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EC-Council

312-50V9 · Question #119

What is the broadcast address for the subnet 190.86.168.0/22?

The correct answer is C. 190.86.171.255. A /22 subnet mask (255.255.252.0) provides 10 host bits, spanning the range 190.86.168.0 to 190.86.171.255, so the broadcast address is 190.86.171.255.

Scanning Networks

Question

What is the broadcast address for the subnet 190.86.168.0/22?

Options

  • A190.86.168.255
  • B190.86.255.255
  • C190.86.171.255
  • D190.86.169.255

How the community answered

(26 responses)
  • A
    15% (4)
  • B
    8% (2)
  • C
    73% (19)
  • D
    4% (1)

Why each option

A /22 subnet mask (255.255.252.0) provides 10 host bits, spanning the range 190.86.168.0 to 190.86.171.255, so the broadcast address is 190.86.171.255.

A190.86.168.255

190.86.168.255 is the broadcast for a /24 subnet beginning at 190.86.168.0, not for the larger /22 block.

B190.86.255.255

190.86.255.255 is the broadcast for a /16 subnet (mask 255.255.0.0), which is far larger than the /22 block in question.

C190.86.171.255Correct

A /22 prefix applies a mask of 255.255.252.0, leaving 10 bits for hosts. Starting from 190.86.168.0, the last two bits of the third octet extend the range from 168 through 171, and all eight bits of the fourth octet extend from 0 to 255, placing the highest (broadcast) address at 190.86.171.255.

D190.86.169.255

190.86.169.255 falls within the valid host range of the /22 subnet but is not the highest address, so it cannot be the broadcast address.

Concept tested: IPv4 subnetting broadcast address calculation

Source: https://learn.microsoft.com/en-us/troubleshoot/windows-client/networking/tcpip-addressing-and-subnetting

Topics

#subnetting#CIDR#broadcast address#IP addressing

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