Question 11 of 13: Question 5 (15 marks, optional)

Question 5 (15 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 its 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.

5.1.2 – Block costs. Each block carries a mining cost (drill/blast/load/haul, usually a function of depth/haul distance from surface) applied to every block regardless of destination, plus – for blocks sent to the plant – a processing cost and, if applicable, downstream smelting/refining and transport costs; waste blocks carry mining cost only (no processing cost), which is exactly why a waste block’s cash flow is simply the negative of its mining cost.

5.1.3 – Grade to revenue. Revenue per block $=$ ore tonnage $\times$ grade $\times$ recovery (the fraction of contained metal actually recovered by the process) $\times$ payable factor $\times$ metal price, less any smelter/refining deductions and transport – i.e. 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 the block 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.

5.1.5 – “Least loss” rule. Where a block could be sent 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 – i.e. compare (a) processing now (cash flow as in 5.1.4), (b) stockpiling for possible future processing (opportunity cost of deferred revenue, discounted, 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 – this is 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 printed 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 (Question 1.5’s 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: 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.