ITPEC FE Morning April 2023 Question 33

62 — a /26 leaves 6 host bits, and two of the 64 addresses are reserved.

• •Host bits: 32 - 26 = 6
• •Addresses: 2⁶ = 64, from 192.168.10.0 to 192.168.10.63
• •Host addresses: 64 - 2 = 62, leaving out the network address 192.168.10.0 and the broadcast address 192.168.10.63

Answer (b)

Why not others:

• •(a) 30 — 2⁵ - 2, the count for a /27 with 5 host bits.
• •(c) 126 — 2⁷ - 2, the count for a /25 with 7 host bits.
• •(d) 254 — 2⁸ - 2, treating the whole last octet as the host part, as in a /24, and ignoring the two extra prefix bits.

Key rule: Host addresses = 2^h - 2, where h is 32 minus the prefix length; the 2 are the network and broadcast addresses.

Networks · AI-written, may contain errors