Given the network 192.168.0.0/24, you want to create subnets that can each support at least 50 usable hosts. Which prefix length (subnet mask) should you use for each subnet?

Introduction / Context:
Subnetting a small private network such as 192.168.0.0/24 is a classic exercise in Cisco exams. The goal here is to choose a subnet size that provides at least 50 usable host addresses per subnet, while using the address space efficiently. The answer requires calculating how many host bits are needed and then converting that to a prefix length.

Given Data / Assumptions:

Base network: 192.168.0.0/24, which has 256 total addresses (0 to 255 in the last octet).
Required usable hosts per subnet: at least 50.
Usable hosts per subnet = 2^h - 2, where h is the number of host bits; 2 addresses are reserved for network and broadcast.

Concept / Approach:
We must find the smallest number of host bits h such that 2^h - 2 ≥ 50. Then we subtract h from 32 to get the prefix length. With a /24 base, we are effectively subnetting the last octet. It is also useful to remember common values: /25 gives 126 usable hosts, /26 gives 62 usable, /27 gives 30 usable, and so on.

Step-by-Step Solution:
Compute candidate host sizes: for h = 5, 2^5 = 32 and 32 - 2 = 30 usable hosts, which is less than 50 and therefore not sufficient. For h = 6, 2^6 = 64 and 64 - 2 = 62 usable hosts, which is greater than or equal to 50 and therefore meets the requirement. With 6 host bits, the prefix length is 32 - 6 = 26, which corresponds to a /26 subnet mask. A /26 mask in dotted decimal is 255.255.255.192, giving four equal subnets within the 192.168.0.0/24 range, each with 62 usable addresses. Checking the options, /26 is the smallest subnet that meets the host requirement and is therefore the best answer.

Verification / Alternative check:
We can list a quick summary from memory or a cheat sheet: /25 → 126 usable, /26 → 62 usable, /27 → 30 usable. Only /25 and /26 satisfy the “at least 50 host” requirement, but /25 gives more hosts than needed and uses larger subnets, while /26 is still sufficient and more granular. The question does not ask for the maximum possible hosts, but for the appropriate mask to satisfy the minimum requirement efficiently, so /26 is preferred.

Why Other Options Are Wrong:
Option a (/24) does not create smaller subnets; it keeps a single subnet with 254 usable hosts, which does not answer the subnetting requirement.
Option b (/25) supports 126 usable hosts per subnet, which is valid but larger than necessary; exam questions usually prefer the smallest mask that meets the requirement unless otherwise stated.
Option d (/27) provides only 30 usable hosts, which is too few for the requirement of 50 hosts.
Option e (/28) is even smaller and provides only 14 usable hosts, far below the requirement.

Common Pitfalls:
Many learners either forget to subtract 2 for network and broadcast addresses or assume that any mask that supports more than the required hosts is equally acceptable. In exam situations, you are typically expected to choose the smallest subnet that meets the requirement, as this conserves address space. Knowing common prefix length and host count pairs by heart saves time and reduces calculation mistakes.

Final Answer:
The correct prefix length is /26 (255.255.255.192), which provides about 62 usable host addresses per subnet.