For the class B network address 172.16.0.0 with a /19 prefix, how many subnets and how many usable hosts per subnet are available?

Introduction / Context:
Subnetting questions are very common in Cisco and networking exams. For a given classful network and prefix length, you must be able to calculate the number of subnets and the number of usable hosts per subnet. This question focuses on a class B network, 172.16.0.0, that has been given a /19 prefix, which is more specific than the default /16. You need to determine how many subnets are created and how many hosts each subnet can support.

Given Data / Assumptions:

Base network is 172.16.0.0, which is a class B network with default mask /16.
Actual prefix is /19, which means additional bits have been borrowed for subnetting.
We assume standard subnetting rules, counting usable hosts as 2^h minus 2.

Concept / Approach:
For a class B network, the default prefix is /16, so there are 16 network bits and 16 host bits. A /19 prefix uses 19 bits for the network and subnet portion, leaving 32 minus 19 equals 13 host bits. The number of bits borrowed for subnetting is 19 minus 16 equals 3 bits. The number of subnets equals 2^s, where s is the number of borrowed bits, so we get 2^3 equals 8 subnets. The number of hosts per subnet equals 2^h minus 2, where h is the number of host bits. With 13 host bits, the host count is 2^13 minus 2, which equals 8192 minus 2, or 8190 usable hosts per subnet.

Step-by-Step Solution:
Step 1: Recognize that the network 172.16.0.0 is a class B address with a default /16 mask.Step 2: Calculate the number of bits borrowed for subnetting: 19 minus 16 equals 3 borrowed bits.Step 3: Compute the number of subnets as 2^3, which equals 8.Step 4: Determine the host bits: 32 total bits minus 19 network bits equals 13 host bits.Step 5: Compute usable hosts per subnet as 2^13 minus 2, which equals 8192 minus 2, or 8190 hosts.

Verification / Alternative check:
As a check, think in terms of the third octet block size. A /19 mask in dotted decimal is 255.255.224.0. The block size in the third octet is 256 minus 224, which equals 32. This gives subnet ranges of 0, 32, 64, 96, 128, 160, 192, and 224 in the third octet, which is 8 distinct subnets. Within each subnet, the combination of the last 13 bits defines host addresses, again leading to 8190 usable hosts. This confirms the previous calculation.

Why Other Options Are Wrong:
Option A is wrong because 7 subnets would imply 2^s minus 1 style counting, which is not used in modern practice, and hosts per subnet would then be incorrect.
Option B is wrong because it combines the wrong number of subnets with the correct host count, which is inconsistent with the number of borrowed bits.
Option C is wrong because 2046 hosts per subnet correspond to 11 host bits, which does not match a /19 configuration on a class B network.

Common Pitfalls:
A common pitfall is to subtract two from the subnet count, using outdated rules that reserved the all zeros and all ones subnets. Modern practice usually allows using all subnets. Another mistake is miscounting host bits or misinterpreting the prefix length relative to the base class. Always carefully subtract the prefix length from 32 to get host bits, and subtract the new prefix from the default prefix to get the number of subnet bits.

Final Answer:
8 subnets, 8190 hosts each