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Difficulty: HardIPv4 Addressing and Subnetting

A network engineer is provisioning subnets from the enterprise address block 10.150.16.0/2010.150.16.0/20. The engineering requirement specifies creating a new subnet that supports a minimum of 115 usable host addresses while minimizing unused IP space. Previously, the block 10.150.16.0/2510.150.16.0/25 was allocated to VLAN 10, and 10.150.16.128/2610.150.16.128/26 was allocated to VLAN 20. Following sequential allocation and proper subnet boundary alignment rules, what is the broadcast address of the next valid subnet that fulfills the target host requirement?

  1. 10.150.17.12710.150.17.127Answer
  2. B
    10.150.16.25510.150.16.255
  3. C
    10.150.16.19110.150.16.191
  4. D
    10.150.17.25510.150.17.255

Answer

The broadcast address of the next valid subnet meeting the host requirement is 10.150.17.12710.150.17.127.
To accommodate 115 hosts, a /25/25 subnet is required (27−2=1262^7 - 2 = 126 usable hosts). A /25/25 subnet has a block size of 128 and must start at either .0.0 or .128.128 in the fourth octet. Since 10.150.16.0/2510.150.16.0/25 occupies .0−.127.0 - .127 and 10.150.16.128/2610.150.16.128/26 occupies .128−.191.128 - .191, the .128.128 block in the 10.150.16.010.150.16.0 third octet is partially used and unavailable. Moving to the next binary-aligned /25/25 boundary yields 10.150.17.0/2510.150.17.0/25, which has a broadcast address of 10.150.17.12710.150.17.127.

Step-by-Step Solution

1
Determine required prefix length for host sizing.
7 host bits needed (27−2=126≥1152^7 - 2 = 126 \ge 115), yielding a prefix of /25/25 (32−7=2532 - 7 = 25).
A /26/26 only yields 62 hosts (insufficient), while /25/25 provides 126 usable host addresses.
2
Identify boundary alignment rules for a /25/25 prefix.
Subnet block size is 128 (272^7). Valid /25/25 network boundaries in any octet must end on multiples of 128 (00 or 128128).
Subnet masks require binary boundaries; /25/25 subnets can only start at .0.0 or .128.128 in the fourth octet.
3
Evaluate existing allocations for address overlap.
10.150.16.0/2510.150.16.0/25 consumes 10.150.16.0−10.150.16.12710.150.16.0 - 10.150.16.127. 10.150.16.128/2610.150.16.128/26 consumes 10.150.16.128−10.150.16.19110.150.16.128 - 10.150.16.191.
The candidate block 10.150.16.128/2510.150.16.128/25 (10.150.16.128−10.150.16.25510.150.16.128 - 10.150.16.255) cannot be used because its first half is partially occupied by 10.150.16.128/2610.150.16.128/26.
4
Find the next available aligned /25/25 network address and calculate its broadcast address.
Next valid boundary is 10.150.17.0/2510.150.17.0/25. Network: 10.150.17.010.150.17.0, Broadcast: 10.150.17.12710.150.17.127.
Incrementing to the next /25/25 boundary gives 10.150.17.0/2510.150.17.0/25, whose broadcast address is 10.150.17.12710.150.17.127.

Key Concept

VLSM Subnet Sizing and Boundary Alignment
Estimated Time:2m 0s
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Topics

IPv4 Addressing and Subnetting

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