Identify the subnet: An Ethernet port has IP 172.16.112.1/25. What is the valid subnet (network) address for this host?

Introduction / Context:
/25 subnets split a /24 into two halves. To find the network address, determine which half the host belongs to by looking at the most significant bit of the last octet under the /25 mask.

Given Data / Assumptions:

• IP: 172.16.112.1/25
• Mask: 255.255.255.128 (/25)

Concept / Approach:
/25 gives two blocks in the fourth octet: 0–127 and 128–255. Since the host's fourth octet is 1, it must be in the 0–127 block, making the network address .0.

Step-by-Step Solution:

Verification / Alternative check:
Compute: 1 - (1 mod 128) = 0; confirms network base is .0.

Why Other Options Are Wrong:

• 172.16.0.0 / 172.16.255.0: Not the correct /25 network for this host.
• 172.16.96.0: Different /24 network.
• 172.16.112.128: That is the next /25 network, not the one containing .1.

Common Pitfalls:
Confusing /24 with /25; failing to realize /25 splits the 4th octet in half.

Final Answer:
172.16.112.0