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Question

Difficulty: MediumIPv4 Addressing and Subnetting

A network administrator is allocating IP addresses for a newly created VoIP segment assigned the IPv4 subnet block 172.31.80.0/21172.31.80.0/21. What is the broadcast address of this subnet?

Answer: 172.31.87.255

Answer

172.31.87.255
For the subnet 172.31.80.0/21172.31.80.0/21, the subnet mask is 255.255.248.0255.255.248.0. The block size in the third octet is 8. The network range extends from 172.31.80.0172.31.80.0 to 172.31.87.255172.31.87.255, making 172.31.87.255172.31.87.255 the broadcast address.

Step-by-Step Solution

1
Determine the subnet mask and block size for a /21 prefix.
A /21 prefix corresponds to a subnet mask of 255.255.248.0255.255.248.0. The block size in the third octet is 256−248=8256 - 248 = 8.
Prefix length /21 leaves 3 subnet bits in the third octet (23=82^3 = 8 block size).
2
Calculate the next subnet network address.
Adding the block size of 8 to the third octet of 172.31.80.0172.31.80.0 yields the next network address: 172.31.88.0172.31.88.0.
The next subnet boundary defines the end of the current subnet range.
3
Subtract 1 from the next network address to find the broadcast address.
172.31.88.0−1=172.31.87.255172.31.88.0 - 1 = 172.31.87.255.
The broadcast address is the highest address within the subnet range where all host bits are set to 1.

Key Concept

Subnet Boundary and Broadcast Address Calculation
Estimated Time:1m 0s
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Topics

IPv4 Addressing and Subnetting

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