Identifying Broadcast Addresses in /25 Subnets
Short answer
A 192.168.1.127/25 is the broadcast address because it is the last address in the 192.168.1.0/25 subnet where all host bits are set to 1.
Question
Key Takeaways
The main trap is confusing the highest host address with the broadcast address; remember that the very last IP in a subnet range is always the broadcast address when all host bits are 1.
This page explains why 192.168.1.127/25 is a broadcast address. Learn to calculate subnet ranges and identify network vs. broadcast IPs for H12-811 certification.
Common mistake: Option B (Host) is incorrect because 192.168.1.127 is not assignable to a device; it is reserved for layer 2/3 broadcast traffic within the subnet. Option D (Network) is incorrect because the network ID is 192.168.1.0.
Expert Analysis
Subnet Mask and Block Size
To determine the type of address, we must first analyze the CIDR notation /25. This prefix length indicates that the first 25 bits are fixed for the network portion, leaving 7 bits (32 - 25) for the host portion. In IPv4 addressing, a block size is determined by the least significant bit of the network mask. For a /25, the last octet mask is 128 (binary 10000000), resulting in a block size of 128 addresses per subnet.
Identifying Network and Broadcast Addresses
We calculate the subnets within the third octet since the first three octets (192.168.1) are part of the network ID. The subnets start at multiples of the block size:
- Subnet 1: 192.168.1.0 to 192.168.1.127
- Subnet 2: 192.168.1.128 to 192.168.1.255
The address in question is 192.168.1.127/25. This falls exactly at the end of the first subnet range.
Address Classification
In any IPv4 subnet:
- The first address (all host bits zero) is the Network Address (192.168.1.0).
- The last address (all host bits one) is the Broadcast Address (192.168.1.127).
- Addresses between them are Host Addresses (192.168.1.1 to 192.168.1.126).
Since 192.168.1.127 has all host bits set to 1 within its /25 subnet, it functions as the directed broadcast address for that specific network segment. It cannot be a unicast host address because it is reserved for broadcasting to all devices on that local network segment.
Conclusion
The address 192.168.1.127/25 is the broadcast address for the 192.168.1.0/25 subnet. Therefore, Option A is correct.
Exam Strategy
When encountering CIDR questions, quickly calculate the block size using the formula 2^(32-prefix). Then, list the start (network) and end (broadcast) addresses of the relevant subnet. If the given IP matches the end address, it is the broadcast; if it matches the start, it is the network. Any IP between these two values is a valid host address.
Frequently Asked Questions
How do I find the broadcast address for any subnet?
Why can't 192.168.1.127 be a host address?
Community Votes
100% of anonymous learners picked answer A. Votes are pick records left by other test-takers — they are not the verified answer.
Discussion 4 contributions
Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.
From public community threads 4
-
Fmoura17 👍 2 Answered AA. Broadcast Explanation With a /25 prefix, the subnet mask is 255.255.255.128. Network address: 192.168.1.0 Broadcast address: 192.168.1.127 Host range: 192.168.1.1 – 192.168.1.126
-
amiie00 👍 1 Answered A192.168.1.0 = Network Address 192.168.1.1 - 192.168.1.126 = Host Address 192.168.1.127 = Broadcast Address
-
ButcherCapybara 👍 1 Answered A127 is the last /25 address, which is an broadcast address.
-
Heidorn 👍 1 Answered AThat is the broadcast bit for that network
Reader comments & corrections
No reader corrections yet — spotted an error or have a note? Share it below.