ITPEC FE Subject A April 2024 Question 24

192.168.11.255 — a /23 network spans two consecutive /24 blocks, and its broadcast address is the last address of the second one.

• •/23 leaves 32 − 23 = 9 host bits, so the network has 2⁹ = 512 addresses
• •The mask is 255.255.254.0: the lowest bit of the third octet is a host bit, so the network covers third octets 10 and 11
• •Range: 192.168.10.0 (network address) to 192.168.11.255
• •Broadcast = all 9 host bits set to 1 = 192.168.11.255

Answer (d)

Why not others:

• •(a) 192.168.10.0 — all host bits 0: that is the network address, not the broadcast address
• •(b) 192.168.10.255 — the broadcast address of a /24; it ignores the ninth host bit in the third octet and is an ordinary host address in this /23
• •(c) 192.168.10.511 — adds the 511 host offset to the last octet as if it could hold it; an octet only goes up to 255, so the carry belongs in the third octet

Key rule: For a prefix shorter than /24, the host bits spill into the third octet: set every host bit to 1 across octets to get the broadcast address.

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