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Difficulty: HardIPv4 Addressing and Subnetting

A network architect is implementing a Variable Length Subnet Mask (VLSM) allocation starting at the beginning of the 10.150.32.0/2010.150.32.0/20 address block for an industrial facility. The design requires three consecutive subnets allocated sequentially in descending order of size:

1. Subnet A (Industrial Controllers): Minimum of 500500 usable host IP addresses
2. Subnet B (HMI Touch Panels): Minimum of 100100 usable host IP addresses
3. Subnet C (Telemetry Sensors): Minimum of 2525 usable host IP addresses

Which of the following statements correctly describe the resulting network subnetting plan? (Select TWO.)

  1. Subnet A is assigned the prefix 10.150.32.0/2310.150.32.0/23 and has a broadcast address of 10.150.33.25510.150.33.255.Answer
  2. B
    Subnet A can be accommodated using a /24/24 prefix because 28=2562^8 = 256 total addresses provides sufficient capacity when combined across adjacent IP boundaries.
  3. Subnet C is assigned the prefix 10.150.34.128/2710.150.34.128/27 with a valid usable host IP range from 10.150.34.12910.150.34.129 through 10.150.34.15810.150.34.158.Answer
  4. D
    Subnet B begins at 10.150.33.0/2510.150.33.0/25 immediately following the first /24/24 boundary of Subnet A.
  5. E
    Subnet C utilizes a subnet mask of 255.255.255.240255.255.255.240 (/28/28) because 25 hosts require a 4-bit host portion.

Answer

The correct statements are that Subnet A is assigned the prefix 10.150.32.0/23 with a broadcast address of 10.150.33.255, and Subnet C is assigned the prefix 10.150.34.128/27 with a valid usable host IP range from 10.150.34.129 through 10.150.34.158.
For VLSM allocations, subnets must be ordered by size and allocated sequentially. Subnet A requires 500 hosts, necessitating 9 host bits (29−2=5102^9 - 2 = 510). A /23/23 subnet beginning at 10.150.32.010.150.32.0 covers 10.150.32.010.150.32.0 to 10.150.33.25510.150.33.255, making its broadcast address 10.150.33.25510.150.33.255. Subnet B uses 10.150.34.0/2510.150.34.0/25 (10.150.34.010.150.34.0 to 10.150.34.12710.150.34.127). Subnet C begins at 10.150.34.12810.150.34.128 and requires 5 host bits (25−2=302^5 - 2 = 30), forming the prefix 10.150.34.128/2710.150.34.128/27. Its usable hosts range from 10.150.34.12910.150.34.129 to 10.150.34.15810.150.34.158.

Step-by-Step Solution

1
Calculate host requirements and prefix length for Subnet A
500 hosts requires 9 host bits (29−2=510≥5002^9 - 2 = 510 \geq 500). Prefix = 32−9=/2332 - 9 = /23. Subnet A range: 10.150.32.0/2310.150.32.0/23 (10.150.32.010.150.32.0 to 10.150.33.25510.150.33.255). Broadcast address: 10.150.33.25510.150.33.255.
Subnets must be sized to accommodate usable host IP counts plus network and broadcast overhead.
2
Determine starting address and prefix length for Subnet B
Next available address is 10.150.34.010.150.34.0. 100 hosts requires 7 host bits (27−2=126≥1002^7 - 2 = 126 \geq 100). Prefix = 32−7=/2532 - 7 = /25. Subnet B range: 10.150.34.0/2510.150.34.0/25 (10.150.34.010.150.34.0 to 10.150.34.12710.150.34.127).
Subnet B is allocated sequentially immediately following Subnet A's address block.
3
Determine starting address, prefix length, and usable range for Subnet C
Next available address is 10.150.34.12810.150.34.128. 25 hosts requires 5 host bits (25−2=30≥252^5 - 2 = 30 \geq 25). Prefix = 32−5=/2732 - 5 = /27. Subnet C range: 10.150.34.128/2710.150.34.128/27. First host: 10.150.34.12910.150.34.129, Last host: 10.150.34.15810.150.34.158, Broadcast: 10.150.34.15910.150.34.159.
Sequential VLSM allocation places Subnet C right after Subnet B, using the minimum required block size.

Key Concept

Variable Length Subnet Masking (VLSM) allocation based on usable host address formulas (2h−22^h - 2).
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Topics

IPv4 Addressing and Subnetting

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