An Introduction to Geostatistics

Presented to Math 216, Spring, 2012

Chris Vanags, Ph.D.Associate Director, Vanderbilt Center for Science Outreach

Instructor, School for Science and Math at Vanderbilt

A brief experiment . . .

Is it hot it here?

Follow up from assigned reading

“Analyzing the Consequences of Chernobyl Using GIS and Spatial Statistics”

Relative to your other course readings was this article. . .

1. Inappropriately simple

2. Easier to understand

3. On par with the level of difficulty

4. More difficult to understand

5. Inappropriately complex

●Did you feel that this article was appropriately informative?

1. The article did not contain enough detail to be interesting

2. The article captured my attention, but was not sufficiently detailed for my level of understanding

3. The article was well matched to the course requirements and my level of understanding

4. The article captured my attention, but was overly detailed for my level of understanding

I found this article to be relevant to what we are studying in this class.

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree

Based on the reading, I can see using geostatistcal tools in the future

Stro

ngly Agree

Agree

Disa

gree

Stro

ngly Disa

gree

25% 25%25%25%1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree

Why am I here?

“How do geostatistics differ from "normal" statistics in terms of determining the probability of given events assuming they have these large, vaguely defined sample sizes?”

A brief history of geostatistics

● $962Billion Global mining industry

Georges Matheron (1930 – 2000)Gold deposits in Witwaterstand, SA

Fundamental concepts: interpolation models

● Nearest neighbor (right)– Exact values

● Inverse-distance weighting– Interpolation based on

distance from known values

● Trend analysis– Interpolation based on

distance and variation

Nearest neighbor approximationFrom: Wikipedia Commons

Fundamental concepts: the (semi)variogram

● Change in distance vs. change in property

● Used to weight estimates of variation between known points

● Key terms:– Nugget– Range– Sill

Semivariogram of topsoil clay content vs. lag distanceFrom: USGS

Fundamental concepts: Kriging

● Interpolation based on the modeled semivariogram

● Provides estimates of properties AND estimates of uncertainty of the prediction (right)

● Multi-dimensional● Computationally

expensive

Fundamental concepts: covariability● “Using information that

is easy to obtain to predict information that is difficult to obtain”

● Trend Kriging● Regression Kriging● Co-Kriging

Where to go from here. . . ?

● Indicator kriging (right)

● Stochastic modeling (below)

Geostatistics in practice:“Predicting the field-scale hydrological impacts of shallow palæochannels in the semi-arid landscape of Northern New South Wales, Australia “

Chris Vanags

Background“Expansion of flood irrigation in the Lower Macquarie Valley of New South Wales, Australia has been suggested as a major cause of increased groundwater recharge”

- Willis et al, 1997

www.boreline.co.uk

“The areas exhibiting the largest probability of excessive DD correspond to permeable soil types associated with a prior stream channel.”

Background

From: Stannard and Kelly (1977)

- Triantafilis et al, 2003

Water BudgetFlow (m3/day) from:

palaeochannel (2)

to water table (4)

Background

Layers 1-5 Layers 6-1011

1 12

2

3 4

y = 8.9302x + 1.1831

R2 = 0.9896

0

5

10

15

20

25

0 0.5 1 1.5 2 2.5

Channel Ksat multiplier

“a two fold increase in the contrast in saturated conductivity between channel sediments and those surrounding the channel increases the predicted deep drainage by 64% in our Modflow simulation”-Vanags and Vervoort, 2004

Study siteMoree, NSW

MethodsDirect

Observation

HydrologicalProperties

ConceptualModel

AncillaryData

GroundwaterFlow

Prediction

Groundwater response to irrigation

Perched watertable during irrigation events

Immediate response to irrigation events

Source of perched water?

769300 769400 769500 769600 769700 7698006751700

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5,6,749

11,12

21,22

Irrigation canalCarroll Creek

-9

-8

-7

-6

-5

-4

-3

25/10/2005 19/11/2005 14/12/2005 8/01/2006 2/02/2006 27/02/2006

Date

wat

erle

vel

in m

be

low

th

e s

urf

ace

A: well 2 (9m) below PC B: well 3 (5m) inside PC C: well 4 (9m) outside PC

A

B

C

Ancillary information“… the high costs and intrinsic features of invasive sampling techniques such as drilling and cone penetrometer technologies limit their use to a finite number of sampling locations and do not allow complete coverage of the area under consideration”

