Nested Spherical Variogram — Gamma Values: 24-MMP-A4 Mine Valuation and Mineral Resource Estimation, December 2018, Q12

Question 3.2: Nested Spherical Variogram — Gamma Values (9 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	0.1
Structure 1: sill C₁ / range a₁	0.5 / 100 m
Structure 2: sill C₂ / range a₂	0.4 / 500 m

Find. The scaled diagram (2 marks) and γ(h) at h = 0, 50, 250, 1000 m (7 marks: 1+2+2+1 for the four values plus 1 for indicating them on the diagram).

Nested spherical model — nugget 0.1, structure 1 (dashed red, sill 0.5/range 100 m), structure 2 (dashed green, sill 0.4/range 500 m), nested total (solid blue). Filled dots mark the four requested distances; the open circle at h=0 marks γ(0)=0 by definition.

Approach. Apply $\gamma(h)=C_0+C_1\,Sph(h/a_1)+C_2\,Sph(h/a_2)$ where $Sph(x)=1.5x-0.5x^3$ for $x<1$ and $Sph(x)=1$ for $x\ge1$, checking whether h has exceeded each structure's own range before substituting.

h = 0 m. By definition γ(0) = 0 exactly — a pair of points at zero separation has zero difference, so the semi-variogram itself starts at the origin; the nugget C₀ is the DISCONTINUOUS JUMP the fitted curve makes for any h>0 (representing micro-scale variability/sampling error below the shortest sampled lag), not a value the curve takes at h=0 itself. $$\boxed{\gamma(0)=0}$$
h = 50 m. Both ranges exceed 50 m, so both structures use the cubic form. x₁ = 50/100 = 0.500, Sph(x₁) = 1.5(0.500) − 0.5(0.500)³ = 0.750 − 0.0625 = 0.6875. x₂ = 50/500 = 0.100, Sph(x₂) = 1.5(0.100) − 0.5(0.100)³ = 0.150 − 0.0005 = 0.1495. Substituting: $$\gamma(50)=0.1+0.5(0.6875)+0.4(0.1495)=0.1+0.34375+0.0598=\boxed{0.504}$$
h = 250 m. h has passed structure 1's range (250 > 100 m, so Sph₁ = 1, fully saturated at its sill) but is still inside structure 2's range (250 < 500 m). x₂ = 250/500 = 0.500, Sph(x₂) = 0.6875 (as computed above). Substituting: $$\gamma(250)=0.1+0.5(1)+0.4(0.6875)=0.1+0.5+0.275=\boxed{0.875}$$
h = 1000 m. h exceeds both ranges (1000 > 100 m and 1000 > 500 m), so both structures are fully saturated and the variogram has reached the TOTAL SILL: $$\gamma(1000)=0.1+0.5(1)+0.4(1)=\boxed{1.00}$$

Distance h (m)	γ(h)
0	0.000
50	0.504
250	0.875
1000	1.000 (= total sill)

Question 12 of 29: Nested Spherical Variogram — Gamma Values

Question 3.2: Nested Spherical Variogram — Gamma Values (9 marks)