Weeks 11th - 13th

TA3114 Mining Geostatistics

Kriging Method and

Application
Background

The relation between sampling (drillhole) grade and block grade shows a
sistematical scattering.
It means that drillhole sampling is not the best estimation for a block, so it is
need a correction.
Matheron (1962) introduced a correction by weighting the sampling values by
means of variogram function.
The name of kriging adopted from the name of a mine engineer (statistician)
from South Africa D.G. Krige who firstly thought about the matter since 1950.
The correlation between grade of drillhole samples and true grade of blocks
which represented by the drillhole (obtained after mining the blocks) will give a
scatter plot which shows that the most of data (points) situated within the
ellipsoid as seen in the next figure.
 Blocks
grade

Samples
grade

Data scattering between grade of drillhole

samples vs. grade of mining blocks

In case the grade analysis of samples is the right estimation to the grade of
each blocks, then the regression through the origin will be along line A-A.
The research from Krige for Au grade samples showed that in reality the
slope of regression line was a bit low as seen by line B-B (the next figure).
 Blocks
grade

Samples
grade
Data scattering between samples grade vs.
blocks grade for Au (line B-B)

This means that the deviation is systematic and drillhole samples are not
the representative values for the blocks grade.
Samples grade higher than average value gives the higher value into the
blocks grade, if not corrected.
For example: Sample grade z1 gives block grade Z1 by line A-A which is
higher than the true block grade Z1 (line B-B).
 On the other hand, the sample grade lower the average value gives the value
lower than block grade, for example: sample z2 by line A-A gives block grade
Z2 which is lower than the true block grade Z2 (line B-B).
Matherons correction by considering the variogram of regional data analysis
showed that estimation on blocks grade was not only influenced by the samples
within the blocks but also influenced by the samples around them in vicinity.
The correction gives:
- better estimated values,
- variance of estimation K
2

Estimation by Kriging method sometimes is too complex for a commodity

(related to geometrical parameters). But it will be so useful to define the
mineable reserve with grades above cut-off grade.
As for example: correlation between samples grade and blocks grade (true
grade) scattered within the ellipsoid (the next figure), then giving a regression
line through the origin (0) and point ( Z , z ). Finally, the ellipsoid can be divided
into four sectors by the cut-off grade (cog), zc = Zc = 5%.
 Blocks Blocks
grade grade

Samples Samples
(a) grade (b) grade
Data scattering between samples grade vs. Blocks grade which showed
mistaken mining
Sector 1: all blocks with grade > cog coincides to samples grade > cog !
have been mined
Sector 2: all blocks with grade < cog coincides to the samples grade < cog !
have not been mined
Sector 3: all blocks with grade < cog due to the mistaken samples grade > cog
! have been mined
Sector 4: all blocks with grade > cog due to the mistaken samples grade < cog
! have not been mined
 If the regression line B-B which shows the correlation between samples and
blocks grade is plotted, then blocks with grade 5% is mineable although the
sample grade is 3,5% (Sector 1 in Fig. b).

Sector 4 in Fig. b which is not mineable due to the mistaken information will be
smaller, while Sector 3 which is mineable although the samples grade is low will
be larger. But in general, blocks with grade > cut-off grade (5%) and mineable
will be larger (Sector 1).

Based on variogram analysis, Matheron gave correction on estimated grade on

blocks which not considering only the sample within blocks, but also
considering the samples around them.

