Difficulty: Medium
Correct Answer: 2,113,662
Explanation:
Introduction: Older, classful IPv4 splits addresses into Classes A, B, and C based on the first octet. A common exam calculation asks for the total number of distinct network IDs available across these three classes, applying standard historical conventions (excluding loopback and the zero network).
Given Data / Assumptions:
Concept / Approach: Compute the number of network IDs in each class and sum them. For classful addressing, the count is strictly positional based on fixed leading bits in the first octet (A: 0xxxxxxx excluding 0 and 127; B: 10xxxxxx; C: 110xxxxx).
Step-by-Step Solution:
1) Class A networks: 1–126 => 126 network IDs.2) Class B networks: 128–191 => 64 first-octet values; each defines a block of 256 second-octet values => 64 * 256 = 16,384 network IDs.3) Class C networks: 192–223 => 32 first-octet values; each defines 256 * 256 network IDs => 32 * 65,536 = 2,097,152.4) Sum total = 126 + 16,384 + 2,097,152 = 2,113,662.Verification / Alternative check: Binary prefixes (A: 0, B: 10, C: 110) yield the same totals when expanded over the subsequent octets dedicated to network ID within each class.
Why Other Options Are Wrong:
Common Pitfalls: Confusing host capacity with the number of networks; excluding additional ranges beyond loopback and zero; or including Class D/E, which are not unicast network-ID pools.
Final Answer: 2,113,662 classful network IDs in A + B + C.
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