ITPEC FE Morning October 2020 Question 33
ITPEC FE Morning October 2020 — Question 33 of 80
Broadcast address with /22 subnet — find the broadcast address for a given IP and CIDR prefix.
Given IP 192.168.57.123/22:
- •
/22means subnet mask255.255.252.0— 22 network bits, 10 host bits - •In the 3rd octet, 6 bits are network bits → block size =
2^(8-6)= 4 - •Network blocks in 3rd octet:
…52, 56, 60… - •
57falls in block 56–59 - •Broadcast = last address in block →
192.168.59.255
Why not others:
- (a) 192.168.55.255 — broadcast of the 52–55 block; 57 is not in that range
- (b) 192.168.57.255 — treats it as /24, but /22 spans 4 values in the 3rd octet
- (d) 192.168.63.255 — broadcast of the next block (60–63)
Key rule: For /22, divide the 3rd octet into blocks of 4. The broadcast is the last address of the block containing the host IP.
AI-generated — may contain errors
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