Pit & Cut-off

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 Mining $ 1.50 Processing (CIP) $17.50 Gold Prices

US$900/oz

Grams Unit Prices

Recoveries (CIP) 92.5%

Value = 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)

31.103 grams per oz2240 lbs per tonnne

Worksheet 2 The Value with Multi-Metal, 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) Costs Mining $ 1.50 Processing (SXEW) $ 5.00 Processing (Floatation) $ 8.00 Gold Copper Prices

US$400/oz US$1.20/lb

Grams Tonnes Unit Prices

Recoveries (SXEW) 65% 30% Recoveries (Floatation) 25% 75%

Value = Revenue - 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) =

31.103 grams per oz2240 lbs per tonnne

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

100 tonnes waste

500 tonnes ore

Bench Level

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 Pit

1 2 3 4 5 6 7 8

Ore 500 1,000 Waste

100 400 900 1,600 2,500 3,600 4,900 6,400

Total 600 1,400 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 values for ore at $2.00/T and waste at -$1.00/T Pit

1 2 3 4 5 6 7 8

Value 900 1,600

Worksheet 4

Floating Cones

Trace around the Optimal Pit

Case A

Case B

+40

+60

+200-70

-80

-30-20

Which Case Overmines Which Case Undermines

-30-80 -80

+100 +100

Worksheet 5

Two Dimensional L-G To see how the Lerchs-Grossmann 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.

6.9 23.923.9

Draw the Lerch-Grossman network and determine the most profitable pit.

Structural arcs for each block

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.

DesiredSlope

1,1 2,1 3,1 etc

1,2

X

Z

How arcs can be referenced.

1, 1 > 1, 2 1, 1 > 2, 2 the third arc from 1,1 is:

>

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

100 tonnes waste

500 tonnes ore

Bench Level

We will use a discount rate of 10%

Top Down Mining Sequence Pit

1 2 3 4 5 6 7 8

Value 900 1,600 2,100 2,400 2,500 2,400 2,100 1,600 Mill Capicty 500 tpp 900 1,530 1,917 2,085 2,057 1,851 1,486 978 Mill Capicty 1000 tpp 900 1,530 1,868 2,011 1,956 1,552 1,049 449

Inner Shell First then Mining out by Shell

Pit

1 2 3 4 5 6 7 8

Value 900 1,600 2,100 2,400 2,500 2,400 2,100 1,600 Mill Capicty 500 tpp 900 1,530 1,935 2,154 2,220 2,161 2,002 978 Mill Capicty 1000 tpp 900 1,530 1,898 2,068 2,105 1,993 1,790 449

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 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 Actual Cash flow

1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 10.00 Simple Discount Factor 10% (1-D/100)

0.90 0.81 0.73 0.66 0.59 0.53 0.48 0.43 0.39 0.35 Discounted Cash flow NPV 0.90 0.81 0.73 0.66 0.59 0.53 0.48 0.43 0.39 0.35 5.86

"Financial" NPV Factor 10% (1/(1+D/100))

0.91 0.83 0.75 0.68 0.62 0.56 0.51 0.47 0.42 0.39 Discounted Cash flow NPV

Worksheet 9 Cut-Off 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) = $15 / ( $12.70* 92.5%)

CUT-OFF GRADE =

Worksheet 10

Cut-Off 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 Year 8 Year 9 Year 10 TOTAL Actual Cashflow 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 10.00 Simple Discount Factor (10%) 0.91 0.83 0.75 0.68 0.62 0.56 0.51 0.47 0.42 0.39 0.35 Discounted Cashflow 0.91 0.83 0.75 0.68 0.62 0.56 0.51 0.47 0.42 0.39 6.14

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

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 10.00 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 6.14 Our Desired CashFlow

1.20 1.20 1.20 1.20 1.20 0.8 0.8 0.8 0.8 0.8 Discount Factor

0.91 0.83 0.75 0.68 0.62 0.56 0.51 0.47 0.42 0.39 Discounted Cashflow

Change in NPV