Question 9 of 27

Question 2.1 optimizing mining truck operations (7 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.

2.1.1 — truck haulage share of operating cost. Truck haulage (fuel, tyres, operating labour, maintenance) is typically the single largest operating-cost line in a conventional truck-shovel open pit, commonly ≈40–50% of total mine operating cost — substantially more than drilling+blasting or loading, which is why match-factor and dispatch optimisation (the rest of this question) has such large leverage on overall mine economics.

2.1.2 — five likely truck delays. (1) Queue/wait time at the shovel (spotting delay behind another truck); (2) queue/wait at the crusher or dump; (3) mechanical breakdown/unscheduled maintenance delay; (4) blast-clearance delay (haul road closed during and after a blast); (5) operational delays — shift change, crew breaks, weather (dust suppression/visibility), or road maintenance/grading blocking a haul segment.

Given (2.1.3–2.1.5). Shovel loading time T _l=6 min; truck full cycle time T _c=45 min (one shovel).

2.1.3 — theoretical trucks. $$N_t = \dfrac{T_c}{T_l} = \dfrac{45}{6}$$ $$\boxed{N_t = 7.5 \text{ trucks}}$$ (in practice rounded to 7 or 8 depending on whether the operator prefers a slight shovel-idle or truck-idle bias).

2.1.4 — match factor definition. Match factor is the ratio of ARRIVING truck loading-capacity to AVAILABLE shovel loading-capacity over the same time window: $$MF = \dfrac{N_t \cdot T_l}{N_s \cdot T_c}$$ where N _t=number of trucks, N _s=number of shovels, T _l=loading time per truck, T _c=truck cycle time. MF=1 means the fleet exactly matches shovel capacity (theoretical trucks, as in 2.1.3); MF<1 means trucks are the constraint (shovel waits); MF>1 means the shovel is the constraint (trucks queue).

1. MF with 5 trucks. $$MF = \dfrac{5 \times 6}{1 \times 45} = \dfrac{30}{45}$$ $$\boxed{MF = 0.667}$$ — trucks are the constraint; the shovel will sit idle roughly a third of the time waiting for a truck.
2. MF with the theoretical 7.5 trucks. $$MF = \dfrac{7.5 \times 6}{1 \times 45} = \dfrac{45}{45}$$ $$\boxed{MF = 1.0}$$ — by construction, the theoretical truck count always gives MF=1 exactly (that is the definition of "theoretical").

2.1.6 — MF above 1. Yes, MF can exceed 1 (more truck capacity arriving than the shovel can absorb); the advantage is that the SHOVEL never waits — shovel utilization is pinned at 100%, maximising the value of the more expensive, less-mobile asset (a shovel is normally far more capital-intensive per unit than one truck), at the cost of truck queueing/idle time. Because MF>1 buys shovel utilization by spending truck utilization, most operations deliberately run a modest MF surplus (1.0–1.2) as a buffer against breakdown/queueing variability rather than targeting exactly 1.0.

2.1.7 — multi-shovel, multi-truck MF. With several dissimilar shovels and trucks, the SAME ratio form is used but with FLEET-WEIGHTED averages: $$MF = \dfrac{N_t \cdot \overline{T_l}}{N_s \cdot \overline{T_c}}$$ using the fleet-weighted-average load time and cycle time (each individual truck/shovel type’s time weighted by its share of fleet hours). For N _t=40, N _s=5, weighted average T _c=45 min, weighted average T _l=4 min: $$MF = \dfrac{40 \times 4}{5 \times 45} = \dfrac{160}{225}$$ $$\boxed{MF = 0.711}$$ — below 1, so this weighted fleet is slightly TRUCK-constrained; a few more trucks (or a modest reduction in average cycle time via dispatch) would bring it to balance.