Difficulty: Medium
Correct Answer: 240
Explanation:
Introduction / Context: Subnet design for a Class C-sized block (default /24) trades the number of subnets against hosts per subnet. Borrowing more bits increases the number of subnets but decreases hosts per subnet. The goal is to satisfy a required minimum number of subnets (here, 10) while leaving the host count as large as possible.
Given Data / Assumptions:
Concept / Approach: For a /24, the host portion is 8 bits. Borrow s bits to form subnets: number_of_subnets = 2^s. We need 2^s ≥ 10 → s = 4 (since 2^3 = 8 is insufficient; 2^4 = 16 meets the need). The new prefix is /28 (24 + 4). In dotted-decimal, /28 corresponds to 255.255.255.240, so the last-octet mask value is 240. Usable hosts per /28 subnet are 2^(8 − 4) − 2 = 14, which is larger than for /29 (6) or /30 (2).
Step-by-Step Solution: Find s such that 2^s ≥ 10 → s = 4.Compute prefix: 24 + 4 = /28.Map /28 to last-octet mask value: 240.Confirm hosts per subnet: 14 usable, maximized among qualifying masks.
Verification / Alternative check: Lesser masks: 224 (/27) → 8 subnets (not enough). Heavier masks: 248 (/29) or 252 (/30) meet subnets but reduce hosts to 6 or 2, violating the “maximize hosts” objective.
Why Other Options Are Wrong: 192 (/26) and 224 (/27): Provide 4 or 8 subnets—insufficient.
248 (/29) and 252 (/30): Provide 32 or 64 subnets but at the cost of too few hosts per subnet.Common Pitfalls: Forgetting that the minimum s must satisfy the current requirement; borrowing more bits than necessary needlessly shrinks subnets.
Final Answer: 240
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