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Difficulty: MediumIPv4 Addressing and Subnetting

A network technician is configuring a dedicated VLAN for a department that requires 3030 desktop computers and 11 IP phone, requiring a total of 3131 usable host IPv4 addresses. Which subnet mask is the most efficient choice that fulfills this requirement without wasting unnecessary address space?

  1. 255.255.255.192Answer
  2. B
    255.255.255.224
  3. C
    255.255.255.240
  4. D
    255.255.255.128

Answer

255.255.255.192
To host 3131 devices, the subnet must offer at least 3131 usable host IP addresses. The formula for usable IPs is 2n−22^n - 2, where nn represents host bits. A /27 subnet (255.255.255.224255.255.255.224) provides 25−2=302^5 - 2 = 30 usable IPs, which is one address short. Therefore, the next larger subnet size is required, which is /26 (255.255.255.192255.255.255.192), giving 26−2=622^6 - 2 = 62 usable host IPs.

Step-by-Step Solution

1
Calculate total host IP addresses required including host overhead.
31 usable IP addresses are required.
The subnet must support 30 computers and 1 IP phone (30 + 1 = 31).
2
Apply the usable host capacity formula 2n−2≥H2^n - 2 \ge H, where nn is the number of host bits and HH is the required usable host count.
For n=5n=5, 25−2=302^5 - 2 = 30 usable addresses (insufficient). For n=6n=6, 26−2=622^6 - 2 = 62 usable addresses (sufficient).
Two addresses in every subnet must be reserved for the network ID and the directed broadcast address.
3
Determine the prefix length and dotted-decimal subnet mask corresponding to 66 host bits.
Prefix length is 32−6=/2632 - 6 = /26. The decimal mask is 255.255.255.192255.255.255.192.
A prefix of /26 leaves 6 bits for hosts (255.255.255.110000002=255.255.255.192255.255.255.11000000_2 = 255.255.255.192).

Key Concept

IPv4 Usable Host Count Calculation
Estimated Time:1m 30s
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Topics

IPv4 Addressing and Subnetting

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