Why the Same 12-Byte Reading Has Two Efficiencies

Why the Same 12-Byte Reading Has Two Efficiencies

Ada re-derives this chapter’s own numbers step by step, at full precision

ADA · CALCULATION AUDIT

Why the Same 12-Byte Reading Has Two Efficiencies

The same 12-byte sensor reading measures 41.4% efficient over LoRaWAN but only 22.2% over MQTT/TCP/IP — nearly a 2x gap for identical data. The difference is entirely the fixed header each stack drags along, from a 13-byte LoRaWAN MAC header to the stacked TCP and IP headers of MQTT. This audit separates payload from overhead to explain why one reading has two efficiencies.

Companion to the chapter Lab: Network Performance — every number here comes from that chapter.

See the relationship before changing it

The figure reads from left to right. The blue card is sensor payload. The middle card applies the page rule. The green card is lorawan efficiency. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

Walk the arrows. Batching lifts efficiency because the fixed header is then shared by more payload.

Derive the baseline in four named moves

1. 1

   Name the input. The chapter baseline is 12 bytes.

2. 2

   Name the relationship. efficiency = payload / (payload + 17 overhead bytes) x 100

3. 3

   Substitute with units. 12 / (12 + 17) x 100 = 41.38%

4. 4

   Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change sensor payload

Try Predict the direction of efficiency = payload / (payload + 17 overhead bytes) x 100. Test another sensor payload, then compare lorawan efficiency.

Sensor payload

12 bytes

Chapter baseline

LoRaWAN efficiency

Observe Batching lifts efficiency because the fixed header is then shared by more payload. Reset sensor payload to 12 and compare lorawan efficiency.

Explain Batching lifts efficiency because the fixed header is then shared by more payload.

Check yourself

What should you do before trusting a moved-control result?

Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.

What does this small model leave out?

Answer: Only sensor payload moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only sensor payload. It is not a field prediction.

TryRecompute Total on air: 12 + 13 + 4 = 29 bytes.

ObserveTrack Goodput efficiency: 12 / 29 = 0.413793, i.e. 41.4%; the overhead is the complement, 17 / 29 = 0.586207 = 58.6%.

ExplainExplain LoRaWAN's advantage: (12 / 29) / (12 / 54) = 54 / 29 = 1.862, so about 1.9x — the "nearly 2x" claim holds.

Ready: use the stated baseline inputs, then compare each displayed result.

Ada: This lab reports that a 12-byte reading is 41.4% efficient over LoRaWAN but only 22.2% over MQTT/TCP/IP, and that LoRaWAN is “nearly 2x” better. Same payload, two answers — the difference is entirely the fixed header each stack drags along, so let me separate payload from overhead.

Over LoRaWAN the reading carries a 13-byte MAC header and a 4-byte MIC:

• Total on air: 12 + 13 + 4 = 29 bytes.
• Goodput efficiency: 12 / 29 = 0.413793, i.e. 41.4%; the overhead is the complement, 17 / 29 = 0.586207 = 58.6%.

Over MQTT the same 12 bytes ride a 2-byte MQTT header, a 20-byte TCP header, and a 20-byte IP header:

• Total: 12 + 2 + 20 + 20 = 54 bytes.
• Efficiency: 12 / 54 = 0.222222, i.e. 22.2%.
• LoRaWAN’s advantage: (12 / 29) / (12 / 54) = 54 / 29 = 1.862, so about 1.9x — the “nearly 2x” claim holds.

The design meaning is that efficiency is a property of the payload-to-header ratio, not of the protocol name. The 17-byte LoRaWAN overhead that swallows 58.6% of a 12-byte reading shrinks to just 17 / (100 + 17) = 0.1453 = 14.5% of a 100-byte reading (100 / 117 = 85.5% efficient). The lesson for a sensor designer is blunt: batching or enlarging the payload lifts efficiency far faster than switching protocols, because the fixed header is paid once per packet no matter how little it carries.

Every number above is taken from the chapter’s own material and re-derived step by step.

Technical boundaries. This efficiency audit deliberately does not simulate contention, retransmissions, aggregation, or variable headers. It divides the same fixed 12-byte application reading by each chapter-defined total frame size.