IPv4 addressing classes (legacy concept): Which class of address provides a maximum of 254 usable host addresses per network ID?
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AClass A
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BClass B
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CClass C
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DClass D
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EClass E
Answer
Correct Answer: Class C
Explanation
Introduction / Context:Although modern networks use Classless Inter-Domain Routing (CIDR), legacy classful ranges still appear in exams and historical documentation. Recognizing the host capacity of each class helps map older designs to current subnetting practice.
Given Data / Assumptions:
- Class C default mask: 255.255.255.0 (/24).
- Usable host count per subnet equals total addresses minus network and broadcast.
- Class D and E are not used for unicast hosts.
Concept / Approach:For a /24 network, there are 2^8 = 256 addresses in the host portion. Subtract 2 for network and broadcast, leaving 254 usable hosts. This aligns with the canonical Class C networks (192.0.0.0–223.255.255.255) under classful rules.
Step-by-Step Solution:
Compute host space for Class C: 8 host bits.Total addresses = 2^8 = 256.Usable hosts = 256 − 2 = 254.Verification / Alternative check:Contrast with Class A (/8) and Class B (/16) which provide far more host addresses. Class D is multicast, not for hosts; Class E is experimental.
Why Other Options Are Wrong:
- Class A and B yield much larger host counts than 254.
- Class D is multicast addressing, not host addressing.
- Class E is reserved/experimental and not used for hosts.
Common Pitfalls:Forgetting to subtract network and broadcast; applying classful logic to CIDR blocks where host counts vary by mask.
Final Answer:Class C