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Difficulty: HardIPv4 Addressing and Subnetting

During a network security audit, an incident response analyst discovers a compromised host with the IPv4 address 172.16.89.138/22172.16.89.138/22. What is the broadcast address for the subnet to which this host belongs?

Answer: 172.16.91.255

Answer

The broadcast address for the subnet is 172.16.91.255.
For an IP address of 172.16.89.138/22172.16.89.138/22, the subnet mask is 255.255.252.0. The third octet increments in multiples of 4 (256−252=4256 - 252 = 4). The network containing the third octet value of 89 starts at 88 (172.16.88.0172.16.88.0). The subnet spans up to 172.16.91.255172.16.91.255, making 172.16.91.255172.16.91.255 the broadcast address where all 10 host bits are set to 1.

Step-by-Step Solution

1
Determine the subnet mask and block size from the CIDR prefix /22.
A /22 prefix corresponds to a subnet mask of 255.255.252.0. The block size in the third octet is 256−252=4256 - 252 = 4.
The prefix /22 leaves 10 host bits across the third and fourth octets, giving 210=10242^{10} = 1024 total IP addresses (44 blocks of 256).
2
Identify the network address for the host IP address 172.16.89.138.
The third octet boundaries increment by 4 (0,4,8,…,84,88,920, 4, 8, \dots, 84, 88, 92). Since 89 lies between 88 and 91, the network address is 172.16.88.0.
The network ID uses the lowest address in the block formed by the multiple of the block size (8888).
3
Calculate the broadcast address of the subnet.
The broadcast address is 172.16.91.255.
The broadcast address has all 10 host bits set to 1, which corresponds to the last IP address in the 172.16.88.0−172.16.91.255172.16.88.0 - 172.16.91.255 range.

Key Concept

IPv4 Non-Octet Boundary Subnetting and Broadcast Address Calculation
Estimated Time:2m 0s
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Topics

IPv4 Addressing and Subnetting

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