hustrulid and kuchta

548zyxwvutsrqponmlkjihgfedcbaYXWVUTSRQPONMLKJIHGFEDCBA Open pit mine planning and design'. Fundamentals

Table 6.18. Incremental capital investments and annual cash flows.

Scenario A Capital investment ($103) ACash flow ($103)

7-7 3332 773

7-6 3334 628

7-5 2666 402

4000 374 7-3

Repeating this process for the others one finds

Scenario i

7-7 7-6 23.1%

7-6 —> 7-5 18.6%

7-5 - » 7-4 14.6%

7 - 4 - * 7-3 8.9%

Now it can be seen that the optimum scenario is 7-4 rather than 7-3. The milling rate would

be 15,300 tpd.

16. Examine all of the other scenarios with respect to the best current one. For pit 7, the

best scenario is 7-4. For pit 6, scenario 6-6 has a higher total ore tonnage than 7-4 with

an acceptable rate of return on the additional capital expenditure. It thus becomes the best

current alternative. Plan 6-5 is better than 6-6 and 6-4 is better than 6-5. Plan 6-3 is not

acceptable. Pit 5 alternatives are compared against 6-4. As can be seen from the table, the

overall best plan is between Plan 5-4 and Plan 5-3. The economic cutoff (mill cutoff) for

the pit is 0.34% Cu.

This leads to the following operation

Ore reserve = 253,000,000 tons

Mill rate = 33,580 tpd

Mining rate = 116,600 tpd

Average ore grade = 0.864% Cu

Mine life = 25 years

Capital investment = $95,500,000

6.7 LANE'S ALGORITHM

6.7.1 Introduction

In 1964, K.F. Lane (Lane, 1964) presented what has become a classic paper entitled 'Choos-

ing the Optimum Cut-off Grade'. This section will describe his approach and illustrate it

with an example. As has been discussed earlier cutoff grade is the criterion normally used in

mining to discriminate between ore and waste in the body of a deposit. Waste may either be

left in place or sent to waste dumps. Ore is sent to the treatment plant for further processing

and eventual sale.

Production planningzyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA 549 yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA

Maximum Unit Capacity Costs

M m

C c

R r

f

s ywusgaI

y

Figure 6.21. The model described by the Lane algorithm (Lane, 1964).

The choice of cutoff grade can directly affect profits. This chapter will examine the

principles which determine the best choice of a cutoff grade under different circumstances.

A mining operation is considered to consist of three stages:

- mining,

- concentrating, and

- refining.

Each stage has its own associated costs and a limiting capacity. The operation as a whole

will incur continuing fixed costs. The three most important economic criteria which can be

applied are:

Case I: Maximum present value.

Case II: Maximum total profits.

Case DOE: Maximum immediate profit.

The maximum present value gives the economic optimum and is that generally applied

lacking special circumstances. It is the one which will be used in this book. As has been

shown by Lane (1964) the second and third correspond to the application of special discount

rates in the first. Case EE, maximum total profits, corresponds to a discount rate of zero percent

whereas Case III is for a high value.

In this chapter, attention is focussed on choosing a cutoff grade to maximize the present

value of the cash flow from the operation.

6.7.2 Model definition

Figure 6.21 is a diagrammatic representation of the elements and symbols used in the

model.

Definitions of the maximum capacities, unit costs and quantities involved in the evaluation

are presented below.

Material

Ore

Concentrate

Product

MINE

CONCENTRATOR

REFINERY

Market

Other Factor;

Fixed costs

Selling prici

Recovery


1. Maximum capacity: M is the maximum amount of material (ore and waste) that the mine

can produce in a given time period (for example 1000 tons/year). It is therefore a restriction

on the maximum rate of progress through the orebody.

C is the maximum amount of ore which can be put through the concentrator in a given

time period (for example 500 tons/year), assuming unrestricted availability of input ore from

the mine. A concentrate of fixed grade is produced.

R is the maximum amount of final product produced in the time period (for example

500 lbs/year), assuming unrestricted availability of concentrate from the concentrator. The

maximum can be due to a restriction on refinery throughput or a market limitation.

2. Costs: m are the mining costs expressed in $/ton of material moved. These are assumed

to be the same irrespective as to whether the material is classified as ore or waste. The unit

mining costs include drilling, blasting, loading, hauling, etc.

c are the concentrating costs expressed in $/ton of material milled. The unit cost c includes

crushing, grinding, floating, leaching, etc. It also includes some haulage if ore is trucked

farther than waste (if not, this can become a credit item in calculating c).

r includes all costs incurred at the product and selling stages such as smelting, refining,

packaging, freight, insurance, etc. These are expressed in terms of $ per unit of product. For

copper it would be $/lb.

/ , the fixed cost, includes all costs such as rent, administration, maintenance of roads

and buildings, etc. which are independent of production levels (within normal limits of

variation) but which would cease were the mine to be closed. It is expressed in terms of a

fixed cost over the production period considered (for example 1 year). Other costs such as

head office charges, depreciation, etc. are not included.

s, the selling price, is expressed in terms of selling price per unit of product. It is a gross

figure provided all selling charges are included in r. If not they must be subtracted from i.

)', the recovery, is an overall figure for the concentrator and the refinery. It is that proportion

of the mineral contained in the original ore feed retained in the final product.

3. Quantities: T is the length of the production period being considered (for example 1 year);

Q,„ is the quantity of material to be mined, Qc is the quantity of ore sent to the concentrator

and Qr is the amount of product actually produced over this production period.

6.7.3 The basic equations

Using the definitions given in the preceding section, the basic equations can be developed.

The total costs Tc are

Tc = inQm + cQc + rQr +fT (6.7)

Since the revenue R is

R = sOr (6.8)

the profit P is given by

P = R — Tc = sOr — (niOm + cQc + rQr + f T ) (6.9)

Combining terms, yields

P = {s-r)Qr-cQc-mQm-fT (6.10)

This is the basic profit expression. It can be used to calculate the profit from the next Qm of

material mined.