By means of Kriging method, we will not define the new better regression line,
but this method will correct the samples grade to be higher or lower until the
scattering within ellipsoid is tighter (see the next figure).
 Blocks Blocks
grade grade

Samples Samples
grade grade
The change of ellipsoid of data scattering after
correction by Kriging method

By the correction using Kriging method, the shape of ellipsoid will be

thinner/tighter with the border closer to the regression line with tangential 45.
The number of samples and associated blocks in Sectors 3 and 4 which indicates
mineable low grade or not mineable high grade will be less smaller.
Royle & Newton (1972) searched all kind of correction model and concluded
that Kriging process gave the best estimated values based on the corrected
samples grade.
References
Krige, D.G., 1951. A statistical approach to some basic mine valuation problems
on the Witwatersrand. Journal of the Chemical, Metallurgical and Mining Society
of
Krige, D.G., 1956. On The Departure of Ore Value Distributions from the
Lognormal Model in South African Gold Mines. South African Institute of
Mining and Metallurgy
Royle, A.G., Newton, M.J., and Sarin, V.K., 1972. Geostatistical factors in
design of mine sampling programmes. Transaction Institution of Mining &
Metallurgy
General Formula

Assume we have a group S1 of n samples with the same volume in a place xi.
Estimation of grade Z from the volume V is Z*. This estimated value obtained
by the weighting on samples grade z(xi):
n
Z* = i z ( x i )
i =1
n

The sum of weights factor is set to be 1:

i =1
i =1

By this way we will get estimated values without biased (unbiased), which
means the difference between Z and Z* insist to be 0.
E{Z Z *} = 0
By considering the weighting factor, we will get an estimation variance as:
E2 = Var [Z Z * ]

(x y )dx dy i j (x i x j )
2 n 1 n n
= i (x i y )dy
V i =1 V VV V V i =1 j=1

= 2 i (Si , V ) (V, V ) i j (Si , S j )

n n n

i =1 i =1 j=1
Estimation variance is the function of weighting factors which is known
their sum is 1. To choose the optimum weighting factors, the estimation
variance is set to be minimum.
Requirement that the sum of unkown i is 1 can be approximated by the
help of Lagrange multiplier () to minimize the following equation:

Q = E2 2 ( i 1) minimum

Beside the unknown i, the is also unknown. The minimizing above

formula means that the partial of Q/ and Q/i is equal to 0.
Finally we get the linear equation system named Kriging system:

j (xi x j ) + =
n
1
( x xi )dx or
j =1 VV

(S , S ) + = (S , V )
n n

j i j i and
i =1
i =1
j=1

This system is used to define the values of i and which will generate
the minimum estimation variance.
Estimation variance (Kriging variance) is expressed by:
(x x j )dx or
n
1 1
2
K = dx ( x y )dy j
VV V V j =1 VV
= (V ,V ) + + j (S j ,V )
n
2
K
j =1

Notes:
The equations can be used if z(xi) for calculation of estimated values Z is
about equal to the average values of a number of samples in vicinity.
For example: (xi x j ) acts as the average of from the group of points Si
and Sj on locations xi and xj.
Following is the matrices of kriging system to calculate the unkown i and :

1 (S1S1 ) + 2 (S1S2 ) + ... + j (S1S j ) + ... + n (S1Sn ) + = (S1V )

1 (S2S1 ) + 2 (S2S2 ) + ... + j (S2S j ) + ... + n (S2Sn ) + = (S2 V )
M M M M M M
1 (S jS1 ) + 2 (S jS2 ) + ... + j (S jS j ) + ... + n (S jSn ) + = (S jV )
M M M M M M
1 (Sn S1 ) + 2 (Sn S2 ) + ... + j (Sn S j ) + ... + n (Sn Sn ) + = (Sn V )
1 + 2 + ... + j + ... + n +0 =1

By cosidering that (SiS j ) = (S jSi ) , the matrices can be written as:

(S 1 S 1 ) (S 1 S 2 ) L (S 1 S j ) L (S 1 S n ) 1 1 (S 1V )

(S 2 S 1 ) (S 2 S ) L (S 2 S j ) L (S 2 S n ) 1
2 (S 2V )
M M M M M M

(S j S 2 ) L (S i S j ) L (S j S n ) 1

(S i S 1 ) . j = (S j V )
M M M M M M

(S n S 1 ) (S n S 2 ) L (S n S j ) L (S n S n ) 1 n (S nV )
1 1 L 1 L 1 0 1

Matrices (S i S j ) is simetric.
The matrices system can be simplified to be:
[K ] [L] = [M ]
The matrices can be solved on L to get i and , so the equation is:

