For a host with IPv4 address 200.10.5.68/28, what is the correct subnetwork (network) address for the subnet to which this host belongs?

Introduction / Context:
This question checks your ability to determine the network address of a subnet when given a host IP address and a prefix length. This skill is fundamental in IPv4 subnetting, routing, and troubleshooting tasks, since routers and administrators think in terms of network addresses rather than individual hosts when designing and summarizing networks.

Given Data / Assumptions:
- Host IP address: 200.10.5.68.
- Prefix length: /28, which corresponds to a subnet mask of 255.255.255.240.
- We must find the network address of the subnet that contains this host.
- The network address is the first address in the subnet and has all host bits set to 0.

Concept / Approach:
A /28 prefix means that 28 bits are used for the network portion, leaving 4 bits for the host portion in the last octet. Four host bits give 2^4 = 16 total addresses per subnet. Therefore, subnets in the last octet increment in blocks of 16. By identifying the block in which 68 falls, we can determine the correct network address. The valid blocks in the last octet are 0, 16, 32, 48, 64, 80, 96, and so on.

Step-by-Step Solution:
Step 1: Convert the /28 prefix to a subnet mask: 255.255.255.240. This confirms there are 16 addresses per subnet in the last octet.Step 2: Determine the block size in the last octet as 256 - 240 = 16.Step 3: List the subnet start values in the last octet: 0, 16, 32, 48, 64, 80, 96, and so on up to 240.Step 4: Identify where the host value 68 falls. It is greater than or equal to 64 and less than 80, so it belongs to the 64 to 79 range.Step 5: The network address for that range is 200.10.5.64, and the broadcast address is 200.10.5.79, with usable hosts from 200.10.5.65 to 200.10.5.78.

Verification / Alternative check:
You can verify by subtracting the nearest lower multiple of 16 from the host address last octet. Dividing 68 by 16 gives 4 with a remainder, and 4 * 16 = 64. This confirms that the network address is 200.10.5.64. Any host with a last octet between 64 and 79 inclusive would be in that same subnet under a /28 mask.

Why Other Options Are Wrong:
200.10.5.56 would be the network address if the block size created a range that contained 56, but 56 is in the previous block (48 to 63), which does not include 68. 200.10.5.32 belongs to the 32 to 47 range, again not containing 68. 200.10.5.0 would be the network address for the first subnet only and does not contain 68. 200.10.5.80 is actually the start of the next subnet range (80 to 95) and is greater than the host address 68.

Common Pitfalls:
A common mistake is to pick the closest lower multiple of 8, 32, or another number without correctly calculating the block size from the mask. Another error is to confuse the broadcast address with the network address. Remember that the network address is always the lowest address in the block and has all host bits set to 0. Practicing with simple block size calculations helps reduce these errors in subnetting questions.

Final Answer:
The correct subnetwork address for host 200.10.5.68 with a /28 prefix is 200.10.5.64.