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Difficulty: MediumIPv4 Addressing and Subnetting

A network technician is provisioning an isolated management subnet for a server rack containing 55 hardware management controllers. The subnet must be carved from the 192.168.45.0/24192.168.45.0/24 network block using the smallest possible prefix length that satisfies the requirement while minimizing unassigned addresses. What is the broadcast address of this newly provisioned subnet?

  1. A
    192.168.45.5192.168.45.5
  2. 192.168.45.7192.168.45.7Answer
  3. C
    192.168.45.8192.168.45.8
  4. D
    192.168.45.15192.168.45.15

Answer

The broadcast address of the subnet is 192.168.45.7192.168.45.7.
To host 5 devices, the formula 2h−2≥52^h - 2 \ge 5 yields h=3h = 3 host bits (66 usable addresses), resulting in a /29/29 prefix (255.255.255.248255.255.255.248). The block size is 23=82^3 = 8. Starting from 192.168.45.0192.168.45.0, the subnet encompasses addresses 192.168.45.0192.168.45.0 through 192.168.45.7192.168.45.7. The highest address in the block, 192.168.45.7192.168.45.7, is the broadcast address.

Step-by-Step Solution

1
Determine the required number of host bits.
3 host bits are required.
Using the usable host formula 2h−2≥hosts2^h - 2 \ge \text{hosts}, 23−2=62^3 - 2 = 6 usable host addresses, which satisfies the 55 host interface requirement.
2
Calculate the CIDR prefix length and block size.
Prefix length is /29/29 and block size is 88.
32−3=2932 - 3 = 29. A /29/29 subnet mask (255.255.255.248255.255.255.248) allocates 232−29=82^{32-29} = 8 total IP addresses per subnet.
3
Find the broadcast address for the subnet starting at 192.168.45.0/29192.168.45.0/29.
Broadcast address is 192.168.45.7192.168.45.7.
The IP range for 192.168.45.0/29192.168.45.0/29 spans from 192.168.45.0192.168.45.0 (network ID) to 192.168.45.7192.168.45.7 (broadcast address).

Key Concept

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

IPv4 Addressing and Subnetting

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