Applied Mathematics

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Tonnage of ore = 92,000 DMT

- Let ;

N = no.of days to do initial stripping

Weight of waste

N = --------------------------

Waste extraction rate

120,000 cu.m x 2.5 WMT/cu.m

N = -------------------------------------

2,500 WMT/day

N = 120 days

c. Waste Extraction rate

Ore Extraction rate = --------------------------- 3

2,500 WMT/day

Ore Extraction rate = --------------------

3

Ore Extraction rate = 833.33 WMT/day x 0.92

Ore Extraction rate = 766.67 DMT/day

d. Life of Mine = Total Reserve

------------------

Ore Extraction rate

Life of Mine = 2,000,000 WMT x 1.1

---------------------------

833.33 WMT/day x 300 days/year

Life of Mine = 8.8 years

A.3 OPEN PIT MINING PROBLEM:

August, 1988. A gold mine extracts ore from several small pits to feed its mill. An orebody was determined to have a tonnage of 14,000 tons and a grade of 3.5 gms Au/ton. Cost of mining is detailed as follows:

Dozing = P460.00/hr

Dozing capacity = 166.4 bcm/hr

Loading = P5.85/lcm (swell factor of 1.64)

Hauling = P13.60/lcm/km

Other services (ore only) = P60.00/ton

Explanation of terms:

Bcm – bank cubic meters – the volume of in-situ material

Lcm – loose cubic meter – the volume of broken material

The in-situ material when blasted becomes 1.64 lcm of muck. Assume the specific gravity of ore and waste is 2.2. The orebody is 0.90 km away from the mill and 0.80 km away from the waste dump. Cost of milling is P168.00/ton milled. Mine General overhead, depreciation, depletion, amortization, interest and other charges, and administration totaling P300.00/ton milled. Gold price is $435/oz and the exchange rate is P21.00 per US $1.00. Historical record shows that ore grade determined at the pit drops by 10% by the time the ore reaches the mill. Recovery of metal at the mill is 85%. Assume that gross metal value equals revenue from sale of metal.

- Determine the cost for every ton of ore extracted. (Ans=P74.74/ton)

- Determine the cost for every ton of waste mined. (Ans=P13.73/ton)

- Using X as the corresponding waste tonnage for every ton of ore mined, present a formula for total cost (all cost items included) of mining the 14,000 tons of ore and the corresponding waste.

Determine the total metal sale considering the two-step drop of grade from ore grade to recovered grade. (Ans=P11,007,675)

- Equate item c to item d and determine the value of X

- What is the breakeven stripping ratio? (Ans=17.74:1)

Solution:

a. Cost /ton of ore extracted:

weight 1 ton

Volume = ---------- = ------------------ = 0.4545 bcm

Density 2.2 tons/bcm

Volume 0.4545 bcm

Dozing Time = ---------------------- = ----------------- = 0.00273 hour

Dozing capacity 166.4 bcm/hr

a.1 Dozing = P460/hr x 0.00273 hr -------------------------------------------- P 1.26

a.2 Loading = P5.85/lcm x 0.4545 bcm x 1 64 lcm/bcm ------------------- P 4.36

a.3 Hauling = P13.60/lcm-km x 0.4545 bcm x 1.64 lcm/bcm x 0.9 km -- P 9.12

a.4 Other Services --------------------------------------------------------------- P60.00

-----------

Total Cost per ton of ore extracted ------------------------ P74.74/ton ore

b. Cost / ton of waste mined:

b.1 Dozing = P460/hr x 0.00273 hr ------------------------------------------- P 1.26

b.2 Loading = P5.85/lcm x 0.4545 bcm x 1.64 lcm/bcm ------------------ P 4.36

b.3 Hauling = P13.60/lcm-km x 0.4545 bcm x 1.64 lcm/bcm x 0.8 km - P 8.11

------------

Total Cost per ton of waste mined ------------------------ P13.73/ton waste

c. Total Cost

Let, X = tons of waste per ton of ore

Total Cost = Cost of mining ore + Cost of mining waste + Milling Cost + Other Costs

c.1 Cost of Mining ore = P74.74/ton x 14,000 tons ------------------------------ P 1,046,360

(P13.73) ( X tons-waste)

c.2 Cost of Mining waste= ------------ x ------------------ x 14,000 tons --------- P 192,220 X

tons-waste tons-ore

c.3 Cost of Milling = P168/ton x 14,000 tons --------------------------------- P 2,352,000

c.4 Other Costs = P300/ton x 14,000 tons --------------------------------- P 4,200,000 c.5 Total Cost = P7,598,360 + P192,220