Channel-Width Reuse
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.
See the relationship before changing it
The figure reads from left to right. The blue card is bonded channel width. The middle card applies this page's rule. The green card is reusable channel widths. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only bonded channel width, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 40 MHz.
- 2
Name the relationship. reuse count = 180 MHz pool / channel width
- 3
Substitute with units. 180 MHz / 40 MHz = 4.5 channel widths
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change bonded channel width
Try Predict the direction of reuse count = 180 MHz pool / channel width. Test another bonded channel width, then compare reusable channel widths.
Observe Wider bonded channels consume more of the same non-DFS spectrum pool. Reset bonded channel width to 40 and compare reusable channel widths.
Explain Wider bonded channels consume more of the same non-DFS spectrum pool.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
Count the nine non-DFS 20 MHz blocks at 5 GHz and form 80 MHz channels, then repeat with all 25 DFS-inclusive blocks.
The non-DFS pool yields floor(9/4) = 2 full 80 MHz channels; the larger pool yields floor(25/4) = 6, with one 20 MHz block left over.
Channel bonding consumes contiguous 20 MHz building blocks, so independent-channel count is integer division rounded down rather than raw spectrum divided optimistically.
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.
Technical boundaries: This block-count model omits channel adjacency and exclusions, local regulation, DFS events, power class, guard bands, interference, client capability, spectral masks, and reuse between cells.