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Mining Revenues and Costs
Assoc/Prof Ali Karrech
School of Civil, Environmental, and mining Engineering
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Outline
• Introduction
•
Life of Mine (LOM) plan
• Financial analysis
• Estimating Revenues
• Estimating costs
Slide 2
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Introduction
•
Evaluation: Determining the numerical values and all the possible
factors that are important in establishing the worth of a project.
! mine evaluation is the assessment of the relative economic viability ofthe mining project or investment opportunity.
Slide 3
Iterative procedure of mine evaluation (source: SME)
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LOM planner’s burden
•
What can be mined profitably? (pit optimisation)
• How to mine it? (mining method selection)
•
How to access it? (pit design)
•
How to dispose of worthless materials? (waste design)
•
How the mine evolves? (scheduling/staging)
• What can be presented to the mill? (scheduling, cut-off grade, dilution)
•
What equipment to be used? (equipment selection)
•
How much equipment are needed? (productivity assessment)•
What is the mining capital cost? (cost model)
•
What is the mining operating cost? (cost model)
Source: modified from T. Elkington , MINE4161o
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Source: cost estimation handbook, 2013
Mining Studies
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Mining Studies - Scoping
•
The potential of an orebody in terms ofsize (tonnage/grade), mining and processingconditions, infrastructure, risks etc.
• Is more drilling needed before proceeding? Whereshould it be?
• Costs taken from prior experiences (within ±30%)
• Decide if further work is justified
Slide 6
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Mining Studies – Pre-feasibility
•
Additional details are developed
• Costs are defined to higher accuracy (15-25%)
• Used to consider mining, processing, sales options for
the deposit
• Focuses on key issues
Slide 7
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Mining Studies – feasibility
•
A full technical study that seeks to optimise all aspectsof the operation from environmental through to mining
•
Used to assess the commercial and technical viabilityof the preferred option(s)
• Used to apply for finance
•
Identifies key issues, risks and management strategies
•
Greater analysis of costs (10-15%)
Slide 8
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Mining Studies –
Banquable Quality Feasibility
•
Details the engineering design, validating the feasibilityor providing amendments
•
Provides specifications and implementation plans forEngineering, Procurement, Construction andManagement (EPCM) and initial operating plan andsystems
• Estimates costs to 5%
Slide 9
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Mining Studies
Source: (T. Elkington , MINE4161)
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MiningStudies
Chapters 9, 10
Chapters 6-7, 8
Chapter 5
Chapters 1-4 Ore Deposits
Grade/PotentialGrade
Net SmelterReturn of
Concentrates
Reserves/ResourcesPotentialTonnage
Optimum of Annual Mine
Capacity
OperatingCosts
InvestmentCosts
Chapter 11Payback Period, Net Present Value ,
Internal Rate of Return
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MiningStudies
Chapters 9, 10
Chapters 6-7, 8
Chapter 5
Chapters 1-4 Ore Deposits
Grade/PotentialGrade
Net SmelterReturn of
Concentrates
Reserves/ResourcesPotentialTonnage
Optimum of Annual Mine
Capacity
OperatingCosts
InvestmentCosts
Chapter 11 Payback Period, Net Present Value ,Internal Rate of Return
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Outline
• Introduction
•
Life of Mine (LOM) plan
• Financial analysis
• Estimating Revenues
• Estimating costs
Slide 14
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Project Value
Future worth
•
If a depositor puts 1$ in a savings account today at a bank paying 10%of simple interest. In one year , he would have $1.1, that is
FW = PV (1+i)FW = Future Worth,
PV=Present Value
i = interest rate
At the end of n years, the accumulated amount would be:
FW = PV (1+i)n
After 5 years, FW = 1(1+0.1)5 = $1.61
Slide 15
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Project Value
Present Value
•
The previous procedure can be reversed. For example if you are toreceive $1.61 in 5 years, and the annual interest rate during this periodis 10% (or 0.10), then the present value of this amount is?
