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Difficulty: HardIPv4 Addressing and Subnetting

A network engineer is provisioning subnets from the assigned address block 10.150.48.0/2210.150.48.0/22 for a branch deployment. The primary VLAN requires capacity for up to 115115 IP phones. The engineer allocates the smallest valid CIDR subnet block that satisfies this requirement starting at 10.150.48.010.150.48.0. What is the broadcast address of the second subnet of the exact same prefix length immediately following the primary VLAN's subnet?

  1. A
    10.150.48.12710.150.48.127
  2. B
    10.150.48.25410.150.48.254
  3. 10.150.48.25510.150.48.255Answer
  4. D
    10.150.49.25510.150.49.255

Answer

The broadcast address of the second /25/25 subnet is 10.150.48.25510.150.48.255.
The primary VLAN requires 115 hosts. Factoring in the network and broadcast addresses (115+2=117115 + 2 = 117), the smallest accommodating block size is 128 addresses, which corresponds to a /25/25 subnet mask. The first /25/25 subnet occupies 10.150.48.010.150.48.0 to 10.150.48.12710.150.48.127. The second /25/25 subnet occupies 10.150.48.12810.150.48.128 to 10.150.48.25510.150.48.255. The last address in this second block (10.150.48.25510.150.48.255) is its broadcast address.

Step-by-Step Solution

1
Determine the required IP address capacity.
115 IP phones require at least 115+2=117115 + 2 = 117 total IP addresses (accounting for network and broadcast IDs).
Each subnet reserves the first IP address as the network ID and the last IP address as the broadcast ID.
2
Select the smallest matching CIDR prefix length.
The smallest power of 2 greater than or equal to 117 is 27=1282^7 = 128. A block size of 128 corresponds to a /25/25 mask (32−7=2532 - 7 = 25).
Using a /25/25 subnet provides 126 usable host IP addresses, which satisfies the 115 host requirement without wasting space.
3
Calculate the IP range of the first subnet block.
Subnet 1 spans from 10.150.48.010.150.48.0 to 10.150.48.12710.150.48.127, with 10.150.48.12710.150.48.127 as its broadcast address.
Adding 128 IP addresses starting at 10.150.48.010.150.48.0 covers 00 through 127127 in the fourth octet.
4
Identify the IP range and broadcast address of the second subnet block.
Subnet 2 begins immediately after Subnet 1 at 10.150.48.128/2510.150.48.128/25 and ends at 10.150.48.25510.150.48.255. The broadcast address is 10.150.48.25510.150.48.255.
The second block of 128 IP addresses spans 128128 through 255255 in the fourth octet.

Key Concept

Subnet Block Sizing and Broadcast Address Calculation
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Topics

IPv4 Addressing and Subnetting

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