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Zorluk: OrtaIPv4 Addressing and Subnetting

An infrastructure team is provisioning a new subnet from the IPv4 address block 172.20.96.0/23172.20.96.0/23 for a server farm that requires support for up to 250250 usable host interfaces. If the team applies the most efficient prefix length that satisfies this requirement, which IPv4 address is the last usable host address in the first subnet created?

  1. 172.20.96.254Cevap
  2. B
    172.20.96.255
  3. C
    172.20.97.254
  4. D
    172.32.96.254

Cevap

172.20.96.254
To support 250250 usable host interfaces efficiently, 88 host bits are required (28−2=2542^8 - 2 = 254 usable hosts), corresponding to a /24/24 subnet prefix. The first /24/24 subnet carved from the parent block 172.20.96.0/23172.20.96.0/23 has a network address of 172.20.96.0172.20.96.0 and a broadcast address of 172.20.96.255172.20.96.255. The last usable host IP address is one position below the broadcast address, which is 172.20.96.254172.20.96.254.

Adım Adım Çözüm

1
Determine the required host bits and prefix length for 250 hosts.
Using 2n−2≥2502^n - 2 \ge 250, n=8n = 8 host bits are needed (28−2=2542^8 - 2 = 254 usable addresses). The corresponding prefix length is 32−8=/2432 - 8 = /24.
Selecting 8 host bits fulfills the requirement of at least 250 usable hosts with minimum wasted IP space.
2
Calculate the boundaries for the first /24 subnet from 172.20.96.0/23.
The first subnet is 172.20.96.0/24, spanning network address 172.20.96.0 to broadcast address 172.20.96.255.
A /24 subnet increments in the third octet by 1, starting at the base network address 172.20.96.0.
3
Identify the last usable host address of the 172.20.96.0/24 subnet.
Subtracting 1 from the broadcast address (172.20.96.255 - 1) gives 172.20.96.254.
Usable host IP addresses lie strictly between the network address and the broadcast address.

Anahtar Kavram

IPv4 Subnetting, Prefix Derivation, and Host Range Boundaries
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Konular

IPv4 Addressing and Subnetting

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