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Zorluk: OrtaIPv4 Addressing and Subnetting

An enterprise network administrator is allocated the IPv4 address block 172.16.48.0/21172.16.48.0/21 to provision subnets for multiple VLANs in a regional branch office. One specific VLAN requires support for at least 100100 host devices while maximizing address efficiency. Which subnetwork address and CIDR prefix pair represents a valid, correctly sized subnet within the assigned parent block?

  1. 172.16.50.0/25172.16.50.0/25Cevap
  2. B
    172.16.50.0/26172.16.50.0/26
  3. C
    172.32.50.0/25172.32.50.0/25
  4. D
    172.16.56.0/25172.16.56.0/25

Cevap

The subnet address 172.16.50.0/25172.16.50.0/25 is correct because it falls within the 172.16.48.0/21172.16.48.0/21 parent block, uses valid RFC 1918 private addressing, and provides 126 usable host addresses.
The subnet 172.16.50.0/25172.16.50.0/25 is correct because a /25/25 prefix provides 27−2=1262^7 - 2 = 126 usable host addresses, which satisfies the 100 host requirement efficiently. Furthermore, 172.16.50.0172.16.50.0 lies well within the assigned parent block of 172.16.48.0/21172.16.48.0/21 (range 172.16.48.0172.16.48.0 - 172.16.55.255172.16.55.255) and belongs to valid RFC 1918 private IP space.

Adım Adım Çözüm

1
Determine the address range of the allocated parent block 172.16.48.0/21172.16.48.0/21.
The block has 21 network bits, leaving 11 host bits. The 3rd octet ranges from 48 (00011000200011000_2) to 55 (00011111200011111_2). The valid IP range is 172.16.48.0172.16.48.0 through 172.16.55.255172.16.55.255.
Any valid subnets must fall strictly within these boundaries.
2
Calculate the minimum prefix length needed for at least 100 host devices.
Using 2h−2≥1002^h - 2 \ge 100, for h=6h=6, 26−2=622^6 - 2 = 62 (insufficient). For h=7h=7, 27−2=1262^7 - 2 = 126 (sufficient). The required prefix length is 32−7=2532 - 7 = 25 (/25/25).
Selecting /25/25 provides enough host IPs while conserving space compared to a larger mask like /24/24.
3
Evaluate candidate /25/25 subnets against the parent block boundaries and RFC 1918 rules.
172.16.50.0/25172.16.50.0/25 starts at a valid /25/25 boundary in the 4th octet (00), resides within the 172.16.48.0−172.16.55.255172.16.48.0 - 172.16.55.255 range, and uses Class B RFC 1918 private space (172.16.0.0/12172.16.0.0/12).
This confirms all sizing, allocation, and RFC standards are met.

Anahtar Kavram

IPv4 Subnet Sizing and RFC 1918 Boundary Analysis
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