-Borchers et al 1997

EM 31

DGPS height-adjustable stabiliser

Easting

Sou

thin

g

769400 769600 769800

6751500

6751700

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6752300

Quad-bike EM survey

Clear delineation of channel inside paddock

No delineation outside paddock

Strongly related to soil wetness

Distance (m)

Var

ianc

e

inside paddock

outside paddock

combined

outside paddock

inside paddock

combined

3 people hours = 2,700 data points

EM Survey: Depth sounding

Tx Rxseparation

½ separation

TxTx RxRxseparation

½ separation

Bi var i at e Fi t of 2 By ydi st Xval ue=2

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_v "

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_10"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_20"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_V"

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ydis t

Bi var i at e Fi t of 1 By ydi st Xval ue=2

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_v "

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_10"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_20"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_V"

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ydis t

1m Rx Tx

Bi var i at e Fi t of 2 By ydi st Xval ue=2

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_v "

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_10"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_20"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_V"

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ydis t

Bi var i at e Fi t of 1 By ydi st Xval ue=2

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_v "

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_10"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_20"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_V"

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0 10 20 30 40 50 60 70 80 90 100

ydis t

Bi var i at e Fi t of 2 By ydi st Xval ue=2

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_v "

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_10"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_20"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_V"

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ydis t

Bi var i at e Fi t of 1 By ydi st Xval ue=2

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_0_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1. 5_v "

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 31_1_V"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_10"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 34_20"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_H"

Sm oot hing Spline Fit , lam bda=0. 1 I D==" EM 38_V"

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ydis t

480 people hours = 1000 data points

1

8

Inverted conductivity profiles

McNeill Discontinuous profiles Channel not delineated

Tikhonov 0th order Laterally smooth profiles Large range in predictions

Tikhonov 1st order Smooth profiles Channel delineated

Tikhonov 2nd order Smooth profiles Channel delineated

McNeill Tikh 0

Tikh 1 Tikh 2

Distance along transect (m)

De

pth

(m

be

low

su

rfa

ce

)

EM results: significant relation with clay content

McNeill

poor correlation with clay content

Tikhonov 0th order

best correlation, high RMSE

Tikhonov 1st order

significant correlation

Tikhonov 2nd order

significant correlation, lowest RMSE

Tikh 2r2=0.19RMSE = 19

Tikh 0r2=0.40RMSE = 42

Clay (g/g)

McNeillr2=0.06RMSE = 77

EC

(m

S/m

)

Tikh 1r2=0.20RMSE = 23

Groundwater flow through palæochannel

heavyclays

coarsesand andgravel

finesand

partic le s ize c lusters a long channel

S outhing

de

pth

be

low

su

rfa

ce (

m)

-15

-10

-5

6751800 6751900 6752000 6752100

1

2

3

coarse gravel (Narrabri Formation)

palæochannel

reduced clays

permanent water table (1-2 m annual variation)

deep drainage

lateralflow

De

pth

(m

be

low

su

rfa

ce

)

Direct characterization

Identify important units

Measure hydrologic properties

Assign reference Ksat values for geologic facies

discontinuous predictions, assumed homogeneity within structure

Testing the scaling methods with 3D regression kriging

Use trend from ancillary data to weight direct observations

Assumptions: Direct observations

are related to ancillary data

Weighting is based on regression analysis

2.5m 1.5m 0.5m

6.5m 5.5m 4.5m 3.5m

10.5m 9.5m 8.5m 7.5m

Improved groundwater model

2D laterally-continuous Ksat from EM data X 20 layers

Simulated input from irrigation channel and deep drainage

Limited temporal prediction (single event)

MODFLOW Simulations

Ksat predictions from 2D EM surveys

Continuous Ksat in slice (projected in 3 dimensions)

Ksat = Kref x λ

Top boundary input

Inversion method accounted for 33% of predicted deep drainage

Tikhonov2m depth

McNeill2m depth

Conclusions

High variability of Ksat within palæochannel

2 orders of magnitude within channel

3 orders between palæochannel and surrounding sediments

Is direct characterization possible for a landscape scale effort?

Palæochannel associated with deep drainage AND lateral flow

In presence of irrigation channel: lateral flow >> deep drainage

31 Ml water lost during a single year of irrigation on one site.

where is this water headed?

what is the water quality?

Future research

Continue groundwater monitoring Calibrate groundwater model

Improve Ksat predictions

Incorporate measured data i.e. Kriging with trend

Incorporate uncertainty from ptf and scaling Generate stochastic groundwater model

Translate to landscape scale Use prediction method for larger data set