Cut-off Grade, Third-Party Ore and a 900 kt Low-Grade Stockpile
A copper concentrator is nameplate 2.5 Mt/y and actually treated 2.45 Mt in the last year. The reserve is 20 Mt at 0.88% Cu, mined at a strip ratio of 1.8. Copper sells for USD 9,000/t, flotation recovery is 88%, concentrate grade 26% Cu, payable 96%, treatment charge USD 90/dmt concentrate, refining USD 0.09/lb. Mining costs USD 4.20/t ore and USD 2.80/t waste, processing USD 13.00/t milled, G&A USD 6.0 M/y, sustaining capex USD 8.0 M/y. Two proposals are on the table. First, a nearby junior offers 400 kt at 0.55% Cu, delivered to the mill for USD 20/t. Second, the operation holds a 900 kt stockpile grading 0.31% Cu built during a previous high-grade campaign. The CFO asks whether the cut-off grade should be lowered to 0.30% "to use up the mill capacity and grow production".
1. Calculate the break-even cut-off grade and show the derivation. Then test its sensitivity to a copper price of USD 7,500/t.
Read the worked answer
Net smelter return per tonne of ore at grade g (fraction).
- Cu in concentrate = 0.88g; payable = 0.88 × 0.96 × g = 0.8448g.
- Revenue = 0.8448g × 9,000 = 7,603g.
- Refining = 0.88g × (0.09 × 2,204.6) = 0.88g × 198.4 = 174.6g.
- Concentrate mass = 0.88g/0.26 = 3.385g; TC = 3.385g × 90 = 304.6g.
- NSR = (7,603 − 174.6 − 304.6)g = 7,124g USD/t ore. At 0.88% Cu, NSR = USD 62.7/t.
Cost per tonne milled.
- Ore mining 4.20 + waste 1.8 × 2.80 = 5.04 → 9.24
- Processing 13.00
- G&A 6.0/2.45 = 2.45
- Sustaining capex 8.0/2.45 = 3.27
- Total = USD 27.96/t
Break-even cut-off = 27.96/7,124 = 0.00392 → 0.39% Cu. Note that omitting the strip ratio and sustaining capex would give 0.28% - a 30% error, and the most common one.
Sensitivity at USD 7,500/t. Revenue = 0.8448 × 7,500 = 6,336g; NSR = (6,336 − 174.6 − 304.6)g = 5,857g. Cut-off = 27.96/5,857 = 0.48% Cu. A 17% price fall moves the cut-off up by almost a quarter, which is why the reserve at 0.88% should be re-classified whenever the price moves materially. Conclude that the CFO's 0.30% target is below break-even at both prices and is not an ore grade under current assumptions.
2. Evaluate the offer to purchase 400 kt at 0.55% Cu for USD 20/t delivered. The mill ran at 98% utilization last year.
Read the worked answer
Stand-alone economics. NSR = 7,124 × 0.0055 = USD 39.2/t. Costs = purchase 20.00 + processing 13.00 + G&A 2.45 = USD 35.45/t. Margin = USD 3.75/t, or about USD 1.5 M over 400 kt. Thin but positive in isolation.
The constraint changes everything. At 2.45 Mt against 2.5 Mt nameplate the mill has no idle hours. Every tonne of purchased ore displaces a tonne of own ore. Contribution per tonne of own ore = NSR 62.7 − mining 9.24 − processing 13.00 − G&A 2.45 = USD 38.0/t. Processing 400 kt of bought ore strips out 400 kt × 38.0 = USD 15.2 M of contribution to gain USD 1.5 M. Net effect on NPV is strongly negative.
Caveats worth stating.
- The displaced reserve is deferred, not destroyed, so the comparison is a timing cost rather than an absolute loss - but at a 10% discount rate over an 8-year reserve life, a one-year deferral of USD 15.2 M is still far more than USD 1.5 M.
- If the mill genuinely had spare hours (a planned shutdown year, a grade shortfall because of pit sequencing, or an expansion), the answer flips. Confirm utilization from throughput records, not from the nameplate.
- Check the metallurgy before anything else: 0.55% Cu ore may carry different hardness, clay or penalty elements, and the USD 13/t processing assumption may not hold.
Counter-proposal. Offer toll milling instead: process the junior's ore and charge a toll fee, capturing margin without buying the material or taking the grade risk. Alternatively, buy only the tonnage that fills hours the mine cannot fill itself.
Answer to the CFO's framing. "Growing production" is not a goal when the constraint is mill capacity; growing cash flow per mill hour is.
3. Advise on the 900 kt stockpile at 0.31% Cu and directly answer the CFO's proposal to cut the cut-off to 0.30%.
Read the worked answer
The stockpile. NSR at 0.31% = 7,124 × 0.0031 = USD 22.1/t. Because the material is already mined, the relevant cost is processing plus G&A = 13.00 + 2.45 = USD 15.45/t, giving a marginal margin of USD 6.6/t, or USD 6.0 M - positive in isolation, and less than the 0.39% break-even, so it is correctly classified as marginal material rather than reserve.
But at 98% mill utilization it displaces 900 kt of 0.88% ore worth USD 38.0/t contribution: USD 34.2 M lost to gain USD 6.0 M. Hold the stockpile.
When it becomes ore. Full-cost basis (including the sunk mining cost, i.e. for reporting): 27.96/0.0031 requires NSR coefficient such that 0.0031 × (0.8448P − 479.2) = 27.96, giving P ≈ USD 11,200/t. Marginal basis (mining sunk): 0.0031 × (0.8448P − 479.2) = 15.45, giving P ≈ USD 6,470/t. Two very different answers depending on the decision being made - the marginal figure governs the operating decision, the full-cost figure governs whether it was ever ore.
Practical uses: mill-fill during a grade shortfall from pit sequencing, blending to stabilize head grade, and processing at the end of mine life when the mill would otherwise be idle.
The 0.30% cut-off proposal. Reject it. Lowering the cut-off when the mill is full does not create capacity; it dilutes head grade, reduces copper output per mill hour and raises cost per pound. Lower cut-off grades are the right response when the mill is under-utilized and the mine is ore-limited, which is the opposite of this operation. If capacity relief is the real objective, the options are debottlenecking, a higher cut-off to maximize cash per hour, or reducing the strip ratio through pit design - not dilution.