indian institute of technology bombay mineral prospectivity mapping alok porwal centre of studies in...

Indian Institute of Technology Bombay


MINERAL PROSPECTIVITY MAPPING

Alok Porwal

Centre of Studies in Resources Engineering,

Indian Institute of Technology (Bombay) India



• Introduction

GIS-based mineral prospectivity mapping

Generalized methodology

• Weights of evidence model

• Fuzzy model

• Other models (if time)



Remote SensingGeophysics

Geochemistry

Geology

Garbage In,

Garbage Out

Mineral potential maps

GIS

Analyse / Combine

Good Data In, Good

Resource Appraisal Out

Mineral potential maps

GIS

Analyse / Combine

Remote SensingGeophysics

Geochemistry

Geology



database

Predictor maps

Continuous-scale favorability map

Binary favorability map

MINERAL POTENTIAL MAP

MODEL

Mineralization processes

Conceptual models

Knowledge-base

Mappable exploration criteria

Spatial proxies

Processing

Overlay

Validation

Systematic GIS-based prospectivity mapping



• Deposits focal points of much larger mass flux and energy systems

• Focus on critical processes that must occur to form a deposit

• Allows identification of mineralization processes at all scales

• Not restricted to particular geologic settings/deposit type

• Amenable to probabilistic analysis

Fluid Discharge

Ligand source

Metal source

Model I

Model II

Model III

Trap Region

Energy(Driving Force)

No Deposits

Mineral System(≤ 500 km) Deposit Halo

Deposit(≤ 10 km)

(≤ 5 km)

1. Energy 2. Ligand 3. Source 4. Transport 5. Trap 6. Outflow

Transporting fluid

Mineral systems approach (Wyborn et al. 1995)



Fluid Discharge

Metal source

Model I

Model II

Model III

Trap Region

Energy(Driving Force)

Mineral System

(≤ 500 km)

Deposit Halo

Deposit(≤ 10 km)

(≤ 5 km)

1. Energy 2. Ligand 3. Source 4. Transport 5. Trap 6. Outflow

Transporting fluid

Mineral systems approach :Processes to Proxies

Deformation Metamorphism

Magmatism

Connate brinesMagmatic fluidsMeteoric fluids

Enriched source rocks

Magmatic fluids

Structures Permeable

zones

Structures Chemical

traps

Structures aquifers

MA

PPA

BLE

C

RIT

ER

IA

Metamorphic grade, igneous

intrusions, sedimentary

thickness

Evaporites, Organics, isotopes

Radiometric anomalies,

geochemical anomalies, whole-rock

geochemistry

Fault/shear zones, folds geophysical/ geochemical anomalies, alteration

Dilational traps, reactive

rocks, geophyiscal/ geochemical anomalies, alteration

magnetic/ radiometric/ geochemical anomalies, alteration, structures

SPA

TIA

L PR

OX

IES

Ligand/brine

source



Broad Regiona

l

Prospect

Scale dependence of exploration criteria

(e.g. orogenic gold system - McCuaig and Beresford, 2009)

‘Camp’ 2-D GIS-based prospectivity mapping






MODEL


Conceptual models

Knowledge-base


Spatial proxies

Overlay

Validation

database

Predictor maps

Processing

Model-based mineral potential mapping


Indian Institute of Technology Bombay Primary

data

solid geology

structural geology

airborne magnetic

gravity

reactivity (Fe2+/(Fetot+ Mg + Ca) contrast

distance to faults

strike of nearest fault

rheological contrast

Empirical

fav. tectonic environment

fav. host rock lithology

Conceptualfav. litho. contact type

Creation of candidate layers in GIS: Feature extraction What are the exploration criteria for the mineralisation?



