IPv4 subnetting design: For network 172.16.0.0, you need 25 subnets for your company now and an additional 30 after a merger (total 55). Each subnet must support about 600 hosts. What single subnet mask will satisfy these requirements?

Difficulty: Medium

Correct Answer: 255.255.252.0

Explanation:


Introduction / Context:
Subnet mask selection must satisfy two constraints simultaneously: number of required subnets and number of hosts per subnet. For a Class B private block (172.16.0.0/16), borrowing bits from the host field increases subnet count while reducing hosts per subnet. The chosen mask must meet or exceed both the subnet and host requirements.


Given Data / Assumptions:

  • Base network: 172.16.0.0 with default mask /16.
  • Required subnets: 25 now + 30 later = 55 total.
  • Hosts per subnet ≈ 600 (usable addresses need to exceed this).
  • Standard IPv4 addressing with CIDR.


Concept / Approach:
Let s = subnet bits borrowed from the /16 host field. Subnets available = 2^s. Hosts per subnet (usable) = 2^(16 - s) - 2. We need 2^s ≥ 55 and 2^(16 - s) - 2 ≥ 600. Find the smallest s that satisfies both.


Step-by-Step Solution:

1) 2^5 = 32 (not enough subnets); 2^6 = 64 (enough).2) With s = 6, host bits = 16 - 6 = 10.3) Usable hosts = 2^10 - 2 = 1024 - 2 = 1022 ≥ 600 (satisfies host requirement).4) New prefix = /16 + 6 = /22 → mask 255.255.252.0.


Verification / Alternative check:
Check neighboring masks: /21 (255.255.248.0) gives 2^5 = 32 subnets (too few). /23 (255.255.254.0) would give more hosts than needed but still only 2^7 = 128 subnets if taken from /16? (That would be /23 = s = 7, but then hosts per subnet = 510, which fails the ~600 requirement.) /22 is the smallest mask that simultaneously satisfies both bounds.


Why Other Options Are Wrong:

255.255.192.0 (/18): only 2^(18-16)=4 subnets; hosts per subnet large, but subnet count fails.255.255.224.0 (/19): 8 subnets; still too few.255.255.240.0 (/20): 16 subnets; still too few.255.255.248.0 (/21): 32 subnets; still too few.


Common Pitfalls:
Forgetting to satisfy both constraints, miscounting usable hosts by ignoring the network and broadcast addresses, and mixing classful intuition with CIDR math.


Final Answer:
255.255.252.0.

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