A network administrator is dividing the IP address block 172.20.0.0/23 into subnets with a… — CompTIA Network+

A network administrator is dividing the IP address block $172.20.0.0/23$ into subnets with a prefix length of $/26$ to accommodate multiple department networks. What is the broadcast address of the third subnet in this allocation?

$172.20.0.191$ Answer
B
$172.20.0.190$
C
$172.20.0.127$
D
$172.20.0.255$

Answer

The broadcast address of the third subnet is

172.20.0.191

With a

/26

subnet mask (

255.255.255.192

), each subnet contains

64

IP addresses. The subnets start at

172.20.0.0

(1st),

172.20.0.64

(2nd),

172.20.0.128

(3rd), and

172.20.0.192

(4th). The broadcast address of the 3rd subnet (

172.20.0.128/26

) is the highest IP address in its block, which is

172.20.0.191

Step-by-Step Solution

Determine the subnet block size for a

/26

prefix length.

/26

prefix uses a subnet mask of

255.255.255.192

. The block size in the fourth octet is

256 - 192 = 64

IP addresses per subnet.

Calculating the block size identifies the boundary increment for each subnet.

List the subnet network boundaries starting from the base address

172.20.0.0/23

Subnet 1 starts at

172.20.0.0/26

.
Subnet 2 starts at

172.20.0.64/26

.
Subnet 3 starts at

172.20.0.128/26

.
Subnet 4 starts at

172.20.0.192/26

Adding the block size of

64

sequentially locates the network ID of each subnet.

Calculate the broadcast address of the third subnet.

The third subnet network ID is

172.20.0.128

. The fourth subnet starts at

172.20.0.192

. Therefore, the broadcast address of the third subnet is one IP address before the fourth subnet's network ID, which is

172.20.0.191

The broadcast address is always the last IP address in a subnet range immediately preceding the next subnet's network ID.

Key Concept

Subnet Masking and Subnet Boundary Calculation

Estimated Time:1m 30s

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