Substituting for FW = $1.61, i=0.1 and n=5
Slide 16
PV =FW
(1+ i)n
PV =1.61
(1+ 0.1)5 = $1
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Project ValuePresent Value of a series of contributions
•
If you are to receive 1$ at the end of each year of 5 consecutive years
and the annual interest rate during this period is 10% (or 0.10), thenwhat is the present value in question?
• One calculates the present value of each of these payments
Slide 17
Year1:PV1 = $1.(1+ 0.1)1
= $0.909
Year 2 : PV2 =$1.
(1+ 0.1)2 = $0.826
Year 3 : PV3
=
$1.
(1+ 0.1)3 = $0.751
Year 4 : PV4 =$1.
(1+ 0.1)4 = $0.683
Year 5 : PV5 =$1.
(1+
0.1)
5 = $0.621
!
The present value of these 5years payments
! In general,
PV = $3.79
PV Ck 1 i k
k 1
n
if Ck C #k, PV C (1 i)
n 1
i(1 i)n
$
%
&
'
(
)
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Economic ConceptsPayback period
•
The period of time required to regain the funds invested, or to reach
the break-even point.• Example: Assume $5 borrowed today to purchase a piece of
equipment and that an interest rate of 10% applies. It is intended torepay the loan in equal yearly payments of $1.
The loan is repaid when the net present value is zero:
! n ~ 7.25 years Slide 18
PV(loan) $5
PV(payments) $1 (1
0.1)n 1
0.1(1 0.1)n
#
$
%
&
'
Net Pr esent Value (NPV)
PV(loan)
PV(payments)
0
NPV $5 $1 (1 0.1)n 1
0.1(1
0.1)n
#
$
%
&
'
0
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Project ValueCash flow•
Net inflow (positive) and outflow (negative) of money, during a specificperiod of time
• Example (calculate the Cash flow)
Slide 21
Gross revenue
- Operating expense = Gross Profit (taxable income)
- Tax = Net Profit
- Capital Costs = Cash flow
Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6
Revenue 170 200 230 260 290
- Op costs -40 -50 -60 -70 -80- Cap costs -200 -100
- Tax costs -30 -40 -50 -60 -70
Cash flow -200 -100 100 110 120 130 140
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Project ValueDepreciation
•
Decrease in value of an asset, e.g.:
•
It can be accounted for as follows:
•
Example: Consider a $100 investment in a 5 years lifetime project. Thesalvage value of the investment is zero. The project dollar income andcapital expenses are estimated to be $(80,30) in year 1, $(84,32) in
year 2, $(88,34) in year 3, $(92,36) in year 4, and $(96,38) in year 5.The effective income tax is 32%.
Calculate the cash flows and the Net Present Value (consider an interestrate of 15%).
Slide 23
Investment ($)
Life (number of year)
Gross revenue
- Operating expense - Depreciation = Gross Profit (taxable income)
- Tax = Net Profit- Capital Costs + Depreciation = Cash flow
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Project ValueDepreciation
•
Solution :
Slide 24
!"#$ & ' ( ) * + ,-.-/#01"!"#"$%" '( ') '' *+ *, ))(
-./"0 12343 -5( -5+ -5) -5, -5' -67(
-8"/0"9:92$ -+( -+( -+( -+( -+( -6((
;4:" 5( 5+ 5) 5, 5' 67(
-?:< @ 5+ A -*B, -6(B+) -6(B'' -66BC+ -6+B6, -C)B)
;D"4 E$12F" +(B) +6B7, +5B6+ +)B)' +CB') 66CB,G 8"/0"1E:92$ +( +( +( +( +( 6((
-H:/E4:> 12343 -6(( -6((
H:3I J2K3 -6(( )(B) )6B7, )5B6+ ))B)' )CB') 66CB,
8E312%4"L HM -6((B(( 5CB65 56BC' +'B5C +CB)5 ++B7* )5B+'
Remark: Depletion, which is a tax consideration given the the owner of amineral deposit, can be considered the same way as Depreciation
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Estimation of the potentialproduction rate
To estimate the Mine Life and production rate Taylor (1977)
proposed the following empirical formulae:Life (years) = 0.2 Tonnage4
Production (tonnes/day)=Tonnage
Life OperatingDays
0.0143 Tonnage0.75
. For a mine operating 350 days/yr
Example: using Taylor’s model estimate the lifetime and dailyproduction of mine knowing that the 9.15 million tonnes are tobe removed. Assume the mine operates 350 days/year.
