Usable hosts: 2^8 - 2 = 254; Broadcast: 172.16.255.255. Only the last two octets are used in a class B
Why C is correct
Subnetting formula: usable hosts = 2^h - 2 where h = host bits = 32 - prefix length. For /16: h = 32 - 16 = 16 host bits. 2^16 = 65,536 total. Minus 2 = 65,534 usable. Broadcast = all host bits set to 1 = 172.16.255.255. Network = all host bits 0 = 172.16.0.0. Quick reference: /24 = 254 usable, /25 = 126, /26 = 62, /27 = 30, /28 = 14, /29 = 6, /30 = 2. For large subnets: /16 = 65,534 usable; /8 = 16,777,214 usable.
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