11Proxies to Predictor Maps: Systematic

Analysis of Exploration Data

GIS

geology(1-of-n coded)

magnetic anomalygamma-ray channels total countdistance to nearest faultKThU

input layer hiddenlayer outputlayer

outputinternal biases = 1

Test/Analyse

Derivative predictor maps

Remote SensingGeophysicsGeochemistry Geology

• Hypothesis testing• Select input layers

• Create candidate layers (Feature extraction based on mineral systems model)

• Input layers



12

For each exploration criteria, define a hypothesis in terms of GIS layer(s)

- Generate all appropriate layers

- Examine each layer’s spatial relationship to the targeted mineral deposits- Quantify results

- Define an experimental plan

Hypothesis incorrect Hypothesis correct

Factor identified,

keep GIS layer for

prospectivity mapping

Discard layer

Red

efi

ne h

yp

oth

esis



Example: Rheology as an exploration criteria for gold trap

4

1

5Basalt

Granite

Komatiite

GRANITE-GREENSTONE

BELTS

UltramaficFelsic

Volcanics

High-Mg

basalts

Gneiss Foliated

monzogranite

Monzogranite

Granodiorite

Tonalite

Dolerite

Granophyric

dolerite

BIF

Volcaniclastic/

clastic sedimentary

rocks

Andesitic

Porphyry

Dolomite-Qtz rock

Tholeiitic basalt

5

3

2

1

0

4

1 2 3 4 5

Reacti

vit

y

(Fe x

[Fe /

Fe+

Mg

+C

a])

Rheological strength

(Provided by David Groves)



14Experiment 1: Host Rock Rheology

Expected Results

0.00

0.20

0.40

0.60

0.80

1.00

1 2 3 4 5Rheology Value

Sp

ati

al associa

tion

Hypothesis: More deposits in high rheological strength rocks

4

1

5Basalt

Granite

Komatiite



15

Experiment 1: Host Rock Rheology

The Results

0.00

0.20

0.40

0.60

0.80

1.00

1 2 3 4 5

Rheology Value

Edjudina

Kalgoorlie

Laverton

Leonora

Yandal

Sp

ati

al associa

tion

Hypothesis rejected!


Indian Institute of Technology Bombay Hypothesis: More deposits adjacent to

lithological contacts that have a higher rheological contrast

Experiment 2: Rheological Contrast

Expected Results

0.00

0.20

0.40

0.60

0.80

1.00

1 2 3 4 5Rheology contrast across geological contacts

Sp

ati

al associa

tion

4

1

5Basalt

Granite

Komatiite



The Results

0.00

0.20

0.40

0.60

0.80

1.00

0 1 2 3 4

Rheology Contrast at lithological contactsSp

ati

al associa

tion

Edjudina

Kalgoorlie

Laverton

Leonora

Yandal

Experiment 2: Rheological Contrast

Hypothesis rejected!



18

Change in Philosophy

Frequent change

in rheology

Increased fluid flow

Deposit localization

Region of increased fluid

flow



19

Hypothesis: Expect deposits to be more common in regions of greater density of rheological contrast.

Experiment 3: Rheological contrast density

Expected Results

0.00

0.20

0.40

0.60

0.80

1.00

1 2 3 4 5Geological contact density weighted by rheology contrast

Sp

ati

al associa

tion



20

0 1 2 3 4

Rheology contrast at lithological contacts Rheology contrast density

LOW HIGH

Experiment 3: Rheological contrast density

Kalgoorlie

Kalgoorlie



21

Edjudina

Kalgoorlie

Laverton

Leonora

Yandal

Experiment 3: Rheology Contrast Density

Results

Rheological Contrast DensityLOW HIGHS

pati

al associa

tion

0.00

0.20

0.40

0.60

0.80

1.00

• Deposits more common in regions of greater rheology contrast density

• Select rheology contrast layer for inputting into mineral potential models



22


Conceptual models

database

Predictor maps




Knowledge-base

MODEL


Spatial proxies

Processing

Overlay

Validation

Model-based mineral potential mapping



23

Structure of a model for mineral resource potential mapping

∫Integrating function

• linear or non-linear

• parameters

Input predictor maps

• Categoric or numeric

• Binary or multi-class

Output mineral potential map

• Grey-scale or binary



24GIS-based mineral resource potential

mapping - Modelling approaches

Exploration datasets with homogenous coverage – required for all models Expert knowledge (a knowledge base) and/or

Model parameters estimated from mineral deposits data(Known deposits required)Brownfields explorationExamples - Weights of evidence, Bayesian classifiers, NN, Logistic Regression