Table 6.19. Initial mineral inventory for the Lane

example. zyxwvutsrqponmlkjihgfedcbaYXWVUTSRQPONMLKJIHGFEDCBA

Production planning 551

Grade (lbs/ton) Quantity (tons)

0.0-0.1 100 0.1-0.2 100

0.2-0.3 100 0.3-0.4 100

0.4-0.5 100

0.5-0.6 100 0.6-0.7 100

0.7-0.8 100

0.8-0.9 100

0.9-1.0 100

Qm = 1000

6.7.4 An illustrative example

To introduce the reader in a soft way to the problem being explored in detail in this section

consider the following example. A final pit has been superimposed on a mineral inventory.

Within the pit outline are contained 1000 tons of material. The grade distribution is shown

in Table 6.19. The associated costs, price, capacities, quantities and recovery are:

Costs

m — mining = $ l / ton

c = concentrating = $2/ton

r — refining = $5/lb

/ = fixed cost = $300/year

Price

s = $25/lb

Capacities

M — lOOtons/year

C = 50tons/year

R = 40 lbs/year

Quantities

Qm — amount to be mined (tons)

Oc = amount sent to the concentrator (tons)

Qr = amount of concentrator product sent for refining (lbs)

Recovery (Yield)

y — 1.0(100%recovery is assumed).


There are a great number of possible mine, concentrator and refinery operating combi-

nations. Which is the optimum? In this section the basic equations will be developed in

addition to demonstrating the process. However, prior to beginning the theoretical treat-

ment, it is considered useful to briefly consider just one of these operating combinations.

The total amount of material to be mined Qm is 1000 tons. If the mine is operated at capacity

(100 tons/year) then the pit would be mined out over a time period of 10 years. Assuming

that the grades (Table 6.19) are equally distributed throughout the pit and a concentrator

cutoff grade of 0.50 lbs/ton is used, then 50 tons of material having an average grade of

0.75 lbs/ton would be sent to the concentrator every year. The other 50 tons would be sent

to the waste dump. Since the concentrator capacity C is 50 tons/year, this is an accept-

able situation. The concentrator product becomes the refinery feed. In this case it would

be 37.5 lbs/year (0.75 lbs/ton x 50 tons/year). Since the refinery can handle 40 lbs/year, it

would be operating at below its rated capacity R. This combination of mining rate and cutoff

grade would yield a yearly profit Py of

Py = (25 - 5)37.5 - 2 x 50 - 1 x 100 - 300 = $250

These profits would continue for 10 years and hence the total profit would be $2500. The

NPV assuming an interest rate of 15% would be

The first question to be asked is whether some other combination of mine production rate

and concentrator cutoff grade would yield a better profit from this deposit? The larger

question is whether the various plant capacities (with their associated costs) are optimum?

The procedure described in this section is a way of determining the combination yielding the

maximum profit for a given set of operating constraints. The constraints may then be changed

(mine, concentrator and refining capacities, for example) and the profit corresponding to

this new combination determined as well as how the various capacities should be utilized

over the life of the pit.

6.7.5 Cutoff grade for maximum profit

Step 1. Determination of the economic cutoff grade - one operation constraining the total

capacity

As indicated, the basic profit expression (6.10) is

P = {s- r)Qr - cOc - mOm - fT

Calculate cutoff grade assuming that the mining rate is the governing constraint. If the

mining capacity M is the applicable constraint, then the time needed to mine material Qm is

Qm xroligfSOMC

M

Equation (6.10) becomes

P = ( s - r)Qr - cOc - (m + Qm (6.12)

To find the grade which maximizes the profit under this constraint one first takes the derivative

of (6.12) with respect to g.

r )d Q r / Q c f n j f \ d Q m

dg dg dg V MJ dg

Production planningzyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA 553

However the quantity to be mined is independent of the grade, hence

d Qm = 0 (6.14)

dg


d P . dQr d Qc T = 0 - r ) ~ c — (6.15) dg d g dg

The quantity refined Qr is related to that sent by the mine for concentration Qc by

Qr —yrmgedcSRQ gyQcywusgaI ( 6. 16)

where g is the average grade sent for concentration, and y is the recovery.

Taking the derivative of Qr with respect to grade one finds that

~r~ = gy-7- (6.17 dg dg

Substituting Equation (6.17) into (6.15) yields

dP r n d Oc

- T = [(s - r)gy ~ c] " f 1 (6-18) dg dg

The lowest acceptable value of g is that which makes

dg

Thus the cutoff grade gm based upon mining constraints is the value of g which makes

(s - r)gy - c = 0

Thus g m = g = 7 C 7 (6.19)

y(s - r)

Substituting the values from the example yields MH

$2 g m = l . Q ( $ 2 5 - $ 5 ) = 0 - 1 0 1 b s / t O n

Calculate cutoff grade assuming that the concentrating rate is the governing constraint. If

the concentrator capacity C is the controlling factor in the system, then the time required to

mine and process a Qc block of material (considering that mining continues simultaneously

with processing) is

Qc

Tc = ~ (6-20)

Substituting Equation (6.20) into (6.10) gives

P = (s-r)Qr-cQc-mQm-f^ (6.21)

Rearranging terms one finds that

P = ( s - r)Qr - U + Q Qc- >nQm


Differentiating with respect to g and setting the result equal to zero yields

— =(s- r)- c + - m—— = 0 dg dg V C) dg dg

As before

dg

dQr

d Qc

Thus

, dQr ( f \ dOc

d Qc s - r The cutoff grade when the concentrator is the constraint is xroligfSOMCC+

fr

gc = — ^ r (6.22)

y(s - r)

For the example, this becomes

ec „.. 0.40 s 1.0(525 - $5)

Calculate cutoff grade assuming that the refining rate is the governing constraint. If the

capacity of the refinery (or the ability to sell the product) is the controlling factor then the

time is given by

Qr Tr = ^ r (6-23)

K Substituting (6.23) into Equation (6.10) yields

P = (s- r)Qr-f%- - cQc - rnQm

K

P= ( s - r - t \ Or-cQc- rnQm (6.24)

or

P= ( x - r -

R

Differentiating with respect to g and setting the result equal to zero gives

d P ( f\dQr d Oc dQm — = j — r — — — c— m—-— = 0 dg V RJ dg dg dg

Simplifying and rearranging gives

gr = 7 ~~ TT (6.25)

