A network administrator is reviewing the network configuration of a server assigned the IP… — Cisco CCNA

A network administrator is reviewing the network configuration of a server assigned the IPv4 address $172.28.143.100$ with a subnet mask of $255.255.252.0$ (/22). Which IPv4 address is the broadcast address for this subnet?

$172.28.143.255$ Answer
B
$172.28.140.0$
C
$172.28.143.254$
D
$172.28.144.255$

Answer

The broadcast address for the subnet containing host

172.28.143.100/22

172.28.143.255

For a /22 subnet mask (

255.255.252.0

), the third octet increments in steps of 4 (

256 - 252 = 4

). For the IP address

172.28.143.100

, the containing network boundary starts at

172.28.140.0/22

and ends at

172.28.143.255

. The last address in the subnet range (

172.28.143.255

) is reserved as the directed broadcast address.

Step-by-Step Solution

Determine the block size in the third octet from the subnet mask

A mask of

255.255.252.0

(/22) leaves

256 - 252 = 4

as the block size in the third octet.

The third octet is the interesting octet where subnetting occurs for a /22 prefix.

Find the network address boundary for the host IP address 172.28.143.100

Multiples of 4 in the third octet are 0, 4, 8, ..., 136, 140, 144. Since 143 falls between 140 and 144, the network address is

172.28.140.0

The subnet boundary begins at the nearest lower multiple of the block size (140).

Calculate the broadcast address of the subnet

The next subnet begins at

172.28.144.0

, so subtracting 1 IP address gives the broadcast address

172.28.143.255

The broadcast address is always one address below the network ID of the next consecutive subnet.

Key Concept

IPv4 Subnet Boundaries and Broadcast Address Calculation

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