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Difficulty: MediumIPv4 Addressing and Subnetting

A network administrator receives an alert that a server with the IPv4 address 172.24.110.99/23172.24.110.99/23 cannot communicate beyond its local segment due to a misconfiguration on the upstream switch interface. What is the correct broadcast address for the subnet to which this server belongs?

  1. A
    172.24.110.255
  2. B
    172.24.111.254
  3. 172.24.111.255Answer
  4. D
    172.24.110.0

Answer

172.24.111.255
For an IP address of 172.24.110.99/23172.24.110.99/23, the prefix length of /23/23 indicates a subnet mask of 255.255.254.0255.255.254.0. The third octet block size is 256−254=2256 - 254 = 2. Because 110110 is an even multiple of 22, the network boundary begins at 172.24.110.0172.24.110.0 and ends at 172.24.111.255172.24.111.255. The broadcast address is the highest address in this range, which is 172.24.111.255172.24.111.255.

Step-by-Step Solution

1
Determine the subnet mask from CIDR notation
A /23 prefix corresponds to a subnet mask of 255.255.254.0255.255.254.0.
Converting CIDR /23 means 23 network bits and 9 host bits (32−23=932 - 23 = 9).
2
Calculate the block size in the third octet
Block size = 256−254=2256 - 254 = 2.
Subnet boundaries in the third octet occur in multiples of 2.
3
Identify the network ID and broadcast address
Network ID = 172.24.110.0172.24.110.0, Broadcast address = 172.24.111.255172.24.111.255.
Since the third octet value is 110110 (a multiple of 22), the subnet range spans 172.24.110.0172.24.110.0 through 172.24.111.255172.24.111.255.

Key Concept

IPv4 Subnet Mask and Broadcast Address Calculation
Estimated Time:1m 30s
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Topics

IPv4 Addressing and Subnetting

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