( s - r - f y y

For this example yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA

$2

( $ 2 5 - $ 5 - ^ ) 1.0 = 0.16 (6.26)

Production planning 555 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA

One can now calculate the amount of material which would be concentrated and refined

under the various constraints as well as the time required. When the mining rate of 100

tons/year is the constraint

1000 tons T m = — 7 — = 1 Q y e a r s

lOOtons/year

Since the cutoff grade gm is 0.10 lbs/ton, a quantity Qc of 900 tons having an average grade

of 0.55 lbs/ton would be sent to the concentrator. The total amount of product refined and

sold Qr is

Q r = 900 x 0.55 = 495 lbs

Substituting these values into the profit equation gives

Pm = ($25 - $5)495 - $2 x 900 - $1 x 1000 - $300 x 10 = $4100

The same procedure can be followed with the other two limiting situations. The results are

given below:

Concentrator limit:

gc = 0.401bs/ton

Qc = 0.60 x 1000 = 600 tons

C = 50 tons/year

^ 6 0 0 , „ Tc = — = 12years

1000 Qm - - ¡ y = 83.3 tons/year

g — 0.71bs/tons

Qr = 600 x 0.7 x 1.0 = 420 lbs

Pc = (25 - 5)420 - 2 x 600 - 1 x 1000 - 300 x 12

= $2600

Refinery limit:

gr = 0.16 lbs/ton

g = 0.581bs/tons

Qc = 0.84 x 1000 = 840 tons

Q r = 840 x 0.58 x 1.0 = 487.2 lbs

487.2 Tr = = 12.18years

40

1000 M — = 82.1 tons/year

12.18 n

840 C = = 69 tons/year

12.18 '

Pr = (25 - 5)487.2 - 2 x 840 - 1 x 1000 - 300 x 12.18

= $ 3 4 1 0


Table 6.20. Total profits as a function of concentrator cutoff with mine

operating at capacity. g

g Profits ($)

Pm Pc Pr

0.0 4000 1000 3250

0.1 4100 1700 3386

0.16 4064 2024 3410

0.2 4000 2200 3400

0.3 3700 2500 3287.50

0.4 3200 2600 3050

0.5 2500 2500 2687.50

0.6 1600 2200 2200

0.7 500 1700 1587.50

0.8 - 8 0 0 1000 850

0.9 - 2 3 0 0 100 -12 .50

0.95 - 3 1 2 5 - 4 2 5 -511.25

In summary, for each operation taken as a single constraint, the optimum cutoff grades are: yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA

g m = 0 . 1 0

gc = 0.40

gr = 0.16

The total profits assuming the single constraint of mining, concentrating or refining are

given as a function of cutoff grade in Table 6.20. The values have been plotted in

Figure 6.22.

Step 2. Determination of the economic cutoff grade by balancing the operations

In the first step, it was assumed that only one of the operations was the limiting factor to

production capacity.

A second type of cutoff is based simply on material balance. To be able to calculate this

one needs to know the distribution of grades of the mined material. The average grade of

the treated material can be found as a function of the chosen cutoff. The average grade, and

the number of units involved are given as a function of cutoff grade in Table 6.21.

For both the mine and mill to be at their respective capacities, then

Qm = 100 tons

Qc = 50 tons

As can be seen from Table 6.21, the cutoff grade should be 0.5 lbs/ton. This balancing cutoff

between mine and concentrator is expressed as gmc. For the concentrator and the refinery to

be at full capacity

Qc = 50 tons

Qr = 40 tons

The relationship between Qc and Or is shown in Table 6.22.

In examining Table 6.22, it can be seen that the required balancing cutoff grade gcr is

gcr = 0.60


Figure 6.22. Total profit as a function of cutoff grade under different constraints.

Table 6.21. Concentrator feed as a function of concentrator cutoff with mine operating at capacity.

Mined amount

(2m) (tons)

Concentrator cutoff

grade (gc) (lbs/ton)

Feed going to the

concentrator (<2c)(tons)

100 0 100

100 0.1 90

100 0.2 80

100 0.3 70

100 0.4 60

100 0.5 50

100 0.6 40

100 0.7 30

100 0.8 20

100 0.9 10

The final balancing cutoff is between the mine and the refinery. As seen in Table 6.23 (assum-

ing 100% concentrator recovery) a cutoff grade of 0.4 yields 42 lbs of product whereas 0.5

yields 37.5 lbs. The desired level of 40 lbs lies between these two. Interpolating one finds that

gmr = 0.456 lbs/ton


Table 6.22. Refinery product as a function of concentrator cutoff with concentrator

operating at capacity.

Amount to be Concentrator Avg. conc. Refinery

concentrated (Qc) cutoff grade (gc) feed grade (gc) product (Qr)

(tons) (lbs/ton) (lb/ton) (lbs)

50 0 0.5 25

50 0.1 0.55 27.5

50 0.2 0.5 30

50 0.3 0.65 32.5

50 0.4 0.7 35

50 0.5 0.75 37.5

50 0.6 0.8 40

50 0.7 0.85 42.5

50 0.8 0.9 45

50 0.9 0.95 47.5

Table 6.23. Refinery feed as a function of mine cutoff with the mine operating at

capacity (assuming 100% concentratory recovery).

Mined amount (Qm)

(tons)

Mine cutoff grade (gm)

(lbs/ton)

Refinery product (Qr)

(lbs)

100 0 50

100 0.1 49.5

100 0.2 48

100 0.3 45.5

100 0.4 42

100 0.5 37.5

100 0.6 32

100 0.7 25.5

100 0.8 18

100 0.9 9.5

In summary, when the operations are taken in combination, the optimum cutoff grades are:

gmc = 0.50

gcr = 0.60

gmr = 0-456

Step 3. Determining the overall optimum of the six cutoff grades

There are six possible cutoff grades. Three (gmc, gcr, and gmr) are based simply upon the grade

distribution of the mined material and capacities. The other three (gm, gc, and gr) are based

upon capacities, costs and the price. The objective is to find the cutoff grade which produces

the overall maximum profit in hght of the mining, concentrating and refining constraints.