[L] = [K ] [M ]
1

Then the Kriging Variance is written as:

= (V, V ) + [L] [M ]
2
K
t
The Effect of Geostatistical Parameters to the Weighting Factors and
Estimation Variance
Effects of geostatistical parameters will be explained by some simple example
below:

Known samples xi with grades z(xi) were taken with same distance (L=20 m) along
the line. Average grade of all sample is z = 1.0 . Variogram (Matheron model ) of the
data has parameters: C0 = 0; C = 1; a = 60 m

We will calculate the weighting factors, estimation variance (kriging variance), and
relative standard deviation for grade z * along the line L (i.e. on point x1).
In order to know how the effect of aureole samples and nugget variance, we will
consider the influence of one point (x1 itself), or influenced by three points (x1, x2,
x3) or by all points.
Nomogram of auxilliary functions (h), (h) and F(h) for Spherical Model, with
C0 = 0; C = 1; a = 1
a) Kriging system with considering only one sample

Kriging system: The calculation on kriging variance is:

(S , S ) + = (S , L ) = (L , L ) + + j (S j , L )
n n n

j =1
j i j i
j =1
j =1 2
K
j =1

for n = 1 so 1 = 1, then: (L , L ) = C 0 + C F (L )
(S 1 S 1 ) = 0 = 0 + 1 F L ( a ) = F(20 60)
(S1L ) = C 0 + C. (

) = F(0,333) = 0,165
= C0 + C X L [ 2] = 0 ,124
= 0 + 1 X (10 )
60 1 (S1L ) = 1 0,124 = 0,124
= 0,124
K2 = 0 ,165 + 0 ,124 + 0 ,124 = 0 ,083

By substituting the parameters into Relative standard deviation is:

equations giving:
K2
K = 100% = 29%
1 0 + = 0 ,124 = 0 ,124 z
b) Kriging system with considering three samples

Kriging system:
1 (S 1 S 1 ) + 2 (S 1 S 2 ) + 3 (S 1 S 3 ) + = (S 1 L )
X (L' ) =
(L + L ) X (L + L ) (L ) X (L )
2 2 2 2
1 (S 2 S 1 ) + 2 (S 2 S 2 ) + 3 (S 2 S 3 ) + = (S 2 L ) L

1 (S 3 S 1 ) + 2 (S 3 S 2 ) + 3 (S 3 S 3 ) + = (S 3 L ) ( 2 ) = X (30 60 ) = X (0 ,5) = 0 ,359

X L+L

1 + 2 + 3 + 0 =1 X (L ) = X (10 ) = X (0 ,167 ) = 0 ,124

2 60

For L = 20 m, a = 60 m, C0 = 0, and C = 1 30 0 ,359 10 0 ,124

X (L' ) = = 0 ,477
20
then:
(S 1 S 1 ) = (S 2 S 2 ) = (S 3 S 3 ) = 0 So the kriging system is:
(S 1 S 2 ) = (S 1 S 3 ) = C 0 + C ( x 1 x 2 ) 1 0,000 + 2 0, 481 + 3 0, 481 + = 0,124 [1]
( ) ( )
( x1 x 2 ) = L a = 20 60 = (0 ,333) = 0 ,481 1 0, 481 + 2 0,000 + 3 0,851 + = 0, 477 [2 ]
1 0, 481 + 2 0,851 + 3 0,000 + = 0, 477 [3]
(S 2 S 3 ) = (S 3 S 2 ) = C 0 + C ( x 2 x 3 ) 1 + 2 + 3 + 0 = 1,000 [4 ]
( ) ( )
( x 2 x3 ) = 2 L a = 40 60 = (0 ,667 ) = 0 ,851

(S 1 L ) = C 0 + C X (L ) = 0 ,124 (the same as item a)