Data-driven Knowledge-driven

Hybrid

Mineral deposit data

Model parameters estimated from both mineral deposits dataand expert knowledge (Known deposits necessary)Semi-brownfields to brownfields explorationExamples – Neuro-fuzzy systems

Training data Expert knowledge

Model parameters estimated from expert knowledge (Known deposits not necessary)Greenfields explorationExamples – Fuzzy systems; Dempster-Shafer belief theory



25

Theory

RichSymbolic

Artificial

Intelligence

Fuzzy

Systems

Neural

Networks/

GA

Poor

Poor Rich

Bayesian

Any of the methods

best: Hybrid Systems

Neuro-fuzzy; fuzzy WofE

Which model is best?

Greenfields Brownfields

Data



26

Probabilistic Model (Weights of Evidence):

- used in the areas where there are already some known deposits

- spatial associations of known deposits/oil well with the geological features are used to determine the probability of occurrence of a mineral deposit (or well) in each unit cell of the study area.

Fuzzy Model:

- used in the areas where there are no known mineral deposits

- each geological feature is assigned a weight based on the expert knowledge, these weights are subsequently combined to determine the probability of occurrence of mineral deposit in each unit cell of the study area.

GIS MODELS FOR MINERAL EXPLORATION


27

Weights of Evidence Model for Mineral Prospectivity Mapping


28

Probabilistic model (Weights of Evidence)

• What is needed for the WofE calculations?– A training point layer –

i.e. known mineral deposits;

– One or more predictor maps in raster format.


29

PROBABILISTIC MODELS (Weights of Evidence or WofE)

Four steps:1. Convert multiclass maps to binary maps2. Calculation of prior probability3. Calculate weights of evidence (conditional probability) for

each predictor map4. Combine weights


30

• The probability of the occurrence of the targeted mineral deposit type when no other geological information about the area is available or considered.

Study area (S)

Target deposits D

Assuming- 1. Unit cell size = 1 sq km2. Each deposit occupies 1 unit cell

Total study area = Area (S) = 10 km x 10 km = 100 sq km = 100 unit cells

Area where deposits are present = Area (D) = 10 unit cells

Prior Probability of occurrence of deposits = P {D} = Area(D)/Area(S)= 10/100 = 0.1

Prior odds of occurrence of deposits = P{D}/(1-P{D}) = 0.1/0.9 = 0.1110k

10k

1k

1k

Calculation of Prior Probability


31

Convert multiclass maps into binary maps

• Define a threshold value, use the threshold for reclassification

Multiclass map

Binary map


• How do we define the threshold?

Use the distance at which there is maximum spatial association as the threshold !



• Spatial association – spatial correlation of deposit locations with geological feature.

A

BC

D

A

BC

D

10km

10km

1km

1km

Gold Deposit (D) Study area (S)


Indian Institute of Technology Bombay A

B C

D

Which polygon has the highest spatial association with D?More importantly, does any polygon has a positive spatial association with D ???

What is the expected distribution of deposits in each polygon, assuming that they were randomly distributed? What is the observed distribution of deposits in each polygon?

Positive spatial association – more deposits in a polygon than you would expect if the deposits were randomly distributed.

If observed >> expected; positive associationIf observed = expected; no association If observed << expected; negative association



A

B C

D

Area (A) = n(A) = 25; n(D|A) = 2Area (B) = n(A) = 21; n(D|B) = 2Area(C) = n(C) = 7; n(D|C) = 2 Area(D) = n(D) = 47; n(D|D) = 4Area (S) = n(S) = 100; n(D) = 10

OBSERVED DISTRIBUTION



A

B C

DArea (A) = n(A) = 25; n(D|A) = 2.5Area (B) = n(A) = 21; n(D|B) = 2.1Area(C) = n(C) = 7; n(D|C) = 0.7Area(D) = n(D) = 47; n(D|D) = 4.7(Area (S) = n(S) = 100; n(D) = 10)