Solution:
Life (years) = 0.2 9e64 = 11 years
Production (tonnes/day)=
0.0143 (9.15e6 )0.75
2349.73 tonnes/day
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Estimation of the potentialproduction rate
does not work well for v. large (>200Mt) deposits, deep, ftatorebodies, steeply dipping tabular orebodies etc.
For open pit mines, the production in mt/yr can be expressed
as
0.416
Tonnage0.5874
. (based on 41 Gold-silver mines (Singer et al. 1998))
0.0236
Tonnage0.74. (based on 45 Copper mines (Singer and Long 2001))
0.123
Tonnage
0.649
. (based on 342 Other mines (Long 2009))
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Taxation and royalties
Taxation regimes differ from a country (or state) to another and oftendepend on commodities.
There are number of means by which governments gain direct benefitsfrom a project by:
•
Royalties.
•
Lease paymentsfor land andmineral rights,
• etc.
Source: Australia’s Future Tax System: Report to the Treasurer p. 47 (Henry Tax Review)
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Influence of risk on NPV
Investors prefer relatively certain profits (“a loaf is better than two in the bush”)
The discount rate, i, used to discount NPV depends on the Weightedaverage cost of capital (WACC) and the level of risk of a project.
It is calculated as a function of three components:
•
i1: WACC (2.5% ~ 5%) .
•
i2: associated to the risk of the project (3% ~ 16%). Depends onthe development level (scoping, pre-feasibility, feasibility etc.)
•
i3: associated with the country (geopolitics) (0% ~ 14%)
•
i = 5.5 ~ 25%
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I fl f i k
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Influence of risk onNPV
i3 associated with thelevel of risk of acountry
http://pages.stern.nyu.edu/~adamodar/New_Home_Page/datafile/ctryprem.html
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Influence of risk on NPV
A company intends to invest in a Copper mine in Indonesia.
The estimated tonnage is 12Mt. Estimate the life time of themine and the discount rate, knowing that the WACC is 5%.
Pr oduction (tonnes/day)
0.0236 Tonnage0.74
Pr oduction (tonnes/day)=Tonnage
Life OperatingDays
Life=Tonnage
Pr oduction (tonnes/day) OperatingDays
Tonnage0.26
0.0236 OperatingDays
8.39 years
i1 = 5%, i2 = 12% (assume a feasibility study), and i3 = 9%for Indonesia! i = 26%
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Estimating RevenuesCurrent prices•
Current prices can be found in several publications such as:
–
Metals Week – Mining Magazine
–
Metal Bulletins
– Industrial Minerals
•
For many minerals, the “ton” is the unit of sale:
1 short ton (st) = 2000 lbs = 0.9072 metric tons
1 long ton (lt) = 2240 lbs = 1.01605 metric tons
1 metric tons = 2204.61 lbs = 1000 kilograms
•
Precious metals (gold, silver, platinum, palladium, rhodium), are
generally sold by troy ounce.
1 troz = 31.1035 grams
1 oz = 28.3495 grams (U.S. standard ounce)
1 carat = 0.2 grams
Slide 33
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Estimating RevenuesMarkets
Principal markets of metals:
• London Metal Exchange (LME), UK: www.lme.com .
• New York Mercantile Exchange (NYME), New York, USA: www.nyme.com .
• KITCO Precious Metals: www.kitco.com
Slide 34
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Slide 35
Estimating RevenuesHistorical price data
Precious metals
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Slide 36
Estimating RevenuesHistorical price data
Base Metals
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Slide 37
Estimating RevenuesHistorical price data
Steel raw materials
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Slide 38
Estimating RevenuesTrend Analysis
•
Idea: replace the actual price-time history with a mathematicalrepresentation which can be used for extrapolation
– A function of the form can be useful. For a
data of size N (N pairs of x and y), the maximum power of such polynomialis m=N-1.
