Question 13 of 13: Open-Pit Capital Cost Estimation – Mular & Poulin Method

Question 7: Open-Pit Capital Cost Estimation – Mular & Poulin Method (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. Soil overburden 23,000 mt/day × 60 days; rock overburden 11,000 mt/day × 300 days; total ore+waste $T=23{,}000$ mt/day; mill (ore) capacity $=9{,}000$ mt/day; all formulae in 1998 dollars.

Find. The stripping ratio, every intermediate cost component, the total fixed capital cost, and its 2014-escalated value.

Approach. Each cost centre is computed directly from its own power-law formula in the given SEQUENCE (shovel size before truck size, before fleet counts, before fleet cost, since each formula's input depends on the previous result), then summed and marked up by the three percentage overhead categories before escalating.

1. 7.1 – Stripping ratio. Waste (ore+waste total less the ore the mill actually takes) over ore: $$SR=\dfrac{T-\text{mill}}{\text{mill}}=\dfrac{23{,}000-9{,}000}{9{,}000}=\boxed{1.56:1}$$ Definition: tonnes of waste removed per tonne of ore delivered to the mill – the same ratio whose upper bound is set by the pit-inclusion rules of Question 1.4.2.
2. 7.2 – Site preparation. $$C_{12}=11410\times23{,}000^{0.5}=11410\times151.66=\boxed{\$1.73\text{M}}$$
3. 7.3 – Pre-production stripping. Total soil $T_S=23{,}000\times60=1{,}380{,}000$ mt; total rock $T_R=11{,}000\times300=3{,}300{,}000$ mt: $$C_{21}=1826\times1{,}380{,}000^{0.5}=\boxed{\$2.15\text{M}}\qquad C_{22}=19395\times3{,}300{,}000^{0.5}=\boxed{\$35.23\text{M}}$$
4. 7.4 – Shovel and truck size. $$S_{raw}=0.1034\times23{,}000^{0.4}=5.74\ \text{m}^3\ \rightarrow\ \text{round up}\ \boxed{S=6.1\ \text{m}^3}$$ $$t_{raw}=9.75\times6.1^{1.1}=71.3\ \text{mt}\ \rightarrow\ \text{round up}\ \boxed{t=77\ \text{mt}}$$
5. 7.5 – Fleet numbers. $$N_{S,raw}=0.0058\times\dfrac{1}{6.1}\times23{,}000^{0.8}=2.93\ \rightarrow\ \boxed{N_S=3}\qquad N_{T,raw}=0.198\times\dfrac{1}{77}\times23{,}000^{0.8}=7.94\ \rightarrow\ \boxed{N_T=8}$$
6. 7.6 – Equipment fleet costs. $$C_{31}=499813\times3\times(6.1\times1.308)^{0.73}=\boxed{\$6.83\text{M}}$$ $$C_{32}=19558\times8\times(77\times1.1023)^{0.85}=\boxed{\$6.82\text{M}}$$ $$C_{33}=1407359\times3\times6.1^{0.73}\times23{,}000^{-0.2}=\boxed{\$2.12\text{M}}$$

7.7 – Unit costs, currency of the estimate, and the drill rule of thumb. Unit shovel cost $=C_{31}/N_S=6.83/3\approx2.28$ ($M) each; unit truck cost $=C_{32}/N_T=6.82/8\approx0.853$ ($M) each. These are 1996–98-dollar figures; today's actual market price for a large hydraulic shovel of this class runs well above $8–12M and an ultra-class truck above $3–5M, which is considerably more than either figure escalated by the generic index of Question 7.12 – consistent with Question 1.3's warning that a single blended index under-predicts fast-escalating, equipment-heavy cost centres. If drills are assumed to cost the same per unit as trucks, the number of drills indicated is $$N_{drills}=\left\lceil\dfrac{C_{33}}{\text{unit truck cost}}\right\rceil=\left\lceil\dfrac{2.12\text{M}}{0.853\text{M}}\right\rceil=\boxed{3}$$ This “equal unit cost” rule of thumb is only a rough cross-check – real drill rigs cost markedly LESS per unit than an ultra-class truck, so applying the rule literally to a fleet of very large trucks indicates an unrealistically LOW drill count. Where the rule gives an inappropriately low number, the estimator should instead size the drill fleet from the actual drilling requirement (blast-hole metres per tonne, pattern burden and spacing, and rig penetration rate) rather than from cost parity with the truck fleet.

1. 7.8 – Maintenance facilities. $$C_4=335629\times23{,}000^{0.3}=\boxed{\$6.83\text{M}}$$

7.9 – Electrical, water, general plant services, access, townsite. As the question states, $C_5$ is estimated during milling/processing infrastructure cost estimation and is not required for this open-pit-only estimate; no calculation is performed for it here.

1. 7.10.1–7.10.3 – Overhead markups. $C_{12}+C_{21}+C_{22}=39.11$ ($M); $C_{31}+C_{32}+C_{33}+C_4=22.60$ ($M); full sum $=61.71$ ($M). $$\text{FEP}=0.05\times39.11\text{M}+0.07\times22.60\text{M}=\boxed{\$3.54\text{M}}$$ $$\text{Supervision/Mgmt/Camp}=0.09\times61.71\text{M}=\boxed{\$5.55\text{M}}$$ $$\text{Admin/Accounting/Legal}=0.055\times61.71\text{M}=\boxed{\$3.39\text{M}}$$
2. 7.11 – Total fixed capital cost. $$\text{Total}=61.71+3.54+5.55+3.39=\boxed{\$74.19\text{M (1998 dollars)}}$$

7.12 – Escalation reasonableness. A blanket “doubles every 12 years” ($\approx6\%$ compound per year) applied over 16 years (1998→2014): $$\text{factor}=2^{16/12}=2.52\qquad\$74.19\text{M}\times2.52=\boxed{\$187.0\text{M (2014 dollars)}}$$ Whether a single blanket rate is reasonable for INDIVIDUAL cost components is doubtful (Question 1.3): labour, steel, energy and specific equipment classes escalate at genuinely different rates, and this particular 16-year window spans the mid-2000s commodity supercycle, during which mining capital cost inflation ran well above a generic 6%/year for several consecutive years. Applied to the Question 7.11 TOTAL, the blanket factor is therefore more likely a CONSERVATIVE (low) estimate of the true 2014 capital cost than an overstatement – consistent with Question 7.7's independent finding that simple escalation under-predicts today's actual heavy-equipment prices for the shovel and truck fleet specifically.