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Zorluk: Çok zorIPv4 Addressing and Subnetting

A network engineer is configuring Variable Length Subnet Masking (VLSM) on a router interface for two internal enterprise subnets using the base block 172.16.10.0/24172.16.10.0/24. Subnet A must support at least 60 usable hosts, while Subnet B must support at least 28 usable hosts. Subnet A is allocated first starting at 172.16.10.0172.16.10.0 using the smallest prefix length that satisfies its requirement. Subnet B is allocated immediately following Subnet A using its smallest required prefix length. Which two statements regarding this subnetting layout are correct? (Select two.)

  1. The broadcast address for Subnet A is 172.16.10.63172.16.10.63.Cevap
  2. The IP address 172.16.10.94172.16.10.94 is a valid, usable host address in Subnet B.Cevap
  3. C
    The assignable IPv4 host address range for Subnet B starts at 172.16.10.64172.16.10.64.
  4. D
    The IP address 172.32.10.65172.32.10.65 is the first usable host address in Subnet B.

Cevap

The broadcast address for Subnet A is 172.16.10.63172.16.10.63, and the IP address 172.16.10.94172.16.10.94 is a valid, usable host address in Subnet B.
For Subnet A, 60 hosts require 6 host bits (26−2=622^6 - 2 = 62 usable), yielding a /26/26 mask. Starting at 172.16.10.0/26172.16.10.0/26, the block ends at broadcast address 172.16.10.63172.16.10.63. Subnet B starts at the next available address, 172.16.10.64172.16.10.64, and requires 5 host bits (25−2=302^5 - 2 = 30 usable) for 28 hosts, giving a /27/27 mask (172.16.10.64/27172.16.10.64/27). The valid usable host range for Subnet B is 172.16.10.65172.16.10.65 through 172.16.10.94172.16.10.94, making 172.16.10.94172.16.10.94 a valid usable host address.

Adım Adım Çözüm

1
Calculate prefix length and host range for Subnet A.
Subnet A prefix is /26/26 (255.255.255.192255.255.255.192), spanning 172.16.10.0/26172.16.10.0/26. Network ID: 172.16.10.0172.16.10.0, Usable range: 172.16.10.1172.16.10.1–172.16.10.62172.16.10.62, Broadcast: 172.16.10.63172.16.10.63.
60 host requirement needs 2h−2≥60  ⟹  h=62^h - 2 \ge 60 \implies h=6 bits. 32−6=2632 - 6 = 26.
2
Determine the starting boundary and prefix length for Subnet B.
Subnet B starts at 172.16.10.64172.16.10.64 with prefix /27/27 (255.255.255.224255.255.255.224). Network ID: 172.16.10.64172.16.10.64, Usable range: 172.16.10.65172.16.10.65–172.16.10.94172.16.10.94, Broadcast: 172.16.10.95172.16.10.95.
Next contiguous subnet after 172.16.10.63172.16.10.63 is 172.16.10.64172.16.10.64. 28 host requirement needs 2h−2≥28  ⟹  h=52^h - 2 \ge 28 \implies h=5 bits. 32−5=2732 - 5 = 27.
3
Evaluate the choices against calculated subnet boundaries and RFC 1918 rules.
Identified 172.16.10.63172.16.10.63 as Subnet A's broadcast address and 172.16.10.94172.16.10.94 as Subnet B's last usable host IP.
Network addresses (172.16.10.64172.16.10.64) cannot be assigned to hosts, and 172.32.x.x172.32.x.x falls outside RFC 1918 private space.

Anahtar Kavram

VLSM Subnet Sizing and Boundary Calculation
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Konular

IPv4 Addressing and Subnetting

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