Spatial decision-making

Why this matters

A beautiful vegetation index map is not yet an answer. The farmer wants to know what to do, where, how much, and whether it worked. This module turns the data from earlier modules into decisions and follows them up honestly. The example shows that looking at the vegetation index alone and fertilising the low zone may treat the wrong problem, because the real cause is water stress.

Water stress

When a plant has enough water, its leaves transpire and cool down, like sweat evaporating from skin. When it is short of water, the stomata close and the leaves warm up. The crop water stress index (CWSI), developed by Idso and colleagues and Jackson and colleagues (1981), compares canopy temperature $T_c$ with two reference temperatures:

$\text{CWSI}=\frac{T_c-T_{\text{wet}}}{T_{\text{dry}}-T_{\text{wet}}}$

$T_{\text{wet}}$ is the temperature of a fully transpiring leaf and $T_{\text{dry}}$ that of a non-transpiring leaf. Maes and Steppe (2012) summarise that a CWSI of 0 means no stress and 1 means fully closed stomata. Both references come from wet and dry reference targets in the field or from a model and depend on the weather at the time, so they must be measured on the same day.

From zones to a prescription map

A prescription map tells the sprayer what rate to apply in each zone. The decision must combine several kinds of evidence, not a single index.

T_WET, T_DRY = 26.0, 36.0                     # °C from reference targets on the same day
zones = {  # zone: (area rai, canopy temperature °C, prescribed liquid fertiliser L/ha)
    "high": (18.33, 28.5, 10.0),
    "medium": (6.67, 31.0, 15.0),
    "low": (5.00, 34.0, 20.0),
}
total = 0.0
for name, (rai, canopy, rate) in zones.items():
    cwsi = (canopy - T_WET) / (T_DRY - T_WET)
    if cwsi >= 0.6:
        action, rate = "check irrigation first, hold fertiliser", 0.0
    else:
        action = f"apply {rate:.0f} L/ha"
    litres = rate * rai * 0.16
    total += litres
    print(f"{name:<6} {rai:5.2f} rai  CWSI {cwsi:.2f}  -> {action} ({litres:.1f} L)")
print(f"total product {total:.1f} L")

high   18.33 rai  CWSI 0.25  -> apply 10 L/ha (29.3 L)
medium  6.67 rai  CWSI 0.50  -> apply 15 L/ha (16.0 L)
low     5.00 rai  CWSI 0.80  -> check irrigation first, hold fertiliser (0.0 L)
total product 45.3 L

Monitoring again: real change or just error?

Two weeks after application, fly again and compare NDVI for each zone. A difference may come from measurement uncertainty (light, calibration, alignment) rather than a change in the crop. Following NIST’s propagation of uncertainty, if the errors of the two surveys are independent:

${\sigma }_{\Delta }=\sqrt{{\sigma }_1^2+{\sigma }_2^2},{\text{LoD}}_{95}\approx 1.96{\sigma }_{\Delta }$

A difference within the LoD95 must be reported as “not detectable”, not “no change”.

import math

SIGMA_1, SIGMA_2 = 0.03, 0.03
lod = 1.96 * math.hypot(SIGMA_1, SIGMA_2)
change = {"Z1": 0.02, "Z2": -0.05, "Z3": -0.12, "Z4": 0.10}   # NDVI survey 2 − survey 1
print(f"LoD95 = {lod:.3f}")
for zone, d in change.items():
    verdict = ("detected increase" if d > lod else "detected decrease" if d < -lod
               else "not detectable")
    print(f"{zone}: dNDVI {d:+.2f} -> {verdict}")

LoD95 = 0.083
Z1: dNDVI +0.02 -> not detectable
Z2: dNDVI -0.05 -> not detectable
Z3: dNDVI -0.12 -> detected decrease
Z4: dNDVI +0.10 -> detected increase

This formula applies when the two errors are independent, unbiased and approximately normal. Otherwise use a method suited to the data. When many pixels are tested at once, some will be “detected” by chance.

Into a decision support system

Results passed into a decision support system or cooperative platform need more than a map: the image dates, camera and calibration method, indices and thresholds, ground-truth results, a recommendation per zone with reasons, and limits such as zones whose cause is still unknown. Deliver GIS layers (such as a GeoPackage) that other systems can read, plus a PDF map for the farmer. The final decision belongs to the farmer and specialists; drone data is supporting evidence.

Module lab

Common mistakes

Summary

• CWSI = (Tc − Twet)/(Tdry − Twet) indicates water stress: 0 is no stress, 1 is full stress
• A prescription map must combine vegetation indices, CWSI and ground truth, not a single index
• A difference must exceed LoD95 to count as detected change
• Pass results into decision support with methods, thresholds and limits

Check your understanding

1. Tc = 30 °C, Twet = 25 °C and Tdry = 35 °C. What is the CWSI?
2. σ₁ = 0.02 and σ₂ = 0.04 (independent). What is LoD95?
3. Using question 2, is an NDVI difference of −0.07 a detected change?
4. How many litres does a 5-rai zone need at 20 L/ha?
5. Why should a zone with low NDVI and high CWSI not simply receive more fertiliser?