– This is a good procedure for interpolation, but cannot be used forextrapolation (the higher orders fluctuate a lot)
– The simplest representation is
–
From the previous curves, it can be noticed that the behavior is non-linear.The following trend could also be tested
– Which can be transformed into an equation of the form (1) such that
y = a0 + a1x+ a2x2+...+ amx
m
y = a 0 + a1x (1)
y = a exp(bx) (2)
y
!
=
a
!
+
a1x such that (.)
!
=
Ln(.)
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Slide 39
Estimating RevenuesTrend Analysis
•
The least square approach shows that:
• The correlation factor can be calculated using:
where
The correlation coefficient is unitless and between +1 and -1. In general,the closer the correlation coefficient is to +1 or -1 the better theassociation between the two variables.
b =xi ln yi! "
1
nxi! lnyi!
xi2! " 1
nxi!( )
2a = exp
1
nlnyi! "
b
nxi!
#
$%
&
'(
r =Sxy
SxxSyy
Sxx = n xi2! " x i!( )
2
Syy = n ln yi( )2
! " ln yi!( )2
Sxy = n xi ln yi! " x i! ln yi!
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Slide 40
Estimating Revenues
N":0
O14%:> /0E1"
2P /"0 %$E4 2P
12FF2LE4Q
6*C5 6' +'B7*'
6*C) 6* +*B,*)
6*CC +( 57B)*6
6*C, +6 )6B'6'
6*C7 ++ +*BC7,
6*C' +5 +CB7,)
6*C* +) 56B6'+
6*,( +C 5+B(C5
6*,6 +, +*B*+6
6*,+ +7 5(B,
6*,5 +' 5(B,
6*,) +* 56B*,
6*,C 5( 5CB(67
6*,, 56 5,B67
6*,7 5+ 5'B++,
6*,' 55 )6B')76*,* 5) )7BC5)
6*7( 5C C7B7
6*76 5, C6B)55
6*7+ 57 C(B,67
6*75 5' C'B'C+
6*7) 5* 7,B,)*
6*7C )( ,5BC5C
6*7, )6 ,'B'+)
Predict the price from 1977to 1988 basedon data from 1955 to 1976
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Estimating Revenues
b xi ln yi
1
n
xi lnyi
xi2
1
nxi
2
a
exp 1
nlnyi
b
nxi
$
&
'
6*77 ++ ,CB'6 ,CB)+6'C5)C
6*7' +5 ,CBC ,'B65+*5'+)
6*7* +) *+B55 7(B*C,57(C+
6*'( +C 6(6B)+ 75B'*,'(C*,
6*'6 +, '5B7) 7,B*C*(*5+
6*'+ +7 7+B*6 '(B6)'+'677
6*'5 +' 77B', '5B),*,5()7
6*') +* ,,B7, ',B*+',6,(+
6*'C 5( ,CBC7 *(BC5(*)+6
6*', 56 ,,B6 *)B+'+C)'7C
6*'7 5+ '+BC *'B6'*,++6,
$ ++
: +,B775
= (B()(),
Slide 41
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Slide 42
Estimating Revenues
Sxx = n x i2! " x i!( )
2
Syy = n ln yi( )2
! " ln yi!( )2
Sxy = n x i ln yi! " xi! ln yi!
r = Sxy
SxxSyy
233 '4*5'
266 *+7)+
236 84'7&'
$9:36;
:> &75*'
• Calculate the correlation coefficient
•
Comment on the accuracy of the proposed method
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Slide 43
Estimating Revenues
•
The average price of a metallic commodity is around 1.5its production cost.
• Prices are imposed by the market rather than controlledby mining companies
• The prices are usually cyclic, in terms of discounteddollars.
