24-MMP-A5 Surface Mining Methods and Design · May 2016
Question 5 of 11: Floating-Cone Excavation on a 2-D Block Section (Grid B)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2016 — 09-MMP-A5, Surface Mining Methods and Design. Three hours, closed book; one hand-written, double-sided 8.5×11″ reference sheet and an approved Sharp or Casio calculator are permitted. Question 1 is compulsory (six parts, 40 marks); candidates then choose three of the five optional questions (2–6, 20 marks each) for a 100-mark paper — only the first three optional answers appearing in the answer book are graded. All six parts of Question 1 and all five optional questions are answered here, because this set is a study resource rather than an exam script.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
W. Hustrulid, M. Kuchta & R. Martin, Open Pit Mine Planning and Design, 3rd ed. — block modelling and record addressing (Ch. 4–5), pit-limit optimization: floating cone and Lerchs–Grossmann (Ch. 12–13), truck–shovel systems, match factor and dispatch (Ch. 15–16), slope design (Ch. 14).
SME Mining Engineering Handbook, 3rd ed. — open-pit design and materials-handling chapters; mine closure and reclamation planning.
H. Lerchs & I. F. Grossmann, “Optimum Design of Open-Pit Mines,” CIM Bulletin (1965).
Y. Lizotte, “The Economics of Computerized Open Pit Design,” International Journal of Surface Mining, Reclamation and Environment, 2:59–78 (1988) — the floating-cone procedure cited directly on this paper's Question 1.3.
BC Health, Safety and Reclamation Code for Mines — mine closure planning and reclamation security (Canadian regulatory context for Question 5).
Question 1.5: Floating-Cone Excavation on a 2-D Block Section (Grid B) (7 marks)
Given. A 5-row × 13-column section carrying three ore blocks: $+4$ at (row 2, col 6), $+10.7$ at (row 4, col 5), and $+19.8$ at (row 5, col 8) — all other cells $-1$. Same 45°-slope cone rule as Question 1.4.
Find. The stage-by-stage floating-cone sequence, the ultimate pit, and (per Question 1.3.4) what happens if the cones are evaluated in the “wrong” order.
Grid B (5×13). Orange = Stage 1 (apex row 5/col 8, the deepest and highest-value block); blue = Stage 2's new cells (apex row 4/col 5) — the two cones overlap at row 3, cols 4–6 and row 2, cols 5–7, which Stage 2 does not re-count.
Approach. Process the three ore blocks in descending block value ($+19.8$, then $+10.7$, then $+4$); at each stage compute the marginal cone value (excluding any cell already mined by an earlier stage) and accept only if positive.
Stage 1 — apex (row 5, col 8), value $+19.8$. Its cone runs row 5 (col 8, 1 cell) up through row 1 (cols 4–12, 9 cells) — 25 cells total, and its row-2 span (cols 5–11) happens to include the $+4$ block at col 6.
$$V_1 = 19.8 + \underbrace{(-3)}_{\text{row 4, cols 7-9}} + \underbrace{(-5)}_{\text{row 3, cols 6-10}} + \underbrace{(4-5)}_{\text{row 2, cols 5-11 (incl.}+4\text{)}} + \underbrace{(-9)}_{\text{row 1, cols 4-12}} = \boxed{+0.8}$$
Accepted: 25 cells mined (this absorbs the $+4$ block at col 6 automatically).
Stage 2 — apex (row 4, col 5), value $+10.7$. Its full cone would be row 4 (col 5), row 3 (cols 4–6), row 2 (cols 3–7), row 1 (cols 2–8) — but row 3/col 6, and row 2/cols 5–7, and row 1/cols 4–8 were already mined in Stage 1, so only the 7 new cells count:
$$V_2 = \underbrace{10.7}_{\text{row 4, col 5 (new)}} + \underbrace{(-1-1)}_{\text{row 3, cols 4-5 (new)}} + \underbrace{(-1-1)}_{\text{row 2, cols 3-4 (new)}} + \underbrace{(-1-1)}_{\text{row 1, cols 2-3 (new)}} = 10.7-2-2-2 = \boxed{+4.7}$$
Accepted: 7 new cells mined.
Stage 3 — check for remaining candidates. All three ore blocks ($+4$, $+10.7$, $+19.8$) are now mined; the algorithm halts. Ultimate pit value $= 0.8+4.7=\boxed{+5.5}$.
Quantity
Result
Stage 1 (apex row 5/col 8)
+0.8, 25 cells
Stage 2 (apex row 4/col 5, new cells only)
+4.7, 7 cells
Ultimate pit — grade-first order
+5.5, 32 cells — matches the Lerchs–Grossmann-equivalent optimum exactly
Sub-optimal counter-example (Q1.3.4)
Evaluating the small $+4$ block first (its own standalone cone is $4-3=+1.0$) strands both larger cones at negative marginal value ($-0.3$ and $-0.2$ respectively, once the shared waste is already “spent”) and the heuristic halts at only +1.0 total — a 4.5-unit shortfall versus the true optimum, from evaluation order alone.
