Question 4 of 11: Floating-Cone Excavation on a 2-D Block Section (Grid A)

Question 1.4: Floating-Cone Excavation on a 2-D Block Section (Grid A) (6 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. A 5-row (row 1 = surface) × 9-column cross-section; every block is $-1$ (waste) except two ore blocks: $+2$ at (row 2, col 3) and $+13$ at (row 4, col 5). The pit-slope constraint is the standard 45° wall: a block at $k$ rows below any given row can only be exposed by also removing every block within $\pm k$ columns of it in that shallower row (the cone widens by one column each side per row toward surface).

Find. The stage-by-stage floating-cone sequence and the resulting ultimate pit for this section.

Approach. Rank the two ore blocks by value (largest first, the standard floating-cone evaluation order); test each unmined apex's cone net value; accept if positive and mark its cells mined; stop when no positive cone remains among unmined ore blocks.

1. Stage 1 — test the larger block, $+13$ at (row 4, col 5), as the apex. The 45° cone reaching from row 4 up to the surface (row 1) spans: row 4 col 5 only (1 cell); row 3 cols 4–6 (3 cells, all $-1$); row 2 cols 3–7 (5 cells — this range captures the $+2$ block at col 3 along with four $-1$ cells); row 1 cols 2–8 (7 cells, all $-1$) — 16 cells in total. $$V_{\text{cone}} = \underbrace{13}_{\text{row 4}} + \underbrace{(-1-1-1)}_{\text{row 3}} + \underbrace{(2-1-1-1-1)}_{\text{row 2 (incl. }+2\text{)}} + \underbrace{(-1\times 7)}_{\text{row 1}} = 13-3-2-7 = \boxed{+1.0}$$
2. Decision. $V_{\text{cone}} = +1.0 > 0$, so the cone is accepted: all 16 blocks are extracted. Because the $+2$ block at (row 2, col 3) lies inside this cone's footprint, it is captured automatically — no separate cone needs to be (or should be) evaluated at it.
3. Stage 2 — check for remaining candidates. Every ore block in the section (both $+2$ and $+13$) is now mined; no unmined positive block remains to seed another cone, so the algorithm halts.