When Two Hops Beat One
When Two Hops Beat One
Ada re-derives this chapter’s own numbers step by step, at full precision
ADA · CALCULATION AUDIT
When Two Hops Beat One
A three-node chain, S → R → GW, loses relay R, and HWMP has to decide whether to heal onto a weaker direct link or find another route. The chapter’s own worked example pits two solid 130 Mbit/s hops against one weak 6 Mbit/s direct link, under its own rule that each extra same-channel hop roughly halves throughput. This audit asks the question that comparison invites: how much longer does the weak direct link actually hold the channel than the two clean hops, and does that gap explain why airtime routing prefers two hops over one?
Companion to the chapter Lab: Wi-Fi Mesh and Self-Healing — every number here comes from that chapter.
See the relationship before changing it
The figure reads from left to right. The blue card is weak-link rate. The middle card applies this page's rule. The green card is time for 1 mbit. 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 weak-link rate, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 6 Mbit/s.
- 2
Name the relationship. airtime = 1 Mbit / link rate x 1,000
- 3
Substitute with units. 1 / 6 s x 1,000 = 166.7 ms
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change weak-link rate
Try Predict the direction of airtime = 1 Mbit / link rate x 1,000. Test another weak-link rate, then compare time for 1 mbit.
Observe A weak direct link can hold the channel far longer than a clean multi-hop path. Reset weak-link rate to 6 and compare time for 1 mbit.
Explain A weak direct link can hold the channel far longer than a clean multi-hop path.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
Compare airtime for one 6 Mbit/s hop with two 130 Mbit/s hops by writing the costs as L/6 and 2L/130 for the same frame length L.
The direct path consumes about 10.8× the channel time of the relay path; two same-channel transmissions also explain the approximate per-hop throughput penalty.
A mesh metric sums transmission airtime rather than hop count, so an extra fast hop can cost far less shared-medium time than one weak low-rate link.
Ada: The worked example says HWMP’s airtime metric prefers two solid 130 Mbit/s hops over one weak 6 Mbit/s direct link, even though hop-counting picks the direct link. Let me put numbers on “airtime” to show the gap is not close.
- Sending a fixed frame of L bits takes
L / rateof channel time (L is illustrative; it cancels below). The direct one-hop path costsL / 6. - The two-hop relay path transmits the same frame once per link:
L / 130 + L / 130 = 2L / 130 = L / 65. - Dividing, the weak direct link occupies the channel
(L / 6) / (L / 65) = 65 / 6 ~= 10.8xlonger than the two clean hops - and that is before its retries, which only widen the gap.
That factor of about 11 is why a shared radio routes for least airtime, not fewest hops. The same store-and-forward accounting explains the chapter’s “each extra same-channel hop roughly halves throughput” rule: one delivery now occupies the channel for two transmissions instead of one. Keeping hop depth shallow and links fast is how self-healing lands on a path that is still fast, not merely still connected.
Every number above is taken from the chapter’s own material and re-derived step by step.
Technical boundaries: This rate-only comparison assumes equal frames and a single shared channel; it excludes retries, contention, interference, half-duplex scheduling, rate adaptation, relay processing, queueing, and hidden nodes.