Question 11 of 11: Mine Design and Scheduling

Question 6 — Mine Design and Scheduling (20 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.

6.1 — using the grade-distribution figure. Figure 6.1 plots the cumulative proportion of blocks (on a normal-probability horizontal scale) whose grade is at or above successive grade values. To find the TONNAGE above a chosen cut-off grade g, read across from g on the vertical (grade) axis to the curve, then down to the cumulative-frequency axis to get the percentage of all blocks at or above g, and multiply that percentage by the total block tonnage in the pit. To find the AVERAGE GRADE of that same above-cut-off tonnage, the curve (or its companion grade-tonnage curve, Fig 6.2) is integrated/read directly at that cumulative-percentage point — Fig 6.2 is in fact the same relationship replotted directly as tonnage- and grade-above-cutoff versus cutoff grade, which is the more directly usable form used for the remaining sub-questions.

6.2 — shape of the two curves. The TONNAGE-above-cutoff curve is a monotonically DECREASING, roughly S-shaped (sigmoid) curve: raising the cut-off progressively excludes more of the lower-grade material, so tonnage falls, slowly at first (while cut-off is still below most of the deposit’s grade), then steeply through the bulk of the grade distribution, then slowly again as only the highest-grade tail remains. The GRADE-above-cutoff curve is the mirror-image INCREASING sigmoid, because excluding progressively more low-grade material necessarily raises the average of what remains. Grade rises RAPIDLY through the middle of the cut-off range specifically because that is where the bulk of the tonnage (and hence the bulk of the low-grade material being progressively excluded) sits — a copper deposit’s grade distribution is typically approximately log-normal, so most of the tonnage clusters in a fairly narrow mid-grade band, and cutting through that band removes a disproportionate share of low-grade tonnes per unit increase in cut-off, pulling the average up steeply.

Given (6.3–6.6). Mining USD 1.75/t (ore or waste); Milling USD 5.00/t (ore); G&A mining USD 0.35/t; G&A milling USD 0.80/t; Smelting/Refining/Sales USD 0.50/lb Cu; recovery 78%; price USD 2.50/lb Cu; 2,204.62 lb/tonne.

6.3.1 — break-even grade.

1. Total cost per tonne of ore. USD 1.75+5.00+0.35+0.80 (mining+milling+both G&A) = USD 7.90/t.
2. Net price per lb Cu recovered. USD 2.50 price − USD 0.50 smelting/refining = USD 2.00/lb.
3. Solve for the grade that exactly covers cost. $$g_{be}\times 2204.62\times 0.78\times 2.00 = 7.90$$ $$\boxed{g_{be} = 0.230\%\ \text{Cu}}$$, closely bracketing the 0.25% cut-off the question itself instructs using for 6.4 (the small gap reflects rounding/estimation in a hand-worked break-even, and possibly a modest contingency the mine has added above the bare-bones break-even).

6.3.2 — using USD 2.50/lb as the long-term price. A single, fixed long-term price assumption ignores the copper price’s well-documented cyclicality; using today’s (or a recent) spot price directly as a LONG-TERM planning price risks materially over- or under-stating the economic pit limit and cut-off grade depending on where in the cycle that price sits — standard practice is to use a trailing multi-year average or a bank/consensus long-term forecast price (deliberately smoothed below current cyclical peaks) for reserve and pit-limit determination, reserving the actual spot price for short-term operating/cut-off decisions only.

Find (6.4). Total tonnes, ore tonnes, average grade, waste tonnes and stripping ratio at the 0.25% cut-off, read from Fig 6.2.

1. Total tonnes (ore + waste). The whole “ultimate pit” tonnage is every block with grade ≥ 0 — i.e. the curve’s own value at cut-off = 0: $$\boxed{Total = 118\ \text{Mt}}$$
2. Ore tonnes at the 0.25% cut-off. Reading the tonnage curve at g=0.25: $$\boxed{Ore = 62\ \text{Mt}}$$
3. Average grade of that ore. Reading the grade curve at g=0.25: $$\boxed{Grade = 0.56\%\ \text{Cu}}$$
4. Waste tonnes. $$Waste = Total - Ore = 118-62=\boxed{56\ \text{Mt}}$$
5. Stripping ratio. $$SR=\dfrac{Waste}{Ore}=\dfrac{56}{62}=\boxed{0.90:1}$$

Find (6.5) — Taylor’s Rule mine life.

1. 6.5.1 State the rule. Taylor (1986) “Rates of working of mines”: an empirical relation between a deposit’s reserve tonnage T and its economically-optimum operating life, $$\text{Life (yr)} = 0.2\times T^{0.25}$$ (T in tonnes) — a purely statistical fit across many operating mines, used for early-stage, pre-feasibility sizing before a detailed schedule exists.
2. 6.5.2 Mine life. Using the ore reserve T = 62,000,000 t: $$\text{Life}=0.2\times(62{,}000{,}000)^{0.25}=0.2\times 88.7$$ $$\boxed{\text{Life} \approx 17.7\ \text{years}}$$
3. 6.5.3 Tonnes milled annually/daily. $$\dfrac{62{,}000{,}000}{17.7}\approx\boxed{3{,}500{,}000\ \text{t/yr}}\ \left(\approx\boxed{9{,}600\ \text{t/day}}\text{ on a 365-day year}\right)$$
4. 6.5.4 Copper produced annually/total. Total Cu = 62,000,000 × 0.0056 × 0.78 ≈ 270,700 t total, or $$\boxed{\approx 15{,}100\ \text{t Cu/yr} \ (\approx 33.3\ \text{million lb Cu/yr})}$$
5. 6.5.5 Equipment/cost estimate. At this early (pre-final-pit-design) stage, equipment and operating costs are sized PARAMETRICALLY off the same annual tonnage (the O’Hara-style Cost=A·T^b approach of Q1.1) — e.g. shovel/truck fleet sized to move ≈9.5 Mt/yr of TOTAL material (ore+waste, using the 0.90:1 stripping ratio ⇒ ≈6.7 Mt/yr waste + 3.5 Mt/yr ore ≈ 10.2 Mt/yr combined), with capital and operating unit costs taken from the same cost-index-updated benchmarks discussed in Question 1.1, refined once a detailed pit design and schedule exist.

Find (6.6) — costs, revenue, profit.

1. Total operating cost (life-of-mine). Sum of mining, milling and both G&A cost streams over 118 Mt total / 62 Mt ore: $$118(1.75+0.35) + 62(5.00+0.80) \Rightarrow \boxed{\$607.4\ \text{million total}\ (\approx \$34.2\ \text{M/yr})}$$
2. Revenue (life-of-mine). Total Cu recovered (270,700 t) converted to pounds and sold at the net price: $$270{,}700\times 2204.62 \times 2.00 \Rightarrow \boxed{\$1{,}183\ \text{million total}\ (\approx \$66.7\ \text{M/yr})}$$
3. Profit (cash flow), total and annual. Revenue less operating cost: $$1183 - 607.4 \Rightarrow \boxed{\$576\ \text{million total}\ (\approx \$32.5\ \text{M/yr})}$$

6.7 — refining the assumed pit (≈10-word answers).