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Difficulty: HardIPv4 Addressing and Subnetting

A network administrator is assigned the IPv4 block 172.31.96.0/22172.31.96.0/22 for a data center expansion. The administrator must allocate the first subnet from this block to support a web server farm requiring at least 110110 usable host IP addresses. Which of the following correctly identifies the subnet ID and broadcast address for this web server subnet using the most space-efficient CIDR prefix?

  1. Subnet ID: 172.31.96.0/25172.31.96.0/25, Broadcast Address: 172.31.96.127172.31.96.127Answer
  2. B
    Subnet ID: 172.31.96.0/25172.31.96.0/25, Broadcast Address: 172.31.96.128172.31.96.128
  3. C
    Subnet ID: 172.31.96.0/26172.31.96.0/26, Broadcast Address: 172.31.96.63172.31.96.63
  4. D
    Subnet ID: 172.31.96.0/24172.31.96.0/24, Broadcast Address: 172.31.96.255172.31.96.255

Answer

Subnet ID: 172.31.96.0/25172.31.96.0/25, Broadcast Address: 172.31.96.127172.31.96.127
To host at least 110 devices, 7 host bits are required because 27−2=1262^7 - 2 = 126 usable hosts. A prefix length of /25/25 (32−732 - 7) creates subnets of size 128. Starting from 172.31.96.0172.31.96.0, the first /25/25 subnet covers addresses 172.31.96.0172.31.96.0 through 172.31.96.127172.31.96.127. Thus, the Subnet ID is 172.31.96.0/25172.31.96.0/25 and the Broadcast Address is 172.31.96.127172.31.96.127.

Step-by-Step Solution

1
Determine the required number of host bits (hh).
To accommodate at least 110110 usable hosts, use the formula 2h−2≥1102^h - 2 \ge 110. For h=6h=6, 26−2=622^6 - 2 = 62 (insufficient). For h=7h=7, 27−2=1262^7 - 2 = 126 usable hosts (sufficient).
Two addresses must be subtracted from the total block size for the network ID and broadcast address.
2
Calculate the CIDR prefix length.
Prefix length = 32−7=/2532 - 7 = /25.
IPv4 addresses consist of 32 bits. Subtracting 7 host bits leaves 25 network/subnet bits.
3
Determine the block size and network range for the first subnet.
Block size = 27=1282^7 = 128. Starting at 172.31.96.0172.31.96.0, the subnet spans from 172.31.96.0172.31.96.0 to 172.31.96.127172.31.96.127.
The first subnet starts at the beginning of the parent block (172.31.96.0172.31.96.0).
4
Identify the Subnet ID and Broadcast Address.
Subnet ID = 172.31.96.0/25172.31.96.0/25, Broadcast Address = 172.31.96.127172.31.96.127. Usable host range is 172.31.96.1172.31.96.1 through 172.31.96.126172.31.96.126.
The lowest address in the block is the Subnet ID, and the highest address in the block is the Broadcast Address.

Key Concept

Calculating host requirements, prefix lengths, and broadcast boundaries in IPv4 subnetting.
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Topics

IPv4 Addressing and Subnetting

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