Subnet range calculation (IPv4 /26): Given IP address 192.168.168.188 with subnet mask 255.255.255.192, what is the valid host range for that subnet (exclude network and broadcast addresses)?

Introduction / Context:
Quickly determining the host range of a subnet is a core networking skill. The mask 255.255.255.192 corresponds to a /26, which creates blocks of 64 addresses. We must locate the correct block containing 192.168.168.188 and then exclude the network and broadcast addresses to get the usable host range.

Given Data / Assumptions:

• IP: 192.168.168.188
• Mask: 255.255.255.192 (that is, /26)
• We want the usable host range only.

Concept / Approach:

With /26, the increment (block size) is 64 in the last octet because 256 - 192 = 64. Subnets therefore start at .0, .64, .128, .192 within the /24 boundary. Determine which block contains .188, then compute network (.128), broadcast (.191), and the host span between them.

Step-by-Step Solution:

Verification / Alternative check:

Binary math confirms: /26 leaves 6 host bits → 2^6 = 64 addresses, subtract 2 for network/broadcast → 62 usable hosts, matching .129–.190 as 62 addresses.

Why Other Options Are Wrong:

.129–.191 includes the broadcast (invalid for hosts).

.128–.190 includes the network address.

.128–.192 mixes network and the next subnet’s network (.192).

Common Pitfalls:

Forgetting to subtract network and broadcast; miscomputing the increment as 32 instead of 64 for /26; ignoring that ranges reset at every block boundary.

Final Answer:

192.168.168.129-190