The local optimums for each pair of operations are first considered. The corresponding

optimum grades for each pair (Gmc , Grc, and Gmr) are selected using the following rules:

gm if gmc < gm

Gmc = go if gmc > gc (6.26a)

gmc otherwise

Production planning 559

gr i f g r c < gr

gc i f g r c > gc zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA(6.26b)

grc otherwise

8m i f gmr < gm

gr if gmr > gr (6.26c)

§mr otherwise

The overall optimum cutoff grade G is just the middle value of Gmc, Gmr, and Grc. In our

example the six possible cutoff grades are:

gm = 0.10

gc = 0.40

gr ~ 0.16

gmc = 0.50

gmr = 0.456

gcr = 0.60

Consider them in groups of three:

g m = 0 . 1 0

gc = 0.40 Choose the middle value Gmc = 0.40

gmc = 0.50

gm = 0 . 1 0

gr = 0.16 Choose the middle value Gmr = 0.16

gmr = 0.456

g ryxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA = 0.10

gc — 0.40 Choose the middle value Gcr = 0.40

gcr = 0.60

Considering the three middle values

Gmc = 0.40

Gmr = 0.16

Gcr = 0.40

one chooses one numerically in the middle

G = 0.401bs/ton

From Table 6.24, the average grade gc of the material sent to the concentrator for a cutoff

of 0.40 lbs/ton would be

gc = 0 . 7 0 lbs/ton

For 100% recovery the quantities are

Qm = 1000 tons

Qc = 6 0 0 tons

Qr — 420 lbs

Group 1

Group2

Group3


Table 6.24. Grade distribution for the first 100 ton parcel mined.

Grade (lb/ton) Quantity (tons)

0.0-0.1 10

0.1-0.2 10

0.2-0.3 10

0.3-0.4 10

0.4-0.5 10

0.5-0.6 10

0.6-0.7 10

0.7-0.8 10

0.8-0.9 10

0.9-1.0 10

Qm = 100

Applying the respective capacities to these quantities one finds that

600

Tc= — = 12 years

4 2 0

Tr = — - = 10.5 years 40 1000

Tm = = 10 years 100 J

Since the concentrator requires the longest time, it controls the production capacity. The

total profit is

P = $2600

and the profit per year Py is

$2600 Px = = $216.70/year

12years

The net present value of these yearly profits assuming an interest rate of 15% is

(1 + f) 12 _ i i 1512 _ 1 N P V = ^ V r T p - = $ 2 1 6 - 7 0 o l 5 a l 5 F = $ 1 1 7 4 - 6 0

In summary: the concentrator is the controlling production limiter; concentrator

feed = 50 tons/year; optimum mining cutoff grade = 0.40 lbs/ton (constant); total con-

centrator feed = 600 tons; average concentrator feed grade = 0.70 lbs/ton; years of

production = 12 years; copper production/year = 35 lbs; total copper produced = 420 lbs;

total profits = $2600.40; net present value = $1174.60.

6.7.6 Net present value maximization

The previous section considered the selection of the cutoff grade with the objective being

to maximize profits. In most mining operations today, the objective is to maximize the net

present value NPV. In this section the Lane approach to selecting cutoff grades maximizing

NPV subject to mining, concentrating and refining constraints will be discussed.


For the example of the previous section a fixed mining cutoff grade of 0.40 lbs/ton was

used. One found that

Qm = 83.3 tons/y ear

Qc = 50tons/year

Q r = 351bs/year

/ = $300/year

Lifetime = 1 2 years

The yearly profit would be

Pj = 35.0 x $20 - 50 x $2 - 83.3 x $1 - $300

= $216.70

when finding the maximum total profit, the profits realized in the various years are simply

summed.

The total profit is, therefore

12

PT = Y^Pj = 1 2 X 216.7 = $2600.40

i= i

The net present value for this uniform series of profits (Chapter 2) is calculated using

NPV = P , ( 1 + I ' ) " - 1

1 ¿(1 + 0"

Assuming an interest (discount) rate of 15% one finds that

The question to be raised is 'Could the NPV be increased using a cutoff grade which, instead

of being fixed, varies throughout the life of the mine?' If so, then 'What should be the cutoff

grades as a function of mine life?' These questions have been addressed by Lane and are

the subject of this section.

Assume that just prior to mining increment Qm (shown to commence at time f = 0 in

Figure 6.23 for simplicity), the present value of all remaining profits is V. This is composed

of two parts. The first, PVp, is from the profit P realized at time T by mining the quantity

Qm. The second, PVW is obtained from profits realized by mining the material remaining

after time T. These profits are indicated as P\ occurring at time Tu P2 occurring at time

7?, etc., in Figure 6.23. The value of all these remaining profits for mining conducted after

t = T as expressed at time T is W. The present values of W and P, respectively, discounted

to time t = 0 are given by

pv~(f = 0 )=(IW (6-27)

pv'(i = 0) = a T ^ (6-28)

where d is the discount rate.


V - - V = PVP + PVW

PV, w

PVp ywusgaI

w

—i N

A y

Figure 6.23. A diagrammatic representation of the NPV calculation.

The present value at time t = 0 is therefore

W P

(6.30)

(6.31)

Since the present value at time t = T of the remaining reserves is W, the difference v

between the present values of the remaining reserves at times t = 0 and t = T is

v = V - W

Equation (6.29) can be rewritten as

W + P = V(l + d)T

Applying the binomial expansion to the term (1 + d)T one finds that

r (T - 1 )d2 T(T - 1 )(T - 2)d

3

(1 + d)T = 1 + Td + T - — + — 1 — + • • • (6.32)

For d small, (1 +d)T can be approximated by

(1 + d)T » 1 + Td (6.33)

Combining Equations (6.31) and (6.33) results in

W + P = V(1 + Td) = V + VTd

or

V -W = P- VTd (6.34)

Comparing Equations (6.30) and (6.34) one finds that

v — P — VTd (6.35)

The profit P obtained through the mining of Qm in time T is given as before by

P = {s — r)Qr — cQc — mQm —fT (6.36)

Combining Equations (6.35) and (6.36) yields

v = (s - r)Qr - cOc - mQm - T(f + Vd) (6.37)


One would now like to schedule the mining in such a way that the decline in remaining

present value takes place as rapidly as possible. This is because later profits get discounted

more than those captured earlier. In examining Equation (6.37), this means that v should

be maximized. As in the previous section one first takes the derivative of v with respect to

grade. Setting the derivative equal to zero, one can solve for the appropriate cutoff grades

subject to mining, concentrating and refining constraints.

