A network administrator is sub-dividing an allocated IPv4 address block of 172.28.16.0/20 … — CompTIA Network+

A network administrator is sub-dividing an allocated IPv4 address block of $172.28.16.0/20$ into contiguous subnets with a $/23$ prefix length for datacenter switches. What is the broadcast address of the second $/23$ subnet?

$172.28.19.255$ Answer
B
$172.28.18.255$
C
$172.28.17.255$
D
$172.28.19.254$

Answer

The broadcast address of the second

/23

subnet is

172.28.19.255

Dividing

172.28.16.0/20

into

/23

subnets yields block increments of

2

in the third octet. The first subnet spans

172.28.16.0

172.28.17.255

. The second subnet starts at

172.28.18.0

and spans through

172.28.19.255

. Therefore,

172.28.19.255

is the broadcast address of the second subnet.

Step-by-Step Solution

Determine the size and block increment of a

/23

subnet in the third octet.

/23

subnet has

32 - 23 = 9

host bits (

2^9 = 512

total addresses). In the third octet, each subnet increments by

512 / 256 = 2

Calculating the block size allows identification of subnet boundaries.

Calculate the network address boundaries for the first and second subnets starting from

172.28.16.0/20

Subnet 1 starts at

172.28.16.0/23

. Subnet 2 starts at

172.28.18.0/23

. Subnet 3 starts at

172.28.20.0/23

Adding the third octet increment of

2

identifies each subnet's network ID.

Identify the broadcast address for the second subnet (

172.28.18.0/23

The broadcast address is one address below the start of Subnet 3 (

172.28.20.0

), which is

172.28.19.255

The highest numerical address in a subnet range is reserved as the broadcast address.

Key Concept

IPv4 Subnet Boundary and Broadcast Address Calculation

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