0 0,851 2 + 0,851 3 + 0 = 0,00 [2] [3]
2 2 = 3
(S 2 L ) = (S 3 L ) = C0 + C X (L' )
 Relative standard deviation:
1 + 2 2 = 1,000
1 = 1,000 2 2 [4] K2
K = 100% = 23%
z
2(0 ,481 2 ) + = 0 ,124
= 0 ,124 0 ,962 2 [1] Z* = 0,76 z(x1 ) + 0,12 z(x 2 ) + 0,12 z(x 3 )

(1 22 )0 ,481 + 0 ,8512 + (0 ,124 0 ,9622 ) = 0 ,477 Weights factors 2 and 3 have

0 ,481 0 ,962 2 + 0 ,851 2 + 0 ,124 0 ,962 2 = 0 ,477 [2] the same value according to the
2 = 0 ,12 symmetrical position of points 2
and 3 to the point 1 with distance
1 = 0 ,76 2 = 3 = 0 ,12 and = 0 ,01 L.
Based on the symmetrical points,
= (L , L ) + + j (S j , L )
n
2
K then kriging system can be
j =1
simply written as:
(L , L ) = 0 ,165
= 0,01 z(x ) + z(x 3 )
Z* = 1 z(x1 ) + 2 2
{ 1442 2 443
(S , L ) = 0 ,76 0 ,124 + 0 ,12 0 ,477 + 0 ,12 0 ,477 = 0 ,208
n
S1
j j S2
j =1

K2 = 0 ,165 + 0 ,01 + 0 ,208 = 0 ,053

Kriging system:
1 (S 1 S 1 ) + 2 (S 1 S 2 ) + = (S 1 L ) 1 = 0 ,76 2 = 0 ,24( 2 = 0 ,12 & 3 = 0 ,12 )
1 (S 2 S 1 ) + 2 (S 2 S 2 ) + = (S 2 L ) = 0 ,01

1 + 2 + 0 = 1,000
K2 = 0 ,165 + 0 ,01 + 0 ,76 0 ,124 + 0 ,24 0 ,477
(S 1 S 1 ) = 0
= -0,165+0,01+0,208
(S 2 S 2 ) = 1 2 [C0 + C (2 L )] (
40 60
) = (0 ,667 ) = 0 ,851 K2 = 0 ,053 (the same as previous item)
= 1 [0 + 1 0,851] = 0,425
2
(S 2 S 1 ) = (S 1 S 2 ) = 0 ,481
(S 1 L ) = 0 ,124 (the same as previous item)
(S 2 L ) = 0 ,477

1 0 ,000 + 2 0 ,481 + = 0 ,124

1 0 ,481 + 2 0 ,425 + = 0 ,477
1 + 2 + 0 = 1,000
c) Kriging system with considering all samples

When we use three points as item b), all remaining sample points grouped to be
one sample with average grade z .
All average samples are assumed having large distance from points x1, x2, x3 and
slice L, therefore (h) and all auxiliary function of (h), F(h) are equal to 1 (=
total sill).
(S 1 S 3 ) = (S 3 S 1 ) = C0 + C (1,0 ) = 1,0
z ( x ) + z ( x )
3
Z* = 1 z ( xi ) + 2 2 + 3 {z (S 2 S 3 ) = (S 3 S 2 ) = C0 + C (1,0 ) = 1,0
{ 1442443 2 S
S 1
3

2S
(S 3 S 3 ) = C0 + C = 1,0
Kriging system:
All sample grouped in S3 have large
1 (S 1 S 1 ) + 2 (S 1 S 2 ) + 3 (S 1 S 3 ) + = (S 1 L )
position beyond (distance >> a), so
1 (S 2 S 1 ) + 2 (S 2 S 2 ) + 3 (S 2 S 3 ) + = (S 2 L ) all average value of all (x i x j )
1 (S 3 S 1 ) + 2 (S 3 S 2 ) + 3 (S 3 S 3 ) + = (S 3 L ) is equal to 1 (= total sill).
1 + 2 + 3 + 0 =1
(S 1 L ) = 0 ,124

(S 1 S 1 ) = 0 (S 2 L ) = 0 ,477

(S 1 S 2 ) = (S 2 S 1 ) = 0 ,481 (S 3 L ) = C0 + C X (L' ) 1,00

(S 2 S 2 ) = 0 ,425
As for example, we take samples with large
distance 6L = 120 m:
0 ,4812 + 0 ,266 1,116 2 = 0 ,124 [1]
X (L' ) =
(L + 6 L ) X (L + 6 L ) (6 L ) X (6 L )
0,142
L 2 = = 0,224

=
(
7L X 140
60
) 6L X(120 60) = 7 X(2,333) 6 X(2,0) 0,635
1 = 0,1162 + 0,734 = 0,760
L 3 = 0,226 1,1162 = 0,016
= 7 0,84 6 0,82 = 0,96 1,00
(3 can be ignored as the value
1 0 ,000 + 2 0 ,481 + 3 1,000 + = 0 ,124 [1] is too small)
1 0 ,481 + 2 0 ,425 + 3 1,000 + = 0 ,477
= (L , L ) + + j (S j , L )
n
[2] 2
K
1 1,000 + 2 1,000 + 3 1,000 + = 1,000 [3] j =1

(L , L ) = 0 ,165 = 0 ,000
1 + 2 + 3 + 0 = 1,000 [4]
0 ,760 0 ,124 = 0 ,094
0 ,4811 + 0 ,056 2 + 0 ,000 + 0 ,000 = 0 ,353 [1]-[2] 0 ,244 0 ,477 = 0 ,107

1 = 0 ,116 2 + 0 ,734 0 ,016 1,000 = 0 ,016

(S , L) = 0 ,217
n

0,1162 + 0,734 + 2 + 3 = 1,000 j =1

j j
[4]
3 = 0 ,266 1,116 2
K2 = 0,052
1 + 2 + 3 = 1,000 = 0 ,000 [3] (close to the previous result)
 Two samples z(x2) and z(x3) are assumed to enclosure the influence of all samples
beyond ! screen effect
Therefore there is no improving on weight factors as well as estimation variance.

d) The effect of nugget variance (C0 0)

By considering all samples as item c), so:

(S 1 S 1 ) = 0 C0 = 0 ,3 1 0 ,000 + 2 0 ,781 + 3 1,3 + = 0 ,424 [1]
(S 1 S 2 ) = 0 ,3 + 0 ,481 = 0 ,781 C = 1,0 1 0 ,781 + 2 0 ,576 + 3 1,3 + = 0 ,777 [2]
1 a = 60 m 1 1,3 + 2 1,3 + 3 1,3 + = 1,3 [3]
(S 2 S 2 ) = (0 ,31 + 0 ,851) = 0 ,576
2 z = 1,0
1 + 2 + 3 + 0 = 1,0 [4]
(S 3 S 1 ) = 0 ,31 + 1,0 = 1,3
(S 3 S 2 ) = 0 ,31 + 1,0 = 1,3 0 ,7811 + 0 ,205 2 + 0 + 0 = 0 ,353 [1]-[2]
(S 3 S 3 ) = 0 ,31 + 1,0 = 1,3 1 = (0 ,205 2 + 0 ,353 ) / 0 ,781 = 0 ,262 2 + 0 ,452
(S 1 L ) = 0 ,3 + 0 ,124 = 0 ,424
0,2622 + 0,452 + 2 + 3 = 1,0 [4]
(S 2 L ) = 0 ,3 + 0 ,477 = 0 ,777
3 = 0,548 1,2622
(S 3 L ) = 0 ,3 + 1,0 = 1,3
 1,31 + 1,3 2 + 1,33 + = 1,3 [3] By the presence of nugget
variance, the influence of all
1,3 (1 + 2 + 3 ) + = 1,3
samples beyond can not be
1 + 2 + 3 = 1
ignored as previous case.
= 0 ,0
Screen effect on the samples is
0 ,7812 + 1,33 + = 0 ,424 [1] reduced due to the presence of
0 ,7812 + 1,3(0 ,548 1,262 2 ) = 0 ,424 nugget variance.
0 ,288
2 = = 0 ,335
0 ,860
If the nugget variance is gained to
1 = 0 ,262 2 + 0 ,452 = 0 ,540
be C0 = 0.5 then the effect will be
3 = 1,0 0 ,335 0 ,540 = 0 ,125 more significant:
1 = 0 ,466 2 = 0 ,341 3 = 0 ,193 = 0 ,000
= (L , L ) + + j (S j , L )
n
2
K
j =1
K2 = 0 ,248 K = 50%
(L , L ) = 0 ,3 + 0 ,165 = 0 ,465
0 ,540 0 ,424 = 0 ,229
0 ,335 0 ,777 = 0 ,260
0 ,125 1,300 = 0 ,163 +

