Mineral Resource Block Modelling and Pit Limits: 24-MMP-B8 Rock Slope Engineering, May 2013, Q3

Question 4: Mineral Resource Block Modelling and Pit Limits (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.

Given. Block dimensions 10 m × 10 m × 10 m; ore and waste density 3000 kg/m³; grade cutoff 1.5%; net mineral value $4200/tonne of recovered mineral; mine recovery 100%, mill recovery 95%; mining cost $20/t, milling cost $15/t (ore only), overhead cost $15/t. The geological block model (percent grade, “nil” = no assay/zero grade) is given as a 5-row × 8-column cross-section:

Fig. 4.1(a) — Geological block model (percent grade), as given. Row 1 = surface bench, row 5 = deepest bench.

Find. (a) the equivalent economic block model (dollar value per block); (b) the optimum 2-D pit outline; (c) the net economic value of that pit.

Approach. Convert each block's grade to a dollar value using the cutoff rule (below cutoff → classified waste, cost only; at/above cutoff → classified ore, net revenue), then solve the 2-D pit-limit problem as a column-envelope optimisation: for every column, choose how many benches deep to mine so that the sum of the included blocks' dollar values is maximised, subject to a 45° wall-slope constraint (adjacent columns' mined depths differ by at most one 10 m bench, since block width = block height). This constrained column optimisation is mathematically identical to a Lerchs-Grossmann graph closure (or an exhaustive floating-cone search) restricted to a single 2-D cross-section, and is solved here by dynamic programming rather than by hand-tracing cones, which is both faster and exactly verifiable.

Check: (1) mining and overhead costs are charged on every tonne mined, ore or waste (both must be moved); milling cost is charged only on tonnes that are actually ore (waste is not sent to the mill). (2) The 45° wall constraint is anchored to zero depth at BOTH edges of the given 8-column section (i.e. the pit must not extend beyond the block model as given) — a reasonable reading since the model is presented as a complete, closed cross-section with no indication the deposit continues past column 1 or column 8.

Block tonnage and value formulas. Each block: $10\times10\times10\ \text{m}^3 = 1000\ \text{m}^3$, so mass $=1000\times3.0\ \text{t/m}^3 = 3000\ \text{t}$. For a WASTE block (grade < 1.5%, including “nil”): $$\text{value}_{waste} = -(20+15)\times3000 = -\text{\$}105{,}000$$ For an ORE block (grade $g\% \ge 1.5\%$), contained mineral $=3000\times(g/100)$ t, recovered $=$ contained $\times1.00\times0.95$, revenue $=$ recovered $\times$ \$4200/t, and mining+milling+overhead cost applies to the whole 3000 t block: $$\text{value}_{ore}(g) = 3000\times\tfrac{g}{100}\times0.95\times4200 - (20+15+15)\times3000 = 119{,}700\,g - 150{,}000$$
Classify and value every block. Applying the cutoff: all “nil” and 1% blocks are WASTE (−\$105,000 each, even the 1% ones — sub-cutoff mineralization still costs money to move). The 2%, 3% and 4% blocks are ORE: $$\text{value}(2\%)=+\text{\$}89{,}400,\quad \text{value}(3\%)=+\text{\$}209{,}100,\quad \text{value}(4\%)=+\text{\$}328{,}800$$ giving the economic block model in Fig. 4.1(b).

Fig. 4.1(b) — Economic block model (US$ per 3000 t block). Green = positive-value ore block, pink = negative-value waste block.

