Pit & Cutoff Optimisation
Work Book
Prepared by
Norm Hanson Whittle Consulting Pty Ltd For
University of Queensland Sept 2008
Worksheet 1
The Value Of Truck Load Of Ore Calculate the value (which is Revenue  Costs) of a single truck load of ore in three example cases 1. 50 tonnes of 2g/t Gold 2. 100 tonnes of 1 g/t Gold 3. 250 tonnes of 0.5% g/t Gold Before you begin the calculation can you guess which is the most valuable truck load? Costs $ 1.50 $17.50
Mining Processing (CIP)
Gold US$900/oz Prices Grams 31.103 grams per oz 2240 lbs per tonnne
Unit Prices Recoveries (CIP)
Value =
92.5%
Revenue

Costs
= [(GRADE* ORE* PRICE*RECOVERY)  (ORE*COSTP)] ROCK*COSTM) Example 1  2 grams per tonne in 50 tonnes = [(2* 50 * PriceAu * 92.5%)  (50 * $17.5)] (50 * $1.50) =
Example 2  1 gram per tonne in 100 tonnes = [(1*100 * PriceAu * 92.5%)  (100 * $17.5)] (100 * $1.50) =
Example 3  0.5 grams per tonne in 250 tonnes = [(0.5*250 * PriceAu * 92.5%)  (250 * $17.5)] (100 * $1.50)
Worksheet 2 The Value with MultiMetal, Destinations If you have more than one metal or more than one possible 4. 250 Tonnes of 0.25% Copper & 0.5 g/t Gold from an Oxide Zone In this example the truck could go to two different processes, a floatation (but since it is oxide ore and has lower recoveries) or SXEW (a form of heap leach with a solvent extraction)
Mining Processing (SXEW) Processing (Floatation)
Costs $ 1.50 $ 5.00 $ 8.00 Gold US$400/oz
Copper US$1.20/lb
Prices Grams Unit Prices Recoveries (SXEW) Recoveries (Floatation)
Value =
Revenue
Tonnes 31.103 grams per oz 2240 lbs per tonnne
65% 25%
30% 75%

Costs
= [(GRADE* ORE* PRICE*RECOVERY)  (ORE*COSTP)] ROCK*COSTM) Example 4a  0.5 grams Au plus 0.25% Cu per tonne in 200 tonnes to SXEW = [(0.5*250 * PriceAu * 65% + 0.25%*250*PriceCu*30%)  (250 * $5)] (250 * $1.50)
Example 44b .5 grams Au plus 0.25% Cu per tonne in 200 tonnes to flotation = [(0.5*250 * PriceAu * 25% + 0.25%*250*PriceCu*75%)  (250 * $8)] (250 * $1.50) =
Worksheet 3
Simple Example Let us consider a very simple ore body which is rectangular, of constant grade, and sits beneath a horizontal topography. Let us further assume that it is of infinite length, so that we do not have to allow for end effects, and need only consider one section. The following diagram shows such an ore body.
Surface
Bench Level 100 tonnes waste
500 tonnes ore Using the quantities indicated in the diagram, we can easily calculate the tonnages for the eight possible pit outlines, and the following table shows these. Tonnages for possible outlines 2 3 4 5
Pit
1
Ore Waste
500 100
1,000 400
Total
600
1,400
900
1,600
2,500
6
7
8
3,600
4,900
6,400
If we assume that ore is worth $2.00 per tonne and that waste costs $1.00 per tonne to mine and remove, then the following table shows the value of each pit.
Pit Value
Pit values for ore at $2.00/T and waste at $1.00/T 1 2 3 4 5 6 900
1,600
7
8
Worksheet 4
Floating Cones Trace around the Optimal Pit
Case A 30 80
80
+100
+100
Case B 20
30 80 70
+200
+40
+60
Which Case Overmines Which Case Undermines
Worksheet 5
Two Dimensional LG To see how the LerchsGrossmann method works, we will use a two dimensional example. To keep the example simple we will use squares and assume a 45o slope.
As an example we can have a two dimensional model 17 blocks long by 5 blocks high. Only three blocks contain ore and they have the values shown (after mining). All other blocks are waste and have the value 1.0.
23.9
6.9
Structural arcs for each block
Draw the LerchGrossman network and determine the most profitable pit.
23.9
Worksheet 6
Handling Variable Slopes Blocks whose centroid is above the desired slope line must be mined, Those below the line do not need to be mined.
