Chapter 7 Numericals – IDW & Suitability

Key Formulas & Process

1. Inverse Distance Weighting (IDW)

IDW estimates the value at an unknown point using values from nearby known points, giving more weight to closer points.

Formula:

Z = Σ(Z_i / d_i^p) / Σ(1 / d_i^p)

Where:

• Z_i = value at known point i

• d_i = distance from unknown point to known point i

• p = power parameter (typically 1 or 2)

Key insight: Higher p → nearest point dominates more. When p=2 (GIS default), the nearest point has much more influence than distant ones.

2. Weighted Suitability Analysis

Combines multiple raster layers using weighted overlay.

Formula:

Score = Σ(Weight_i × Rating_i)

Where:

• Weight_i = importance of factor i (must sum to 1.0)

• Rating_i = suitability rating of the cell for factor i (typically 1–10)

• Suitable area = count of suitable cells × cell area

Process Steps

IDW: Calculate weight w_i = 1/d_i^p for each known point
IDW: Multiply each weight by its value: w_i × Z_i
IDW: Sum weighted values and divide by sum of weights: Z = Σ(w_iZ_i) / Σw_i
Suitability: Reclassify each input layer to a common scale (1–10)
Suitability: Multiply each reclassified layer by its weight
Suitability: Sum weighted layers, apply threshold to find suitable cells

Q1. IDW Interpolation VERIFIED

Exam: 2074 Bhadra Q3b [4 Marks]

Question: Given four known elevation values at different distances from an unknown point, estimate the elevation using IDW with p=1 and p=2.

Point	Elevation (Z)	Distance (d)
1	160	1000 m
2	170	400 m
3	180	600 m
4	90	120 m

Solution (a) — p = 1 (Linear)

With p=1, weights are inversely proportional to distance: w = 1/d

⚙ IDW_p1.calc

// IDW with p = 1 (Linear weighting)
// Formula: Z = Σ(Zi/di) / Σ(1/di)
//  Point  Z       d      w = 1/d        w × Z
Pt 1: Z=160  d=1000  w = 1/1000 = 0.00100  wZ = 0.160
Pt 2: Z=170  d= 400  w = 1/400  = 0.00250  wZ = 0.425
Pt 3: Z=180  d= 600  w = 1/600  = 0.00167  wZ = 0.300
Pt 4: Z= 90  d= 120  w = 1/120  = 0.00833  wZ = 0.750
// Sum of weights
Σw = 0.00100 + 0.00250 + 0.00167 + 0.00833 = 0.01350
// Sum of weighted values
ΣwZ = 0.160 + 0.425 + 0.300 + 0.750 = 1.635
Z(p=1) = 1.635 / 0.01350 = 121.11

GIS CalculationUTF-8

Solution (b) — p = 2 (Quadratic — GIS Default)

With p=2, weights are inversely proportional to the square of distance: w = 1/d². This gives much more influence to nearby points.

⚙ IDW_p2.calc

// IDW with p = 2 (Quadratic weighting)
// Formula: Z = Σ(Zi/di²) / Σ(1/di²)
//  Point  Z       d       d²          w = 1/d²        w × Z
Pt 1: Z=160  d=1000  d²=1,000,000  w=0.0000010  wZ=0.000160
Pt 2: Z=170  d= 400  d²=  160,000  w=0.0000063  wZ=0.001063
Pt 3: Z=180  d= 600  d²=  360,000  w=0.0000028  wZ=0.000500
Pt 4: Z= 90  d= 120  d²=   14,400  w=0.0000694  wZ=0.006250
// Sum of weights
Σw = 0.0000010 + 0.0000063 + 0.0000028 + 0.0000694
   = 0.0000795
// Sum of weighted values
ΣwZ = 0.000160 + 0.001063 + 0.000500 + 0.006250
    = 0.007973
Z(p=2) = 0.007973 / 0.0000795 = 100.32
// Why is p=2 closer to 90? Because Point 4 (Z=90, d=120)
// is MUCH closer than others, and p=2 squares the weight
// making it 87% of total weight vs 62% with p=1

GIS CalculationUTF-8

Key Takeaway:
• p=1: Z = 121.11 — moderate influence from all points
• p=2: Z = 100.32 — nearest point (Z=90 at 120m) dominates
Higher power ⇒ result is pulled more toward the nearest point's value

Q2. Rice Crop Suitability Analysis NEW

Exam: 2074 Magh Q6 [14 Marks]

Question: Using the given 5×5 raster with coded landuse, slope, and soil ratings, calculate the weighted suitability score for rice cultivation. Cell size = 25 m. Weights: Slope 50%, Soil 30%, Landuse 20%. Suitable if score > 6.

Step 1 — Rating Table

Code	Slope Rating	Soil Rating	Landuse Rating
1	10	8	8
2	8	4	1
3	6	10	4
4	4	1	8
5	1	1	10

Step 2 — Weighted Score Calculation

⚙ Rice_Suitability.calc

// Weighted Suitability Score
// Formula: Score = Slope×0.5 + Soil×0.3 + LU×0.2
// Suitable if Score > 6
//  Code   Slope×0.5    Soil×0.3    LU×0.2    Score
Code 1: 10 × 0.5 = 5.0
         8 × 0.3 = 2.4
         8 × 0.2 = 1.6
        Total = 5.0 + 2.4 + 1.6 = 9.0  ✓ SUITABLE
Code 2:  8 × 0.5 = 4.0
         4 × 0.3 = 1.2
         1 × 0.2 = 0.2
        Total = 4.0 + 1.2 + 0.2 = 5.4  ✗ NOT suitable
Code 3:  6 × 0.5 = 3.0
        10 × 0.3 = 3.0
         4 × 0.2 = 0.8
        Total = 3.0 + 3.0 + 0.8 = 6.8  ✓ SUITABLE
Code 4:  4 × 0.5 = 2.0
         1 × 0.3 = 0.3
         8 × 0.2 = 1.6
        Total = 2.0 + 0.3 + 1.6 = 3.9  ✗ NOT suitable
Code 5:  1 × 0.5 = 0.5
         1 × 0.3 = 0.3
        10 × 0.2 = 2.0
        Total = 0.5 + 0.3 + 2.0 = 2.8  ✗ NOT suitable

GIS CalculationUTF-8

Step 3 — Suitability Summary

Code	Score	Result
1	9.0	✓ Suitable
2	5.4	✗ Not suitable
3	6.8	✓ Suitable
4	3.9	✗ Not suitable
5	2.8	✗ Not suitable

Step 4 — Area Calculation

⚙ Rice_Area.calc

// Count cells with Codes 1 or 3 (suitable)
Suitable cells = 5 // count from the coded raster grid
// Cell area
Cell area = 25 m × 25 m = 625 m²
// Total suitable area
Total area = 5 × 625 = 3,125 m²
Suitable area = 3,125 m² = 0.3125 hectares

GIS CalculationUTF-8

Answer: Only Codes 1 (score 9.0) and 3 (score 6.8) are suitable for rice cultivation. The total suitable area is 3,125 m² (0.3125 ha). Code 1 has the highest score because it has the best slope (10) and good soil/landuse. Code 5 is worst because its steep slope (rating 1) outweighs even good landuse (rating 10) due to slope's 50% weight.