Examkin

Question

Difficulty: MediumIPv4 Addressing and Subnetting

A network engineer is configuring a core router interface assigned the IPv4 address 10.16.42.138/2210.16.42.138/22. What is the broadcast address for this subnet?

Answer: 10.16.43.255

Answer

The broadcast address for the 10.16.42.138/2210.16.42.138/22 subnet is 10.16.43.25510.16.43.255.
For the host IP 10.16.42.138/2210.16.42.138/22, the subnet mask is 255.255.252.0255.255.252.0. In the third octet, subnets increment by 4 (0,4,8,…,40,440, 4, 8, \dots, 40, 44). The network address containing 10.16.42.13810.16.42.138 is 10.16.40.0/2210.16.40.0/22, and the subnet range spans from 10.16.40.010.16.40.0 to 10.16.43.25510.16.43.255. Therefore, the broadcast address for this subnet is 10.16.43.25510.16.43.255.

Step-by-Step Solution

1
Determine the subnet mask and block size from the CIDR prefix.
A /22/22 CIDR prefix corresponds to a subnet mask of 255.255.252.0255.255.252.0. The block size in the third octet is 256−252=4256 - 252 = 4.
The CIDR prefix /22/22 allocates 22 bits to the network portion, leaving 2 network bits in the third octet.
2
Calculate the network boundary (Network ID) for the host IP address 10.16.42.13810.16.42.138.
The third octet network boundaries in increments of 4 are 0,4,8,…,40,440, 4, 8, \dots, 40, 44. Since 42 lies between 40 and 44, the Network ID is 10.16.40.010.16.40.0.
The network ID is determined by rounding down to the nearest multiple of the block size in the third octet.
3
Calculate the broadcast address of the subnet.
The next network boundary is 10.16.44.010.16.44.0. Subtracting 1 from this address gives the broadcast address 10.16.43.25510.16.43.255.
The broadcast address is the highest possible IP address within the subnet range, setting all 10 host bits to 1.

Key Concept

IPv4 Subnetting and Broadcast Address Calculation
Rate this question

Topics

IPv4 Addressing and Subnetting

More questions from this topic