Mining
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MiningStudies
Chapters 9, 10
Chapters 6-7, 8
Chapter 5
Chapters 1-4 Ore Deposits
Grade/Potential
Grade
Net SmelterReturn of
Concentrates
Reserves/
ResourcesPotentialTonnage
Optimum of Annual Mine
Capacity
OperatingCosts
InvestmentCosts
Chapter 11Payback Period, Net Present Value ,
Internal Rate of Return
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Slide 45
Estimating RevenuesNet Smelter Return
• For base metals (eg. copper, lead, zinc), prices are not quoted for
concentrates. Prices are given for refined metals.•
Net Smelter Return: the payment received by the company from thesmelter for their concentrates.
•
A mill produces a copper concentrate containing G % of metal. Thecontained copper in one ton of concentrate is:
CM = G x 2000 / 100 (lbs of metal /ton of concetrate)
– G: concentrate Grade (% metal)
– 2000: lbs/ton
•
The contained copper value is:
CV = G x 2000 x P / 100 ($/ton)
– P: current market price ($/lb)
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Slide 46
Estimating RevenuesNet Smelter Return
• It is not possible for smelting and refining to recover 100% of the
contained metal. Some of it is lost in the slag. •
To account for this, the smelter pays only for a portion of the metalcontent in the concentrate.
•
The deductions may take one of the following forms:
–
Percentage deduction: the smelter pays only for a concentration C of thecontained metal
– Unit deduction: the concentrate grade is reduced by a certain fixed amountcalled unit reduction
– A combination of the above.
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Slide 47
Estimating RevenuesNet Smelter Return
• Therefore, the effective concentrate grade is:
(%)
–
u: fixed unit deduction (%)
– C: Credited percentage of the metal content (%)
•
The payable (accountable) metal content in one ton of concentrate is:
(lbs/ton)
•
Smelters often pay only a certain proportion of the current market price.The relation between the actual price to the market price is the “pricefactor”, f.
•
The gross value of one ton of concentrates is:
($/ton of concentrate)
Ge = C
100G!u( )
Me =
C
100
G!u( )
1002000
GV=MePf
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Slide 48
Estimating RevenuesNet Smelter Return
• The basic smelter return (BSR) can be obtained by taking into account
the charges incurred during treatment, refining, and selling:($/ ton of concentrate)
– r: refining and selling cost ($/ (lb of payable metal))
– T: treatment charge ($/ton of concentrate)
•
The net smelter return (NSR) accounts for the presence of other metalswhether it is advantageous (resulting in a by-product credit Y) ordisadvantageous (resulting in a penalty charge X):
($/ ton of concentrate)
•
Pe=Pf-r is also known as the effective metal price.
•
The At Mine Revenue (AMR): net value of the concentrate to the mineis expressed as: ($/ ton of concentrate)
–
R :is the realisation cost (e.g. Freight, Insurance, sales items commission)
BSR =Me Pf ! r ( )!T
NSR =Me Pf ! r ( )!T !X+Y
AMR =NSR!R
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Slide 49
Estimating RevenuesNet Smelter Return
• According to O’Hara (1980), Freight cost F in Canadian $ (1979) per
ton of concentrate can be expressed as:($/ ton of concentrate)
– Tm: miles per road (truck)
– Rm: miles per railroad
– D0: days of loading, ocean travel and unloading on a 15000-ton freighter
• The percent payment (PP) is expressed as:
–
CV: value of the metal contained in the concentrate
F = 0.17Tm
0.9+ 0.26R
m
0.7+ 0.8D
0
PP =AMR
CV
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Slide 50
Estimating RevenuesNet Smelter Return
• According to O’Hara (1980), Freight cost F in Canadian $ (1979) per
ton of concentrate can be expressed as:($/ ton of concentrate)
– Tm: miles per road (truck)
– Rm: miles per railroad
– D0: days of loading, ocean travel and unloading on a 15000-ton freighter
• The percent payment (PP) is expressed as:
–
CV: value of the metal contained in the concentrate
F = 0.17Tm
0.9+ 0.26R
m
0.7+ 0.8D
0
PP =AMR
CV
