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.
Segment ii — ramp, 10% grade, 125 m elevation change
1,250 m (= 125/0.10)
Segment iii — level, to crusher/mill
1,950 m
Rolling resistance, all surfaces
6%
Posted downhill speed limit
30 km/h
Loading time histogram (Fig. 3.1), counts (0–30/31–60/61–90/91–120/121–150 s)
5/42/62/30/20, n=159
Dumping time histogram (Fig. 3.1), counts (same bins)
25/47/60/17/8, n=157
Find. (a) the expected cycle time and its variability range; (b) the optimum number of trucks per shovel.
Approach. Read a representative sustained speed off Fig. 3.2 for each segment's total resistance (rolling resistance ± grade), sum the loaded and empty travel times; add the loading and dumping times, whose mean and spread are read from the Fig. 3.1 histograms; convert the resulting cycle-time distribution into a truck count that keeps the shovel continuously fed.
Fig. 3.3 — Haul route profile and total resistance (TR = rolling resistance ± grade) used to read Fig. 3.2.
Check: Fig. 3.2's total-resistance curves are straight lines radiating from a common near-origin point for TR = 0/4/6/8/10/15%, i.e. a constant sustained speed per curve. the same representative sustained speeds are reused here for consistency: loaded 30 km/h at TR=6%, 12 km/h at TR=15% (the chart's highest plotted curve, used to cap the 16% climb — slightly overestimates speed, so cycle time is slightly underestimated); empty 40 km/h at TR=6%. Segment ii empty descends at TR = 6%−10% = −4% (favourable/negative, off the bottom of the chart), so it is governed instead by the posted 30 km/h downhill speed limit, not an extrapolated curve. Each of the three haul segments is treated as accelerating from rest independently (no velocity splicing between segments).
Loaded travel time (shovel → crusher). Segments i and iii run level at TR=6% (30 km/h); segment ii climbs at TR=16%, capped at the 15% curve (12 km/h): $$t_i=\frac{0.100}{30}\times60=0.20\ \text{min},\quad t_{ii}=\frac{1.250}{12}\times60=6.25\ \text{min},\quad t_{iii}=\frac{1.950}{30}\times60=3.90\ \text{min}$$ $$\boxed{t_{loaded}=0.20+6.25+3.90=10.35\ \text{min}}$$
Empty travel time (crusher → shovel). Segments iii and i run level at TR=6% (empty, 40 km/h); segment ii descends and is governed by the 30 km/h posted limit: $$t_{iii}=\frac{1.950}{40}\times60=2.925\ \text{min},\quad t_{ii}=\frac{1.250}{30}\times60=2.50\ \text{min},\quad t_i=\frac{0.100}{40}\times60=0.15\ \text{min}$$ $$\boxed{t_{empty}=2.925+2.50+0.15=5.575\ \text{min},\qquad t_{travel}=t_{loaded}+t_{empty}=15.925\ \text{min}}$$
Loading and dumping statistics from Fig. 3.1. Treating each histogram bin's midpoint as its representative value, the weighted mean and standard deviation are $$\mu_{load}=\frac{\sum n_ix_i}{\sum n_i}=78.9\ \text{s},\ \sigma_{load}=31.1\ \text{s}\qquad \mu_{dump}=63.2\ \text{s},\ \sigma_{dump}=31.3\ \text{s}$$ mean load+dump time $=78.9+63.2=142.1\ \text{s}=2.368\ \text{min}$, combined standard deviation (independent draws) $\sqrt{31.1^2+31.3^2}=44.1\ \text{s}=0.735\ \text{min}$.
Part (a) — expected cycle time and its range. Travel time is fixed by the road profile; load+dump time is the variable component: $$\bar t_{cycle}=t_{travel}+\mu_{load+dump}=15.925+2.368=18.29\ \text{min}$$ $$\boxed{\bar t_{cycle}\approx18.3\ \text{min},\quad \text{expected range (}\pm1\sigma\text{)}\approx17.6\text{--}19.0\ \text{min}}$$
Part (b) — optimum number of trucks per shovel. A shovel is kept continuously fed when the fleet completes one truck-load every $t_{load}$ on average, i.e. the match number is cycle time divided by loading time: $$N=\frac{\bar t_{cycle}}{\mu_{load}}=\frac{18.29\times60}{78.9}=13.9$$ $$\boxed{N\approx14\ \text{trucks (rounded up, to avoid starving the shovel)}}$$ Rounding down to 13 leaves the shovel idle a small fraction of the time each cycle (the more economic choice, fewer trucks); rounding up to 14 fully saturates the shovel at the cost of one extra truck's capital and operating cost, with correspondingly more truck queuing at the shovel. Given the long (3,300 m one-way, 10%-grade) haul already drives a double-digit fleet, 14 is recommended to guarantee the shovel — the more capital-intensive, harder-to-substitute asset — is never starved; 13 is a defensible lower-cost alternative if a small amount of shovel idle time is acceptable.