Step 1. Deteimination of the economic cutoff grades - one operation constraining the total

capacity

(a) Calculate cutoff grade assuming that the mining rate is the governing constraint.

The time Tm is given by

ryrmgedcSRQ -9m M

Substituting this into Equation (6.37) yields

f+Vd vm = ( s - r)Qr - cQc - m + "

M Q,r

Differentiating (6.39) with respect to grade g gives

dv , d Qr

dg dg

d 2 c

dg yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA772 +

dQrr f + Vd

M J dg

However the quantity mined Qm does not depend upon the grade:

d Qn

dg = 0

Hence

dvm d Qr

— = (J - 7-) — dg dg

d2c

' d g

(6.38)

(6.39)

(6.40)

(6.41)

(6.42)

The relationship between the quantities refined Qr and those sent for concentration Oc is

Qr = Qcgcy (6.43)

where gc is the average grade of ore sent for concentration and y is the recovery in

concentration.

Thus

d2 r _ d2c -r~ = gcy-j-dg dg

Substituting Equation (6.44) into (6.42) yields

dv,„ r -, dOc — = [(i - r)gcy - cj —— dg L dg

The average grade gc is defined as the mining cutoff (breakeven) grade g„, when

dg

Setting Equation (6.45) equal to zero and solving for gc = gm one finds that

c gm — (s - r)y

(6.44)

(6.45)

(6.46)

(6.47)


Substituting the values from the example yields

$2 gm = = 0.10 lbs/ton S m ( $ 2 5 - $ 5 ) 1 . 0 1

(b) Calculate cutoff grade assuming that the concentrating rate is the governing constraint.

If the concentrator throughput rate is the limiting factor then the time T is controlled by the

concentrator.

Oc T = = £ (6.48)

where Qc is the total number of tons which will be sent to the concentrator, and C is the

tons/year capacity.


vc = (yxutsronljiebXVUTSONLKJIHFECAJ - r)Or - cQc - rnQm - ( / + dV)^ (6.49)

c+f + dV vc = ( s - r)Or

J— Qc-mQm-rn (6.50)

Since the total amount of material Om is fixed,

rnQm — const

Thus the cutoff grade affects only Qr and Qc.

Substituting as before

Qr = Qc'gcy

one finds that

f + dV

C

To make vc as large as possible the term

Vr = (s - r)gcy - c + Qc-mQm (6.51)

C

should be as large as possible. At breakeven (the cutoff grade), the term is zero. Thus

(s - r)gcy - c +

c + f+dV

gc = ' ; (6.52) y(s - r)

(c) Calculate cutoff grade assuming the refitting rate is the governing constraint. If the

refinery output is the limiting factor then the time T is controlled by the refinery,

O r

T = y (6.53)

where Qr is output of the refinery and R is the refining/sales capacity per year.

Substituting into Equation (6.37) yields yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA

v r = (J - r)Qr - cQc - mQm - (/ + dV) % (6.54) K

( f + dV\ v r = i s - r - 1 Or - cQc - rnQm (6.55)


As before

Qr = grl'Qc

Thus

/ f + dV\ vr = I s - r — I gryQc - cQc - mQm

The total amount of material in the pit is fixed, therefore

mQm = const

Maximizing the expression for Vr one finds that

f + dV\ s~r — I gryQc = cQc

Solving yields

^ = 7 j ^ y - (6.56)

In summary, this first type of cutoff grade determination is based upon finding the grade for

which the net increase in overall present value is zero. The expressions are as in formulas

(6.47), (6.52) and (6.56):

8m yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA(5 - r)y

gc =

ctreUSODC + f-±f

y(s - r)

c gr =

As can be seen, the expressions for gc and gr contain the unknown value of V.

Step 2. Determination of the economic cutoff grade by balancing the operations

This step is exactly the same as that discussed in the previous section. Hence only the results

will be presented here.

gmo = 0.50

gar = 0.60

gmr = 0.456

Step 3. Deteimining the optimum of the six cutoff grades

There are six possible cutoff grades. Three are based simply upon the grade distribution of

the mined material and capacities.

gmc = 0.50

gcr = 0.60

gmr = 0.456

566zyxwvutsrqponmlkjihgfedcbaYXWVUTSRQPONMLKJIHGFEDCBA Open pit inine planning and design-. Fundamentals

The other three are based upon capacities and cost/price. Substituting in the known values

one finds that

gm — 0.10

_ I f + dV O , 300 +0.15V = c _ 50

8c y(s-r) 1 . 0 ( 2 5 - 5 )

_ 8 +0 .003V

~~ 2 0

( 2 5 - 5 -300 + 0.15y

40 ) 1.0 12.5 — 0.003757

Of these, two of the limiting economic cut-off grades are not known initially since they

depend upon knowing the overall present value. This in turn depends upon the cutoff grade.

Since the unknown V appears in the equations an iterative process must be used.

An optimum grade will be determined for each of the three pairs of operations. This will

be followed by finding the optimum of the three final candidates. For the mine and the

concentrator considered as a pair, there are three possible candidates for the optimum cutoff

grade Gmc. These are gm, gc, and gnw. The following rules are used to select Gmc.

Gmc =

gm if gmc — gm

gc if gmc > gc

gmc otherwise

This simple sorting algorithm yields the middle value. Treating the concentrator and refinery

as a pair, the optimum Grc is found from

gr

gc

grc

if grc < gr

if grc > gc

otherwise

Finally the optimum cutoff grade Gmr when the mine and refinery are treated as a pair is

gm

Gmr =

if gmr < gm

if gmr > gr

otherwise

As the first step in the iteration process, it will be assumed that V = 0.

