Which subnet mask supports 500 hosts with fewest unused IP addresses?
A network technician is working to deploy a new subnet to support 500 hosts. The technician needs to limit the number of unused IP addresses. Which of the following subnets should the technician choose?
Community Votes
100% of anonymous learners picked answer D. Votes are pick records left by other test-takers — they are not the verified answer.
Community Insight
This question tests your ability to calculate the smallest subnet that meets a host requirement without leaving excessive unused IPs, and the trap is confusing total addresses with usable addresses or choosing a larger subnet for expansion.
To support exactly 500 hosts while minimizing unused IP addresses, a /23 subnet mask (255.255.254.0) is required because it provides 512 total addresses and 510 usable addresses. Community experts agree that D is correct because any larger subnet would waste more addresses.
Choosing a larger subnet such as C (255.255.252.0 /22) because it comfortably supports 500 hosts, while forgetting that the question specifically asks to limit unused IP addresses; /22 wastes 522 usable addresses.
Community Discussion (6 comments)
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Expert Analysis
Why the Answer Is Correct
A /23 subnet has the mask 255.255.254.0, providing 9 host bits. With 9 host bits, 2^9 = 512 total addresses, of which 2 are reserved for network and broadcast, leaving 510 usable addresses - enough for 500 hosts with only 10 unused usable addresses. This is the smallest subnet that satisfies the 500-host requirement, so it best meets the goal of limiting unused IPs. Comments [1], [3], [5], and [6] all confirm that 255.255.254.0 /23 yields 512 total and 510 usable hosts, making it the correct match.
Why the Other Options Are Wrong
Option A (255.255.224.0 /19) provides 8,190 usable addresses, leaving thousands unused. Option B (255.255.248.0 /21) provides 2,046 usable addresses, wasting over 1,500 addresses compared to the requirement. Option C (255.255.252.0 /22) provides 1,022 usable addresses, still wasting 522 addresses; although it would support 500 hosts, it does not limit unused addresses as tightly as D. As comment [2] explains, the other subnets provide more addresses but result in a higher number of unused addresses, which violates the stated requirement.
Community Comment Notes
Multiple comments, including [1], [3], and [5], correctly identify D as the answer by calculating 9 host bits and 2^9=512 total IPs. Comment [3] makes an important point: the question does not mention expandability, so you should not assume a larger subnet is needed - the goal is the fewest unused addresses. Comment [2] explicitly compares the subnets and notes that options like /19 and /21 create many more unused addresses. The community is unanimous with 100% votes for D, reinforcing that the right method is exact-fit subnetting rather than over-provisioning.
Official Reference
Exam Strategy
Calculate the required host bits using 2^n - 2 >= required hosts. For 500 hosts, n must be at least 9, so the subnet prefix is /23 (32 - 9), which equates to 255.255.254.0. Avoid the trap of adding extra bits for future growth unless the question explicitly mentions expansion - always read for the 'limit unused addresses' requirement.