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Question

Difficulty: HardIPv4 Addressing and Subnetting

A security analyst is investigating a potential incident involving a server assigned the IP address 10.80.35.138/2610.80.35.138/26. What is the broadcast IPv4 address of the subnet to which this server belongs?

Answer: 10.80.35.191

Answer

10.80.35.191
The CIDR prefix /26/26 indicates a subnet mask of 255.255.255.192255.255.255.192, creating subnet increments of 6464 addresses in the fourth octet. The subnets start at .0.0, .64.64, .128.128, and .192.192. The IP address 10.80.35.13810.80.35.138 lies within the .128.128 subnet block (10.80.35.128/2610.80.35.128/26). The broadcast address is the final IP in this block, which is 10.80.35.19110.80.35.191.

Step-by-Step Solution

1
Determine the subnet mask and block size from the CIDR prefix length.
A /26/26 prefix corresponds to a subnet mask of 255.255.255.192255.255.255.192. The block size in the fourth octet is 256−192=64256 - 192 = 64.
The CIDR notation /26/26 uses 26 network bits, leaving 6 host bits (26=642^6 = 64 addresses per subnet).
2
Identify the network boundaries for the fourth octet.
Subnet boundaries increment by 6464: 00, 6464, 128128, and 192192.
Each subnet span covers 64 continuous IP addresses.
3
Locate the given IP address within its subnet range.
The IP address 10.80.35.13810.80.35.138 falls within the range 10.80.35.12810.80.35.128 through 10.80.35.19110.80.35.191.
Since 128≤138<192128 \le 138 < 192, the network address for this subnet is 10.80.35.12810.80.35.128.
4
Determine the broadcast address for the subnet.
The broadcast address is 10.80.35.19110.80.35.191.
The broadcast address is the last address in the subnet range, immediately preceding the next subnet network ID (10.80.35.192−1=10.80.35.19110.80.35.192 - 1 = 10.80.35.191).

Key Concept

IPv4 Subnetting and Broadcast Address Calculation
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Topics

IPv4 Addressing and Subnetting

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