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Difficulty: HardIPv4 Addressing and Subnetting

A network administrator is carving out a new subnet from the IP address block 10.150.0.0/1610.150.0.0/16 for a high-density virtualization cluster. The cluster requires static IP address assignments for 250250 virtual machine hosts, 33 redundant default gateway router interfaces, and 22 out-of-band management switches. To minimize wasted address space, the administrator allocates the smallest CIDR subnet starting at network address 10.150.32.010.150.32.0 that satisfies all host requirements. Which of the following statements regarding this newly created subnet are correct? (Select TWO).

  1. The subnet requires a /23 CIDR prefix (subnet mask 255.255.254.0) to accommodate all 255 required IP endpoints.Answer
  2. B
    A /24 CIDR prefix (subnet mask 255.255.255.0) is sufficient because 2^8 equals 256 total IP addresses, which satisfies the 255 required host endpoints.
  3. The IPv4 address 10.150.33.254 is a valid usable host IP address within this subnet.Answer
  4. D
    The broadcast address for this subnet is 10.150.32.255.
  5. E
    The maximum number of usable host IP addresses available in this subnet is 512.

Answer

The subnet requires a /23 CIDR prefix (subnet mask 255.255.254.0) to accommodate all 255 required IP endpoints, and the IPv4 address 10.150.33.254 is a valid usable host IP address within this subnet.
The scenario demands 255 usable IP addresses (250+3+2=255250 + 3 + 2 = 255). A /24 subnet provides 28=2562^8 = 256 total addresses, but subtracting the network ID and broadcast address leaves only 254254 usable host IPs, which fails to accommodate the 255 endpoints. Therefore, a /23 subnet (29−2=5102^9 - 2 = 510 usable hosts) with subnet mask 255.255.254.0 is required. Within the 10.150.32.0/2310.150.32.0/23 subnet, the address space spans from 10.150.32.010.150.32.0 (network ID) to 10.150.33.25510.150.33.255 (broadcast ID). Thus, 10.150.33.25410.150.33.254 is the last usable IP address in the range.

Step-by-Step Solution

1
Calculate total host endpoints required
250 virtual machines + 3 router interfaces + 2 switches = 255 IP addresses required
Every active host and interface on the subnet requires a distinct usable IPv4 address.
2
Determine the minimum host bits (h) and smallest CIDR prefix
A /24 prefix yields 28−2=2542^8 - 2 = 254 usable hosts (insufficient). A /23 prefix yields 29−2=5102^9 - 2 = 510 usable hosts (sufficient). Minimum CIDR is /23.
Subnet design requires 2h−2≥required hosts2^h - 2 \ge \text{required hosts}. For 255255 hosts, h=9h=9 host bits are needed.
3
Calculate network boundaries for 10.150.32.0/23
Network ID: 10.150.32.010.150.32.0, First usable IP: 10.150.32.110.150.32.1, Last usable IP: 10.150.33.25410.150.33.254, Broadcast IP: 10.150.33.25510.150.33.255
A /23 subnet covers two consecutive /24 blocks (10.150.32.010.150.32.0 and 10.150.33.010.150.33.0).

Key Concept

IPv4 Subnetting & Usable Host Calculation (2^h - 2 rule)
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Topics

IPv4 Addressing and Subnetting

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