ITPEC FE Subject A October 2024 Question 24

192.168.131.255 — set all 10 host bits of the /22 network to 1.

• •/22 leaves 32 − 22 = 10 host bits: the last 2 bits of the third octet and all 8 bits of the fourth
• •Third octet of the network: 128 = 10000000; the network spans 2² = 4 values of it, 128 to 131
• •Host bits all 1: third octet 10000011 = 131, fourth octet 11111111 = 255
• •Broadcast address: 192.168.131.255 (the next network would start at 192.168.132.0)

Answer (c)

Why not others:

• •(a) 192.168.128.127 — the broadcast of 192.168.128.0/25: only 7 host bits set, as if the prefix were /25
• •(b) 192.168.128.255 — the broadcast of 192.168.128.0/24: sets only the fourth octet and ignores the 2 host bits in the third octet
• •(d) 192.168.255.255 — sets every bit of the last two octets, which is right only for a /17 or /16 prefix; it treats the network bits of the third octet as host bits

Key rule: Broadcast = network address with every host bit set to 1; when the prefix ends inside an octet, the block size in that octet is 2ⁿ for its n host bits, and the broadcast is the next block's start minus 1.

Networks · AI-written, may contain errors