Question 13 of 13: Capital-Cost Estimation – O’Hara Method

Question 7: Capital-Cost Estimation – O’Hara Method (15 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.

Find. Stripping ratio; each cost centre C12…C4; shovel/truck size and fleet counts; indirect costs; total fixed capital cost of the open-pit section (mine only, excluding milling infrastructure per the question’s own instruction).

Approach. This is a sequential, published parametric cost-estimating chain (O’Hara 1980 / Mular & Poulin CANCOST 1998): compute each cost centre’s own power-law formula from the given tonnages in the stated order, round equipment sizes and counts up to the next standard/integer unit exactly as instructed, then sum direct costs and apply the percentage indirect-cost factors to reach the total.

1. 7.1 – quick order-of-magnitude estimate (0.5 marks). A ±40%, one-day turnaround estimate is a Class 5 / order-of-magnitude (conceptual) estimate in AACE terms – the right contact is an in-house senior mining/cost engineer (or a mine-equipment vendor’s applications engineer for a quick equipment-based check) working from parametric formulae exactly like the O’Hara method below, not a detailed engineering firm, since a ±10–15% definitive estimate needs weeks of detailed take-off, not a day.
2. 7.2 – the O’Hara method (1 mark). O’Hara’s method (CIM Bulletin, Feb. 1980, later extended as Mular & Poulin’s CANCOST, CIM Special Volume 47, 1998) is a parametric (power-law regression) capital and operating cost estimator: each major cost centre of a mine/mill (site prep, stripping, mobile equipment, maintenance facilities, etc.) is fit to a simple formula of the form $C = a\,X^{b}$ against a single scale variable (throughput tonnage, equipment size), calibrated from a database of actual completed mine capital costs; summing the cost-centre formulae with the standard indirect-cost percentages (7.4.9) gives a rapid, order-of-magnitude total project capital cost from only a handful of basic design parameters, exactly as exercised in 7.4 below.
3. 7.3 – escalating to present-day dollars (0.5 marks). Convert using a published mining capital-cost index (e.g. the Marshall & Swift Mine/Mill index of Q1.7, or an equivalent CE/Nelson-Farrar-type index) ratioed between the base year and the target year: $$C_{2012} = C_{1980}\times\frac{Index_{2012}}{Index_{1980}}, \qquad C_{2015}=C_{1998}\times\frac{Index_{2015}}{Index_{1998}}$$ using the 1980-base index series for the O’Hara-year costs and the 1998-base series for the CANCOST-year costs, since each formula set was calibrated in its own base-year dollars.
4. 7.4.1 – stripping ratio. Ore feed equals mill capacity (9,000 mt/day); waste is the remainder of the 23,000 mt/day total: $$\text{ore}=9{,}000\ \text{mt/d}, \quad \text{waste}=23{,}000-9{,}000=14{,}000\ \text{mt/d}$$ $$SR = \frac{\text{waste}}{\text{ore}} = \frac{14{,}000}{9{,}000} = \boxed{1.56:1}$$ Stripping ratio is the tonnes of waste that must be removed for every tonne of ore mined – the single most common measure of a pit’s overall economic burden.
5. 7.4.2 – site-preparation cost C12. $$C_{12} = 11410\,T^{0.5} = 11410\sqrt{23{,}000} = \boxed{\$1{,}730{,}412}$$
6. 7.4.3 – pre-production stripping costs. Total soil $T_S = 23{,}000\times60 = 1{,}380{,}000$ mt; total rock $T_R=11{,}000\times300=3{,}300{,}000$ mt: $$C_{21} = 1826\sqrt{1{,}380{,}000} = \boxed{\$2{,}145{,}064} \qquad C_{22} = 19395\sqrt{3{,}300{,}000} = \boxed{\$35{,}232{,}767}$$
