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Zorluk: Çok zorIPv4 Addressing and Subnetting

An enterprise network architect is designing an IP addressing plan within the RFC 1918 private address space using the allocated 172.16.0.0/12172.16.0.0/12 block. The design requires server LAN subnets that can each support at least 1,0001,000 usable host IP addresses, as well as point-to-point infrastructure subnets requiring /30/30 prefixes. Which two statements regarding this subnetting design and IPv4 address allocation are correct?

  1. A /22/22 subnet prefix provides 1,0221,022 usable host addresses, with 172.24.4.0/22172.24.4.0/22 spanning from network ID 172.24.4.0172.24.4.0 to broadcast address 172.24.7.255172.24.7.255.Cevap
  2. B
    A /22/22 subnet prefix provides 1,0241,024 usable host addresses because 210=10242^{10} = 1024 host addresses are available per subnet.
  3. The subnet 172.31.255.252/30172.31.255.252/30 represents the final valid /30/30 point-to-point subnet available within the assigned RFC 1918 172.16.0.0/12172.16.0.0/12 private block.Cevap
  4. D
    The subnet 172.32.0.0/30172.32.0.0/30 is a valid point-to-point subnet within the allocated RFC 1918 Class B private block.

Cevap

The two correct statements are the assertion that a /22 prefix provides 1,022 usable host addresses with 172.24.4.0/22 spanning up to broadcast address 172.24.7.255, and the assertion that 172.31.255.252/30 is the final valid /30 subnet within the RFC 1918 172.16.0.0/12 private block.
A /22 subnet provides 10 host bits, which allows for 1,022 usable hosts (2^10 - 2) and spans 172.24.4.0 through 172.24.7.255. Furthermore, 172.31.255.252/30 is the final /30 subnet within the 172.16.0.0/12 RFC 1918 private block, which terminates at 172.31.255.255.

Adım Adım Çözüm

1
Calculate host capacity and range for a /22 subnet prefix
Host bits = 32 - 22 = 10 bits. Total IP addresses = 2^10 = 1,024. Usable host addresses = 2^10 - 2 = 1,022. The 3rd octet mask value is 252 (255.255.252.0), creating a 3rd octet block increment of 256 - 252 = 4. Therefore, subnet 172.24.4.0/22 spans from network ID 172.24.4.0 to broadcast address 172.24.7.255.
Determining the required prefix length for 1,000 hosts requires at least 10 host bits (1,022 usable), and validating the subnet boundaries ensures proper addressing without overlap.
2
Evaluate the RFC 1918 Class B private range boundaries
RFC 1918 defines Class B private space as 172.16.0.0/12, which ranges from 172.16.0.0 to 172.31.255.255.
Verifying private IPv4 address boundaries prevents using public addresses (such as 172.32.0.0/30) internally.
3
Verify the last /30 subnet within the RFC 1918 Class B block
A /30 subnet uses a mask of 255.255.255.252 with a block size of 4 in the fourth octet. The last block in the 172.31.255.0/24 subnet starts at 172.31.255.252 and ends at 172.31.255.255 (broadcast).
Confirming the upper boundary of 172.16.0.0/12 validates that 172.31.255.252/30 is the last valid /30 private subnet.

Anahtar Kavram

Subnetting host capacity formulas (2^n - 2) and RFC 1918 private IPv4 address range boundaries (172.16.0.0/12).
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