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(Prog. Concept Question) Subnetting**
Which of the following is a valid IP host address given the network ID of 191.254.0.0 while using 11 bits for subnetting?
Pick ONE option
Show answer & explanation
The network ID 191.254.0.0 with 11 bits for subnetting creates a /22 classless network (Class B defaults to /16, plus 11 subnet bits = /27 effective subnetting within the third octet). With 11 subnet bits in the third and fourth octets, valid host ranges exist between subnet boundaries. Option C (191.254.1.29) falls within a valid host range, while A and B are subnet/boundary addresses (powers of 2), and D has an incorrect network octet (54 instead of 254).
Step-by-step Derivation:
Step 1: Network ID is 191.254.0.0 (Class B, /16 base). With 11 subnet bits, we use /27 addressing (16 + 11 = 27).
Step 2: With /27, each subnet has 2^(32-27) = 2^5 = 32 addresses. Subnets occur at boundaries: 0, 32, 64, 96, 128, etc. in the combined third-fourth octets.
Step 3: Convert to 16-bit representation of third-fourth octets:
- 191.254.0.32 = third octet = 0, fourth octet = 32 (subnet boundary, not a host)
- 191.254.0.96 = third octet = 0, fourth octet = 96 (subnet boundary, not a host)
- 191.254.1.29 = third octet = 1, fourth octet = 29 (within subnet range 1.0–1.31, valid host)
- 191.54.1.64 = network octet wrong (54 ≠ 254, invalid)
Step 4: Option C is the only valid host address in a legitimate subnet range with correct network address.