A network administrator is assigning an IP address to a core switch interface with the IPv… — Cisco CCNA

A network administrator is assigning an IP address to a core switch interface with the IPv4 address $172.16.138.200$ and a subnet mask of $255.255.248.0$ . What is the broadcast address of this subnetwork?

$172.16.143.255$ Answer
B
$172.16.143.254$
C
$172.16.139.255$
D
$172.32.143.255$

Answer

The broadcast address for the subnetwork containing

172.16.138.200/21

172.16.143.255

For a subnet mask of

255.255.248.0

(/21), the third octet block size is

256 - 248 = 8

. Subnet boundaries in the third octet occur at multiples of 8 (

0, 8, 16, \dots, 136, 144

). Since the host IP is

172.16.138.200

, it falls in the subnet starting at

172.16.136.0

. The broadcast address is one less than the next subnet boundary (

172.16.144.0

), resulting in

172.16.143.255

Step-by-Step Solution

Determine the prefix length and block size from the subnet mask

255.255.248.0

The subnet mask

255.255.248.0

corresponds to a /21 prefix. The interesting octet is the third octet, giving a block size of

256 - 248 = 8

Identifying the block size determines the increment between consecutive subnet boundaries.

Calculate the network ID containing the host address

172.16.138.200

Dividing the third octet value

138

8

yields

17.25

. Multiplying

17 \times 8 = 136

establishes the network boundary at

172.16.136.0/21

Finds the starting network address of the subnet.

Calculate the broadcast address by subtracting 1 from the next subnet boundary.

The next subnet boundary is

172.16.136.0 + 8 = 172.16.144.0

. Subtracting 1 gives

172.16.143.255

The broadcast address is the highest address in the subnet with all host bits set to 1.

Key Concept

Subnet Block Size and Broadcast Address Derivation

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