You administer the Class B network 134.57.0.0 and must create exactly 8 subnets while maximizing the number of usable host IDs per subnet. Which subnet mask should you assign?

Introduction / Context: Subnetting a Class B (/16) network involves borrowing bits from the host portion to create subnets. The design goal here is to produce exactly 8 subnets and keep each subnet as large as possible (maximum hosts per subnet).

Given Data / Assumptions:

• Base network: Class B 134.57.0.0 with default /16.
• Required subnets: 8.
• Objective: largest possible hosts per subnet.
• Use contemporary practice that allows all-zeros and all-ones subnets.

Concept / Approach: Number of subnets = 2^(borrowed bits). To get exactly 8 subnets, borrow 3 bits (since 2^3 = 8). New prefix = /16 + 3 = /19. The /19 mask in dotted decimal is 255.255.224.0 (since 224 = 11100000 in the third octet). Hosts per /19 subnet = 2^(32 − 19) − 2 = 8192 − 2 = 8190 usable addresses.

Step-by-Step Solution: Borrow 3 bits from the host field: /16 → /19.Convert /19 to dotted decimal: 255.255.224.0.Confirm subnets: 2^3 = 8 distinct subnets.Confirm host capacity: 2^(13) − 2 = 8190 usable per subnet.

Verification / Alternative check: Subnet increments occur every 32 in the third octet (.0, .32, .64, .96, .128, .160, .192, .224), yielding exactly eight /19 blocks—matching the requirement.

Why Other Options Are Wrong: 255.255.240.0 (/20), 255.255.248.0 (/21), 255.255.252.0 (/22): Borrow 4, 5, or 6 bits → 16, 32, or 64 subnets (more than needed) and smaller host ranges.

255.255.255.255: Host mask, not usable for subnets.

Common Pitfalls: Forgetting to maximize hosts (thus borrowing too many bits) or using outdated “no all-zeros/all-ones subnet” rules which would alter counts.

Final Answer: 255.255.224.0