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Slide 51
Estimating RevenuesNet Smelter Return
Gangue
M inera l Metal
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Slide 52
Estimating RevenuesNet Smelter Return
Gangue
M inera l Metal
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Slide 54
Estimating RevenuesNet Smelter Return
• Solution: For copper, the Basic Smelter Return is:
–
The by-product credit for silver is
– The penalty for the excess of lead is:
X = (units present – units allowable) x charge per ton of concentrate == (2-1)$10. = $10 /ton of concentrate
BSR
Me Pf r T C
100
G u
100 2000 (Pf r) T
98
100
30 1
100 2000 ($1 1 $0.1) 75
$436.56 / ton of concentrate
Y
C
100(G u) (Pf r)
95
10030 1
($6 1 $0.35)
$155.66 / ton of concentrate
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Slide 55
Estimating RevenuesNet Smelter Return
• Solution: the Net Smelter Return is:
•
Assume that concentrates are shipped 500 miles by rail to the smelterand that $1 (CAN, 1979) = $3.6 (AU, current). The transport cost is:
F = 3.6 x 0.26 x Rm0.7 = 0.936 (500)0.7= $72.53
•
The at-mine-revenue is: AMR = NSR-R = 582.22 – 72.53 = $509.7
• The contained Value in one tonne of concentrates is:
•
The percent payment is
PP = AMR/CV = 509.7/800= 63%
NSR
$436.56
$10.
$155.66
$582.22 / ton of concentrate
CV 30
100
2000
$1
copper
! "## $##
30
$6silver
!"$
2
100
2000
$0.5
lead
! "### $###
$800
Outline
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Outline
• Introduction
•
Life of Mine (LOM) plan
• Estimating Revenues
• Estimating costs
–
Types of costs – Approach I (order 0 experience)
" Fleet requirement
"
Mining cost
– Approach II (empirical - statistics)
Slide 56
E ti ti C t
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Slide 57
Estimating CostsTypes of costs
• Cost categories may be:
–
Capital cost (Capex)
–
Operating cost (Opex)
– General and administration cost (G&A).
•
The capital is the investment required to bring a project to acommercially operable status.
• The operating costs are used by an organization just to maintain itsexistence. They would reflect labour, drilling, blasting, loading, hauling,dozing, maintenance (some companies include maintenance in G&A),etc.
•
The G&A expenses include all of the costs of operating the businessother than the costs of preparing the product for sale,
(e.g. managerial salaries, area supervision, mine/head office expense,mine surveying, state/local taxes, insurance.)
E ti ti C t
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Slide 58
Estimating CostsEscalation of older costs
• Is there a simple way to
update costs to makeestimation today?
– This can be achieved byescalating costs through theapplication of various
published indexes
–
e.g. construction cost, buildingcost, skilled labour, commonlabour, materials.
• Question: Assume that the costof the mine maintenance
building was $100000 in 1978,estimate its cost in 2011.
•
Answer:
$100k x 5058/1664 = $303.96k
Outline
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Outline
• Introduction
•
Life of Mine (LOM) plan
• Estimating Revenues
• Estimating costs
–
Types of costs – Approach I (order 0 experience)
" Capital cost
"
Mining cost
–
Approach II (empirical - statistics)
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Capital costs
•
Once equipment replacement and purchase has been determined, it is asimple process to determine the capital costs of the equipment
•
Need to know all up cost!
– Freight
–
Tyres –
Insurance – Buckets/Ground Engaging Tools (GET) – Additional extras
• Capital cost per unit (rules of thumb)
– Drills = $47,000 x Weight (t) – Shovel = $21,000 x Weight (t) – Truck = $37,000 x Weight (t)
–
Water cart = $37,000 x Weight (t) – Wheel loader = $37,000 x Weight (t) – Track dozer = $28,000 x Weight (t) – Wheel dozer = $32,000 x Weight (t) – Grader = $51,000 x Weight (t)
61
Source: T. Elkington , MINE4161o
Mining cost estimate
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Mining cost estimate
62
Mining cost estimate (cont )
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Mining cost estimate (cont.)
