A network administrator is auditing interface configurations across an enterprise network.… — Cisco CCNA

A network administrator is auditing interface configurations across an enterprise network. A host interface is assigned the IPv4 address $172.20.108.90$ with a subnet mask of $255.255.255.224$ . Which IPv4 address represents the broadcast address for this subnetwork?

$172.20.108.95$ Answer
B
$172.20.108.94$
C
$172.20.108.96$
D
$172.20.108.127$

Answer

The broadcast address for the subnetwork is

172.20.108.95

For a

/27

subnet mask (

255.255.255.224

), the subnet block size is

256 - 224 = 32

. The subnets increment by

32

in the fourth octet (

0, 32, 64, 96, 128

). The host address

172.20.108.90

falls into the subnet starting at

172.20.108.64

. The broadcast address is the highest IP address within this range, which is

172.20.108.95

(one less than the next subnet ID of

172.20.108.96

Step-by-Step Solution

Determine the CIDR prefix length and block size from the subnet mask.

The mask

255.255.255.224

corresponds to a

/27

prefix (

24 + 3

bits). The block size in the fourth octet is

256 - 224 = 32

The subnet mask defines the boundary between the network and host portions of the IP address.

Identify the network ID for the given IP address.

Subnet multiples of

32

in the fourth octet are

0, 32, 64, 96, 128...

Since

90

lies between

64

and

95

, the network ID is

172.20.108.64

The network ID is the lowest address in the subnet block.

Calculate the broadcast address.

The next subnet starts at

172.20.108.96

. Subtracting

1

from the next network ID yields the broadcast address

172.20.108.95

The broadcast address is always the last address in the subnet range where all host bits are set to 1.

Key Concept

IPv4 Subnet Boundaries and Broadcast Address Calculation

Estimated Time:1m 0s

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