Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
09-MMP-B8, Mine Management & Systems Analysis — May 2013 sitting. 3-hour closed-book exam, answer all questions, 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); 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. Also: only Questions 1, 3, 4 and 5 exist anywhere in the 6-page exam booklet — the cover sheet instructs “ANSWER ALL 5 QUESTIONS FOR A TOTAL OF 100 MARKS” and each question is marked out of 20, but no Question 2 appears on any page between Question 1 (ending “2 of 6”) and Question 3 (starting on page 3). This is a genuine gap in the original exam booklet — all four questions that DO exist are answered in full below (80 of the stated 100 marks).
Given. Haul route (one-way, shovel → crusher): 300 m level, then a ramp climbing 175 m of elevation on a 10% grade ($175/0.10=1750$ m along the ramp), then 850 m level to the crusher; rolling resistance 6% on all surfaces; downhill speed limit 30 km/h. Fig. 5.1 gives histograms of shovel loading time and truck dumping time (bin counts read from the source figure):
Fig. 5.1 — time-study histogram bin counts (as read from the source charts)
Time bin (s)
Loading — occurrences
Dumping — occurrences
0–30
5
26
31–60
43
47
61–90
62
60
91–120
29
16
121–150
18
8
Fig. 5.2 gives EMPTY and LOADED truck performance charts (distance one-way, m, vs. time, min) for total-resistance curves of 0%, 4%, 6%, 8%, 10% and 15% (grade + rolling resistance); read directly for this problem's 6% (level) segments, with the 16% loaded-ramp figure obtained by linear extrapolation beyond the chart's 15% curve (see the check note).
Find. (a) the expected range of truck cycle times; (b) the optimum number of trucks to assign to a single shovel.
Approach. Take the loading and dumping times as the frequency-weighted mean of each histogram; read each haul segment's travel time from the appropriate (loaded/empty) performance chart at its own distance and total resistance, using the imposed 30 km/h speed limit for the downhill ramp descent (the chart has no negative-grade curve); sum load + haul + dump + return for the mean cycle time, and use the histograms' modal bin to bound the expected range; then apply the shovel-truck match-factor method to size the fleet.
Check: (1) each haul segment's travel time is read from the performance chart independently, i.e. assuming the truck accelerates from rest at the START of that segment, rather than carrying forward the speed reached at the end of the previous segment (the equivalent-distance splicing method) — a reasonable simplification at this course's level given the chart's coarse (200 m / 0.4 min grid) resolution. (2) The loaded uphill ramp's total resistance is $10\%\ \text{grade}+6\%\ \text{rolling}=16\%$, one point beyond the chart's steepest plotted curve (15%); the 16% travel-time line is obtained by linearly extrapolating the trend from the given 10% and 15% curves, which is a small, well-behaved extrapolation (curve slope is smoothly decreasing with resistance). (3) The empty return's ramp segment is governed by the posted 30 km/h downhill speed limit, not by the performance chart (which plots only positive-resistance, accelerating curves).
Fig. 5.3 — haul-route profile: 300 m level, 1750 m @ 10% ramp (175 m climb), 850 m level.
Mean loading and dumping times. Using the bin midpoints (15, 45.5, 75.5, 105.5, 135.5 s) weighted by occurrence count: $$\bar{t}_{load} = \frac{5(15)+43(45.5)+62(75.5)+29(105.5)+18(135.5)}{157} = 77.8\ \text{s} = 1.30\ \text{min}$$ $$\bar{t}_{dump} = \frac{26(15)+47(45.5)+60(75.5)+16(105.5)+8(135.5)}{157} = 62.6\ \text{s} = 1.04\ \text{min}$$
Loaded haul time (shovel → crusher). Reading each segment from the LOADED chart at its own resistance (300 m and 850 m level segments at 6% total resistance; the 1750 m ramp at the extrapolated 16%): $$t_{300,6\%}=0.57\ \text{min},\quad t_{1750,16\%}=11.31\ \text{min},\quad t_{850,6\%}=1.91\ \text{min}$$ $$\boxed{t_{haul,loaded} = 0.57+11.31+1.91 = 13.78\ \text{min}}$$
Empty return time (crusher → shovel). The two level segments are read from the EMPTY chart at 6% total resistance; the ramp descent is governed by the 30 km/h speed limit ($=500\ \text{m/min}$): $$t_{850,6\%}=1.05\ \text{min},\quad t_{1750,\text{limit}}=\frac{1750}{500}=3.50\ \text{min},\quad t_{300,6\%}=0.38\ \text{min}$$ $$\boxed{t_{return,empty} = 1.05+3.50+0.38 = 4.94\ \text{min}}$$
Mean cycle time and expected range. Total haul (loaded + empty) $=13.78+4.94=18.72\ \text{min}$; adding the mean load and dump times: $$\boxed{\bar{t}_{cycle} = 18.72+1.30+1.04 = 21.06\ \text{min}}$$ Using the histograms' shared modal bin (61–90 s) as the fastest/slowest representative load+dump combination instead of the mean gives the expected range: $$t_{cycle} = 18.72 + \tfrac{2\times61}{60}\ \text{to}\ 18.72+\tfrac{2\times90}{60} = \boxed{20.8\ \text{to}\ 21.7\ \text{minutes}}$$ — consistent with (bracketing) the mean-based estimate.
Optimum number of trucks (match-factor method). A shovel is kept continuously busy when $N\times t_{load} = t_{cycle}$: $$N = \frac{\bar{t}_{cycle}}{\bar{t}_{load}} = \frac{21.06}{1.30} = 16.2$$ Sixteen trucks give a match factor of $16\times1.30/21.06=0.985$ (the shovel is idle about 1.5% of the time, no truck queuing); seventeen trucks give $17\times1.30/21.06=1.05$ (shovel fully utilised, but trucks occasionally queue). $$\boxed{N_{optimum} = 16\ \text{trucks}\ (\text{match factor} \approx 0.99,\ \text{no truck queuing})}$$
