NivaarExam PrepOfficial exam papers ↗

24-MMP-B8 Rock Slope Engineering · Undated paper

Question 5 of 5: Open Pit Limits

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

09-MMP-B8, Mine Management & Systems Analysis — May 2019 sitting. 3-hour closed-book exam, answer all 5 questions for a total of 100 marks, Appendix A (discounted cash-flow factor tables) attached.

Reference texts. Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design (pit optimization, truck/shovel matching, mine scheduling); Hartman & Mutmansky (eds.), SME Mining Engineering Handbook (mine life-cycle, project economics, haulage systems); Blank & Tarquin, Engineering Economy (DCF/NPV/IRR/payback); Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK Guide) (Critical Path Method).

Check: every page of the examination is headed “09-MMP-B8 Mine Management & Systems Analysis”. The content below solves the paper as printed.
Check: the data used below are as printed in the exam. (1) Table 1's LoM totals reconcile exactly against their own row sums (LoM ore 16,497 kt, contained 485.0 koz, recovered 397.7 koz, waste 86,468 kt). (2) The Mining unit cost in Table 3 is $11.75/t. (3) Question 2(a) asks for the gross and net value of ore per tonne. (4) The rolling resistance for Question 3's haul route is 6%. (5) Question 4's task table includes the task “Expand u/g diesel powered equipment fleet”. (6) Question 5's 2-D block model is a 5-row×8-column grid, the net processed mineral value is $2,800/tonne, and a 1.5% cutoff grade is stated.

Question 5: Open Pit Limits (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.

Given.

Block model parameters
QuantityValue
Block volume1,000 m³
Density (ore and waste)2.5 t/m³ (tonnage/block = 2,500 t)
Cutoff grade1.5%
Combined mining + milling + overhead cost$50/t
Net processed mineral value$2,800/t contained metal
Block model size5 rows (depth) × 8 columns
Geological Block Model (% grade)0%0%2%3%4%3%1%1%0%0%1%4%4%0%1%0%0%0%1%4%2%3%0%0%1%1%1%2%3%1%0%0%0%1%1%2%2%1%0%0%
Fig. 5.1 — Geological block model (top row = surface bench; darker green = higher grade).

Find. The 2-D pit outline that maximizes total profit, obeying a 45° maximum wall slope (adjacent columns' pit floors may differ in depth by at most one block).

Approach. Convert every grade block to a dollar value, then find the pit outline (one mined depth per column, changing by at most one block between adjacent columns, anchored to zero depth at both edges of the model) that maximizes total profit — the 2-D equivalent of Lerchs-Grossman graph closure, solved here by an envelope dynamic program over cumulative column profit.

Check: block economic value equals tonnage times (grade fraction times $2,800 minus $50), i.e. the $2,800/tonne figure is a metal price applied to the tonnes of contained metal in each block (tonnage × grade fraction), not a flat dollar-per-tonne-of-rock revenue — this reading is confirmed by the resulting block values landing on clean round numbers (−$125,000, −$55,000, $15,000, $85,000, $155,000 for grades 0–4), which would not happen under an alternative unit interpretation. Waste-grade blocks (0% and 1%, below the 1.5% cutoff) are assigned the SAME cost-minus-revenue formula rather than a separate waste-only mining cost, the standard simplified convention for a 2-D Lerchs-Grossman teaching exercise absent a stated separate waste mining rate — a 0%-grade block naturally values at −$125,000 (pure cost, zero revenue) under this formula, so the cutoff grade is used to LABEL the optimal pit's ore/waste tonnage after solving, not to change the optimization itself.
  1. Block economic value. Each block is $1{,}000\ \text{m}^3\times2.5\ \text{t/m}^3=2{,}500$ t. $$v(g) = 2{,}500\times\left[\frac{g}{100}\times2800 - 50\right] = 2{,}500(28g - 50)\ \text{\$}$$ $$\boxed{v(0)=-\$125{,}000,\ v(1)=-\$55{,}000,\ v(2)=\$15{,}000,\ v(3)=\$85{,}000,\ v(4)=\$155{,}000}$$
  2. Cumulative column profit. For each of the 8 columns, sum block values from the surface (row 1) down to depth $d$ (0–5 blocks); e.g. column 5 (grades 4,4,2,3,2 top-to-bottom) gives $P(1)=155\text{k}$, $P(2)=310\text{k}$, $P(3)=325\text{k}$, $P(4)=410\text{k}$, $P(5)=425\text{k}$ (all $\times\$1000$).
  3. Envelope DP for the maximum-profit pit outline. Choosing a mined depth $d_c\in\{0,\ldots,5\}$ per column $c$, subject to $|d_c-d_{c-1}|\le1$ (45° slope) and $d_0,d_7\le1$ at the model's outer edges, maximize $\sum_c P_c(d_c)$ by dynamic programming across columns (analogous to the floating-cone/envelope DP already used for this discipline's other pit-limit questions): $$\boxed{d = [0,\,0,\,1,\,2,\,2,\,1,\,0,\,0]\ \text{(columns 1--8, blocks mined from surface)}}$$
  4. Optimal pit profit. Summing $P_c(d_c)$ over all 8 columns: $$\boxed{\text{Total pit profit} = 15{,}000+85{,}000+155{,}000+155{,}000+155{,}000+85{,}000 = \$650{,}000}$$
  5. Ore/waste tonnage within the optimal pit. The pit mines 6 blocks (columns 3–6, depths 1/2/2/1), all at grade $\ge2\%$ (above the 1.5% cutoff), so every mined block is ore: $$\boxed{\text{Ore} = 6\times2{,}500 = 15{,}000\ \text{t},\quad \text{Waste} = 0\ \text{t},\quad \text{Overall strip ratio} = 0}$$ The optimizer naturally excludes every column whose best-achievable cumulative profit is negative (columns 1, 2, 7, 8 stay at $d=0$) and stops each ore column exactly where the next block's marginal value would turn the cumulative sum down (e.g. column 5 stops at $d=2$ even though extending it further would itself still be profitable in isolation ($P(3)=325\text{k}$, $P(4)=410\text{k}$) — it is the slope constraint linking column 5 to column 6, whose own best cumulative profit is only $P(1)=85\text{k}$ (its $P(2)=-40\text{k}$ turns negative), that holds column 5 back to $d=2$, not any weakness in column 5's own economics).
Profit Block Model ($, x1000) with optimal pit outline-125k-125k15k85k155k85k-55k-55k-125k-125k-55k155k155k-125k-125k-125k-125k-125k-55k155k15k85k-125k-125k-55k-55k-55k15k85k-55k-125k-125k-125k-55k-55k15k15k-55k-125k-125k
Fig. 5.2 — Profit block model ($, ×1000) with the optimal pit outline (red) from the envelope DP: green = mined ore, pink = left in place (negative-value waste).
Question 5 — final results
ItemResult
Optimal pit floor depth per column (1–8)0, 0, 1, 2, 2, 1, 0, 0
Total pit profit$650,000
Ore mined / waste mined15,000 t / 0 t
Blocks mined6 (columns 3–6)
Back to the paper →