24-MMP-A4 Mine Valuation and Mineral Resource Estimation · December 2014
Question 12 of 27: Nested Spherical Model – Gamma Values
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Mining and Mineral Processing Engineering, 09-Mmp-A4 Mine Valuation and Mineral Resource Estimation, 2014-Dec. 3 hours duration; one handwritten 8.5×11 in reference sheet permitted (not an open-book exam); only approved Sharp or Casio calculators allowed. Question 1 is compulsory (40 marks, parts 1.1–1.7); candidates then select FOUR of the six optional Questions 2–7 (15 marks each) to complete the paper.
Reference texts: Isaaks & Srivastava, An Introduction to Applied Geostatistics (variogram modelling, kriging estimators, volume–variance relations); Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design (mine valuation, NPV and cut-off grade methodology, mineable reserves); Gentry & O'Neil, Mine Investment Analysis (Canadian mining taxation, smelter/refining contract terms, net smelter return); SME Mining Engineering Handbook, 3rd ed. (mineral exploration/evaluation stages, ore reserve classification); CIM Best Practice Guidelines and NI 43-101 (Canadian Securities Administrators).
Question 3.3: Nested Spherical Model – Gamma Values (6 marks)
Nested spherical model (nugget 0.1, structure 1 sill 0.5/range 100 m, structure 2 sill 0.4/range 500 m); dots mark the four requested distances (h=1000 m is off-chart, at the total sill).
Approach. Apply $$\gamma(h)=C_0+C_1\,Sph(h/a_1)+C_2\,Sph(h/a_2),\ Sph(x)=1.5x-0.5x^3\ (x\lt1),\ Sph(x)=1\ (x\ge1)$$ structure by structure, checking whether h has exceeded each structure's own range.
h = 0 m. By definition γ(0) = 0 exactly (zero separation, no variance) — the nugget is the jump the curve makes for any h>0, not a value 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, so Sph1 = 1, fully saturated) but is still inside structure 2's range (250 < 500). 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 and 1000 > 500), so both structures are fully saturated and the variogram has reached the total sill: $$\gamma(1000)=0.1+0.5(1)+0.4(1)=\boxed{1.00=\text{Sill}}$$
