Channel-Width Reuse
Ada re-derives this chapter’s own numbers step by step, at full precision
ADA · CALCULATION AUDIT
Channel-Width Reuse
Bonded channels, DFS timing, and 6 GHz room, ~3 minutes
An installer wants maximum per-link speed and sets every access point in a 5 GHz plan to 80 MHz, then finds only two or three non-DFS 80 MHz channels exist because UNII-1 and UNII-3 together give just 9 twenty-megahertz blocks. Dropping to 40 MHz roughly doubles the independent-channel count and restores reuse in the dense site. This audit asks the question that reuse trade-off invites: exactly how many non-overlapping channels does each channel width leave in the 2.4 GHz, non-DFS 5 GHz, DFS-inclusive 5 GHz, and 6 GHz pools?
Companion to the chapter Wi-Fi Bands and Channels — every number here comes from that chapter.
2.4 GHz reuse
The planning set has three non-overlapping 20 MHz channels: 1, 6, and 11.
40 MHz plan: floor(3 / 2) = 1 full bonded channel
5 GHz non-DFS pool
UNII-1 contributes 4 channels and UNII-3 contributes 5, so the non-DFS pool has 9 twenty-megahertz blocks.
40 MHz plan: floor(9 / 2) = 4
5 GHz with DFS
Adding UNII-2A and UNII-2C gives 4 + 4 + 12 + 5 = 25 twenty-megahertz blocks, but DFS channels require the availability check first.
DFS check: about 60 s before transmit
6 GHz room
The chapter’s ~1200 MHz figure gives the upper-bound width math before local rules, power class, and client support narrow it.
1200 / 160 = 7.5, so 7 full 160 MHz channels
Width planning is integer arithmetic: count the 20 MHz building blocks, divide by the bonded width, then round down because a leftover block cannot form another independent wide channel.
Every number above is taken from the chapter’s own channel-width example and re-derived step by step.