IPv4 subnet design: For the private network 192.168.10.0, choose a subnet mask that supports up to 25 hosts per subnet while maximizing the number of subnets.

Difficulty: Medium

Correct Answer: 255.255.255.224

Explanation:


Introduction / Context:
Choosing a subnet mask is a balancing act between hosts per subnet and number of subnets. The requirement here is to support at least 25 hosts while maximizing subnet count, so we should pick the smallest mask that still yields ≥25 usable addresses.



Given Data / Assumptions:

  • Network: 192.168.10.0 (a class C private block, but we use CIDR math).
  • Need at least 25 usable hosts per subnet.
  • Want the maximum number of subnets (choose the most restrictive mask that still meets hosts requirement).


Concept / Approach:

Usable hosts = 2^(host_bits) - 2. Evaluate candidate masks and find the smallest number of host bits that gives ≥25. The fewer host bits used, the more network (subnet) bits available, which increases the number of subnets.



Step-by-Step Solution:

Test /27 (255.255.255.224): host_bits = 5 → usable hosts = 2^5 - 2 = 30 (meets 25).Test /28 (255.255.255.240): host_bits = 4 → usable hosts = 2^4 - 2 = 14 (too small).Test /26 (255.255.255.192): host_bits = 6 → usable hosts = 62 (meets, but wastes addresses and yields fewer subnets than /27).Therefore, /27 (255.255.255.224) is the tightest mask that satisfies ≥25 hosts, maximizing subnet count.Choose 255.255.255.224.


Verification / Alternative check:

Subnet tables show /27 yields 8 subnets in a /24 space and 30 hosts each, which aligns with the requirement. A quick test block: 192.168.10.0/27 ranges 192.168.10.1–.30 with .31 broadcast.



Why Other Options Are Wrong:

a: /26 allows 62 hosts; it meets the host count but does not maximize subnets.

c, d, e: /28, /29, /30 allow 14, 6, and 2 hosts respectively—insufficient for 25 hosts.



Common Pitfalls:

Forgetting to subtract network and broadcast addresses, or picking a larger host pool than needed, which reduces the number of subnets and wastes address space.



Final Answer:

255.255.255.224

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