High-Density WLAN Channel Planning for Throughput
Short answer
A 5 channels are required to provide 100 Mbps of raw throughput capacity for 100 users at 1 Mbps each, given 20 Mbps per channel.
Question
How many channels must the design use to provide 1 Mbps per user prior to RF overhead?
Key Takeaways
The problem tests the ability to translate user throughput needs into channel capacity, specifically requiring the exclusion of RF overhead from the initial bandwidth allocation to determine the minimum cell count and thus channel count.
This question addresses the calculation of required non-overlapping channels in a high-density wireless LAN design to meet specific per-user throughput requirements before RF overhead.
Common mistake: Many candidates mistakenly include RF overhead (typically 20-30%) in the per-cell capacity calculation or fail to recognize that 'prior to RF overhead' implies using the raw 1 Mbps figure directly against the available channel capacity, leading to an underestimation of required channels if they overcomplicate the math.
Expert Analysis
Why the Answer Is Correct
To support 100 users with 1 Mbps each prior to overhead, we need 100 Mbps of raw throughput capacity. The design allows 20 Mbps per cell/channel. Dividing the total required throughput by the per-channel capacity (100 Mbps / 20 Mbps) yields exactly 5 cells. In Wi-Fi design, each cell typically corresponds to a unique non-overlapping channel to avoid interference in high-density scenarios. Therefore, 5 distinct channels are required.
Why the Other Options Are Wrong
Options B, C, and D represent higher numbers of channels which would provide more capacity than necessary for the stated pre-overhead requirement. While adding more channels is often good practice to handle real-world overhead and density spikes, the question specifically asks for the number required to provide 1 Mbps prior to overhead, making 5 the precise mathematical answer based on the given constraints.
Community Comment Notes
As noted by leonel591, the calculation is straightforward: 100 users require 100 Mbps total. With 20 Mbps per cell, 5 cells are needed. This aligns perfectly with the logic that each cell operates on its own channel in this specific high-density planning model.
Exam Strategy
When calculating channel counts for high-density designs, always check if the question asks for 'pre-overhead' or 'post-overhead' requirements. If it specifies pre-overhead, use the raw throughput figures provided without applying typical efficiency loss factors.
Frequently Asked Questions
Why do we not subtract RF overhead first?
Does one channel equal one cell?
Discussion 1 contributions
Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.
From public community threads 1
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leonel591 👍 1 Answered A100 usuarios / 100 Mbps = 1 (usuario/Mbps) 20 (Mbps/celda ) 1 Usuario/Mbps = 20 (usuarios/celda) 100 (usuarios) 1/20 (celda/usuario)= 5 (celdas)
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