EXPECTED DISTRIBUTION

Expected number of deposits in A = (Area (A)/Area(S))*Total number of deposits



A

B C

D

Area (A) = n(A) = 25; n(D|A) = 2.5

Area (B) = n(A) = 21; n(D|B) = 2.1

Area(C) = n(C) = 7; n(D|C) = 0.7

Area(D) = n(D) = 47; n(D|D) = 4.7

(Area (S) = n(S) = 100; n(D) = 10)

EXPECTED DISTRIBUTIONArea (A) = n(A) = 25; n(D|A) = 2

Area (B) = n(A) = 21; n(D|B) = 2

Area(C) = n(C) = 7; n(D|C) = 2

Area(D) = n(D) = 47; n(D|D) = 4

Area (S) = n(S) = 100; n(D) = 10

OBSERVED DISTRIBUTION

Only C has positive association!So, A, B and D are classified as 0; C is classified as 1.

Another way of calculating the spatial association : = Observed proportion of deposits/ Expected proportion of deposits= Proportion of deposits in the polygon/Proportion of the area of the polygon= [n(D|A)/n(D)]/[n(A)/n(S)] • Positive if this ratio >1• Nil if this ratio = 1 • Negative if this ratio is < 1



LA

BC

D

10km

10km

1km

1km

Gold Deposit (D) Study area (S)

Convert multiclass maps into binary maps – Line features


1km

1km

Gold Deposit (D)

10

0

000

0

000

11

112

11

1

1111

111

111

1

11

2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8

2 3 4 5 6 7 2 3 4 5 6 2 3 4 5 6

2 3 4 5 2 3 4 2 3 4 2 3 4

3 2 3 2 4 3 2 4 3 2

5 4 3 2 5 4 3 2

Distance from

the fault

No. of

pixels

No of deposit

s

Ratio (Observe

d to Expected

)

0 9 1 1.1

1 21 3 1.4

2 17 0 0.0

3 16 3 1.9

4 14 2 1.4

5 9 0 0.0

6 6 0 0.0

7 4 0 0.0

8 3 1 3.3

9 1 0 0.0



• Calculate observed vs expected distribution of deposits for cumulative distances

Gold Deposit (D)

10

0

000

0

000

11

112

11

1

1111

111

111

1

11

2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8

2 3 4 5 6 7 2 3 4 5 6 2 3 4 5 6

2 3 4 5 2 3 4 2 3 4 2 3 4

3 2 3 2 4 3 2 4 3 2

5 4 3 2 5 4 3 2

Distance from the fault

No. of pixels

Cumul No. of pixels

No of deposits

Cumul No. of deposits

Ratio (Observed to Expected)

0 9 9 1 1 1.1

1 21 30 3 4 1.3

2 17 47 0 4 0.9

3 16 63 3 7 1.1

4 14 77 2 9 1.2

5 9 86 0 9 1.0

6 6 92 0 9 1.0

7 4 96 0 9 0.9

8 3 99 1 10 1.0

9 1 100 0 10 1.0

=< 1 – positive association (Reclassified into 1)> 1– negative association (Reclassified into 0)



Weights of evidence ~ quantified spatial associations of the resource with predictor maps

Calculation of Weights of Evidence

Study area (S)

10k

10k

Unit cell

1k1k

Objective: To estimate the probability of occurrence of D in each unit cell of the study area

Approach: Use BAYES’ THEOREM for updating the prior probability of the occurrence of deposits to posterior probability based on the conditional probabilities (or weights of evidence) of the predictor features.

Deposit locationsPredictor feature (B1)

Predictor Feature (B2)


42

P{D|B} = P{D& B}

P{B}= P{D} P{B|D}

P{B}

P{D|B} = P{D & B}

P{B}= P{D} P{B|D}

P{B}

Posterior probability of D given the presence of B

Posterior probability of D given the absence of B

Bayes’ theorem:

D- Well

B- Predictor Feature

THE BAYES EQUATION ESTIMATES THE PROBABILTY OF A WELL GIVEN THE PREDICTOR FEATURE FROM THE

PROBABILITY OF THE FEATURE GIVEN A WELL

ObservationInference



43

Using odds (P/(1-P)) formulation:

