Mineral Economics and Mine Planning: Resource Classification, Discounted Cash Flow, Valuation, Taxation, Sampling, Reserve Estimation, Geostatistics, Mine Size, Pit Design and Cut-off Grade
1. Mineral resource classification
A mineral resource is a concentration in or on the earth with reasonable prospects of eventual economic extraction; a mineral reserve (ore reserve) is the part of a resource that has been shown, by studies that include mining, processing, economic, legal, environmental and social factors, to be minable at a profit now. The CRIRSCO family of codes (JORC and its equivalents) divides resources by increasing geological confidence into inferred, indicated and measured, and converts only indicated and measured resources into probable and proved reserves respectively — an inferred resource can never be reported as a reserve.
India classifies deposits by the United Nations Framework Classification (UNFC), which gives each quantity a three-digit code on three axes: E, economic viability; F, feasibility (the project’s status); and G, geological knowledge — each graded 1 (highest) downward. 111 is a proved mineral reserve (economically viable, backed by a feasibility study, from detailed exploration); 121 and 122 are probable reserves; codes whose first digit is 2 or 3 are resources that are not, or not yet, economic, and the G4 end is reconnaissance.
2. Discounted cash flow, mine valuation and mineral taxation
Money has a time value: ₹1 received n years from now is worth 1/(1 + i)ⁿ today at discount rate i. The net present value of a project is NPV = Σ CFₜ/(1 + i)ᵗ − I₀; accept if NPV > 0. A uniform annual cash flow A for n years has present value A × [1 − (1 + i)⁻ⁿ]/i (the annuity factor), so ₹40 a year for four years at 10% is worth 40 × 3.1699 = 126.79, and an investment of ₹100 has an NPV of 26.79. The internal rate of return is the discount rate that makes NPV zero: ₹100 now for ₹50 and ₹66 in years one and two gives 50/1.1 + 66/1.21 = 45.45 + 54.55 = 100, so IRR = 10%. The payback period ignores the time value of money and everything after payback; the profitability index is PV of inflows over the investment. For mutually exclusive projects NPV, not IRR, decides.
A mine is a wasting asset: its income stops when the reserve is exhausted, so its value must return the capital as well as a profit. The Hoskold (dual-rate) formula values a mine earning a constant annual income A for n years as V = A/[r + s/((1 + s)ⁿ − 1)], where r is the speculative (risk) rate of return and s the safe rate at which a sinking fund redeems the capital. With A = ₹10 crore, r = 15%, s = 5% and n = 10, the sinking-fund factor is 0.05/(1.05¹⁰ − 1) = 0.0795, so V = 10/0.2295 = ₹43.57 crore. The other approaches to mine valuation are the income approach by DCF, the market approach (comparable transactions, value per tonne of reserve) and the cost approach (for exploration properties).
- Royalty is the payment to the owner of the mineral — in India the State — for its removal, at rates set in the schedules of the Mines and Minerals (Development and Regulation) Act, 1957, mostly ad valorem as a percentage of the average sale price, and for some minerals per tonne.
- Dead rent is a fixed annual charge per hectare of the lease area; the lessee pays either the royalty or the dead rent, whichever is greater, never both, so dead rent is the floor that stops a lease being held idle.
- Since the 2015 amendment of the Act, mineral concessions are granted by auction, with an auction premium on top of royalty, and lessees contribute a prescribed share of royalty to the District Mineral Foundation (for the people of mining-affected areas) and to the national trust for mineral exploration. Beyond these the mine pays the ordinary taxes — income tax, GST, cesses — and depreciation and depletion allowances reduce taxable income.
3. Sampling, reserve estimation, geostatistics and quality control
Sampling is the base of every estimate. Underground and in pits: channel samples cut across the ore at right angles to the lode, chip samples, grab samples from broken ore, and bulk samples for metallurgical testing; in exploration, diamond-drill core (split, one half assayed) and reverse-circulation chips. A sample must represent its interval, so channels are cut to a constant width and depth, and assays of unequal lengths are combined as a length-weighted average grade Σlᵢgᵢ/Σlᵢ: intervals of 1.2, 0.8 and 2.0 m at 2.5, 4.0 and 1.5% average (3.0 + 3.2 + 3.0)/4 = 2.3%, not the plain mean 2.67%. Tonnage is volume × in-situ density: 50 000 m² × 4 m × 2.8 t/m³ = 560 000 t.
The classical reserve estimation methods assign each sample an area of influence: polygons (each drill hole’s grade applied to the polygon of points nearer it than any other), triangles, cross-sections with the prismoidal or end-area rule between them, and inverse distance weighting (IDW), which weights each sample by 1/dᵖ, usually p = 2: samples of 2% at 10 m and 5% at 20 m estimate (2/100 + 5/400)/(1/100 + 1/400) = 2.6%. Geostatistics treats grade as a regionalised variable whose continuity is described by the semivariogram γ(h) = [1/(2N(h))] Σ [z(x) − z(x + h)]². Its nugget (the jump at the origin) is short-scale randomness and sampling error; it rises to a sill, about the sample variance, at the range beyond which samples are uncorrelated; the spherical model is the usual fit. Kriging uses the variogram to find the weights that give the best linear unbiased estimate, with a kriging variance that depends only on the sample geometry and the variogram, not on the sample values — which is what lets a drill pattern be designed before it is drilled.
Quality control (QA/QC) keeps the database honest: blanks detect contamination, duplicates at each stage (field, coarse reject, pulp) measure precision, certified reference materials check assay accuracy, and a proportion of pulps is sent to an umpire laboratory. Grade control in operating mines uses blasthole or dedicated grade-control drilling to mark ore and waste boundaries before loading.
4. Mine planning: size and life, pit design, cut-off grade, dilution, facility location and equipment
A mine plan sets the method, the layout of access and infrastructure, the production rate and schedule, the equipment, the manpower, the processing route, the environmental management and closure, and the economics that tie them together. Mine size and life trade against each other: a larger rate earns revenue sooner (good for NPV) but costs more capital and shortens life. Taylor’s rule gives a first estimate from the expected ore tonnage T in tonnes: life ≈ 0.2 T^0.25 years, so a 100 Mt deposit supports about 0.2 × 100 = 20 years, i.e. about 5 Mt a year.
Ultimate pit design starts from a block model: each block’s economic value is its recoverable revenue minus mining and processing costs (or minus mining cost alone if it is waste). The Lerchs–Grossmann algorithm finds, by graph theory, the pit outline that maximises the total undiscounted value while respecting the slope constraints; the floating (moving) cone method is a faster heuristic that can miss combinations of blocks that pay only together. Running the optimiser at a range of prices gives nested pits, which guide the sequence of pushbacks, and the pit is then designed with ramps and benches.
The grade–tonnage curve of a deposit shows, for each cut-off, the tonnage above it and its average grade: as the cut-off rises the tonnage falls and the average grade rises. The break-even cut-off grade is the grade at which a tonne of ore just pays its costs: g = cost per tonne/(price × recovery), in consistent units — with costs of ₹2550 per tonne of ore, a metal price of ₹6 lakh per tonne and 85% recovery, g = 2550/(600 000 × 0.85) = 0.5%. Lane’s theory sets cut-offs that maximise NPV by balancing the mine, mill and market capacities, and raises the cut-off early in life. Dilution is waste mined with the ore: 100 t of 2% ore with 15 t of barren waste gives (100 × 0.02)/115 = 1.74% over 115 t, a dilution of 15%. Recovery (mining recovery) is the fraction of the ore reserve actually extracted, and processing recovery the fraction of the metal the plant recovers.
- Facility location — a washery, a crusher, a central workshop — is optimised by minimising the load-weighted transport distance. The centre-of-gravity method puts it at x̄ = Σwᵢxᵢ/Σwᵢ (and likewise ȳ): mines producing 200 t/d at x = 0 and 300 t/d at x = 10 km place it at 3000/500 = 6 km. Where transport is along roads at right angles, the weighted median minimises the rectilinear distance exactly.
- Equipment selection matches the machine to the method, the production target, the ground, the haul profile and the other machines, and compares candidates on cost per tonne (owning plus operating cost over output). Loaders and trucks are matched in bucket passes and by the match factor (trucks × time to load one truck)/(loaders × truck cycle time), 1 at a perfect match.
Key takeaways
- Only indicated and measured resources become probable and proved reserves; UNFC 111 is a proved reserve, codes starting 2 or 3 are resources.
- NPV = ΣCFₜ/(1 + i)ᵗ − I₀; IRR makes NPV zero; Hoskold V = A/[r + s/((1 + s)ⁿ − 1)] returns capital through a sinking fund.
- The lessee pays royalty or dead rent, whichever is greater; concessions are now auctioned, with shares of royalty to the DMF and the exploration trust.
- Weight grades by length; IDW weights by 1/d²; the variogram’s nugget, sill and range drive kriging, whose variance does not depend on the sample values.
- Taylor: life ≈ 0.2T^0.25 years; break-even cut-off = cost/(price × recovery); diluted grade = metal/(ore + waste); centre of gravity x̄ = Σwx/Σw.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Under the United Nations Framework Classification used in India, the code 111 denotes:
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Answer: B — a proved mineral reserve
The three digits are E (economic viability), F (feasibility) and G (geological knowledge), each 1 at its highest: 111 is economically viable, backed by a feasibility study and based on detailed exploration — a proved reserve. Probable reserves are 121 and 122, and reconnaissance quantities sit at the G4 end of the resource codes.Under CRIRSCO-type reporting codes such as JORC, an inferred mineral resource:
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Answer: C — cannot be converted into any category of reserve
Geological confidence in an inferred resource is too low to support mine planning, so it must first be upgraded by further drilling. Indicated resources convert to probable reserves and measured resources to proved (or probable) reserves; measured is the highest-confidence resource class, not the same as inferred.A mining project costs ₹100 crore now and returns ₹40 crore at the end of each of the next four years. At a discount rate of 10%, the NPV, in ₹ crore, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 26.79
Annuity factor = (1 − 1.1⁻⁴)/0.1 = (1 − 0.68301)/0.1 = 3.16987, so PV = 40 × 3.16987 = 126.79 and NPV = 126.79 − 100 = 26.79. The undiscounted surplus is 60; discounting every flow as if it came in year four gives 40 × 4 × 0.683 − 100 = 9.28.An investment of ₹100 lakh returns ₹50 lakh at the end of year 1 and ₹66 lakh at the end of year 2. Its internal rate of return, in %, is ____.
Numerical answer — type the value.
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Answer: 10
At 10%, PV = 50/1.1 + 66/1.21 = 45.45 + 54.55 = 100.00, which equals the outlay, so NPV = 0 and IRR = 10%. The simple average return (116 − 100)/100/2 = 8% ignores timing; at 8% the NPV is +2.88, at 12% it is −2.74, bracketing 10%.A mine is expected to earn a net income of ₹10 crore a year for 10 years. Using Hoskold’s formula with a speculative rate of 15% and a safe (sinking fund) rate of 5%, its present value, in ₹ crore, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 43.57
Sinking-fund factor = s/((1 + s)ⁿ − 1) = 0.05/(1.62889 − 1) = 0.079505; V = A/(r + factor) = 10/(0.15 + 0.079505) = 10/0.229505 = 43.57. A single-rate annuity at 15% gives 50.19, which does not provide for replacing the capital at the safe rate.Under the Mines and Minerals (Development and Regulation) Act, 1957, the holder of a mining lease pays in a year:
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Answer: B — royalty or dead rent, whichever is greater
The lessee is liable for royalty on the mineral removed or dead rent on the lease area, whichever is greater, so an idle lease still pays dead rent and a productive one pays royalty. Paying both would double-charge the same right to mine, and the smaller of the two would reward holding a lease idle.Three consecutive channel samples across a lode are 1.2 m at 2.5%, 0.8 m at 4.0% and 2.0 m at 1.5% metal. The length-weighted average grade, in %, is ____.
Numerical answer — type the value.
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Answer: 2.3
Σlg/Σl = (1.2 × 2.5 + 0.8 × 4.0 + 2.0 × 1.5)/(1.2 + 0.8 + 2.0) = (3.0 + 3.2 + 3.0)/4.0 = 9.2/4 = 2.3%. The plain mean of the three assays, 2.67%, over-weights the short high-grade sample.An orebody has a horizontal area of 50 000 m², an average thickness of 4 m and an in-situ density of 2.8 t/m³. Its tonnage, in tonnes, is ____.
Numerical answer — type the value.
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Answer: 560000
Tonnage = area × thickness × density = 50000 × 4 × 2.8 = 560000 t, typed as 560000. Omitting the density gives the volume, 200000 m³, and using a loose (broken) density instead of the in-situ one would understate the tonnage.A block is estimated by inverse distance squared weighting from two samples: 2% at 10 m and 5% at 20 m. The estimated grade, in %, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 2.6
Weights 1/100 = 0.01 and 1/400 = 0.0025: grade = (2 × 0.01 + 5 × 0.0025)/(0.01 + 0.0025) = 0.0325/0.0125 = 2.6%. Inverse distance (power 1) gives 3.0%, and the plain average 3.5% ignores distance altogether.Which of the following statements about the semivariogram and kriging are correct? (Select all that apply.)
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Answer: B — The nugget effect is the discontinuity of the semivariogram at the origin; C — Beyond the range, samples are uncorrelated; D — Kriging gives the best linear unbiased estimate for the given variogram
The nugget is the jump at h → 0 from short-scale variability and sampling error; past the range γ(h) stays at the sill and samples carry no information about each other; kriging is the BLUE estimator. Its variance depends only on the variogram and on where the samples lie, not on their values — which is why drill spacing can be designed in advance.By Taylor’s rule, the life in years of a mine with an expected ore reserve of 100 million tonnes is ____.
Numerical answer — type the value.
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Answer: 20
Life = 0.2 × T^0.25 = 0.2 × (10⁸)^0.25 = 0.2 × 100 = 20 years, so the annual output is about 10⁸/20 = 5 Mt. Using T in million tonnes (100^0.25 = 3.16) gives 0.63 years, because the rule is written for tonnes.Mining, processing and overhead costs are ₹2550 per tonne of ore. The metal sells at ₹6 lakh per tonne and the overall metallurgical recovery is 85%. The break-even cut-off grade, in %, is ____.
Numerical answer — type the value.
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Answer: 0.5
g = cost/(price × recovery) = 2550/(600 000 × 0.85) = 2550/510 000 = 0.005 = 0.5%. Leaving out the recovery gives 0.425%, and forgetting to convert the fraction to a percentage gives 0.005.A stope contains 100 t of ore at 2% metal. Mining it brings in 15 t of barren wall rock. The grade of the ore delivered to the mill, in %, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.74
Metal = 100 × 0.02 = 2 t, now in 115 t, so the diluted grade = 2/115 = 1.74%. Reducing the grade by 15% (2 × 0.85 = 1.70%) treats dilution as a fraction of the diluted tonnage rather than of the ore; the delivered tonnage rises by 15%.Two mines on a straight road produce 200 t/day at x = 0 km and 300 t/day at x = 10 km. By the centre-of-gravity method, a common coal washery should be located at x (in km) = ____.
Numerical answer — type the value.
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Answer: 6
x̄ = Σwx/Σw = (200 × 0 + 300 × 10)/(200 + 300) = 3000/500 = 6 km, pulled towards the larger producer. The midpoint, 5 km, ignores the loads; minimising total tonne-km exactly on a line would put it at the weighted median, the 300 t/day mine itself.The Lerchs–Grossmann algorithm for ultimate pit design finds the pit outline that:
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Answer: B — maximises the total undiscounted value subject to slope constraints
Lerchs–Grossmann is a graph-theory method that finds the closure of the block model with the greatest total block value, honouring the slope precedences exactly; it does not schedule or discount. A pit containing every ore block would take waste that the ore cannot pay for, and minimising the stripping ratio would mean mining almost nothing.As the cut-off grade applied to a deposit is raised, which of the following happen? (Select all that apply.)
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Answer: A — The tonnage of ore above the cut-off decreases; B — The average grade of the ore above the cut-off increases
Raising the cut-off drops the lowest-grade blocks, so tonnage falls and the average grade of what remains rises — the two branches of the grade–tonnage curve. Less ore at the same rate means a shorter life, and the metal contained can only fall, because the blocks removed carried some metal.