IPv4 subnetting design: for the private Class B network 172.16.0.0, you want each subnet to support approximately 600 usable hosts. Which single subnet mask should you assign?

Difficulty: Medium

Correct Answer: 255.255.252.0

Explanation:


Introduction / Context:
When choosing a subnet mask, two requirements must be met: the number of hosts per subnet and the number of subnets available. Here, we focus on host capacity per subnet for 172.16.0.0 (a Class B private block with default /16). We must select a mask that provides at least ~600 usable addresses per subnet.


Given Data / Assumptions:

  • Base network: 172.16.0.0/16.
  • Target: ≈600 usable hosts per subnet.
  • Usable hosts per subnet = 2^(host_bits) - 2.


Concept / Approach:
Borrow s bits from the /16 host space, leaving (16 - s) host bits. The number of usable addresses is 2^(16 - s) - 2. We need this to be ≥ 600. Compute for successive masks until the requirement is satisfied, choosing the smallest mask length that still meets the host count to conserve address space.


Step-by-Step Solution:

Try /23 (borrow 7 bits): host_bits = 9 → usable = 2^9 - 2 = 510 (insufficient).Try /22 (borrow 6 bits): host_bits = 10 → usable = 2^10 - 2 = 1022 (meets ≥ 600).Corresponding dotted-decimal mask for /22 is 255.255.252.0.Therefore, /22 is the smallest prefix length that satisfies the requirement.


Verification / Alternative check:
Cross-check neighboring options: /21 (255.255.248.0) yields 2^11 - 2 = 2046 usable (more than enough but wastes addresses). /23 is too small (only 510). Hence /22 is the optimal choice.


Why Other Options Are Wrong:

255.255.192.0 (/18): far too many hosts; not targeted and wastes space.255.255.224.0 (/19) and 255.255.240.0 (/20): still much larger than required and do not represent the minimal mask meeting 600.255.255.248.0 (/21): meets hosts but is not the minimal that satisfies ≥ 600.


Common Pitfalls:
Forgetting to subtract 2 addresses for network/broadcast, or misreading the host requirement as ‘‘exactly 600’’ rather than ‘‘at least 600’’.


Final Answer:
255.255.252.0.

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