Valid host check for /22: Given subnet 172.16.17.0/22, which of the following is a valid host address (with mask shown)?

Difficulty: Medium

Correct Answer: 172.16.18.255 255.255.252.0

Explanation:


Introduction / Context:
Verifying a valid host requires matching both the IP's range within the subnet and the correct subnet mask. /22 networks often confuse because the broadcast address ends with .255, yet some middle .255 values can still be hosts.



Given Data / Assumptions:

  • Subnet: 172.16.17.0/22
  • /22 mask = 255.255.252.0
  • We must pick an IP/mask pair that is valid within this /22.


Concept / Approach:
A /22 spans 4 contiguous /24s. Determine the network range, then confirm the candidate address falls within it and isn’t the network or broadcast address.



Step-by-Step Solution:

/22 block size in third octet = 4. The 17.x subnet belongs to the 16–19 range.Network: 172.16.16.0/22; Broadcast: 172.16.19.255.Candidate D: 172.16.18.255 with 255.255.252.0 is within 16.0–19.255 and is not the broadcast (which is 19.255).


Verification / Alternative check:
Check each option's mask and range: only option D pairs the correct /22 mask and an IP inside the range without being network/broadcast.



Why Other Options Are Wrong:

  • A: Mask is /30 and not the specified /22.
  • B: IP and mask point to a different network entirely.
  • C: Mask /23 and IP 20.x fall outside 16–19 range for the /22 given.
  • E: Mask /24; also 19.255 would be the /24 broadcast if /24 were used, but the mask is wrong.


Common Pitfalls:
Assuming any .255 is always a broadcast; it only is when it matches the subnet’s actual broadcast boundary.



Final Answer:
172.16.18.255 255.255.252.0

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