Column profit-vs-depth (prefix sums). For each of the 8 columns, sum block values from the surface (row 1) down to depth $d$ (0–5 rows), $P(d,\text{col})$. Columns 1, 2, 7 and 8 are all-waste and best left unmined ($P$ decreases monotonically with $d$, maximum at $d=0$). Column 4 (the richest) increases every step to $P(5)=+\text{\$}1{,}045{,}500$. Column 5 dips at $d=3$ (a buried “nil” block) but recovers by $d=5$. Column 3 peaks early at $d=1$ ($+\text{\$}89{,}400$) then turns negative. Column 6 peaks at $d=3$ ($+\text{\$}313{,}200$).
Envelope optimisation with the 45° slope constraint. A pure “mine every column to its own local best depth” choice (col 3 to depth 1, col 4 to depth 5, col 5 to depth 5…) would violate the slope constraint between neighbouring columns and is not a buildable pit wall. Solving the constrained dynamic program (mined depth per column $d_j$, $|d_j-d_{j-1}|\le1$, anchored to $d=0$ at both section edges) for the depth sequence that maximises the total of the included column prefix-sums gives: $$\boxed{d_1..d_8 = 0,\ 1,\ 2,\ 3,\ 2,\ 1,\ 0,\ 0\ \text{benches}}$$ Column 2 is taken one bench deep (a losing, $-\text{\$}105{,}000$ waste block) purely to satisfy the slope constraint into column 3's valuable material below — a smaller local loss that unlocks a much larger gain two columns over, exactly the trade-off Lerchs-Grossmann graph closure is designed to find.

Fig. 4.1(c) — Optimum pit outline. Shaded blocks are mined (green = ore, amber = waste taken inside the pit envelope to hold a buildable 45° wall); grey blocks are left in place.

Net economic value of the optimum pit. Summing the included blocks' values column by column ($0 + (-105{,}000) + (-15{,}600) + 866{,}700 + 657{,}600 + 209{,}100 + 0 + 0$, using each column's own prefix sum at its chosen depth): $$\boxed{\text{Net pit value} = \text{\$}1{,}612{,}800}$$ The pit contains 7 ore blocks and 2 waste blocks (9 blocks total, 27,000 t moved), a strip ratio of $2/7\approx0.29{:}1$ waste-to-ore — low, because the slope constraint only forces two extra waste blocks into an otherwise compact, rich pit.

Question 4 — final results
Item	Result
(a) Block value formulas	Waste: −$105,000/block; Ore: $119,700×grade% − $150,000/block
(b) Optimum mined depth per column (1–8)	0, 1, 2, 3, 2, 1, 0, 0 benches
Blocks mined (ore / waste / total)	7 / 2 / 9 (27,000 t)
Strip ratio (waste:ore)	0.29 : 1
(c) Net economic value of the optimum pit	$1,612,800

Topic. Economic block modelling and 2-D optimum open-pit limit determination (Lerchs-Grossmann / floating cone).

Key relations. Block value: waste $=-(\text{mining}+\text{overhead})\times\text{tonnage}$; ore $=(\text{recovered mineral})\times\text{price} - (\text{mining}+\text{milling}+\text{overhead})\times\text{tonnage}$. Column-envelope DP (2-D equivalent of Lerchs-Grossmann graph closure): maximise $\sum_j P(d_j,j)$ subject to $|d_j-d_{j-1}|\le1$.

Why this works. The pit-limit problem is not "mine every positive-value block" — a pit wall must stand at a stable slope, so reaching a deep, valuable block often requires mining shallower, uneconomic blocks above and beside it purely to keep the wall buildable. The envelope/graph-closure formulation directly maximises total value under exactly that geometric constraint, which is why it (and not simple per-block selection) is the industry-standard method.

Common pitfall. Forgetting that sub-cutoff mineralized blocks ("nil" or 1% here) still cost real money to mine if they fall inside the pit envelope — treating them as zero-cost rather than as waste blocks understates the true pit cost; and ignoring the wall-slope constraint entirely, which would select only the four richest cells in isolation and produce a physically unbuildable, checkerboard "pit".

Source. Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design, 3rd ed. (Lerchs-Grossmann algorithm, floating cone method, economic block modelling).

Question 3 of 4: Mineral Resource Block Modelling and Pit Limits

Question 4: Mineral Resource Block Modelling and Pit Limits (20 marks)