Question 10 of 13: Power-Law Mine Operating Cost Model — Cost per Day and per Tonne

Question 4: Power-Law Mine Operating Cost Model — Cost per Day and per Tonne (15 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.

4.1 Comment on the claims. This power-law form, Cost/day = K·T^x, is a standard parametric (factored) mining cost-estimating relationship, fitted by regressing ln(cost) on ln(T) across a population of operating mines of different scale – exactly the same statistical philosophy as Taylor's Rule in Q1.1. Dividing by T gives cost/tonne = K·T^(x−1) = K/T^(1−x), confirming the algebra in the claim. Because most such components exhibit economies of scale, the fitted exponent x is normally less than 1 (cost grows sub-linearly with throughput, reflecting fixed-cost components, bulk purchasing and mechanization efficiencies spread over more tonnes); since (x−1) is then negative, cost/tonne decreases as T increases, which is the physically sensible behaviour the claim implies. Two cautions are warranted: (i) K and x are only valid within the range of mine sizes actually used to calibrate them – extrapolating far outside that range risks missing real diseconomies (haul-distance/pit-geometry limits, logistics bottlenecks) that a pure power law cannot capture; and (ii) different cost components legitimately have different exponents, because they scale with different physical drivers – here Haulage correctly carries a shallower exponent (0.6) than Drilling/Blasting/Loading/General (0.7), consistent with haul fleet requirements scaling somewhat differently with throughput than drill-and-blast pattern intensity.

Given.

Find. Individual component and total costs, in $/day and $/tonne.

Approach. Evaluate T^0.7 and T^0.6 once, then substitute into each component's own K and x; cost/tonne is cost/day divided by T (equivalently K·T^x−1).

1. Common powers of T. $$T^{0.7}=50{,}000^{0.7}=1946.61 \qquad T^{0.6}=50{,}000^{0.6}=659.75$$
2. Drilling. Cost/day = 1.90(1946.61) = $\$3,698.56$; cost/tonne = 3,698.56/50,000 = $\$0.0740$/t.
3. Blasting. Cost/day = 3.17(1946.61) = $\$6,170.75$; cost/tonne = $\$0.1234$/t.
4. Loading. Cost/day = 2.67(1946.61) = $\$5,197.45$; cost/tonne = $\$0.1039$/t.
5. Haulage. Cost/day = 18.07(659.75) = $\$11,921.75$; cost/tonne = $\$0.2384$/t.
6. General. Cost/day = 6.65(1946.61) = $\$12,944.96$; cost/tonne = $\$0.2589$/t.
7. Totals. Summing all five components: $$\text{Total cost/day} = \boxed{\$39{,}933.48}\qquad \text{Total cost/tonne} = \boxed{\$0.7987/\text{t}}$$ (check: $\$39,933.48$ / 50,000 t = $\$0.7987$/t, consistent).