Question 9 of 13: Truck-Shovel Dispatch – Closed Out vs. Dispatched

Question 3: Truck-Shovel Dispatch – Closed Out vs. Dispatched (20 marks, optional)

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 (incl. positioning) 3.0 min; loaded haul 12.0 min (both ore and waste); backup+dump 1.0 min; empty return 8.0 min. Full routing set: Shovel 1 (ore)→Crusher 12 min, Crusher→Shovel 1 8 min; Shovel 2 (waste)→Waste Dump 12 min, Waste Dump→Shovel 2 8 min; Crusher→Shovel 2 4 min; Waste Dump→Shovel 1 3 min; load at Shovel 1 3 min, dump ore at Crusher 1 min; load waste at Shovel 2 3 min, dump at Waste Dump 1 min.

Find. The theoretical cycle time and match factor (3.1), the closed-out and dispatched truck requirements (3.4), which is more efficient (3.5), and the resulting loads delivered in an 8-hour shift (3.6).

Approach. A closed-out loop sizes independently off its own cycle time; a dispatched fleet instead shares the whole pool around ONE combined circuit built from the SHORTER cross-legs (Crusher→Shovel 2, Waste Dump→Shovel 1), so both are found from cycle-time÷load-time, just applied to a different loop.

1. 3.1 – Theoretical cycle time and match factor. $$T_c=3.0+12.0+1.0+8.0=\boxed{24.0\text{ min}}$$ The theoretical (matched) truck count for one shovel is $N=T_c/T_{load}=24/3=8$; substituting into $MF=N\,T_{load}/(N_{shovels}T_c)$ at that count gives $$MF=\dfrac{8\times3.0}{1\times24.0}=\boxed{1.0}$$ – the textbook definition of a theoretically matched fleet.

1. 3.4 – Closed-out truck requirement. Each dedicated loop uses the SAME per-loop cycle time as 3.1 (load+haul+dump+return, here 24 min for either shovel): trucks per shovel $=24/3=8$; with two independent loops, $$N_{closed}=8+8=\boxed{16\text{ trucks}}$$
2. Dispatched supercircuit. Using the cross-links, one shared circuit delivers ONE ore load and ONE waste load: Load@S1(3)+S1→Crusher(12)+DumpOre(1)+Crusher→S2(4)+LoadWaste@S2(3)+S2→WasteDump(12)+DumpWaste(1)+WasteDump→S1(3): $$T_{circuit}=3+12+1+4+3+12+1+3=\boxed{39\text{ min for 2 loads}}$$
3. 3.4 – Dispatched truck requirement. A truck must arrive at whichever shovel is due every $T_{load}=3$ min to keep BOTH shovels saturated: $$N_{dispatched}=T_{circuit}/T_{load}=39/3=\boxed{13\text{ trucks}}$$

3.5 – Which is more efficient. Dispatched needs only 13 trucks against 16 for closed out to keep both shovels at maximum production – a saving of $$\left(1-\dfrac{13}{16}\right)\times100=\boxed{18.75\%\text{ fewer trucks}}$$ so the dispatched configuration is more efficient: it exploits the shorter cross-legs (4 and 3 min) in place of a second full dedicated loop, sharing the whole fleet across both shovels instead of splitting it 8/8.

3.6 – Loads per 8-hour shift, dispatched configuration. $$\text{circuits per truck}=\left\lfloor\dfrac{8\times60}{39}\right\rfloor=\lfloor12.31\rfloor=12$$ Each circuit delivers one ore load and one waste load, so with 13 trucks: $$\text{total loads}=12\times13\times2/2=\boxed{156\text{ loads/shift (78 to the crusher, 78 to the dump)}}$$ This is a theoretical ceiling that assumes zero queuing delay, perfectly timed arrivals and no breakdowns, shift-change, blast or weather delays; a real 8-hour shift typically achieves 60–85% of the theoretical figure once those losses are included, so roughly 95–130 combined loads (not 156) is a more realistic expectation, though the theoretical value remains the correct basis for fleet sizing.