Examkin

Question

Difficulty: HardIPv4 Addressing and Subnetting

A network administrator is allocating subnets from the enterprise address block 172.16.0.0/12172.16.0.0/12 to provision a new datacenter VLAN that must support at least 500500 usable host interfaces. The design requires using the most efficient subnet mask possible to conserve IP address space, while ensuring all assigned host addresses strictly adhere to RFC 1918 private IPv4 specifications. Which statements regarding this subnetting design are correct? (Select TWO.)

  1. The most efficient subnet mask to accommodate the host requirement is 255.255.254.0255.255.254.0 (/23/23).Answer
  2. The IP address 172.31.255.250172.31.255.250 is a valid private IPv4 host address within the allocated 172.16.0.0/12172.16.0.0/12 address space.Answer
  3. C
    A subnet mask of 255.255.255.0255.255.255.0 (/24/24) provides 512512 total host addresses, satisfying the requirement with minimal wasted addresses.
  4. D
    The IP address 172.32.10.15172.32.10.15 is a valid private IPv4 host address within the 172.16.0.0/12172.16.0.0/12 block.

Answer

The correct answers are the statement identifying 255.255.254.0 (/23) as the most efficient subnet mask and the statement confirming that 172.31.255.250 is a valid private IPv4 address within the 172.16.0.0/12 range.
The subnetting calculation requires 9 host bits (29−2=5102^9 - 2 = 510) to support 500 hosts, making /23/23 (255.255.254.0255.255.254.0) the most efficient choice. Additionally, RFC 1918 defines Class B private space as 172.16.0.0172.16.0.0 to 172.31.255.255172.31.255.255, which includes 172.31.255.250172.31.255.250.

Step-by-Step Solution

1
Calculate the host bits required for 500 usable host interfaces.
Using the formula 2h−2≥5002^h - 2 \ge 500, h=9h = 9 host bits are needed because 29−2=5102^9 - 2 = 510 usable addresses (28−2=2542^8 - 2 = 254 is insufficient).
Two addresses in every subnet are reserved for network identity and subnet broadcast.
2
Determine prefix length and dotted-decimal subnet mask.
Prefix length =32−9=/23= 32 - 9 = /23. In dotted-decimal format, /23 translates to 255.255.254.0255.255.254.0.
A 23-bit network prefix leaves 9 bits for host addressing.
3
Verify RFC 1918 private IPv4 range boundaries for Class B.
The RFC 1918 Class B private address space spans 172.16.0.0172.16.0.0 to 172.31.255.255172.31.255.255 (172.16.0.0/12172.16.0.0/12).
Address 172.31.255.250172.31.255.250 falls inside this range, whereas 172.32.10.15172.32.10.15 is in public IP space.

Key Concept

Subnet Host Calculation and RFC 1918 Private Addressing
Rate this question

Topics

IPv4 Addressing and Subnetting

More questions from this topic