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Hourly costs of equipment (trucks/loaders/
drills)• Fuel
–
Fuel burn (L/hr) x Fuel cost ($/L) –
Fuel burn rule of thumb – (L/hr = Z x Diesel power in kW)• Where Z = 0.19 for loaders and drills, varies for trucks
•
Salary cost ($/year) / Hours worked (exc leave) x manning ratio –
Salary cost includes direct pay, other benefits (social security), payroll tax,accommodation, meals, training etc.
–
Manning ratio rule of thumb – 2.8 per machine (including maintenance labour)
• Maintenance (includes GET, tyres, parts, oil, accidental damage)
–
Rule of thumb $/hr = Y x Weight of equipment in t
•
Where Y = 2.12 for drills, 1.05 for loaders, 1.06 for trucks• Lease cost (spreads capital cost across equipment life)
–
Rule of thumb $/hr = K x Weight of equipment in t• Where K = 0.68 for loaders, 1.19 for trucks, 1.52 for drills
64
Source: T. Elkington , MINE4161o
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•
Fuel burn – Calculated in segments (as per time) – Fuel burn (L/hr)
– Where EngineLoading
•
Uphill or level
• Downhill/Idle = 0
– Fuel burn
65
Hourly costs of equipment
Source: T. Elkington , MINE4161o
Hourly fuel burn = 0.233 engine power 0.96 Engine loading +0.04
H l t f i t (t k /l d /
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•
Fuel burn – Calculated in segments (as per time) – Fuel burn (L/hr) = 0.233 x GrossEnginePower x
(EngineLoading*0.96+0.04)
–
Where EngineLoading•
Uphill or level = UpRampTruckSpeed x TotalTruckWeight x9.81 x (RampGrade+RollingResistance)/(3.6 xGrossEnginePower x TransmissionEfficiency x(1+RampGrade^2)^0.5)
•
Downhill/Idle = 0
66
Hourly costs of equipment (trucks/loaders/drills)
Source: T. Elkington , MINE4161o
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Other costs
• Blasting
–
Powder factor ~ 0.6 kg/m3 is common – Explosive costs ~ $1,000/t for Emulsion and $800/t for ANFO
• Ancillary
–
Rule of thumb – 25% of combined loading and truck cost ($/t)
•
Management on-cost/contractor margin – Rule of thumb – 20% of all total cost
• Admin & dayworks
–
Rule of thumb – 10% of total cost67
Source: T. Elkington , MINE4161o
cost($/t) PF explosive cost
Rock density Labour adjust. (1.3) consumable adjust. (1.1)
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Some equipment data
68
Source: T. Elkington , MINE4161o
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Estimating Costs
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Slide 70
Estimating CostsO’Hara (1980) cost estimator - CAPEX
• Daily tonnage:
–
T = tons of ore milled/day – T0 = tons of ore mined/day
– Tw = tons of waste mined/day
– Tc = tons of passing the primary crasher/day
– Tp = Tc+Tw =total material minded/day
Assume the mill operates three 8-hour shifts, 5 days/week. Many open pit mines 7 days/week others operate 5 days/week. In the second case, T = 5 T0/7.
•
The combined mine/mill capital cost is
– Assume that $1 (CAN, 1979) = $3.6 (AUD, today)
C $400,000T0.6
C $1, 440, 000T0.6
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Slide 71
Estimating CostsO’Hara (1980) cost estimator - CAPEX
• Personnel numbers
–
Number of personnel required in open pit (trucks and shovels)
– Number of personnel required to operate mills treating T tons of low grade ore:
–
Number of service personnel:
– Number of administrative and technical personnel:
Nop =0.034Tp
0.8 for hard rock
0.024Tp0.8 for competent soft rock
!"#
$#
Nml =
5.9T0.3 for cyanidation of precious metal ores
5.7T0.3 for flotation of low-grade base metal ores
7.2T0.3 for gravity concentration of iron ores
!
