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Difficulty: HardIPv4 Addressing and Subnetting

A network engineer is configuring a point-to-point connection and internal VLANs within an enterprise infrastructure. An interface is assigned the IPv4 address 10.144.67.100/2110.144.67.100/21. What is the decimal value of the third octet of the broadcast address for the subnet to which this host belongs?

Answer: 71

Answer

The third octet of the broadcast address is 71 (broadcast address is 10.144.71.255).
For the prefix /21 (subnet mask 255.255.248.0), the third octet operates with a block size of 8 (256 - 248 = 8). The third octet value 67 falls into the subnet starting at 64 (10.144.64.0). The next subnet starts at 72 (10.144.72.0), making the broadcast address for the current subnet 10.144.71.255. The third octet of this broadcast address is 71.

Step-by-Step Solution

1
Convert CIDR prefix /21 to dotted-decimal mask.
Mask is 255.255.248.0 (21 network bits: 8 + 8 + 5).
The third octet contains 5 subnet bits and 3 host bits.
2
Calculate the block size (increment) for the third octet.
256 - 248 = 8.
Subnets increment in steps of 8 in the third octet.
3
Find the third octet network boundary for IP 10.144.67.100.
Network address is 10.144.64.0.
The greatest multiple of 8 less than or equal to 67 is 64.
4
Determine the broadcast address of the subnet.
Broadcast address is 10.144.71.255.
The subnet spans from 10.144.64.0 to 10.144.71.255, one value prior to the next subnet 10.144.72.0.

Key Concept

Subnet Boundary and Broadcast Address Calculation
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Topics

IPv4 Addressing and Subnetting

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