Question 10 of 11: Block-Model Economics and Pit-Limit Optimization

Question 5: Block-Model Economics and Pit-Limit Optimization (20 marks, optional)

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.

5.1.1 – Block model and block numbering. A block model discretises the deposit and surrounding waste into a regular 3-D array of fixed-size rectangular blocks, each assigned its own grade (and hence value) estimated from drill-hole data (kriging or another interpolation method). Each block is given a single unique index by fixing an origin and block dimensions and computing $\text{ID}=i+j\cdot n_x+k\cdot n_x n_y$ from its column/row/level indices $(i,j,k)$; the same formula run in reverse ($k=\lfloor ID/(n_xn_y)\rfloor$, etc.) recovers $(i,j,k)$, and real-world $(X,Y,Z)$ follow directly from the origin plus $(i,j,k)\times$ block size – the SAME indexing logic used to address the R-column/depth-row grid of Figure 5.2 below.

5.1.2 – Block costs (ore vs. waste). Every block, ore or waste, carries a MINING cost (drill/blast/load/haul, typically a function of depth or haul distance from surface) simply because it must be excavated and moved regardless of destination. An ORE block additionally carries a PROCESSING cost (crushing/grinding/concentration) and, if applicable, downstream smelting/refining and transport costs, since it is sent to the plant rather than the dump. A WASTE block therefore has mining cost only, which is exactly why its cash flow (5.1.4) reduces to the negative of that mining cost.

5.1.3 – Grade to revenue. Revenue per block $=$ ore tonnage $\times$ grade $\times$ recovery (fraction of contained metal actually recovered by the process) $\times$ payable factor $\times$ metal price, less any smelter/refining deductions and transport – only the PAYABLE, RECOVERED metal is monetised, not the block's in-situ contained metal.

5.1.4 – Block cash flow and the waste threshold. A block's cash flow is (revenue if processed) $-$ (mining cost) $-$ (processing cost), evaluated as if it were sent to the plant; if that value is negative, the block is redefined as WASTE (mining cost only, sent to the dump) because processing it would destroy value. The break-even (cut-off) grade is exactly the grade at which processed cash flow equals the waste-only cash flow – the mechanism by which the positive/negative values in Figure 5.2 were themselves derived from grade.

5.1.5 – "Least loss" rule. Where a block could go to the plant, a low-grade stockpile, or the waste dump, "least loss" sends it to whichever destination MINIMISES the loss relative to the best possible outcome for that block: compare (a) processing now (cash flow as in 5.1.4), (b) stockpiling for possible future processing (opportunity cost of deferred, discounted revenue – Question 1.2's discounting logic applied at the block level – less rehandle cost), and (c) waste (sunk mining cost, zero further loss). The block is routed to the option with the smallest net loss, which is why marginal-grade material often goes to a stockpile rather than straight to waste or straight to the mill.

Given. Figure 5.2 block cash flows (table below), 15×15 m blocks, 45° wall slope (one block of horizontal step per row of depth).

Find. The optimal 45°-constrained expansion envelope, its ore/waste block count and total cash flow, and whether it is optimal.

Approach. Apply the 1.5.1 moving-cone rule mechanically as a 1-D envelope dynamic program: for each column find the mining depth $d(R)$ that maximises the column's cumulative cash flow $P(d,R)=\sum_{r=1}^{d}V(r,R)$, subject to the 45° constraint $|d(R)-d(R{+}1)|\le1$ between adjacent columns – the exact numerical equivalent of testing every candidate cone in 1.5.1, done exhaustively rather than block-by-block.

1. 5.2.1 – Optimal envelope. Running the depth-profile dynamic program column by column (right-anchored $R=0\ldots11$) against the cash-flow grid gives the depth profile $$d(R):\ R_{2}{=}1,\ R_{3}{=}2,\ R_{4}{=}3,\ R_{5}{=}4,\ R_{6}{=}3,\ \text{tapering to }0\text{ by }R_{9}$$ – a single apex at $R5$ (depth 4, capturing the richest block cluster, values 4 and 3 at rows 3–4) tapering symmetrically at 45° on both sides, exactly the shape 1.5.1's cone-acceptance rule predicts around the highest-value block.
2. 5.2.2–5.2.3 – Blocks mined and total cash flow. Summing the envelope column by column (table below): 4 ore blocks (values 2, 4, 3, 4 – positive cash flow) and 7 waste blocks (all negative cash flow), plus one break-even block (value 0); columns $R7$–$R8$ only step down to satisfy the 45° tie-back to the existing (unshaded) pit wall and contain no real candidate material at that shallow depth. $$\text{Total cash flow}=-2-3+0+6+2=\boxed{+3\text{ units}}$$

5.2.4 – Is this optimal? Yes – the depth-profile dynamic program evaluates every 45°-constrained envelope simultaneously (the Question 1.5 Lerchs–Grossmann argument applied to this single cross-section) and returns the maximum-value one by construction, so no alternative outline respecting the same slope constraint can score higher on this cash-flow grid. The margin is thin (+3 units against individual block swings of ±4–5), which is itself the pedagogical point (Question 1.2's front-loading logic applies at project scale, but block-model sensitivity matters at THIS scale too): a small error in block value, cost, or price assumption could flip this expansion from marginally profitable to a net loss, so a sensitivity check on the ore-block values would normally precede a final go/no-go decision.

5.2.5 – Outline on Figure 5.2. The optimal outline is the staircase apex at $R5$ (depth 4) tapering one column per row on each side, exactly as tabulated above – drawn directly onto the candidate expansion area of the supplied Figure 5.2 as a heavy boundary line separating the mined (shaded) blocks from the unmined remainder.