Apply /23 logic — valid hosts on 172.16.2.1/23 LAN A router interface E0 is 172.16.2.1/23 (mask 255.255.254.0). Which of the following are valid host addresses on that same LAN segment? Consider: 1) 172.16.1.100 2) 172.16.1.198 3) 172.16.2.255 4) 172.16.3.0.

Introduction / Context:Hosts must share the same subnet to communicate without routing. With a /23 mask, two contiguous /24 blocks form one larger subnet. Determining whether candidate addresses fall inside that range is a common exam and field task.

Given Data / Assumptions:

• Interface E0: 172.16.2.1/23.
• Mask: 255.255.254.0 (block size 2 in the third octet).
• We must select all addresses within 172.16.2.0/23 usable host range.

Concept / Approach:A /23 covering 172.16.2.0 spans from 172.16.2.0 to 172.16.3.255, with usable hosts 172.16.2.1 through 172.16.3.254. Therefore, any address in the 172.16.2.x or 172.16.3.x ranges—except the network (.2.0) and broadcast (.3.255)—is valid.

Step-by-Step Solution:List usable range: 172.16.2.1–172.16.3.254.Evaluate #1 (172.16.1.100): outside the /23, invalid.Evaluate #2 (172.16.1.198): also outside, invalid.Evaluate #3 (172.16.2.255): within usable range (not the broadcast), valid.Evaluate #4 (172.16.3.0): within usable range (not the network of the /23), valid.

Verification / Alternative check:Compute network by ANDing 172.16.2.1 with mask 255.255.254.0 → 172.16.2.0; broadcast = next /23 start (172.16.4.0) − 1 = 172.16.3.255. #3 and #4 sit between these values.

Why Other Options Are Wrong:

• A/B/E: Include addresses from the 172.16.0.0/23 below, not this LAN.
• D: Incorrect because two candidates are valid.

Common Pitfalls:Assuming any .255 is always a broadcast; in /23, 172.16.2.255 is a host. Also mistaking 172.16.3.0 for a network boundary; the network is 172.16.2.0.

Final Answer:3 and 4 only