Truck–Shovel Cycle Time, Match Factor and Dispatch: 24-MMP-A5 Surface Mining Methods and Design, May 2016, Q9

Question 4: Truck–Shovel Cycle Time, Match Factor and Dispatch (20 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.

Given. Load time 3.0 min, haul (loaded, same-side) 12.0 min, dump 1.0 min, empty return (same-side) 8.0 min; the detailed dispatch network additionally offers two short cross-routes: Crusher→Shovel 2 (4.0 min) and Waste Dump→Shovel 1 (3.0 min).

Find. 4.1 theoretical closed-out cycle time and match factor; 4.3 trucks required, closed-out vs. dispatched; 4.4 the more efficient configuration; 4.5 truckloads per 8-hour shift.

The two short cross-routes (green) that make “dispatched” operation possible: a truck dumping ore at the Crusher can cross directly to Shovel 2 (4 min) instead of returning empty to Shovel 1 (8 min); a truck dumping waste at the Waste Dump can cross directly to Shovel 1 (3 min) instead of returning to Shovel 2 (8 min).

Approach. 4.1: sum the four closed-out cycle legs, then use the match-factor identity (number of trucks needed to keep one shovel continuously loaded) $= T_c/T_{\text{load}}$. 4.3: closed-out totals two independent single-shovel fleets; dispatched instead forms one continuous circulating loop through both shovels and both dump points, using the short cross-routes as the return legs, and the number of trucks needed to keep both shovels continuously fed is the loop time divided by the loading interval. 4.5: convert the dispatched loop's per-truck delivery rate to a per-shift total.

4.1 — closed-out cycle time and match factor. $$T_c = 3.0+12.0+1.0+8.0 = \boxed{24.0\ \text{min}}$$ Match factor $\text{MF}=\dfrac{N_{\text{trucks}}\times T_{\text{load}}}{N_{\text{shovels}}\times T_c}$; for a single shovel, $\text{MF}=1$ (trucks exactly matched to shovel capacity, no truck or shovel queuing) requires $$N_{\text{trucks/shovel}} = \frac{T_c}{T_{\text{load}}} = \frac{24.0}{3.0} = \boxed{8\ \text{trucks per shovel}}$$
4.3 — closed-out truck requirement. Two independent shovel/dump pairs, each needing 8 trucks at MF$=1$: $$N_{\text{closed-out}} = 2\times 8 = \boxed{16\ \text{trucks}}$$
4.3 — dispatched loop and truck requirement. Using the short cross-routes, a single truck can serve both shovels in one continuous loop: load ore at S1 (3) → haul to Crusher (12) → dump (1) → cross to S2 (4) → load waste (3) → haul to Waste Dump (12) → dump (1) → cross back to S1 (3). $$T_{\text{loop}} = 3+12+1+4+3+12+1+3 = \boxed{39.0\ \text{min}}$$ With $N$ trucks evenly spaced around this shared loop, the interval between successive trucks passing any fixed point (in particular, each shovel) is $T_{\text{loop}}/N$; setting this equal to the 3.0-min load time (so neither shovel ever waits) gives $$N_{\text{dispatched}} = \frac{T_{\text{loop}}}{T_{\text{load}}} = \frac{39.0}{3.0} = \boxed{13\ \text{trucks}}$$
4.4 — which is more efficient. Dispatched needs only 13 trucks (vs. 16 closed-out) to keep both shovels continuously loaded, because the two short cross-routes (4 and 3 min) replace the long same-side empty returns (8 min each) that closed-out trucks would otherwise run empty — a 19 % fleet reduction for the same production.
4.5 — truckloads per 8-hour shift (dispatched, the more efficient configuration). Each of the 13 trucks completes one full 39-min loop delivering exactly one ore load (to the crusher) and one waste load (to the dump): $$\text{loads/min} = \frac{13}{39} = 0.3333, \qquad \text{loads in an 8-h shift} = 0.3333\times(8\times60) = \boxed{160\ \text{truckloads}}$$ (160 to the crusher and 160 to the waste dump, since every loop delivers one of each.)

Quantity	Result
4.1 Closed-out cycle time	24.0 min
4.1 Trucks/shovel at MF = 1	8
4.3 Closed-out fleet	16 trucks
4.3 Dispatched loop time / fleet	39.0 min / 13 trucks
4.4 More efficient configuration	Dispatched (19% fewer trucks for equal production)
4.5 Truckloads per 8-h shift	160 to crusher, 160 to waste dump

Check: the 160-loads-per-shift figure assumes zero queuing, breakdown or shift-change loss and perfectly even truck spacing around the loop — a real operation typically achieves 80–90 % of this theoretical maximum once spotting variability, weather and maintenance are included, and would carry a small spare-truck allowance above the theoretical 13.

Concept & Theory

Topic. Truck–shovel match factor and the productivity gain from open (“dispatched”) versus dedicated (“closed-out”) truck routing.

Key relations. $T_c = T_{\text{load}}+T_{\text{haul}}+T_{\text{dump}}+T_{\text{return}}$; $\text{MF}=N_{\text{trucks}}T_{\text{load}}/(N_{\text{shovels}}T_c)$; loop-sharing trucks required $=T_{\text{loop}}/T_{\text{load}}$.

Why this works. A dedicated (closed-out) truck idles for its entire 8-minute empty-return leg every cycle; routing it instead through whichever shovel is geometrically closer after dumping converts that idle leg into productive travel toward the next load, which is exactly what the short cross-routes in this network make possible.

Common pitfall. Computing the dispatched truck requirement as if it were two independent MF$=1$ calculations (one per shovel) using each shovel's own shorter cross-route return time — this double-serves the fleet and understates the requirement; the loop must be treated as one shared cycle because the same trucks serve both shovels.

Source. Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design, Ch. 15 (truck–shovel systems, match factor and dispatch).

Question 9 of 11: Truck–Shovel Cycle Time, Match Factor and Dispatch

Question 4: Truck–Shovel Cycle Time, Match Factor and Dispatch (20 marks)