A network administrator configures a core switch interface with the IPv4 address 10.50.180… — Cisco CCNA

A network administrator configures a core switch interface with the IPv4 address $10.50.180.75/21$ . What is the broadcast address for the subnetwork to which this interface belongs?

$10.50.183.255$ Answer
B
$10.50.180.255$
C
$10.50.183.254$
D
$10.50.191.255$

Answer

The broadcast address for the subnetwork is

10.50.183.255

The prefix /21 designates 21 subnet bits, yielding a subnet mask of

255.255.248.0

. In the third octet, subnets increment in steps of 8. For the IP address

10.50.180.75

, the containing subnet boundary starts at

10.50.176.0

and spans up to

10.50.183.255

. Since the broadcast address requires all host bits set to 1,

10.50.183.255

is the correct subnet broadcast address.

Step-by-Step Solution

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

A /21 prefix provides 21 network bits and 11 host bits. In dotted-decimal form, this is

255.255.248.0

. The host portion spans into the third octet. The block size in the third octet is

256 - 248 = 8

Prefix length identifies where the network portion ends and the host portion begins.

Calculate the network address for the given IP address

10.50.180.75

Dividing the third octet value (180) by the block size (8) gives

180 / 8 = 22.5

. Multiplying

22 \times 8 = 176

. Thus, the network address is

10.50.176.0/21

The network address is the lowest boundary of the subnet block.

Calculate the broadcast address.

Adding the block size minus 1 (

8 - 1 = 7

) to the third octet network boundary gives

176 + 7 = 183

. Setting all host bits in the fourth octet to binary 1s gives 255. The broadcast address is

10.50.183.255

The broadcast address is the highest possible IP address within the subnet range.

Key Concept

IPv4 Subnet Boundary and Broadcast Address Derivation

Estimated Time:1m 30s

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