7. 7.4.4 – shovel and truck size. $$S_{calc} = 0.1034\,T^{0.4} = 0.1034\times23{,}000^{0.4} = 5.74\ \text{m}^3 \;\Rightarrow\; \boxed{6.1\ \text{m}^3\ (8\ \text{yd})}\ \text{(next standard size up)}$$ $$t_{calc} = 9.75\,S^{1.1} = 9.75\times6.1^{1.1} = 71.3\ \text{mt} \;\Rightarrow\; \boxed{77\ \text{mt}\ (85\ \text{st})}\ \text{(next standard size up)}$$
8. 7.4.5 – fleet counts. $$N_S = \frac{0.0058\,T^{0.8}}{S^{0.8}} = \frac{0.0058\times23{,}000^{0.8}}{6.1^{0.8}} = 4.21 \;\Rightarrow\; \boxed{5\ \text{shovels}}$$ $$N_T = \frac{0.198\,T^{0.8}}{t^{0.8}} = \frac{0.198\times23{,}000^{0.8}}{77^{0.8}} = 18.9 \;\Rightarrow\; \boxed{19\ \text{trucks}}$$
9. 7.4.6.1 – fleet and drilling costs. $$C_{31} = 499813\,N_S(1.308S)^{0.73} = 499813\times5\times(1.308\times6.1)^{0.73} = \boxed{\$11{,}381{,}303}$$ $$C_{32} = 19558\,N_T(1.1023t)^{0.85} = 19558\times19\times(1.1023\times77)^{0.85} = \boxed{\$16{,}201{,}305}$$ $$C_{33} = 1407359\,N_S\,S^{0.73}\,T^{-0.2} = 1407359\times5\times6.1^{0.73}\times23{,}000^{-0.2} = \boxed{\$3{,}534{,}436}$$
10. 7.4.6.2 – unit costs, index check, and drill count. $$\text{cost/shovel} = C_{31}/N_S = \boxed{\$2{,}276{,}261} \qquad \text{cost/truck} = C_{32}/N_T = \boxed{\$852{,}700}$$ These are plausible present-day large-shovel/haul-truck unit costs, so the formula set (once escalated per 7.3) still tracks real equipment pricing reasonably well – the main risk in relying on 1978–80-vintage cost data is that it under-represents the disproportionate cost growth of emissions controls, electronics/automation content and ultra-class truck sizes that have entered the market since, so the escalated total should be treated as a lower-bound sanity check, not a definitive estimate. If drills cost approximately the same as trucks, the “rule of thumb” number of drills follows from the drilling-cost total divided by the unit truck cost: $$N_{drills} \approx \frac{C_{33}}{\text{cost/truck}} = \frac{3{,}534{,}436}{852{,}700} = 4.14 \;\Rightarrow\; \boxed{5\ \text{drills}}$$ This rule of thumb becomes unreliable for very large trucks, since drill cost scales far more slowly with pit throughput than ultra-class truck cost does – at very large truck sizes the rule can imply an implausibly low drill count; the estimator should cross-check against a drills-per-shovel operating ratio (typically 1–2 drills per shovel) rather than accept the cost-ratio result blindly when it is inconsistent with normal operating fleet ratios.
11. 7.4.7 – maintenance facilities. $$C_4 = 335629\,T^{0.3} = 335629\times23{,}000^{0.3} = \boxed{\$6{,}829{,}326}$$
12. 7.4.9 – indirect costs. Let $\Sigma_{prep}=C_{12}+C_{21}+C_{22}$ and $\Sigma_{equip}=C_{31}+C_{32}+C_{33}+C_4$: $$\Sigma_{prep} = 1{,}730{,}412+2{,}145{,}064+35{,}232{,}767 = \$39{,}108{,}243$$ $$\Sigma_{equip} = 11{,}381{,}303+16{,}201{,}305+3{,}534{,}436+6{,}829{,}326 = \$37{,}946{,}370$$ $$C_{7.4.9.1}\, (\text{Feasibility/Eng/Planning}) = 0.05\,\Sigma_{prep}+0.07\,\Sigma_{equip} = \boxed{\$4{,}611{,}658}$$ $$C_{7.4.9.2}\, (\text{Supervision/Mgmt/Camp}) = 0.09(\Sigma_{prep}+\Sigma_{equip}) = \boxed{\$6{,}934{,}915}$$ $$C_{7.4.9.3}\, (\text{Admin/Accounting/Legal}) = 0.055(\Sigma_{prep}+\Sigma_{equip}) = \boxed{\$4{,}238{,}004}$$
13. 7.4.10 – total fixed capital cost (mine only). $$Total = \Sigma_{prep}+\Sigma_{equip}+C_{7.4.9.1}+C_{7.4.9.2}+C_{7.4.9.3} = \boxed{\$92{,}839{,}190}$$ At roughly $92.8M for a 23,000 mt/day open-pit mine (excluding the mill, per the question’s own instruction to exclude 7.4.8), this is a reasonable order-of-magnitude fixed capital figure once escalated by the appropriate index (7.3) from the formula set’s 1978–80 base to the target estimate year – the answer is reasonable at Class-5 (±40%) precision, consistent with the intent of a same-day parametric estimate.