18-Geol-B8 Resource Economics & Valuation · May 2016
Question 5 of 6: Mining Dilution, Diluted Grade, and Copper-Equivalent Grade
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geol-B8, Resource Economics and Valuation. Three-hour, open-book exam; any non-communicating calculator permitted. Six questions are printed; the exam's own cover notes state that only the first four questions in the answer book are marked, and each of the six is of equal value (25 marks) — all six are answered here as a complete study resource. Most questions require mathematical solutions, and clarity and organization of the steps involved are explicitly graded.
Reference texts: Torries, Evaluating Mineral Projects: Applications and Misconceptions (SME, 1998) — discounted cash flow valuation of mine projects, net smelter return economics, royalty valuation, and cut-off grade theory; Rudenno, The Mining Valuation Handbook, 4th ed. (Wrightbooks, 2012) — comparable-transaction and appraised-value (Kilburn) methods, copper-equivalent grade, and resource/reserve-stage valuation; EGBC Geoscience Professional Practice Guidelines for assumption-disclosure conventions on open-book calculations.
Find. (i) The mineable (diluted) tonnage and Cu/Au/Ag grades. (ii) The grade of the mineable material expressed as a single copper-equivalent grade.
Approach. Recover 90% of the block as ore, then dilute that recovered tonnage with waste at the stated 15% dilution rate to get the total mined tonnage and its metal-weighted diluted grades; convert every metal's payable value per tonne of that diluted material to dollars using its own mill recovery, net smelter return and price, sum the three, and express the total as an equivalent grade of copper alone by dividing by copper's own dollar value per percentage point of grade.
Mineable tonnage. Ore recovered from the block is $10{,}000\times0.90=9{,}000$ t. With dilution defined as the waste fraction of the total mined mass ($\text{waste}=0.15\times T$, $\text{ore}=0.85\times T=9{,}000$):
$$T_{\text{mined}}=\frac{9{,}000}{0.85}=\boxed{10{,}588\ \text{t}},\qquad \text{waste}=10{,}588-9{,}000=1{,}588\ \text{t}.$$
Diluted grades. Each metal's diluted grade is the ore contribution plus the waste contribution, divided by the total mined tonnage:
$$\text{Cu}=\frac{9{,}000(1.4\%)+1{,}588(0.20\%)}{10{,}588}=\boxed{1.22\%},\qquad \text{Au}=\frac{9{,}000(3.0)+1{,}588(0.25)}{10{,}588}=\boxed{2.59\ \text{g/t}},\qquad \text{Ag}=\frac{9{,}000(50)+1{,}588(2.0)}{10{,}588}=\boxed{42.8\ \text{g/t}}.$$
Payable value per tonne of mined material, by metal. Each metal's value is (diluted grade) × (mill recovery) × (net smelter return) × (price):
$$V_{\text{Cu}}=0.0122\times0.92\times0.75\times5500=\$46.30/\text{t},\quad V_{\text{Au}}=2.59\times0.50\times0.80\times35=\$36.23/\text{t},\quad V_{\text{Ag}}=42.8\times0.40\times0.70\times0.50=\$5.99/\text{t}.$$
$$\text{Total value}=46.30+36.23+5.99=\boxed{\$88.52/\text{t mined material}}.$$
Copper-equivalent grade. One percentage point of copper grade is worth $0.01\times0.92\times0.75\times5500=\$37.95/\text{t}$; dividing the total value by that unit value converts the whole block's worth into an equivalent copper percentage:
$$\text{Cu}_{\text{eq}}=\frac{88.52}{37.95}=\boxed{2.33\%\ \text{CuEq}}.$$
Check: “15% dilution” is taken here as waste = 15% of the total mined tonnage. Some operations instead quote dilution as waste tonnes / ore tonnes; under that convention the waste is $0.15 imes9{,}000=1{,}350$ t, the mined tonnage is 10,350 t, the diluted grades are 1.24% Cu, 2.64 g/t Au and 43.7 g/t Ag, the payable value is USD 90.29/t and the copper-equivalent grade is 2.38% CuEq. The two conventions differ by only about 2% in the final CuEq, and under exam note 1 either is acceptable if it is stated clearly, as done above.
Dilution mass balance: 9,000 t of recovered ore is mixed with 1,588 t of waste (15% dilution) to give 10,588 t of mined material at lower grade than the in-situ block.
