ITPEC FE Morning April 2022 Question 34

192.168.128.63 — the broadcast address sets all 6 host bits of the /26 network to 1.

• •Host bits: 32 − 26 = 6, a block of 2⁶ = 64 addresses
• •The network 192.168.128.0/26 covers the last-octet values 0 to 63
• •Last octet with the host bits all 1: 00 + 111111 = 00111111 = 63

Answer (a)

Why not others:

• •(b) 192.168.128.127 — the broadcast address of a /25, 7 host bits and 128 addresses; it lies outside this /26.
• •(c) 192.168.128.252 — 11111100 is not in the range 0 to 63 at all; it belongs to the /26 network starting at .192, and even there its host bits 111100 are not all 1. (252 is the last octet of a /30 subnet mask.)
• •(d) 192.168.128.255 — the broadcast address of the /24 network 192.168.128.0/24.

Key rule: The broadcast address is the network address plus 2^h − 1, where h is the number of host bits; the network address is the same with all host bits 0.

Networks · AI-written, may contain errors