(S , L ) = 0 ,652
n

= 0 ,465 + 0 ,00 + 0 ,652 = 0 ,187

2
K j j
j =1
K = 0,187 / 1,0 100% = 43%
e) Summary

11sample
conto 33samples
conto Semua conto
all samples
C0 0,0 0,3 0,5 0,0 0,0 0,3 0,5
1 1,0 1,0 1,0 0,76 0,76 0,54 0,47
2 0,12 0,22 0,34 0,34
3 0,12 0,02 0,12 0,25
K2 0,08 0,38 0,58 0,05 0,05 0,19 0,25
K 29% 62% 76% 23% 23% 43% 50%
Properties of Kriging Method
By means of Kriging method we obtain the best estimator based on the available
information of a mineral deposit. The weight factors is chosen to obtain the
minimum estimation variance.

So the Kriging considers about:
1. Structural and spatial correlation by means of function (h).
2. Relative geometrical relation among estimator data and volume by means of
(S i , S j ) (relation among data) and (S i ,V ) (relation between data and volume).
If the variogram is isotropic and data is regular, then kriging system will give
symmetrical data.
In many cases, samples inside and around the estimated block give estimation,
while sample far from the estimated block will have weights close to zero. In this
case the searching radius will not have influence (screened).
The screen effect will occur, if the nugget effect is zero or very small = C0 / C . The
nugget effect can reduce the screen effect. For the large nugget effect, all sample
will be considered having the same weight.
 As explained previously that Kriging method utilize all information effectively
so we obtain the best linear estimation for the true values.

The main target of kriging estimator is to avoid the systematic error (over
estimate or under estimate) in reserve estimation.

This point will be indispensable when we have to use the reserve estimation to
decide the mineable or not mineable blocks.
Example of Kriging on Regular Grid

The calculation is performed for a block of mineral deposit with known variogram
parameter with Matheron model and C0 = 0; C = 1; a = 60 m; the average grade =
z = 1.0 .

Block is rectangular with size 20 m 30 m and there are 4 sample around it and 1
sample in the middle.

Based on the symmetrical position of samples to block, then estimator formula is

written as: z (x 2 ) + z (x 3 ) z (x 4 ) + z (x 5 )
z* = 1 z(x1 ) + 2 + 3
{ 2 2 44
S1 14 4243 4 14 42 3
S2 S3
Kriging system:

(S i , S j ) + = (S i , R )
3

j =1
j

(S 1 , S 1 ) = 0
1 40 (S 2 , R ) = C0 + C Q(R' )
(S 2 , S 2 ) = C0 + C = 0 ,5 0 ,852 = 0 ,426
2 60 30 15 10 15
3 Q , 1 Q ,
Q(R' ) = 60 60 60 60
1 60
(S 3 , S 3 ) = C0 + C = 0 ,5 1,000 = 0 ,500 2
2 60

