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Difficulty: HardIPv4 Addressing and Subnetting

A network administrator is designing an IP addressing scheme for a new branch office location. The department requires static IP assignments for 60 user workstations, 2 IP phone gateways, and 2 redundant default gateway router interfaces. The administrator must assign the smallest possible subnet block from `10.150.0.0/16` that will successfully accommodate all required host devices. Which CIDR prefix and subnet mask should the administrator configure?

  1. /25 (255.255.255.128)Answer
  2. B
    /26 (255.255.255.192)
  3. C
    /27 (255.255.255.224)
  4. D
    /24 (255.255.255.0)

Answer

/25 (255.255.255.128)
The scenario requires 64 usable IPv4 addresses (60 workstations + 2 phone gateways + 2 router interfaces). Using the formula for usable hosts 2h−22^h - 2, a /26 prefix has 6 host bits (26−2=622^6 - 2 = 62), which falls short by 2 addresses. Therefore, the network administrator must allocate a /25 prefix with 7 host bits (27−2=1262^7 - 2 = 126), which is the smallest subnet block capable of supporting all 64 host IP addresses.

Step-by-Step Solution

1
Calculate the total number of host IP addresses required.
60 workstations + 2 IP phone gateways + 2 router interfaces = 64 host IP addresses.
Every active device and router gateway interface on the subnet requires a unique usable IPv4 address.
2
Apply the usable host formula 2h−2≥642^h - 2 \ge 64 to determine the required host bits (hh).
For h=6h = 6, 26−2=622^6 - 2 = 62 usable hosts (insufficient). For h=7h = 7, 27−2=1262^7 - 2 = 126 usable hosts (sufficient).
Subnetting calculations require subtracting 2 reserved addresses (the Network ID and Broadcast ID) from total addresses (2h2^h).
3
Determine the prefix length and dotted-decimal subnet mask.
Prefix length is 32−7=/2532 - 7 = /25, which corresponds to the subnet mask `255.255.255.128`.
Subtracting 7 host bits from 32 total IPv4 bits leaves 25 network bits.

Key Concept

IPv4 Subnetting and Usable Host Calculation
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Topics

IPv4 Addressing and Subnetting

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