A network engineer is partitioning the address block 10.45.160.0/19 to provision a new VoI… — CompTIA Network+

A network engineer is partitioning the address block $10.45.160.0/19$ to provision a new VoIP network segment that must accommodate at least $500$ IP phones. To prevent address space depletion, the engineer allocates the smallest valid subnet starting at the beginning of the block that satisfies this host capacity requirement. What is the last usable host IP address of this newly allocated subnet?

A
10.45.161.255
B
10.45.160.254
10.45.161.254Answer
D
10.45.163.254

Answer

10.45.161.254

To accommodate at least

500

usable host addresses with minimal wastage,

9

host bits are needed (

2^9 - 2 = 510

usable addresses), establishing a

/23

subnet mask. The first

/23

subnet within

10.45.160.0/19

ranges from

10.45.160.0

(Network ID) to

10.45.161.255

(Broadcast Address). Subtracting

1

from the broadcast address yields

10.45.161.254

as the final usable host IP address.

Step-by-Step Solution

Determine the required host bits and CIDR prefix length

9 host bits are required, yielding a /23 subnet prefix

The formula for usable hosts is

2^h - 2 \ge 500

. For

h = 8

2^8 - 2 = 254

(insufficient). For

h = 9

2^9 - 2 = 510

(sufficient). The prefix length is

32 - 9 = 23

Calculate the network block boundaries for the first /23 subnet

Network ID: 10.45.160.0/23, Broadcast Address: 10.45.161.255

A /23 subnet has a block size of

2

in the third octet (

256 - 254 = 2

). Starting at

10.45.160.0

, the network spans from

10.45.160.0

through

10.45.161.255

Identify the last usable host IP address within the subnet boundary

10.45.161.254

The last usable host IP address is always one numeric value below the broadcast address (

10.45.161.255 - 1 = 10.45.161.254

Key Concept

Subnet Host Capacity and Subnet Boundary Calculation

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic