BAB 3 Geostat

8/8/2019 BAB 3 Geostat

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KELOMPK 3

ANGGOTA:

HARTANTO

HERU FIANISAL

ARNOL

MUHAMMAD FARHAD

MUHAMAD RIYAN KAMIL

MOHAMMAD ANDHIKA BUDIAWAN

ISRA KHOIRI

ADE IVUNG REVELES ELYSAMBA

GADHANG HERI WIBOWO

AKBAR HADI MUNAF

IMAM FADLI

BETMAN HUTABART

ASEP FIRDAUS

GEDE ABDI DHARMA PRIBADI

DYAH FIRGIANI

STEPHANIE OCTORINA SAING

RIKI FIRDAUS


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3.1 The Object of Kriging

In mining terms, the problem of kriging is to find the best linear estimator

possible of the grade of a panel, taking into account all the available information,

i.e. the assay values of the different samples that have been taken, either inside or

outside the panel we want to estimate. In order to solve a problem of kriging, i.e. to

compute effectively the proper optimal weight which is to be assigned to each

sample, it is necessary to make certain assumptions about the geostatistical

characteristics of the orebody under study, i.e., essentially to give oneself to

covariance function or the variogram of the R.F. of which the punctual assays are

supposed to constitute a realization.

The first interesting thing about kriging comes from the definition itself. By

minimazing the estimation variance, we ae sure to make the best use of the

available information or, in the other words, to obtain the most precise possible

estimation of the panel concerned. By far, the most important practical point of

kriging is not that it provides the best estimation possible, but that in addition it

avoids any systematic errors. In most ore-bearing deposits, we select for

exploitation a certain number of panels which are juged payable, and abandon

others deemed unpayable.

The reason for this general phenomenon is that the variance of the panelstrue grade is always lower than the variance of the inside samples. In the other

words, the histogram of the actual grades of the panels always contains fewer

extreme grades (rich or poor ) and more intermediate values than the histograms

deduced from the inside samples.


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For the very reason that we have selected a rich panel, the aureole of

outside samples will have, in general, a lower grade tah the inside ones ; yet its

influence on the panel to be estimated is not negligible since kriging has assigned a

non-zero weight to it. To explain how this occurs, let us imagine a vein-type deposit

developed by two drives AA and BB

The grades of BB are deemed payable, while those of AA are payable only

over a segment CC. If we merely take into account the mean grades of BB and CC,

we are sure to make an error to overestimation : the two segments AC and CA

although poor have a not negligible effect on the grade of trapezium BB, CC

selected for mining. If it were possible to draw the precise boundary between

payable and unpayable ore, the real boundary would not lie along the straight lines

BC or BC but would be, at general, a very irregular curve like B D E F.

In the case of a homogenous orebody, and the existence of a precise

boundary as defined by the grades seems doubtful, we shall speak of kriging. The

word 'smudging' is used when rich and poor parts, geologically heterogenous, are

separated by a continuous boundary which can be represented by a fairly regular

curve. The essential effect of such a weighting procedure is to eliminate - on

average -any systematic error of over-evaluation, which is particularly dreadful.

Compare with this prime objective, the improvement of the precision itself appears

as a relatively minor result.

Krige re-determine the empirical coefficients applied by South African miners,

by hypothesize that the expectation of the sample assays taken inside the panel

was equal to the true mean grade of this panel. Hence, if the variable are Gaussian,

the regression line giving the conditional expectation of the sample as a function of

the panel grade is identical to the first bisector. The other regression line, giving the

expectation of the panel as a function of the sample has thus a slope less than 1. If

the sample assay x is greater than the general average m, the panel expectation yis thus inferior to x, and conversely. Hence Krige corrected this systematic error by

using the equation y = m + (x m) of this second regression line, with a

regression coefficient < 1. More precisely, if is the correlation coefficient, x

and y the standard deviations of the samples and the panels, the two regression

coefficients (respectively 1 and ) are:


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, ;

hence: =

The coefficient is equal to the ratio of the variances (within the whole orebody) of

the samples and the panels.

In the case of punctual kriging at a point X0, this system becomes

In the continous case, there always exists a unique element , the

limit in the mean square of finite linear combinations verifying condition, and such

that:

This unique element is the projection of z0. Z* has not necessarily a representation

of the form

With a measure lamda with support in S verifying


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3-4-2 Optimal estimation of m

We can try to estimate m=[z(x)], we take the linier combination

We next impose on the coefficients the universality condition

The coefficients are chosen to minimize E(m-m*)=D2(m*) and take this to the

account. From

There appears a Lagrange factor :

At the optimum, we get so that the lagrange

factor is equal to the variance of the optimal estimator

STATIONARY R.F. WITH ZERO OR PRIOR KNOWN EXPECTATION

Let Y(x) be a R.F with zero expectation, (x,y) its covariance, S = (X) the set of

experimental points (supposed at the starts to be finite), Y0 = the

variable to be estimated, As an estimator, we shall use the linear combination

Yk = Z

Equating to Zero the partial derivatives of this quadran form we get this system:

= Y0 . . .(3-2)


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At the optimum, the cross and quadratic terms are th us equal, and we get:

= D2(Y0) - Y0. . .(3-3)

In the case of punctual kriging, the system is reduced to:

= , x0

And (3-4)

= x0,x0 - ,x0

When the set S is infinite, we try to estimate Y0 with the help of an estimator of the

form:

Yk = . . . (3-5)

And the system (3-2) become

(x,y) = y, Yo

(3-6)

= D2(Y0) - x, Y0

Condition is given as its same with

With computing step, we can get

And the corresponding variance is :


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In the continuous case, we have likewise:

Finally, from equation above, we can concluding that in the punctual case, we still

obtain an exact interpolar

From linear character of the right hand side of equation above

the solutionof

There is superimposition, or linear combination of the punctual krigging : this

relationship applies also the lagrange factor :

On the other hand, the variance cannot be obtained by linear combination of the

variances of the punctual krigings.

Punctual kriging at a point x0, this system becomes :

= ,x0 +

D2[Z(x0) Z*] = x0x0 + ,Z0

So, it can be verified that puctual kriging constitutes an exact interpolator


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3-4-2 Optimal estimation of m.

We can try to estimate the expectation m = E[Z(x)] itself.

To esltimate m, we take the linear combination

M* =

Impose on the coefficients the universality condition

= 1

m* is without bias, whatever the unknown value of m, and the coefficients are

chosen so as to minimize E(m-m*)2 = D2 (m*) by taking condition into account;.

From

D2 (m-m*) =

We deduce that the constitute the unique solution of the following system, in

which there appears a Lagrange factor 0 :

= 0

So that the Lagrange factor 0 is equal to the variance of the optimal estamator.

D2(M*) = 0

The measure k=e-a[h] R(direct placed at R) is indeed the solution of (3-6) for any

yR and can be computed directly as well.

a) K = e-a[h] R for y R and K2


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b) let Z(x) be a R.P. with unknown expectation a and e-a[h] its centered

covariance. Optimal estimator n of a(when the realization is known on)

verifies :

(

( n= aR + ZR+Z-R (Z = 1/2R R Zx (dx)

( 1+aR 2 -R

(

(

( D2 (n2)=

[establish relationships (R+-R)e-a[x-y]=e-aR ) chay and R e-a[x-y] dx= 2/a 2/a-aR

-R

chay for -R y R . deduce from this that the measure v0= a/2 dx + (R+-R)

verifies R C(dx) a[x-y]= 1 and that the measure 0= v0 is the

-R

solution required with the lagrange parameter 0= D2 (n2)= ]

c). with the same condition as in (b), krige the point x 0=R+h (h>0) by applying the

additivity relationship.

Z=e-a[h] ZR+(1- e-a[h])n; D2(Z x0-Z)=2 e-2ah+

From the linear character of the the right p(dx) Z(x) is :

There is linear combination of thr punctual kriging (Lagrange Factor) : =

in the other hand, the variance (3-17) cannot br obtained by linear

combination of the variances (x0) of punctual krigings.

aR

1+aR

1

1+aR

1

1+aR

11+aR

(1- e-2ah )2

1+aR


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Condition gives - , i.e the same universality condition as in the

preceding paragraph :

When this condition is verified, we know from the mechanism of paragraph

(2-2-1) that the variance of Z Z0 can be computed as if there existed a

covariance equal to y.Thus the system ( 3-11) can be transposted directly : the

coefficients of the optimal estimator constitute the unique solution of the

system .

=

And the corresponding variance is :

Finally the punctual case , we obtain an exact interpolator.

(

From the linier character of the right hand side of (3-17), the solution of the kriging

of is :


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There is superimposition, or linear combination of the punctual krigings : this

relationship applies also to Larange factor :

On the other hand, the variance (3-17) cannot be obtained by linear combination of

the variences of the punctual krigings.

EXERCISES ON KRIGING.

Exercise 1

(Kriging of a segment of length ) a/ Consider, on the real line R, an I.R.F.

without variance with semi-variogram and four sampling point x1, x2 = x1 + ,

x3 = x2 + , x4 = x3 + . We want to krige the segment (x2,x3) of length from the

values z1, z2, z3, z4 of the realization at the point x1, x2,x3, x4. Write the system (3-17)

with the help of the auxiliary function X and F of paragraph 2-5-2. [show that

, , which is obvious by symmetry, with the solution of :

Eliminate the lagrange parameter , hence :

b/ Application to the case and interpretation.

Solution : use Exercise 5 of chapter 2. We get


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for = 0, = (pure nugget effect, the optimal estimator is the

arithmetic mean of the four samples). When increase, decrease , is equal to 0

when = 1 (markovian property of the linear variogram), and becomes negative

when 1


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a) show that the optimal estimator (kriging) of Z is Z* = ZM + (1- ) ZT with

=

(weighting by the reciprocal of the respective estimation

variances) and that the corresponding variance is

(the coefficients are of the form and -1 because the universality

condition. Start with

and minimize with respect to ).

b) Application to the De Wijaian case when the length of the drifts and raises are

greater than their equidistance.

[Call LM and LT the (total) length of the drifts and raises, and use the relationship E2

= /2 S/L2 of 2-8-2. Then (weighting by the squares of the

lengths, and ) .

Exercise 3 (Markovian property of the exponential covariance and of the linear

variogram) This preparatory exercise may help to understand the following

exercise.

a. A R.F. Z(x) on the real line is Markovian if, for any x0, the Z(x), x > x0 and the

Z(x), x < x0 are conditionally independent when Z(x0) is given. Show that a


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stationary R.F. of order 2 with Gaussian distribution is Markovian if and only if

its centred covarianced is of the form C e-a|h|.

If x1, x2, x3 are Gaussian, the correlation coeffitient of x1 and x3 for x2 fixed

is:

Deduce therefrom that the R.F. is Markovian if and only if C(h+h) = C(h)

C(h)

(h, h positive).

b. Let Z(x) be an I.R.F. on the real line. Z(x) is said to have independent

increments if the Z(x1) Z(y1) are independent when the intervals (yi, xi) are

disjoint or have a single common point. Show that an I.R.F. with Gaussian

increments has independent increments if and only if its variogram is linear.

Exercise 4 (exponential covariance, continuous case)

a. On a line, let Y(x) be a R.F. with zero expectation and C e-a|h| be its covariance.

The realization of Y(x) is known in the interval (-R, +R). Krige the point x0 = R

+ h (h > 0).

If Y(x) is Gaussian, we can deduce from the Markovian property of exercise 3

that the optimal estimator, is e-a|h| Y(r) with variance 1 - e-2a|h|. This property,

related to the ony covariance, remains if Y(x) is not Gaussian


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The measure K = e -a|h| R (dirac placed at R) is indeed the solution of (3-6) for any y

R, and can be computed directly as well.

b/ let Z(x) be a R.F. with unkown expectation m and e -a|h| its centered covariance.

Show that the optimal estimator m* of m (whwn the realization is known on (-R, +R)verifies:

[establish the relationships chay and

chay for R y R. Deduce from this that the measure

verifies and that the measure

is the solution required with the lagrange parameter ]

c/ with the same condition as in b/, krige the point by applying the

additivity relationship.

Exercise 5 (exponential covariance, discrete case).- a/ Let Y(x) be a R.F. distributed

along a line with zero expectation and a stationary covariance e -a|h| . the realization

is known at the n+1 points of abscissae 0, 1, . . . n. Compute the punctual kriging of

the point

xo = i + ( 0 < < 1 ) (i < n) .


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[from the Markovian property, we must look for a solution of the form YK = Y i +

Yi+1 . show that the system (3-4) is the true for j = 0 ( j i , j i+1 ) ,

]

b/ With the same condition, let Z(x) be a R.F. of covariance e -a|h| and unknown

expectation m. Show that the optimal estimator is of the form

Let Y (x) be a R.F. distributed along a line with a zero expectation and a stationery

covariance

The realizationis known at the points of abscissae 0, 1, ..., n. Compute the punctual

kriging of the point

Solution :

From the markovian property, we must look for a solution of the form

Show that the system :

is true for :


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Exercise 6 (linear program) :

Consider along a line, an I.R.F. with no drift and its variogram.

The realization is known at the point X = R, and at a finite or infinite number of

points < R.

Show that the kriging of itself with

This solution is suggested by the markovian property of . Verify that the

measure is the trus solution of :

For any Y < R.

Exercice 7 (linear variogram or nugget effect)

On segment (-R, +R) a realization of an I.R.F. is known, with a variogram which

comprises a linear term and a nugget term. As we are in the continuous case, we

will take h| - C nugget effect represented by a Dirac Measure). Krige the point x = R+h (h>0). To do this :


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Establish the relationships chax sha chay, shax dx

=n . Hence deduce that with :

Show that the function (x) = verifies the system :

. Hence deduce that the optimal

estimator is Z*(Xo) = , with the lagrange parameter and

.

Exerxcise 8

When the covariance is very regular, it may be the case that the kriging solution

cannot be represented by a measured. For example, consider on a line, a stationary

R.F of covariance C(h) = and zero expectation. We want to krige the point Xo =

R + h, h >0, knowing the realization on (-R,+R). If the opyimal estimator is of the

type Yk = , the measureverifies

Show that a measure obeying this relationship cannot exist.

Exercise 9 ( Des Wijsian kriging in R2)

in R2 the wijsian scheme has a property of markovian character : when the

realization is known on a closed contour C, there is no correlation between the

inside and the outside of C : when kriging an element inside C, the knowledge

information outside C does not change the solution. There is a total screen effect,

which an electrical analogy allows us to understand (log r is the harmonic potential


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of the plane, and plays the same role in R2 as the Newtonian potential 1/R in R3)

a. Z(x) defines as a De Wijsian I.R.F. in R2 and C as closed contour is known. s is

the curvilinear abscissa of a point of C. r (s, s) is the function of distance

between two points of abscissa s and s on C. To

krige a small circle centres at point xo inside C,

define R (s) as the distance between xo and a

moving point on C. The optimal estimator will be

and the function (s) such that :

So, (s) is the density induced by a mass -1 placed at xo (the potential

is log r). it can be stated that (C) is an equipotential curve when a

mass -1 is placed at xo and the (s) on C.

The potential is constant in around C (inside and outside). So, if y is

any point outside C, its still

the density (s) still gives the kriging solution if the realization is

known on C and on any set S inside of C.

b. The green function G(x) relative to the point xo if G(x) is constant on C and G

(x) log r (xo,x) is a regular and harmonic function at any point x inside C,

including xo. The induced density (x) is given by (to positive normal).

So, the kriging of a point xo will be known the realization on a straight line.


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The Green function is log and

c. Same question for an inside point xo, knowing the realization on a circle of

radius R.

[ here G = log with X0 the invers of X0

In an inversion of modulus R2]. Use polar

coordinates

We get the density

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