O{D|B} = O{D} P{B|D}

P{B|D}Odds of D given the presence of B

O{D|B} = O{D} P{B|D}

P{B|D}Odds of D given the absence of B

Taking logs on both sides:

Loge (O{D|B}) = Loge(O{D}) + Log of odds of D given the presence of BP{B|D}

P{B|D}loge

Loge (O{D|B}) = Loge(O{D}) + Log of odds of D given the absence of BP{B|D}

P{B|D}loge

+ive weight of evidence (W+)

-ive weight of evidence (W-)



Contrast (C) measures the net strength of spatial association between the

geological feature and mineral deposits

Contrast = W+ – W-

+ ive Contrast – net positive spatial association

-ive Contrast – net negative spatial association

zero Contrast – no spatial association

Can be used to test spatial associations

44

Calculation of contrast


2121

2

2

1

121

2

2

1

1

21

21

2

2

1

1

2

2

1

1

21

21

21

21

21

2121

2

2

1

121

2

2

1

121

21

2121

)()/&(log(

))/(

)/(log()

)/(

)/(log()()/&(log(

)/(

)/(.

)/(

)/().(

)/()./(

)/()./().(

)()/(

.)(

)/().(

)()/(

.)(

)/().(

)&()/&(

).(

)&()/&(

).(

)&/(

)&/()/&(

)(

)/(.

)(

)/().()&/(

,)(

)/(.

)(

)/()/&(

is, that ce,independen lconditiona assume We

)&(

)/&().()&/(

)(

)/().()/(

WWDOddsDBBOdds

DBP

DBP

DBP

DBPDOddsDBBOdds

DBP

DBP

DBP

DBPDOdds

DBPDBP

DBPDBPDOdds

BPDBP

BPDBP

DP

BPDBP

BPDBP

DP

BBPDBBP

DP

BBPDBBP

DP

BBDP

BBDPDBBOdds

BP

DBP

BP

DBPDPBBDP

BP

DBP

BP

DBPDBBP

BBP

DBBPDPBBDP

BP

DBPDPBDP

WEIGHTS OF EVIDENCE FOR MULTIPLE MAPS


2121

2121

2121

2121

)()/&(log(

)()/&(log(

)()/&(log(

)()/&(log(

WWDOddsDBBOdds

WWDOddsDBBOdds

WWDOddsDBBOdds

WWDOddsDBBOdds

WEIGHTS OF EVIDENCE FOR MULTIPLE MAPS


47

Loge (O{D|B1, B2})

=

Loge(O{D}) + W+B1 + W+B2

Loge (O{D|B1, B2})

=

Loge(O{D}) + W-B1 + W+B2

Loge (O{D|B1, B2})

=

Loge(O{D}) + W+B1 + W-B2

Loge (O{D|B1, B2})

=

Loge(O{D}) + W-B1 + W-B2

Log of odds of D given the presence of B1 and B2

Log of odds of D given the absence of B1 and presence B2

Log of odds of D given the presence of B1 and absence B2

Log of odds of D given the absence of B1 and B2

Loge (O{D|B1, B2, … Bn})

=

Loge(O{D}) + ∑W+/-Bii=1

n

Or in general, for n predictor features,

The sign of W is +ive or -ive, depending on whether the feature is absent or present

The odds are converted back to posterior probability using the relation 0 = P/(1+P)

Feature B2

Feature B1

Well D


48

Loge (O{D|B1, B2})

=

Loge(O{D}) + ∑W+/-Bii=1

n

Calculation of posterior probability (or odds) require:• Calculation of pr prob (or odds) of occurrence of wells in the study area• Calculation of weights of evidence of all predictor features, i.e,

Implementation

P{B|D}

P{B|D}log

e

P{B|D}

P{B|D}log

e

W+ =

W- =

&


49

Exercise-2Layer – B1 (Fold Axes buffers

1.25k)

There are 54 base-metal deposits in the study area. Out of these, 35 deposits occur

within 1.25 km buffers of fold axes (B1), and 42 deposits are associated with mafic

volcanic rocks (B2). The total area in the fold axes buffers is 7132 sq km, while the

mafic volcanic rocks occupy 4374 sq km area. If the total study area is 55987 sq km,

calculate the following (assuming 1 sq km unit area size):

