ITPEC FE Subject A April 2024 Question 24
Source exam: ITPEC FE Subject A April 2024Topic: Networks
ITPEC FE Subject A April 2024 — Question 24 of 60
Broadcast address — find the broadcast address for a given CIDR network.
Given 192.168.10.0/23:
- •/23 → subnet mask =
255.255.254.0→ 9 host bits - •Network spans 2 consecutive /24 blocks:
192.168.10.0–192.168.11.255 - •Broadcast = all host bits set to
1→192.168.11.255
Why not others:
- (a) 192.168.10.0 — this is the network address itself
- (b) 192.168.10.255 — would be broadcast for /24, but /23 covers a wider range
- (c) 192.168.10.511 — 511 is impossible in an octet (max 255)
Key rule: For /23, the network spans 2 × /24 blocks. Broadcast = last address of the second block.
AI-generated — may contain errors
The original exam layout is preserved in the image so diagrams, formulas, tables, and code remain accurate.
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.