Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
09-MMP-B8, Mine Management & Systems Analysis — May 2018 sitting. 3-hour closed-book exam, answer all 4 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/PVR/payback); Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK Guide) (Critical Path Method).
Check: the exam booklet is headed “09-MMP-B8 Mine Management & Systems Analysis”, not Rock Slope Engineering. The content below solves the paper as printed (mine-life/DCF economics, shovel-truck fleet analysis, CPM project scheduling, 2-D pit-limit design), not rock-slope-stability content.
Segment ii — ramp, 10% grade, 175 m elevation change
1,750 m (= 175/0.10)
Segment iii — level, to crusher
1,850 m
Rolling resistance, all surfaces
6%
Posted downhill speed limit
30 km/h
Loading time histogram (Fig. 2.1)
bins 0–150 s, counts 6/42/62/30/19, n=159
Dumping time histogram (Fig. 2.1)
bins 0–150 s, counts 26/47/60/16/8, n=157
Fig. 2.2 LOADED, sustained speed read off the chart
23.7 km/h at TR 6%; 13.2 km/h at TR 10%; 9.5 km/h at TR 15%
Fig. 2.2 EMPTY, sustained speed read off the chart
53.0 km/h at TR 6% (about 64 km/h at TR 0%)
Find. (a) the expected cycle time and its variability range; (b) the optimum number of trucks per shovel.
Approach. Read each segment's travel time off Fig. 2.2 on the line for its total resistance (rolling resistance ± grade), treating each segment as starting from rest — every chart line begins at a small common offset near the origin that represents the acceleration allowance — and sum the loaded and empty legs. Add the loading and dumping times, whose mean and spread come from the Fig. 2.1 histograms, then convert the cycle time into the truck count that keeps the shovel continuously fed.
Fig. 2.3 — Haul route profile and total resistance (TR = rolling resistance ± grade) used to read Fig. 2.2.
Every line is straight after a small common start offset, i.e. a constant sustained speed $v$ plus a fixed start allowance $t_0$: loaded 394.7 m/min (23.7 km/h) at TR 6% with $t_0=0.150$ min, 13.2 km/h at TR 10%, 9.5 km/h at TR 15% ($t_0=0.086$ min); empty 884.4 m/min (53.0 km/h) at TR 6% with $t_0=0.220$ min, and about 64 km/h at TR 0%. The loaded ramp runs at TR = 10% + 6% = 16%, one point past the chart's last (15%) line, so its speed is extrapolated: 146 m/min linearly in TR and 148 m/min at constant power ($v\cdot TR$ constant); their mean, 146.9 m/min (8.8 km/h), is used. Running the ramp on the 15% line itself would give 11.2 min, about 0.8 min shorter. The empty descent runs at TR = −10% + 6% = −4% (favourable), where the truck could exceed even the 0% line's 64 km/h, so the posted 30 km/h limit governs. Chart readings are good to roughly ±3%.
Loaded travel time (shovel → crusher). On a chart line the time to cover distance $d$ from rest is $t=t_0+d/v$. Segments i and iii run on the TR 6% line, segment ii at TR 16% (extrapolated speed, start offset of the 15% line): $$t_i = 0.150+\frac{400}{394.7}=1.16\ \text{min},\quad t_{ii}=0.086+\frac{1750}{146.9}=12.00\ \text{min},\quad t_{iii}=0.150+\frac{1850}{394.7}=4.84\ \text{min}$$ $$\boxed{t_{loaded} = 1.16+12.00+4.84 = 18.00\ \text{min}}$$
Empty travel time (crusher → shovel). Segments iii and i run empty on the TR 6% line; segment ii descends at the posted 30 km/h (500 m/min): $$t_{iii}=0.220+\frac{1850}{884.4}=2.31\ \text{min},\quad t_{ii}=\frac{1750}{500}=3.50\ \text{min},\quad t_i=0.220+\frac{400}{884.4}=0.67\ \text{min}$$ $$\boxed{t_{empty} = 2.31+3.50+0.67 = 6.48\ \text{min}, \qquad t_{travel}=t_{loaded}+t_{empty}=24.48\ \text{min}}$$
Loading and dumping statistics from Fig. 2.1. Treating each histogram bin's midpoint (15, 45, 75, 105, 135 s) as its representative value, the weighted mean and standard deviation are $$\mu_{load}=\frac{\sum n_i x_i}{\sum n_i}=\frac{12{,}345}{159}=77.64\ \text{s},\ \ \sigma_{load}=31.09\ \text{s} \qquad \mu_{dump}=\frac{9{,}765}{157}=62.20\ \text{s},\ \ \sigma_{dump}=31.25\ \text{s}$$ so the mean load+dump time is $77.64+62.20=139.84\ \text{s}=2.33\ \text{min}$, with combined standard deviation (independent draws) $\sqrt{31.09^2+31.25^2}=44.08\ \text{s}=0.73\ \text{min}$.
Expected cycle time and its range. Travel time is fixed by the route and the chart; load+dump time is the variable component: $$\bar{t}_{cycle}=t_{travel}+\mu_{load+dump}=24.48+2.33=26.81\ \text{min}$$ $$\boxed{\bar{t}_{cycle}\approx 26.8\ \text{min}, \quad \text{expected range (}\pm1\sigma\text{)}\approx 26.1\text{--}27.6\ \text{min}}$$ The absolute limits implied by the histograms (load and dump each anywhere from 0 to 150 s) are 24.5–29.5 min.
Optimum number of trucks per shovel. The shovel is kept continuously fed when the fleet returns one truck every $\mu_{load}$, so the match number is the cycle time divided by the loading time: $$N = \frac{\bar{t}_{cycle}}{\mu_{load}} = \frac{26.81\times 60}{77.64} = 20.72$$ With 20 trucks the fleet delivers $20\times60/26.81=44.8$ loads/h against the shovel's $3600/77.64=46.4$ loads/h, so the shovel stands idle 3.5% of the time; with 21 trucks it delivers 47.0 loads/h, the shovel is never starved and the trucks are 98.7% utilized (about 1.3% queueing). $$\boxed{N = 21\ \text{trucks (rounded up)}}$$ Rounding up is preferred because the shovel is the more expensive, production-setting unit, and the load/dump variability in Fig. 2.1 causes bunching that would starve a fleet sized exactly at the match number.
