A host is configured with the address 172.20.35.77/20. What is the network address of its subnet?
LPIC-1 Exam 102-500, objective 109. Networking fundamentals hard
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The options
Not correct 172.20.0.0
Wrong. 172.20.0.0 would be the network address for a /16 prefix, where the whole third octet is host space. A /20 splits the third octet.
Correct 172.20.32.0
Correct. A /20 is the mask 255.255.240.0. ANDing the third octet, 35 & 240 = 32, and the fourth octet becomes 0, giving 172.20.32.0.
Not correct 172.20.35.0
Wrong. 172.20.35.0 assumes a /24, where only the last octet is host space. With a /20 the boundary falls inside the third octet.
Not correct 172.20.16.0
Wrong. 172.20.16.0 is the previous /20 block; it covers 172.20.16.0 through 172.20.31.255, and 35 falls above that range.
Why
A /20 mask is 255.255.240.0, so /20 blocks step by 16 in the third octet: 0, 16, 32, 48 and so on. 35 falls in the block that starts at 32, so the network address is 172.20.32.0 and the broadcast address is 172.20.47.255. Bitwise AND of the address with the mask always yields the network address.
Where this comes from
- Cited
- LPI exam objective 109.1
- What it says
- Calculate the network address of a host given its address and subnet mask.
Practise this
Reading one question is not practice. The trainer will draw a set from objective 109 and space the ones you get wrong.