You are working in a data center environment and are assigned the address range 10.188.31.0/23. You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30

The correct answer is D. 10.188.31.0/27. The /23 network (10.188.30.0–10.188.31.255) contains 512 total IP addresses. To support up to 30 hosts per subnet, you need at least 5 host bits: 2^5 − 2 = 30 usable addresses. A /27 prefix leaves exactly 5 host bits (32 − 27 = 5), producing 30 usable host addresses per subnet. T

Troubleshoot Basic Connectivity

Question

You are working in a data center environment and are assigned the address range 10.188.31.0/23. You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30 hosts each. Which IP address range meets these requirements?

Options

A10.188.31.0/26
B10.188.31.0/25
C10.188.31.0/28
D10.188.31.0/27
E10.188.31.0/29

How the community answered

(58 responses)

A
7% (4)
B
22% (13)
C
2% (1)
D
66% (38)
E
3% (2)

Explanation

The /23 network (10.188.30.0–10.188.31.255) contains 512 total IP addresses. To support up to 30 hosts per subnet, you need at least 5 host bits: 2^5 − 2 = 30 usable addresses. A /27 prefix leaves exactly 5 host bits (32 − 27 = 5), producing 30 usable host addresses per subnet. This is the longest (most specific) prefix that still meets the 30-host requirement, which maximizes the number of subnets from the /23 space: 512 ÷ 32 = 16 subnets. A /28 yields only 14 usable hosts (insufficient), while /26 and /25 support more hosts per subnet but produce fewer subnets overall - the opposite of the goal.

Topics

#subnetting#VLSM#host calculation#IP address planning

Community Discussion

No community discussion yet for this question.

Full 200-101 Practice