ITPEC FE Morning April 2022 Question 33

510 — a /23 prefix leaves 9 host bits, and the host parts of all 0s and all 1s cannot be assigned.

• •Host bits: 32 − 23 = 9
• •Addresses in the network: 2⁹ = 512, from 172.16.64.0 to 172.16.65.255
• •The network address 172.16.64.0 (host bits all 0) and the broadcast address 172.16.65.255 (host bits all 1) are reserved: 512 − 2 = 510

Answer (c)

Why not others:

• •(a) 254 — the assignable hosts of a /24, 2⁸ − 2: treats the network as ending at the third octet, although a /23 takes one host bit from it.
• •(b) 256 — 2⁸, all the addresses of a /24, with the same wrong host-bit count and nothing subtracted.
• •(d) 512 — the right total, but the network and broadcast addresses are counted as assignable.

Key rule: The number of assignable host addresses is 2^h − 2, where h is 32 minus the prefix length.

Networks · AI-written, may contain errors