IPv4 subnetting practice: What is the subnetwork (network) address for a host with IP address 200.10.5.68/28?

Difficulty: Easy

Correct Answer: 200.10.5.64

Explanation:


Introduction / Context:
Subnetting with Classless Inter-Domain Routing (CIDR) requires identifying the block size and then finding the exact network address that contains a given host IP. Here we are working with a /28 prefix, a common exam scenario.



Given Data / Assumptions:

  • Host IP: 200.10.5.68/28
  • /28 corresponds to mask 255.255.255.240
  • We seek the network (subnetwork) address for that host


Concept / Approach:
The block size is determined by 256 minus the last-octet mask value. For /28, the last octet mask is 240, so the block size is 256 - 240 = 16. Valid network boundaries occur at increments of 16 in the fourth octet: 0, 16, 32, 48, 64, 80, etc.



Step-by-Step Solution:

Mask /28 => 255.255.255.240 (block size 16).List boundaries in the 4th octet: 0, 16, 32, 48, 64, 80, 96...Locate 68 between 64 and 80.Therefore, the network address is 200.10.5.64 and the broadcast is 200.10.5.79.


Verification / Alternative check:
Compute: 68 - (68 mod 16) = 68 - 4 = 64, confirming the network base.



Why Other Options Are Wrong:

  • 200.10.5.56: 56 is not a /28 boundary (16-multiple) for this range.
  • 200.10.5.32 and 200.10.5.0: Different /28 networks; do not contain 68.
  • 200.10.5.80: Next network boundary above 68, not the one containing it.


Common Pitfalls:
Mixing up /28 with /29 block sizes; forgetting boundaries must be exact multiples of the block size.



Final Answer:
200.10.5.64

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