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Difficulty: EasyIPv4 Addressing and Subnetting

A system administrator is configuring a subnet for a new department containing 45 desktop computers and 3 network printers. Which of the following CIDR prefix lengths provides the minimum necessary capacity to accommodate all 48 devices while minimizing wasted IP addresses?

  1. /26Answer
  2. B
    /27
  3. C
    /25
  4. D
    /28

Answer

The prefix length /26 is the correct choice as it provides 62 usable host addresses, accommodating the required 48 devices with minimal address waste.
To support 48 host devices (45 PCs + 3 printers), a subnet must accommodate at least 48 usable addresses. A /26 subnet provides 26−2=622^6 - 2 = 62 usable IP addresses, which satisfies the host requirement while allocating the smallest possible subnet block.

Step-by-Step Solution

1
Calculate the total host IP addresses required.
45 workstations + 3 printers = 48 total host devices required.
Every active host interface requires a unique usable IP address on the subnet.
2
Determine the required host bits (h) using the formula 2h−2≥required_hosts2^h - 2 \ge \text{required\_hosts}.
For h=5h = 5: 25−2=302^5 - 2 = 30 usable hosts (insufficient). For h=6h = 6: 26−2=622^6 - 2 = 62 usable hosts (sufficient).
Two IP addresses in every subnet are reserved for the network ID and broadcast address.
3
Convert host bits (h) to CIDR prefix length.
32 total IPv4 bits - 6 host bits = /26 prefix length.
Subtracting host bits from 32 yields the network prefix length.

Key Concept

Determining minimum CIDR prefix length based on host capacity requirements
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Topics

IPv4 Addressing and Subnetting

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