Applying these rules to the example values

g m = 0 . 1 0

gc = 0.40

gr = 0.16

gmc = 0.50

grc = 0.60

gmr = 0.456


Table 6.25. Concentrator product as a function of concentrator cutoff grade with mine operating at

capacity.

Concentrator Concentrator Average feed Concentrator/refinery cutoff feed grade product (lbs/ton) (tons) (lbs/ton) (lbs)

0.0 100 0.50 50 0.1 90 0.55 49.5 0.2 80 0.60 48

0.3 70 0.65 45.5 0.4 60 0.70 42 0.5 50 0.75 37.5

0.6 40 0.80 32

0.7 30 0.85 25.5

0.8 20 0.90 18

0.9 10 0.95 9.5

1.0 0 1.00 0

one finds that

Gmc = 0.40

Grc = 0.40

Gmr = 0.16

The overall optimum cutoff grade G is the middle value of Gmc, Gmr, and Grc.

G — middle value (Gmc, Gmr, Grc)

In this case

G = 0.40

Step 4. Calculation of quantities

The next step in the procedure is to determine the maximum quantities Qm, Qc and Qr

which could be produced and not violate the capacities. Assume that 100 tons are mined

(Qm — 100). The grade distribution of this material is as shown in Table 6.24.

From Table 6.25, a cutoff grade of 0.40 would yield an average feed grade of 0.70. Each of

the capacities must, however, be considered. A mining capacity (Qm = M) of 100 tons would

mean 60 tons to the concentrator and 42 product units. Both the concentrator and refinery

capacities are exceeded. In meeting the concentrator capacity (Qc = 50), the required mining

and refinery capacities are:

5 Qm = - X 100 = 83.3

6

Qr = 50 x 0.70 = 35

These are both less than the maximum values. In meeting the refinery capacity of Qr = 40,

the required concentrating and mining capacities are:

40 Qc = — = 57.1

0.7

57.1 Qm = — 1 0 0 = 95.2


Thus the concentrating capacity is violated. The result is that the concentrator is the

bottleneck.

In the further calculations

Qm = 83.3

Qc = 50

Qr = 35

The profit from time period T is expressed as

P = (s — r)Qr — cQc — mQm —fT

For T = 1 year one finds that

P = (25 - 5) x 35 - 2 x 50 - 1 x 83.3 - 300 x 1 = $216.7

Since the total amount of material to be mined from the pit is Q — 1000 units, the number

n of years required is

1000 n = = 12 years

83.3

The present value V corresponding to 12 equally spaced payments of P — $216.7 using an

interest rate of 15 percent is

216.7[1.1512 — 1]

0 .15(1.15)" = $ 1 1 7 4 - 6

This value of V becomes the second approximation of V (the first was V = 0) for use in the

formulas to calculate gc and gr.

c | f+dV 2 | 300+0.15x1174.6

c ' 50 = 0.576 oc

y(s - r) 2 5 - 5

c 2 _ S r = 7 7 = on 300+0.15x1174.6 =

S ~ ryrmgedcSRQ -R-y 40

The new six choices become

gm = 0.10

gc = 0.576

= 0.247

gmc = 0.50

grc =zywvutsrqponmlkjihgfedcbaZYXVUTSRQPONMLKJIHFEDCBA 0.60

gmr = 0.456

Applying the rules to select the overall pair optimum yields

Gmc = 0.50

Grc --- 0.576

Gmr = 0.247


The overall optimum is the middle value

G = 0.50

Returning to the grade distribution Table 6.24 one finds that the average grade is 0.75. If the

mining rate Qm = 100, then Oc = 50 and Or = 37.5. Both the mine and the concentrator are

at their rated capacities.

The profit in a given year is

P ~ (s — r)Qr — cQc — mQm —fT

= 20 x 37.5 - 2 x 50 - 1 x 100 - 300

= $250

The number of years is

Q 1000 n = —— = — — = 10 years

Qm 100

The present value becomes

1.1510 — 1 V = 2 5 ° 0 . 1 5 x 1 . 1 5 - 0 - $ 1 2 5 4 - 7

This becomes the third estimate for V to be used in calculating gc and gr.

2 300+0.15x1254.7

= ¿ f = 0.588 zywvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLJIHGFEDCBA

2

2 0 - ™ 8 r 300+0.15x1254.7 — 0 . 2 5 7 40

The six possible values are:

Em = 0.10

gc = 0.588

g r = 0.257

gmc = 0.50


gmr = 0.456

The optimum pairs are:

Gmc = 0.50

Grc = 0-588

Gmr - 0.257

and the overall optimum (G = 0.50) is the same as found with the previous estimate. Hence

in year 1

Optimum cutoff grade = 0.501bs/ton

Quantity mined = 100 tons

Quantity concentrated = 50 tons

Quantity refined = 37.5 lbs

Profit = $250


Table 6.26. Reserve distribution at the end of year 1.


0.0-0.1 90

0.1-0.2 90

0.2-0.3 90

0.3-0.4 90

0.4-0.5 90

0.5-0.6 90

0.6-0.7 90

0.7-0.8 90

0.8-0.9 90

0.9-1.0 90

Total = 900

Table 6.27. Reserve distribution at the start of year 8.


0.0-0.1 30

0.1-0.2 30

0.2-0.3 30

0.3-0.4 30

0.4-0.5 30

0.5-0.6 30

0.6-0.7 30

0.7-0.8 30

0.8-0.9 30

0.9-1.0 30

Total = 300

The reserves must now be adjusted to those given in Table 6.25 and the process is repeated

assuming V = 0, calculating gc and gr, etc.