1. Prior probability and odds of base-metal deposits in the study area;

2. W+B1, W-B1, W+B2, W-B2;

3. Contrast values for B1 and B2; and

4. Posterior Log(odds) and probability of base-metal deposits in the study area, where

i. Both B1 and B2 are present; ii. B1 present but B2 absent; iii. B2 present but B1 absent; and iv. Both B1 and B2 are absent

Layer – B2 (Mafic volcanic rocks)

ARAVALLI STUDY AREA


EXERCISE

Feature B2

Feature B1

Well D

Calculate posterior probability for each cell.


51

• The probability of the occurrence of the wells when no other information about the area is available or considered.

Study area (S)

Wells D

Assuming- 1. Unit cell size = 1 sq km2. Each deposit occupies 1 unit cell

Total study area = Area (S) = 10 km x 10 km = 100 sq km = 100 unit cells

Area where wells are present = Area (D) = 10 unit cells

Prior Probability of occurrence of wells = P {D} = Area(D)/Area(S)= 10/100 = 0.1

Prior odds of occurrence of wells = P{D}/(1-P{D}) = 0.1/0.9 = 0.11

10k

10k

1k

1k

Calculation of Prior Probability


52

Basic quantities for estimating weights of evidenceTotal number of cells in study area: n(S)

Total number of cells occupied by wells (D): n(D)

Total number of cells occupied by the feature (B): n(B)

Total number of cells occupied by both feature and wells : n(B&D)

Derivative quantities for estimating weights of evidence

Total number of cells not occupied by D: n( ) = n(S) – n(D)

Total number of cells not occupied by B: n( ) = n(S) – n(B)

Total number of cells occupied by B but not D: n( B & D) = n(B) – n( B & D)

Total number of cells occupied by D but not B: n(B & D) = n(D) – n(B & D)

Total number of cells occupied by neither D but nor B: n( B & D) = n(S) – n(B) – n(D) + n( B & D)

D

B

Probabilities are estimated as area (or no. of unit cells) proportions


P{B|D}

P{B|D}log

e

W+ = P{B|D}

P{B|D}log

e

W- =

= n( )/n(D)

= n( )/

= n( )/n(D)

= n( )/

B & D

B & DP{B|D}

P{B|D} B & D

P{B|D}

P{B|D} B & D

n(D)

n(D)

Where,

B1

D

B2

B2 D B2 D

B2 D

B2 D

B1 D B1 D

B1 D

B1 D

B1

S

D

B2

S

D


53

Exercise

B2 D B2 D

B2 D

B2 DB1 D B1 D

B1 D

B1 D

10k

10k

B1

B2

S

B1

SD

B2

S

D

Unit cell size = 1 sq km & each well occupies 1 unit cell

n(S) = 100

n(D) = 10

n(B1) = 16

n(B2) = 25

n(B1 & D) = 4

n(B2 & D) = 3

Calculate the weights of evidence (W+ and W-) and Contrast values for B1 and B2

= n( )/n(D)

= n( )/ = [n(B) – n( )]/[n(S) –n(D)]

= n( )/n(D) = [n(D) – n( )]/n(D)

= n( )/ = [n(S) – n(B) – n(D) + n( )]/[n(S) –n(D)]

B & D

B & DP{B|D}

P{B|D} B & D

P{B|D}

P{B|D} B & D

n(D)

n(D)

Where,

B & D

B & D

B & D

P{B|D}

P{B|D}log

e

W+ = P{B|D}

P{B|D}log

e

W- =


54

Solution

B2 D B2 D

B2 D

B2 DB1 D B1 D

B1 D

B1 D

10k

10k

B1

B2

S

B1

S

DB

2

S

D

W-B1 = loge = loge(0.6000/0.8667)

= loge(0.6923) = -0.3678

= n( )/n(D) = 4/10 = 0.4000

= n( )/ = 12/90 = 0.1333

= n( )/n(D) = 6/10 = 0.6000

= n( )/ = 78/90 = 0.8667

B1& D

B1&D P{B1|D}P{B1|D}

B1&D

P{B1|D}

P{B1|D}

B1& D

n(D)n(D)

