Question 12 of 13: Question 6 (15 marks, optional)

Question 6 (15 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.

6.1.1 – “Closed out” vs. “dispatched”. A closed-out circuit dedicates a fixed group of trucks permanently to one shovel-to-dump route (the truck always returns to the same shovel) – simple to plan but, as Question 6.2 shows numerically, wasteful of fleet capacity because it cannot share trucks between routes with spare capacity. A dispatched fleet is assigned dynamically, truck by truck, to whichever shovel/dump combination the dispatch system currently judges best (by BP, LP or DP logic, Question 1.6), sharing the whole fleet across every active shovel and route.

6.1.2 – Maximized production. The dispatcher continuously compares each shovel’s current queue against its digging rate and re-routes the next arriving empty truck to whichever active shovel is closest to running out of trucks (i.e. would otherwise sit idle), so total tonnage moved is maximised by keeping every shovel loading continuously rather than by optimising any single truck’s cycle.

6.1.3 – Minimized equipment / optimal mill head grade / stripping ratio. Because a shared, dispatched fleet is not tied to any one route, the SAME truck count can service more shovels than a closed-out fleet would need (Question 6.2.2 quantifies this directly) – minimizing the number of trucks purchased for a given production target. For head grade, the dispatcher can weight assignments toward higher-grade shovels/faces when the mill needs a higher feed grade in a given shift (or blend two faces in a set ratio by proportioning truck assignments), and for stripping ratio it can throttle waste-truck assignments up or down relative to ore-truck assignments to keep the realised ore:waste ratio on the planned schedule even when individual faces are running ahead or behind.

6.1.4 – Maintaining the stripping ratio. Same dispatch logic as 6.1.3: the system tracks cumulative ore and waste tonnes moved against the period target ratio and biases the next several truck assignments toward whichever material (ore or waste) is currently running behind schedule, correcting drift continuously rather than waiting for a shift-end reconciliation.

6.1.5 – Computerised database. The same GPS/vehicle-tracking backbone that drives real-time dispatch also logs every load (tonnes, grade, source, destination, time), every truck’s engine/ hour-meter and fault codes, and consumable usage (tyres, fuel, filters) – feeding production reporting, condition-based maintenance scheduling, and a running fleet-utilisation record that directly informs the “buy vs. keep” replacement/purchase decisions referenced in 1.1 and 6.2.3.

Given. 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. Theoretical trucks needed closed-out vs. dispatched, the more efficient configuration, and the resulting loads per 8-hour shift.

Approach. For a dedicated (closed-out) route, the number of trucks needed to keep one shovel saturated is the full cycle time divided by the load time (a truck departs every load-time interval, so that many trucks are in the cycle at once); for the dispatched routing, the fleet instead runs one shared loop that uses the shorter cross-links between the crusher and Shovel 2, and the Waste Dump and Shovel 1.

1. 6.2.2 – Theoretical trucks, closed out. Each dedicated loop: load (3) + haul (12) + dump (1) + empty return (8): $$T_c = 3+12+1+8=\boxed{24\text{ min}}$$ Trucks to keep one shovel saturated $=T_c/\text{load time}=24/3=8$ per shovel; two shovels (closed out, no sharing): $$N_{closed}=8+8=\boxed{16\text{ trucks}}$$
2. For the dispatched configuration the fleet runs one shared circuit using the cross-links: Shovel 1→Crusher (12) → dump ore (1) → Crusher→Shovel 2 (4) → load waste (3) → Shovel 2→Waste Dump (12) → dump waste (1) → Waste Dump→Shovel 1 (3): $$T_{circuit}=12+1+4+3+12+1+3=\boxed{36\text{ min}}$$ $$N_{dispatched}=T_{circuit}/\text{load time}=36/3=\boxed{12\text{ trucks}}$$
3. 6.2.3 – Which is more efficient. Dispatched needs only 12 trucks against 16 for closed out to achieve the same maximum production – a saving of $$\left(1-\dfrac{12}{16}\right)\times100=\boxed{25\%}\text{ fewer trucks}$$ so the dispatched configuration is more efficient, because it uses the shorter cross-links (4 and 3 min) in place of the longer dedicated empty-return legs (8 min each) and shares the whole fleet across both shovels instead of splitting it 8/8.
4. 6.2.4 – Loads per 8-hour shift. $$\text{circuits per truck}=\left\lfloor\dfrac{8\times60}{36}\right\rfloor=\lfloor13.33\rfloor=13$$ $$\text{total loads}=13\times12=\boxed{156\text{ loads/shift}}$$ This is a realistic order of magnitude for an 8-hour shift once allowance is made for shift-change, delays and queuing (the theoretical figure ignores those, so the practical number achieved would be somewhat below 156, but the same order of magnitude).