Through year 7, it will be found that the optimum cutoff grade remains at 0.50 with the

quantities mined, concentrated and refined being 100,50 and 37.5, respectively. The annual

profit is $250. The reserves going into year 8 are those given in Table 6.27. The balancing

grades remain at

gmc = 0.50 zywvutsrqponmlkjihgfedcbaZYXVUTSRQPONMLKJIHFEDCBA

= 0.60

gmr = 0.456

The first approximation for the economic cutoff grades (V = 0) is

gm

8c

gr

= 0.10

= 0.40

= 0.16


The optimum values of the pairs are

Gmc = GW = 0.40

Gmr — 0.16

The overall optimum is 0.40, and the quantities are

Qm = 83.3

Qc = 50

Qr = 35

The profit is $216.7 as before. The number of years becomes 3 0 0

n = 8 3 3 = y e a r s

The present value V becomes

f l 15-)3-6 - 1 V = 216 .7—— — = 571.2

0.15(1.15)3-6

Substituting this into the formulas for gc and gr yields

gc = 0.486

gr = 0.193

Combining them with the others

gm = 0.10

gmc = 0.50


gmr = 0.456

yields

Gmc = 0.486

Grc = 0.486

Gmr = 0.193

The overall optimum cutoff is G — 0.486 and the average grade above cutoff drops to 0.743:

Tons Grade Tons x Grade

4.2 0.493 2.07

30 0.55 16.50

30 0.65 19.50

30 0.75 22.50

30 0.85 25.50

30 0.95 28.50

Total =154 .2 Avg=: 0.743 Sum =114.57

There are 154.2 ore tons out of the 300 tons remaining to be mined. Since the concentrator

capacity is 50 tons/year, the mine life would be

154.2

50 = 3.08 years

572zyxwvutsrqponmlkjihgfedcbaYXWVUTSRQPONMLKJIHGFEDCBA Open pit mine planning and design: Fundamentals



0.0-0.1 20.27

0.1-0.2 20.27

0.2-0.3 20.27

0.3-0.4 20.27

0.4-0.5 20.27

0.5-0.6 20.27

0.6-0.7 20.27

0.7-0.8 20.27

0.8-0.9 20.27

0.9-1.0 20.27

Total = 202.27

The yearly mine production becomes

300 Om - = 97.3

3.08

and Qr — 37.15. Calculating the profit one finds that

P = 20 x 37.15 - 2 x 50 - 1 x 97.3 - 300 = 245.7

The corresponding present value is

(1 15)3 08 - 1 V = 245.7— r— = $572.96

0.15 x (1.15)3 08

Repeating the process with this new estimate of V yields

gc = 0.486

g r = 0.193

These are the same as before. Hence the values for year 8 are

G = 0.486

2m = 9 7

Qc — 50

a - = 37.1

Profit = $245.7

The reserves are those given in Table 6.28. In year 9, the initial values for a cutoff of 0.4 are

Qm = 83.3

Qc = 50

Qr — 35

The profit would be $216.7.

Based upon this mining rate, the reserves would last

n = ^ y = 2.43 years


and the present value is

(1 15)2-43 - 1 V = 216.7——— — = $416.0

0.15 x (1.15)2-43

Recomputing gr and gc we obtain

gc = 0.462

gr = 0.183

The other possible values are:

gmyxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA — 0.10

gmc = 0.50

grczywvutsrqponmlkjihgfedcbaZYXVUTSRQPONMLKJIHFEDCBA = 0.60

gmr = 0.456

The optimum pair values are:

Gmc - 0.462

G^ = 0.462

Gmr = 0.183

The middle value of these is

G = 0.462

Examining the reserve distribution suggests that there are 109.05 tons out of the total

202.7 tons which are above cutoff. The average grade of this remaining ore is 0.731:

Tons Grade Tons x Grade

7.70 0.48 3.70

20.27 0.55 11.15

20.27 0.65 13.18

20.27 0.75 15.20

20.27 0.85 17.23

20.27 0.95 19.26

Total =109.05 Avg = 0.731 Sum = 79.72

Since the maximum concentrating rate is 50 tons/year the life is

109.05 „ i n = 2.18 years

50 J

The amount of product is

Qr = 50 x 0.731 = 36.55

202.7 _ Qm = = 93 * 2.18

The profit becomes

P = 20 x 36.55 - 2 x 50 - 1 x 93 - 300 = $238




0.0-0.1 11

0.1-0.2 11

0.2-0.3 11

0.3-0.4 11

0.4-0.5 11

0.5-0.6 11

0.6-0.7 11

0.7-0.8 11

0.8-0.9 11

0.9-1.0 11

Total =110

The present value is

(1.15)2'18 — 1 „ V = 2 3 8 0 . 1 5 x l . l 5 ^ X $ 4 1 7

Iterating again does not change the values. The new distribution is shown in Table 6.29.

In year 10, the initial values for a cutoff of 0.4 yields:

Qm = 83.3

Qc = 50

Q r = 35

Profit = $216.7

Based upon this mining rate, the reserves would last yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA

1 1 0 1 OO n = = 1.32 years 83.3


(1 15)1-32 - 1 V = 2 1 6 - 7 0 T 5 x ( 1 . 1 5 ) . - 3 2 - $ 2 4 3 - 4

Calculating gc and gr and using this approximation for V yields

gc = 0.437

gr = 0.172

The other possible values are:

g m = 0 . 1 0

gmc = 0.50

grc - 0.60

gmr = 0.456


The optimum pair values are:

Gmc = 0.437

Grc = 0.437

Gmr = 0.172

The optimum value is G — 0.437. Examining the reserve distribution suggests that there are

62 tons of the 110 tons remaining which are above this cutoff.

Tons

7

Grade

0.469

0.55

0.65

0.75

0.85

0.95

Tons x Grade

3.28

6.05

7.15

8.25

9.35

10.45

Total = 62 Avg = 0.718 Sum = 44.53

The average grade is 0.718. The number of years would be

62 72 = — = 1.24 years

50 '

The rate of mining and refining would be

110 Qr, 89

1.24

Qr = 35.9

and the profit would become

P = 20 x 35.9 - 2 x 50 - 1 x 89 - 300

= $229


(1.15)124 — 1 y = $229- = $243

0.15 x (1.15)1-24

Further iteration yields no change. In year 11, the grade distribution is shown in Table 6.30.