Z Desired Slope
1,2 1,1
2,1
3,1
etc
How arcs can be referenced.
1, 1 1, 1
> >
1, 2 2, 2
the third arc from 1,1 is:
>
X
Worksheet 8
The Sequence of Mining If we have another look at our simple example pit design, we can look at the discounted cash flow and its interaction with the sequence of mining and production rates
Surface
Bench Level 100 tonnes waste
500 tonnes ore
We will use a discount rate of 10% Top Down Mining Sequence Pit
1
Value Mill Capicty 500 tpp Mill Capicty 1000 tpp
900 900 900
2
3
4
5
6
7
8
1,600 1,530 1,530
2,100 1,917 1,868
2,400 2,085 2,011
2,500 2,057 1,956
2,400 1,851 1,552
2,100 1,486 1,049
1,600 978 449
Inner Shell First then Mining out by Shell Pit
1
Value Mill Capicty 500 tpp Mill Capicty 1000 tpp
900 900 900
2
3
4
5
6
7
8
1,600 1,530 1,530
2,100 1,935 1,898
2,400 2,154 2,068
2,500 2,220 2,105
2,400 2,161 1,993
2,100 2,002 1,790
1,600 978 449
Top Down Mining is the
WORST CASE NPV
BEST CASE NPV
Inner Shell out Mining is the
WORST CASE NPV
BEST CASE NPV
Worksheet 7
Discounted Cash Flow Analysis Suppose we have a cash flow of $1 million for ten years. Year 1 Year 2 Year 3 Actual Cash flow 1.00 1.00 1.00
Year 4
Year 5
Year 6
Year 7
Year 8
Year 9
Year 10
1.00
1.00
1.00
1.00
1.00
1.00
1.00
0.53
0.48
0.43
0.39
0.35
Simple Discount Factor 10% 0.90 0.81 0.73 Discounted Cash flow 0.90 0.81 0.73
(1D/100) 0.66 0.66
"Financial" NPV Factor 10% 0.91 0.83 0.75 Discounted Cash flow
10.00
0.59 0.59
0.53
0.48
0.43
0.39
0.35
0.51
0.47
0.42
0.39
NPV 5.86
(1/(1+D/100)) 0.68
0.62
0.56
NPV
Worksheet 9
CutOff Grades Value = [(METAL* PRICE * REC)  (ORE*COSTP)] (ROCK*COSTM) Breakeven situation is when Revenue from Recovered Metal = The Cost of Processing
METAL* PRICE*REC
= ORE*COSTP
(GRADE * ORE )* PRICE*REC
= ORE*COSTP
GRADE
=
COSTP * ORE / (PRICE* REC) * ORE
GRADE
=
COSTP / (PRICE* REC)
CUTOFF GRADE
=
=
$15 / ( $12.70* 92.5%)
Worksheet 10
CutOff Grades Label this graph
Worksheet 11
Delay Cost We again have a cash flow of $1 million for ten years. Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Actual Cashflow 1.00 1.00 1.00 1.00 1.00 1.00 1.00 Simple Discount Factor (10%) 0.91 0.83 0.75 Discounted Cashflow 0.91 0.83 0.75
Year 8
Year 9
Year 10
TOTAL
1.00
1.00
1.00
10.00
0.68
0.62
0.56
0.51
0.47
0.42
0.39
0.68
0.62
0.56
0.51
0.47
0.42
0.39
0.35
Now consider what happens to NPV if we delay production Delay in First Year 0.83 0.75 Cost of Delay Delay in Seventh Year 0.91 0.83
0.47
0.42
0.39 0.35 Cost of Delay
6.14
Worksheet 12
Change (Cost) If we could somehow receive a higher cashflow in the first five years (say $1.2 million) and then accepting a lower cash flow later (eg $0.8 million), what happens to the NPV? This is like starting with a high price which falls over time.
Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 Actual Cashflow 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 Discount Factor 0.91 0.83 0.75 0.68 0.62 0.56 0.51 0.47 0.42 0.39 Discounted Cashflow 0.91 0.83 0.75 0.68 0.62 0.56 0.51 0.47 0.42 0.39 Our Desired CashFlow 1.20 1.20 1.20 Discount Factor 0.91 0.83 0.75 Discounted Cashflow
1.20
1.20
0.8
0.8
0.8
0.8
0.8
0.68
0.62
0.56
0.51
0.47
0.42
0.39
Change in NPV
10.00
6.14