Question 8 of 13: Question 2 (15 marks, optional)

Question 2 (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.

Given. Ore and waste tonnage by 15 m bench elevation and phase (tables below); ore capped at 7 Mt/yr starting year 1; waste capped at 6 Mt in year −1, 12 Mt/yr from year 0 onward; Phase 1 ore/waste must be completed before Phase 2 of the same material starts; on any bench, waste must not be mined ahead of the ore one bench below it.

Find. The year-by-year, bench-by-bench NW-corner ore and waste schedules, the shape of the head-grade curve, and how the resulting truck-hour profile is converted into a truck fleet size.

Approach. The North-West-Corner method fills demand “cells” (years) from supply “rows” (benches, taken top-down, Phase 1 before Phase 2) strictly in order, saturating each year’s tonnage cap before moving to the next year – mechanically identical to the transportation-problem NW-corner rule, just applied to a mine schedule instead of a shipping table.

1. 2.1.1 – Ore schedule. Working down the bench list (1970→1880) for Phase 1 (24 Mt total), then Phase 2 (26 Mt total), and filling each year to its 7 Mt/yr cap before advancing: $$\text{Year total} = \min(7,\ \text{remaining bench supply},\ \text{remaining year capacity})$$ gives the schedule in the results table below. Phase 1 completes partway through year 4 (21 Mt in years 1–3, 3 Mt more in year 4); Phase 2 fills the rest of year 4 (4 Mt) and continues through year 8, where only 1 Mt remains – matching the source figure’s own year ranges (Phase 1: years −1…4; Phase 2: years 4…8). Total mined $=24+26=\boxed{50\text{ Mt over 8 years}}$ (years −1, 0 pre-production; 1–8 at 7 Mt/yr except the partial final year).
2. 2.2.1 – Waste schedule. The same NW-corner sweep is applied to the waste tonnages, but the year −1 cap is fixed at 6 Mt and year 0 onward at 12 Mt/yr (both given directly, not derived), and Phase 1 waste (43 Mt) completes before Phase 2 waste (41 Mt) begins – giving the schedule in the results table, with Phase 1 finishing 1 Mt into year 3 and Phase 2 finishing with a partial 6 Mt in year 6. Total waste $=43+41=\boxed{84\text{ Mt}}$. The “waste never ahead of the ore one bench below” rule is automatically satisfied by this top-down bench order, since the schedule always exposes a bench’s waste before or in the same period as the ore bench immediately beneath it becomes active.

2.1.2 – Head-grade curve shape. A porphyry-copper/epithermal-gold deposit typically develops a near-surface supergene-enriched blanket (secondary copper sulphides/oxides concentrated by decades of surface leaching and downward re-precipitation) sitting above lower-grade, more uniform primary (hypogene) ore at depth. Mining the upper benches first therefore usually shows head grade RISING through the first 1–3 years of production as the schedule works down into the richest part of the enrichment blanket, peaking roughly a third of the way through Phase 1, then DECLINING steadily for the remainder of Phase 1 as mining passes below the blanket into primary ore, and declining further (a visible step down) at the Phase 1→Phase 2 transition if Phase 2’s primary ore carries a lower average grade than Phase 1’s – a classic asymmetric curve (fast rise, slow tail) rather than a simple straight-line decline.

2.2.2 – Truck-hours curve. Truck-hours in any given year scale with total material moved (ore+waste) in that year, not ore alone – since ore is fixed at 7 Mt/yr from year 1 but waste varies with the schedule (6, then 12 Mt/yr, occasionally less in a partial final year), the truck-hour curve tracks the (ore+waste) total: it starts low in years −1/0 (waste-only, 6 and 12 Mt), jumps to its highest, sustained plateau through years 1–7 (7+12 = 19 Mt/yr total, close to the source figure’s own stated 19 Mt maximum-year total), and drops sharply in the final partial year (only the 1 Mt of ore and 6 Mt of waste remaining).

2.2.3 – Converting truck-hours to a fleet size, and minimizing purchases. The number of trucks required in any period is the peak-year truck-hour demand divided by the productive hours available per truck in that period: $N_{trucks} = \dfrac{\text{truck-hours required per year}}{\text{hours available per truck per year (after allowing for scheduled maintenance, weather and shift-change downtime)}}$, rounded up to a whole number of trucks. Because the truck-hour curve in 2.2.2 has a single sustained plateau rather than a sequence of ever-increasing peaks, the fleet only needs to be SIZED to that plateau – the number of trucks purchased is minimized by (i) buying to the plateau demand rather than to the highest instantaneous peak, absorbing short peaks with overtime/rented trucks instead of owned fleet; (ii) staging purchases to match the ramp-up in years −1/0/1 rather than buying the full fleet up front; and (iii) reviewing the fleet size at each phase transition (here, the Phase 1/2 boundary) since a change in haul distance or grade can shift the truck-hour requirement even at constant tonnage.