The initial values (V = 0), yield a cutoff of 0.4 and

Qm =zywvutsrqponmlkjihgfedcbaZYXVUTSRQPONMLKJIHFEDCBA 21

Qc = 12.6

QroUTSONMLIDCA = 8.8

The time would be the largest of

Tm = — = 0.21 yxwvutsrponmlkjihgfedcbaYWVUTSRQPONMLKJIHGFEDCBA100

12.6 T = = 0 25

c 50

576zyxwvutsrqponmlkjihgfedcbaYXWVUTSRQPONMLKJIHGFEDCBA Open pit mine planning and design: Fundamentals



0.0-0.1 2.1

0.1-0.2 2.1

0.2-0.3 2.1

0.3-0.4 2.1

0.4-0.5 2.1

0.5-0.6 2.1

0.6-0.7 2.1

0.7-0.8 2.1

0.8-0.9 2.1

0.9-1.0 2.1

Total = 21.0

which is again controlled by the concentrator. The profit is

300 P = 20 x 8.8 - 12.6 x 2 - 21 x 1

4

= $54.8


d 15)0-25 - 1 V = $ 5 4 - 8 0 C T 5 ^ = $ 1 2 - 5

Solving for gc and gr yields

= 0.402

gr = 0.161

Combining with the others

gmzywvutsrqponmlkjihgfedcbaZYXVUTSRQPONMLKJIHFEDCBA = 0.16

gmc = 0.50

g t c = 0.60

gmr = 0.456

one finds that

Gmc = 0.402

Grc = 0.402

Gmr = 0 . 1 6 1

The cutoff grade is

G = 0.402

The distribution is only slightly changed and farther iteration is not warranted.


Table 6.31. The production schedule determined by the first pass.

Year Optimum Quantity Quantity Quantity Profit Net present

cutoff grade mined concentrated refined zywvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLJIHGFEDCBA($) value

(lbs/ton) (tons) (tons) (lbs)

1 0.5 100 50 37.5 250 1255 2 0.5 100 50 37.5 250 1193 3 0.5 100 50 37.5 250 1122 4 0.5 100 50 37.5 250 1040

5 0.5 100 50 37.5 250 946

6 0.5 100 50 37.5 250 838 7 0.5 100 50 37.5 250 714

8 0.486 97 50 37.1 245.7 574

9 0.462 93 50 36.55 238 417

10 0.437 89 50 35.9 229 243

11 0.40 21 12.6 8.8 55 53

Table 6.32. The final schedule for the manual example.

Year Optimum Quantity Quantity Quantity Profit Net present

cutoff grade mined concentrated refined ($) value

(lbs/ton) (tons) (tons) (lbs)

1 0.50 100.0 50.0 37.5 250.0 1257.8

2 0.50 100.0 50.0 37.5 250.0 1196.5

3 0.50 100.0 50.0 37.5 250.0 1126.0

4 0.50 100.0 50.0 37.5 250.0 1044.9

5 0.50 100.0 50.0 37.5 250.0 951.7

6 0.50 100.0 50.0 37.5 250.0 844.5

7 0.50 100.0 50.0 37.5 250.0 721.1

8 0.49 97.2 50.0 37.1 245.6 579.3

9 0.46 93.0 50.0 36.6 238.2 420.6

10 0.44 88.7 50.0 35.9 229.5 245.5

11 0.41 21.0 12.5 8.8 54.7 52.8

The net present value is calculated using the yearly profits.

_ (1.15)7 - 1 245.7 238 229 55 NPV — 250— 1 1 1 1

0.15 x (1.15)7 (1.15)8 (1.15)9 ^ ( L 1 5 ) i o ^ (i.i5)io.25

= 1040 + 80.32 + 67.70 + 56.61 + 13.12

= $ 1 2 5 8

Step 5. Repetition of the iteration process

In Table 6.31, the present value column reflects the current approximation to V as each years

cutoff grade was calculated. The present value of $1258 obtained using the yearly profits

should be the same as that shown in the table for year 1. Since the values are not the same

($1258 versus $1255), the process is repeated from the beginning using V = $1258 as the

initial estimate for V. Using a computer this iterative procedure is completed in fractions

of a second. The final results are shown in Table 6.32. The NPV is slightly higher than the

$1255 which would have been obtained by maintaining a constant cutoff grade of 0.5.


In summary:

- Initially the mine and the concentrator are in balance, both operating at capacity. In the

last few years, the concentrator is the limiter.

- The cutoff grade begins at 0.50 lbs/ton and drops to 0.41 lbs/ton at the end of mine life.

- Mine life is slightly more than 10 years.

- Total copper produced = 380.9 lbs.

-Tota l profits = $2518.

- Net present value = $1257.80.

This net present value should be compared to that of $1174.60 obtained with the fixed

cutoff grade.

6.8 MATERIAL DESTINATION CONSIDERATIONS

6.8.1 Introduction

The term 'cutoff grade' is a rather poorly defined term in the mining literature. A major

reason for this is that there are many different cutoff grades. Furthermore the values change

with time, mining progress, etc. A cutoff grade is simply a grade used to assign a destination

label to a parcel of material.

The destination can change. During the evaluation of final pit limits, the destinations to

be assigned are:

- to the surface, and

- left in the ground.

Once the destination 'to the surface' has been assigned, then the destination label 'where

on the surface' must be assigned as well. In the distant past there were really only two

surface destinations:

- to the mill, and

- to the waste dump.

A grade was used to assign the location. The distinction between destinations was called

the mill cutoff grade. In more recent times, the potential future value of material carrying

values has been recognized. Hence the lean (low grade) ore dump has become a destination.

Thus the 3 destinations require 2 distinguishing grades:

Destination

- to the mill

- to the lean ore dump

- to the waste dump

Today there are many more possible destinations as our ability to handle and treat materials

have improved. Leach dumps/leach pads are a common destination. An active stockpile is

a less common destination.

This section will deal with alternate destinations to the mill and waste dump. These will

be discussed with respect to cutoff grade. However the reader should remember that these

simply are a way of assigning material destinations.

Assignment

mill cutoff grade

waste cutoff grade

hustrulid and kuchta

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