W+B1 = loge = loge(0.4000/0.1333)

= loge(3.0007) = 1.0988

P{B1|D}P{B1|D}

P{B1|D}P{B1|D}

Contrast = W+B1 – W-B1 = 1.0988 – (-0.3678) = 1.4666

W-B2 = loge = loge(0.7000/0.7555)

= loge(0.9265) = -0.0763

= n( )/n(D) = 3/10 = 0.3000

= n( )/ = 22/90 = 0.2444

= n( )/n(D) = 7/10 = 0.7000

= n( )/ = 68/90 = 0.7555

B2&D

B2& DP{B2|D}

P{B2|D}

B2&D

P{B2|D}

P{B2|D}

B2& D

n(D)

n(D)

W+B2 = loge = loge(0.3000/0.2444)

= loge(1.2275) = 0.2050

P{B2|D}P{B2|D}

P{B2|D}P{B2|D}

Contrast = W+B2 – W-B2 = 0.2050 – (-0.0763) = 0.2812

Assuming Unit cell size = 1 sq km and each well occupies 1 unit cell

n(S) = 100

n(D) = 10

n(B1) = 16

n(B2) = 25

n(B1 & D) = 4

n(B2 & D) = 3

n( ) = - n(B2) - n(D) + n( ) =68

B2& D n(S) B2&D

n( ) = n(B2) – n( ) = 22

B2& D

n( ) = n(D) – ( ) = 7

B2& D

B2& D B2&

D

n(D) = n(S) – n(D) =90

n(B2) = n(S) – n(B2) = 75

Cells occupied by

deposits

n( ) = n(B1) – n( ) =12

B1& D

n( ) = n(D) – n( ) = 6

B1&D

B1& DB1& D

n(D) = n(S) – n(D) =90

n(B1) = n(S) – n(B1) = 84

n( ) = - n(B1) - n(D) + n( ) = 78

B1& D n(S) B1& D


55

Loge (O{D|B1, B2})

=

Loge(O{D}) + W+/-B1 + W+/-B2

Loge(O{D}) = Loge(0.11) = -2.2073

Exercise - Calculation of Posterior Probability

Calculate posterior probability given:1. Presence of B1 and B2;2. Presence of B1 and absence of B2;3. Absence of B1 and presence of B2;4. Absence of both B1 and B2

B1

B2

S

Prior Prb = 0.10

Prior Odds = 0.11


56

Loge (O{D|B1, B2})

=

Loge(O{D}) + W+/-B1 + W+/-B2

Loge (O{D|B1, B2}) = -2.2073 + 1.0988 + 0.2050 = -0.8585

O{D|B1, B2} = Antiloge (-0.8585) = 0.4238

P = O/(1+O) = (0.4238)/(1.4238) = 0.2968

For the areas where both B1 and B2 are present

Loge (O{D|B1, B2}) = -2.2073 + 1.0988 - 0.0763 = -1.1848

O{D|B1, B2} = Antiloge (- 1.1848) = 0.3058

P = O/(1+O) = (0.3058)/(1.3058) = 0.2342

For the areas where B1 is present but B2 is absent

Loge (O{D|B1, B2}) = -2.2073 - 0.3678 + 0.2050 = -2.3701

O{D|B1, B2} = Antiloge (-2.3701) = 0.0934

P = O/(1+O) = (0.0934)/(1.0934) = 0.0854

Loge (O{D|B1, B2}) = -2.2073 - 0.3678 - 0.0763 = -2.6514

O{D|B1, B2} = Antiloge (-2.6514) = 0.0705

P = O/(1+O) = (0.0705)/(1.0705) = 0.0658

For the areas where both B1 and B2 are absent

For the areas where B1 is absent but B2 is present

Loge(O{D}) = Loge(0.11) = -2.2073

Posterior probability

0.2968

0.2342

0.0854

0.0658

Ground Water potential Map

Calculation of Posterior Probability

B1

B2

S

Prior Prb = 0.10

indian institute of technology bombay mineral prospectivity mapping alok porwal centre of studies in...

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