ITPEC FE Subject A October 2024 Question 24
ITPEC FE Subject A October 2024 — Question 24 of 60
Broadcast Address Calculation — find the broadcast address for a given CIDR network.
Given 192.168.128.0/22:
- •/22 mask =
255.255.252.0→ 10 host bits - •Block size in 3rd octet:
256 - 252=4 - •Network starts at
192.168.128.0, next network =192.168.132.0 - •Broadcast = next network − 1 =
192.168.131.255
Quick method:
- Find block size: 256 - subnet_octet_mask
- Next network = network_start + block_size
- Broadcast = next_network − 1, with last octet = 255
Why not others:
- (a) 192.168.128.127 — broadcast for /25 (block size 128 in 4th octet)
- (b) 192.168.128.255 — broadcast for /24 (block size 1 in 3rd octet)
- (d) 192.168.255.255 — broadcast for /16
Key rule: Broadcast address = all host bits set to 1. For /22, that means 10 host bits across the 3rd and 4th octets.
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.