Broadcast address determination: An interface has IP 192.168.192.10/29. What broadcast address will hosts on this LAN use?
Networking
Subnetting
Difficulty: Easy
Choose an option
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A192.168.192.15
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B192.168.192.31
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C192.168.192.63
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D192.168.192.127
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E192.168.192.11
Answer
Correct Answer: 192.168.192.15
Explanation
Introduction / Context:For any subnet, the broadcast address is the highest address in that block. With /29, determining the 8-address block is straightforward and frequently tested.
Given Data / Assumptions:
- Host: 192.168.192.10/29
- /29 mask = 255.255.255.248 (block size 8)
Concept / Approach:Identify the /29 block containing .10 by finding the nearest lower multiple of 8 and then adding 7 for the broadcast address.
Step-by-Step Solution:
Block boundaries (4th octet) every 8: 0, 8, 16, 24, ....10 lies in the 8–15 block.Network = .8, Broadcast = .15, Hosts = .9–.14.Verification / Alternative check:Compute: 10 - (10 mod 8) = 8; broadcast = 8 + 7 = 15.
Why Other Options Are Wrong:
- .31/.63/.127: Broadcasts for larger subnets (/27, /26, /25), not /29.
- .11: A valid host, not the broadcast.
Common Pitfalls:Forgetting that /29 uses 8-address blocks; misidentifying the correct block around the host.
Final Answer:192.168.192.15