A network administrator is provisioning a new VLAN for an enterprise VoIP deployment that … — Cisco CCNA

A network administrator is provisioning a new VLAN for an enterprise VoIP deployment that requires support for up to $500$ usable IP phones. The network block allocated for this subnetwork starts at $172.25.100.0$ . To accommodate the required number of host devices while minimizing unallocated address space, what is the broadcast address of this subnetwork?

$172.25.101.255$ Answer
B
$172.25.100.255$
C
$172.25.101.254$
D
$172.32.101.255$

Answer

The broadcast address of the subnetwork is

172.25.101.255

To support 500 usable host devices, 9 host bits are required (

2^9 - 2 = 510

hosts). This yields a prefix length of

/23

(

32 - 9 = 23

). Starting at network address

172.25.100.0

, a

/23

subnet encompasses the range

172.25.100.0

through

172.25.101.255

, making

172.25.101.255

the correct broadcast address.

Step-by-Step Solution

Determine the minimum number of host bits required.

n = 9

host bits because

2^9 - 2 = 510 \ge 500

A host bit count of 8 (

2^8 - 2 = 254

) is insufficient for 500 hosts.

Calculate the prefix length and subnet mask.

Prefix length is

/23

(

32 - 9 = 23

), corresponding to subnet mask

255.255.252.0

Subtracting 9 host bits from 32 total bits yields a 23-bit network prefix.

Determine the subnet boundaries for

172.25.100.0/23

Network address is

172.25.100.0

, usable host range is

172.25.100.1

172.25.101.254

, and the broadcast address is

172.25.101.255

The block size in the third octet for a /23 subnet is

256 - 252 = 4

2^1 = 2

, spanning

172.25.100.0

through

172.25.101.255

Key Concept

Subnet Mask Derivation and Broadcast Address Calculation

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic