A subnet is defined as 192.168.10.0/26. How many addresses in it can actually be assigned to hosts?
LPIC-1 Exam 102-500, objective 109. Networking fundamentals medium
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The options
Not correct 64
Wrong. 64 is the total size of the block (2^(32-26) = 64 addresses), but the first address is the network address and the last is the broadcast address, so two of the 64 are not assignable.
Not correct 30
Wrong. 30 usable hosts is what a /27 gives you (32 addresses minus network and broadcast).
Not correct 126
Wrong. 126 usable hosts is a /25 (128 addresses minus two).
Correct 62
Correct. A /26 leaves 6 host bits, so 2^6 = 64 addresses; subtracting the network address (192.168.10.0) and the broadcast address (192.168.10.63) leaves 62 assignable host addresses.
Why
For an IPv4 subnet with a prefix length of n, the block holds 2^(32-n) addresses and 2^(32-n) - 2 of them are usable by hosts, because the all-zeros host part is the network address and the all-ones host part is the broadcast address. /26 -> 64 total, 62 usable. The two special addresses are why a /30 gives only 2 usable addresses, which is the classic point-to-point link size.
Where this comes from
- Cited
- LPI exam objective 109.1
- What it says
- Understand network masks and CIDR notation, and how they determine the size of a subnet.
Practise this
Reading one question is not practice. The trainer will draw a set from objective 109 and space the ones you get wrong.