"#
$#
Nsv = 25% of (Nop + Nml )
Nat =11% of (Nop + Nml + Nsv )
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Slide 72
Estimating CostsO’Hara (1980) cost estimator - CAPEX
• Mine site clearing
–
For the pit, the required area in acre is:
–
The clearing cost depends upon the topography, type of cover and area:
•
Pre-production waste stripping
where Ts: tons of soil and Tws: tons of waste
Total clearing cost =
$300Ap0.9 for flat land with no shrubs and no trees
$1600Ap0.9 for 20% slopes with light tree growth
$2000Ap0.9 for 30% slopes with heavy trees
!
"##
$##
Ap = 0.0173Tp0.9
soil stripping costs= $3.2Ts0.8 for soil not more than 20 ft deep
waste stripping costs= $340Tws0.6 for rock requiring blasting loading and haulage
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Slide 73
Estimating CostsO’Hara (1980) cost estimator - CAPEX
• Drilling: number of drills Nd should be
–
The cost of drilling equipment can be estimated by , where d
is the hole diameter.• Shovels: the optimum size is estimated by (dipper size yd3)
– The number of shovels can be estimated by
–
The cost of the fleet of shovel (and auxiliaries) will be
• Trucks: the optimum truck size is matched to the size of the shovel’s
bucket size: truck size t (tons) = 9.0 S1.1
–
The number of trucks required:
– Haulage equipment costs:
Nd =2 if tonnage ! 25000tpd3 if 25000tpd < tonnage ! 60000tpd
4 if tonnage > 60000tpd
"#$
%$
$20000Ndd1.8
S= 0.145!Tp0.5
NS =
0.011!Tp0.8 / S
$510000!Ns!S
0.8
N t 0.25 Tp0.8
/ t
$20000!Nt ! t
0.9
Estimating Costs
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Slide 74
Estimating CostsO’Hara (1980) cost estimator - CAPEX
•
Mill associated costs
cost of concentrator building =$27,000T0.6
cost of gyratory crusher = $63 Tc
0.9
cost of primary crushing plant = $15,000 Tc0.7
cost of fine crushing plant = $18,000 T0.7
cost of grinding
$18, 700T0.7for hard ore ground to 70% passing 200 mesh
$12, 700T0.7for soft ore ground to 55% passing 200 mesh
$22, 500T0.7for hard ore ground to 85% passing 200 mesh
!
"#
$#
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Slide 75
Estimating CostsO’Hara (1980) cost estimator - OPEX
• Pit operating costs
•
Processing costs
gold ores :$40, 000 to 105,000 T0.5
base metal ores: $13,700 to 20,000 T0.6
uranium ores ! $150,000 to 200,000 T0.5
drilling cost per day = $1.9Tp0.7; blasting cost per day = $3.17Tp
0.7
loading cost per day = $2.67Tp0.7; haulage cost per day = $18.07Tp
0.7
general services cost per day = $6.65Tp0.7
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Slide 76
Estimating CostsO’Hara (1980) cost estimator - OPEX
• Concentrator operating costs
crushing cost per day =$7.9T0.6
fine crushing cost per day = $12.6 T0.6
grinding cost per day = $4.9 T0.8
processing cost per day
$65T0.6 for cyanidation of glod/silver ores
$54T0.6 for flotation of simple base metal ores
$34 to $41T0.7 for complex base metal ores
$65T0.7 for Uranium ores by leaching
$45T0.7 for nonfloatable nonsulfide ores responding
to gravity separation
!
"
####
$
###
#
Tailings costs per day = $0.92 T0.8
Assaying costs per day = $1.27 T0.8
Supervision, maintenance and general costs per day = $40.80 T0.8
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Estimating CostsO’Hara (1980) cost estimator (G&A)
•
Mine project overhead costs (D: direct costs)
engineering costs =$2.3D0.8;
general site costs =$0.31D0.9
project supervision costs =$1.8D0.9;
administration costs = $1.5D0.8