Question 11 of 11: Pit-limit optimization by Lerchs–Grossmann (20 marks)

Question 6: Pit-limit optimization by Lerchs–Grossmann (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.

6.1.1 Estimating per-block cash flow

Each block's cash flow is estimated the same way regardless of sign: (revenue the block would generate if it is ore, from its assayed grade × tonnage × metal price × recovery, less that block's OWN processing/refining/selling costs) MINUS (mining cost to drill, blast, load and haul that block, whether it turns out to be ore or waste). A genuine ORE block nets out positive once its process/selling costs are covered by its metal revenue; a WASTE block has no revenue at all, so its cash flow is simply the negative of its mining cost — which is exactly why every waste value in Figure 6.1 is negative and grows (more negative) with depth, since deeper waste generally costs more to haul to the dump.

6.1.2 / 6.2.1 True Lerchs–Grossmann pit outline (not moving cone)

Given. The shaded expansion in Figure 6.1 forms a 45° staircase of up to 4 side-by-side columns (labelled A–D below), read directly off the figure from the pit bottom (level 1) up to the crest (level 8); each column's cash flow by level (deep→shallow) is tabulated below.

Find. The maximum-value 45°-constrained outline. Approach. Lerchs–Grossmann's graph-closure result, specialised to a single 2-D section, is EXACTLY equivalent to a 1-D envelope dynamic program: choose the deepest level $d(k)\in\{0,\dots,\text{max}_k\}$ mined in each column $k$ (0 = not mined; mining is always contiguous from the column's own shallow top down to $d(k)$) to maximise the sum of each column's own cumulative cash flow, subject to the 45° precedence constraint that adjacent mined columns cannot differ in mined depth by more than one level: $|d(k)-d(k+1)|\le1$ whenever both are mined. This was solved by exhaustive search over all feasible $(d_A,d_B,d_C,d_D)$ — the same closure the L–G max-flow/min-cut algorithm would return for this section.

1. Column A (levels 1–8, deep→shallow $-1,-2,-2,-3,2,3,4,2$): mining down to level 5 (i.e. levels 5,6,7,8) sums $2+3+4+2=\boxed{11}$, the best contiguous-from-the-top slice.
2. Column B (levels 1–7, deep→shallow $-1,0,4,-3,-3,2,2$): mining down to level 6 (levels 6,7) sums $2+2=\boxed{4}$; extending down to level 3 to capture the buried $+4$ would cost $-3-3+4=-2$ net on top, so it is correctly excluded.
3. Column C (levels 1–7, deep→shallow $-1,-1,4,-1,-3,-1,2$): every contiguous slice from the shallow top is checked; level 7 alone nets $+2$, but the 45° constraint (below) forces C to track its neighbours, and the search finds C is best left unmined ($d_C=0$) once that constraint with columns B and D is enforced.
4. Column D (levels 1–4, deep→shallow $-2,-2,2,3$): mining down to level 3 (levels 3,4) sums $2+3=\boxed{5}$.
5. Enforce the 45° adjacency constraint. The exhaustive search over $(d_A,d_B,d_C,d_D)$ subject to $|d_A-d_B|\le1$, $|d_B-d_C|\le1$, $|d_C-d_D|\le1$ (only where both neighbours are mined) confirms $(d_A,d_B,d_C,d_D)=(5,6,0,3)$ is the GLOBAL optimum: $|5-6|=1$ OK; C unmined removes its constraint with both neighbours; D at 3 has no mined neighbour on its far side. $$\text{Total} = 11+4+0+5=\boxed{20}$$

6.1.3 Where the optimal-pit value is recorded

In a true L–G solve, the optimal pit VALUE appears at the ROOT/source node of the closure graph once max-flow has been computed (equivalently, in the moving-cone-style hand method, at the deepest/last block added whose cumulative cone value is still positive) — here, that is the cumulative total carried at column A's own deepest mined level (level 5, the block that "closes" the search once no further level extension anywhere in the section increases the total), which is the running total of $\boxed{20}$ tabulated in the final-results row below.

6.2.2–6.2.4 Blocks mined, cash flow, and optimality

1. Block count (6.2.2). Column A contributes 4 blocks (levels 5–8, ALL positive — ore), column B contributes 2 blocks (levels 6–7, both positive), column D contributes 2 blocks (levels 3–4, both positive); column C contributes none. $$\text{Total blocks mined} = 4+2+0+2=\boxed{8},\quad \text{ore}=\boxed{8},\quad\text{waste}=\boxed{0}$$ every block the optimal outline selects happens to be a positive-value (ore) block here — a direct consequence of the search discarding any column extension whose cumulative sum would need to dip negative to reach a deeper positive pocket.
2. Total cash flow (6.2.3). $$\boxed{+20\ \text{(cash-flow units, as printed on the block figure)}}$$
3. Is the FULL shaded outline optimal (6.2.4)? No. The shaded region in Figure 6.1 shows all 26 candidate expansion blocks (4+4+4+4+3+3+3+1) as a single "outlined" area, but mining every shaded block would drag in columns A's and B's negative upper levels, C's negative levels entirely, and D's negative levels — a large net loss. The TRUE L–G-optimal expansion mines only the 8 positive-value blocks identified above ($+20$ total), leaving column C unmined altogether and columns A, B and D each cut off well short of their full shaded extent. This is exactly the teaching point of 1.6.b/1.6.c: a hand-drawn "proposed expansion outline" is a moving-cone-style FIRST GUESS, not a proof of optimality — only the graph-closure (or its 1-D envelope-DP equivalent used here) certifies the true optimum.