An infrastructure team is provisioning a new subnet from the IPv4 address block 172.20.96.… — Cisco CCNA

An infrastructure team is provisioning a new subnet from the IPv4 address block $172.20.96.0/23$ for a server farm that requires support for up to $250$ 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?

172.20.96.254Answer
B
172.20.96.255
C
172.20.97.254
D
172.32.96.254

Answer

172.20.96.254

To support

250

usable host interfaces efficiently,

8

host bits are required (

2^8 - 2 = 254

usable hosts), corresponding to a

/24

subnet prefix. The first

/24

subnet carved from the parent block

172.20.96.0/23

has a network address of

172.20.96.0

and a broadcast address of

172.20.96.255

. The last usable host IP address is one position below the broadcast address, which is

172.20.96.254

Step-by-Step Solution

Determine the required host bits and prefix length for 250 hosts.

Using

2^n - 2 \ge 250

n = 8

host bits are needed (

2^8 - 2 = 254

usable addresses). The corresponding prefix length is

32 - 8 = /24

Selecting 8 host bits fulfills the requirement of at least 250 usable hosts with minimum wasted IP space.

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.

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.

Key Concept

IPv4 Subnetting, Prefix Derivation, and Host Range Boundaries

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