6.2.5 – Shovel production vs. number of trucks. Production rises roughly linearly with truck count while the shovel is starved (few trucks, shovel frequently waits idle for the next arrival), then the curve bends over and FLATTENS once the truck count reaches the “match” fleet size (Question 6.3.2) – beyond that point the shovel is already saturated, so adding more trucks produces no further gain in tonnes/shift, only growing truck queuing/idle time (Question 6.2.6).

6.2.6 – Over-trucking (double the theoretical fleet). The shovel is already saturated at the theoretical truck count, so doubling the fleet adds essentially zero additional production – the surplus trucks simply queue at the shovel (rising truck-side idle/wait time, congestion on the haul road, and unnecessarily higher owning & operating cost for no tonnage benefit); it is a pure loss of capital efficiency, not a production gain.

6.3.1 – Spotting, double back-up, drive-by. Spotting is manoeuvring an empty truck into the single loading position beside the shovel/loader, requiring the shovel to pause digging while the truck backs in and positions; double back-up uses two truck spots on either side of the shovel so a second truck can back in and be ready the instant the first pulls away loaded, minimising shovel idle time between trucks at the cost of a wider working pad; drive-by (drive-through) loading positions trucks so the next truck can pull directly into the loading position without reversing at all (a through, one-way traffic pattern), the fastest spotting method but requiring a blast layout/muck pile shaped to allow a drive-through pattern rather than a single dead-end face.

6.3.2 – Match factor. Match factor (MF) is the ratio of truck arrival capacity to shovel loading capacity for a given fleet: $$MF=\dfrac{N_{trucks}\times T_{load}}{N_{shovels}\times T_{cycle}}$$ $MF<1$ means the shovel is under-trucked (waits for trucks); $MF=1$ means the fleet exactly matches the shovel’s loading rate (the theoretical, most cost-efficient point); $MF>1$ means trucks queue and wait for the shovel (Question 6.2.6’s over-trucked case).

1. 6.3.3 – Theoretical cycle time and match factor. Load 3.0 + haul 12.0 + backup/dump 1.0 + empty return 8.0 min: $$T_c=3.0+12.0+1.0+8.0=\boxed{24.0\text{ min}}$$ The “theoretical” truck count for one shovel is by definition $N_{trucks}=T_c/T_{load}=24/3=8$ (Question 6.2.2), so substituting back: $$MF=\dfrac{8\times3.0}{1\times24.0}=\boxed{1.0}$$ A match factor of exactly 1.0 is the textbook definition of a “theoretically matched” fleet – neither the shovel nor the trucks wait on each other, on average.

6.3.4 – Computer hardware for maximum trucking efficiency.

• GPS/vehicle tracking units on every truck and shovel – real-time position for dispatch and collision avoidance.
• On-board computer/display in each truck cab – receives dispatch assignments, shows routing/queue instructions to the operator.
• Payload weighing system (suspension or strut-pressure based) on each truck – measures actual load for production reporting and to flag under/over-loading.
• Radio/wireless network (or private LTE) – carries dispatch commands and telemetry between the fleet and the control room.
• Central dispatch server running the BP/LP/DP optimisation (Question 1.6) – ingests all truck/shovel state and issues assignments.
• Condition-monitoring sensors (engine, hydraulics, tyre pressure/temperature) feeding the maintenance database referenced in 6.1.5.