Difficulty: Easy
Correct Answer: 6
Explanation:
Introduction / Context:
Small subnets such as /30 and /29 are common on WAN links and access segments. You must know exactly how many usable host addresses each mask provides to avoid over-subscribing a subnet or wasting address space.
Given Data / Assumptions:
Concept / Approach:
For /29, host bits h = 32 − 29 = 3. Therefore, usable hosts = 2^3 − 2 = 8 − 2 = 6. This total includes any router interface addresses and all attached host devices. The network and broadcast consume the two excluded addresses in each /29 block of 8 addresses.
Step-by-Step Solution:
Compute host bits: h = 3.Compute usable hosts: 2^3 − 2 = 6.Confirm that 192.168.192.8/29 spans .8–.15 with usable .9–.14, which contains .10.Therefore, total usable addresses on that LAN (including the router) = 6.
Verification / Alternative check:
Block size for /29 is 256 − 248 = 8 addresses per subnet. Usable = 8 − 2 = 6 corroborates the formula above.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing “hosts total” with “usable hosts,” and forgetting that the question includes the router interface in the same count.
Final Answer:
6
Discussion & Comments