CIDR capacity — how many subnets and hosts does 172.16.0.0/19 provide? For the private Class B network 172.16.0.0 with a /19 mask, determine the number of subnets created and the number of usable hosts per subnet.

Introduction / Context:Mask length determines both the number of subnets and host capacity. Quickly computing these values is essential for scalable design, especially when planning allocations inside private ranges like 172.16.0.0/12 and its sub-blocks.

Given Data / Assumptions:

• Base classful reference: Class B default /16.
• Given mask: /19 → 3 more bits beyond /16.
• Usable hosts per subnet exclude network and broadcast addresses.

Concept / Approach:Borrowed subnet bits = 19 − 16 = 3 → number of subnets = 2^3 = 8. Host bits per subnet = 32 − 19 = 13 → usable hosts = 2^13 − 2 = 8192 − 2 = 8190. These computations do not depend on classful “zero subnet” restrictions in modern CIDR practice.

Step-by-Step Solution:Compute subnet count: 2^(borrowed) = 2^3 = 8.Compute host bits: 32 − 19 = 13.Compute usable hosts: 2^13 − 2 = 8190.Report as “8 subnets, 8,190 hosts each.”

Verification / Alternative check:Block size per subnet in the third octet is 256 − 224 = 32; subnets progress 0, 32, 64, … which corroborates 8 distinct subnets across the /16 space (256/32 = 8).

Why Other Options Are Wrong:

• A/D: 7 subnets is invalid; with 3 borrowed bits you get 8.
• C/E: Host counts correspond to /21 and /22-style calculations, not /19.

Common Pitfalls:Mixing classful defaults with CIDR arithmetic or forgetting to subtract 2 for network and broadcast when host addressing is required.

Final Answer:8 subnets, 8,190 hosts each