Nested Spherical Variogram – Numeric Evaluation: 24-MMP-A4 Mine Valuation and Mineral Resource Estimation, May 2017, Q7

Question 2.1: Nested Spherical Variogram – Numeric Evaluation (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Parameter	Value
Nugget, C₀	0.05 %Cu²
Structure 1: sill contribution C₁, range a₁	0.10 %Cu², 15 m
Structure 2: sill contribution C₂, range a₂	0.5 %Cu², 100 m

Find. γ(h) at h = 0, 10, 50 and 150 m, and a to-scale sketch of the nested model.

Fig. 2.1 – The nested two-structure spherical model, to scale, with the four requested evaluation points marked: (0, 0), (10, 0.210), (50, 0.494), (150, 0.650 = sill).

Approach. Evaluate each structure's spherical function Sph(h/a_i)=1.5(h/a_i)−0.5(h/a_i)³ while h≤a_i, clamped at 1 once h exceeds that structure's own range, then sum γ(h)=C₀+C₁·Sph(h/a₁)+C₂·Sph(h/a₂) for h>0 (with γ(0)≡0 by definition).

Total sill. $$\text{Sill} = C_0+C_1+C_2 = 0.05+0.10+0.5 = \boxed{0.65\ \%\text{Cu}^2}$$
γ(0). By definition γ(0)=0 exactly (no separation → no variance); C₀ is the discontinuous jump the model makes as h→0+, not a value AT h=0. $$\boxed{\gamma(0)=0}$$
γ(10). h=10 lies inside both ranges. Structure 1: x₁=10/15=0.6667, Sph(x₁)=1.5(0.6667)−0.5(0.6667)³=1.0000−0.1481=0.8519. Structure 2: x₂=10/100=0.100, Sph(x₂)=1.5(0.1)−0.5(0.1)³=0.150−0.0005=0.1495. $$\gamma(10)=0.05+0.10(0.8519)+0.5(0.1495)=0.05+0.08519+0.07475=\boxed{0.210\ \%\text{Cu}^2}$$
γ(50). h=50 has passed structure 1's range (50>15, Sph₁=1, fully saturated) but is inside structure 2's range (50<100). Structure 2: x₂=50/100=0.500, Sph(x₂)=1.5(0.5)−0.5(0.5)³=0.6875. $$\gamma(50)=0.05+0.10(1)+0.5(0.6875)=0.05+0.10+0.34375=\boxed{0.494\ \%\text{Cu}^2}$$
γ(150). h=150 exceeds both ranges (150>15 and 150>100), so both structures are fully saturated and the variogram has reached the total sill. $$\gamma(150)=0.05+0.10(1)+0.5(1)=\boxed{0.65\ \%\text{Cu}^2 = \text{Sill}}$$

Quantity	Value
Sill (C₀+C₁+C₂)	0.65 %Cu²
γ(0)	0.000
γ(10)	0.210 %Cu²
γ(50)	0.494 %Cu²
γ(150)	0.650 %Cu² (sill reached)

Concept & Theory

Topic: Nested spherical variogram models.

Key relations: Sph(x)=1.5x−0.5x³ for x≤1, else 1; total sill = ΣC_i (i=0..n).

Why this works: Each spherical structure is itself a valid, positive-definite covariance model, and a sum of valid variogram models is always itself a valid variogram model, so nesting structures at different ranges is a mathematically safe way to add fitting flexibility – here a short-range (15 m, grain/vein-scale) and a longer-range (100 m, zone-scale) continuity structure – without breaking the kriging system's solvability.

Common pitfall: Forgetting that once h exceeds a structure's OWN range, that structure's Sph term is clamped at 1 rather than continuing to apply the cubic formula (which would overshoot past the sill).

Source: Isaaks & Srivastava, Ch. 7–8 (variogram models, nested structures).

Question 7 of 18: Nested Spherical Variogram – Numeric Evaluation

Question 2.1: Nested Spherical Variogram – Numeric Evaluation (5 marks)