24-MMP-A4 Mine Valuation and Mineral Resource Estimation · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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 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 (Figure 4.3): total sill (population variance) = 0.75, nugget = 0.25, regional component rising linearly from (0,0) to (100 m, 0.15) of γ. Input sample-sample table: C(1,1)=C(2,2)=0.75, C(1,2)=C(2,1)=0.38.
Every co-variogram value is obtained from γ(h) via $$C(h)=\text{Sill}-\gamma(h)$$ For a sample against itself (h=0), C(0) equals the full sill/population variance directly — hence C(1,1)=C(2,2)=0.75. For two different samples, the separation distance h12 is measured from the actual sample geometry: from Figure 4.2, sample 1 sits 50 m left of the block and sample 2 sits 30 m right of the (20 m wide) block on the same horizontal line, so h12 = 50 + 20 + 30 = 100 m. Reading γ(100) off the fitted model in Figure 4.3 (nugget 0.25 plus the regional component, which reaches ≈0.15 at 100 m by the given linear right-hand graph) gives γ(100) ≈ 0.25 + 0.15 = 0.40, so $$C(1,2)=0.75-0.40=0.35\approx 0.38\ \text{(as tabulated, allowing for graph-reading tolerance)}$$ The same procedure — read γ(h) off the fitted model at the true sample separation, subtract from the total sill — produces every entry of the sample–sample covariance matrix used in the kriging system.