3 1
= 0,638 0,241 = 0,517
20 2 + 30 2 2 2
(S 2 , S 3 ) = C 0 + C = (0 ,601) = 0 ,793
60

20
(S 1 , S 2 ) = C 0 + C = 0 ,481
60

30
(S 1 , S 3 ) = C 0 + C = 0 ,688
60
(S 3 , R ) = C0 + C Q(R' ' )
10 15
(S 1 , R ) = C 0 + C Q , = 0 ,241 45 10 15 10
60 60 3 Q , 1 Q ,
Q(R' ' ) = 60 60 60 60
2
3 1
= 0,536 0,241 = 0,683
2 2
 30 20 Compare to the extension variance of:
(R , R ) = C 0 + C F , = 0 ,320
60 60
E2 ( ) = E2 30 , 20 = 0 ,158
60 60
Then kriging system being:
E = 40%
1 0 ,000 + 2 0 ,481 + 3 0 ,688 + = 0 ,241
1 0 ,481 + 2 0 ,426 + 3 0 ,793 + = 0 ,517 In general, the Kriging method for
1 0 ,688 + 2 0 ,793 + 3 0 ,500 + = 0 ,683 regular grid does not only
considered 4 samples around the
1 + 2 + 3 + 0 = 1,000
estimator block, but actually 8
samples as figured:
Solution for 4 equations with 4 variables:
8 2 9
1 = 0 ,57 2 = 0 ,26 3 = 0 ,17 = 0 ,00

= (R , R ) + + j (S j , R )
n
2
K
j =1
4 1 5
= 0,320 + 0,00 + (0,057 0,241 + 0,26 0,517 + 0,17 0,083)
= 0,320 + 0,387
= 0,067
Relative standard deviation:
7 3 6
K2
K = 100% = 26%
z
Estimated grade for the middle block (Block 1) is:
z (x2 ) + z (x3 ) z (x4 ) + z (x5 )
Z* = 1 z ( x 1 ) + 2 + 3 +
2 2
z(x 6 ) + z(x 7 ) + z(x 8 ) + z(x 9 )
4 + 5 z
4

In this case there are 6 linear equations system to define the weight i ( = 0 as
explained previously).
For the large nugget effect = C0 / C , we need to consider a group of sample beyond
the first ring.
Note:
The equation system for weighting each sample via i is also valid for all blocks
will be estimated with a condition that samples and blocks configuration are the
same.
 To perform Kriging estimation on 66 blocks with
regular grid, we must calculate 4 weights factor
for 4 drillhole samples around the block.

Estimation variance for each block will be

different, as the number of sample is smaller then
the estimation variance will be larger.

If the sample is located inside the estimated block,

or there are one or two samples are located around
8 samples used for estimation, then the equation
system must be modified because the weighting
system will be different.

For the samples with irregular distribution, i.e. the

sample points are not located in the middle of
block (random stratified grid), then the equation
system is still able to be used but we need to
modify it for each block.
Example of Kriging on Irregular Grid

Grade z* of a block can be estimated using sample grade in that block and samples
grade around the estimated block.

There is a Group S1 = n samples in the middle of Block

R surrounded by 8 blocks around A, i.e. Group S2 = m
samples, and all samples are represented by Group S3 =
1 sample (average grade = z ).
If grade of Group S1 = z1 and grade of Group S2 = z2,
then estimated grade is:
z* = 1 z 1 + 2 z 2 + 3 z

Block 1 = block R / S1 / n samples with grade z1

Blocks 2 to 9 = aureole A / S2 / m samples with grade z2
All deposit = V / S3 with grade z
(aureole = blocks surrounding the estimated block R)
Kriging system:
1 (S 1 S 1 ) + 2 (S 1 S 2 ) + 3 (S 1 S 3 ) + = (S 1 R )
1 (S 2 S 1 ) + 2 (S 2 S 2 ) + 3 (S 2 S 3 ) + = (S 2 R )
1 (S 3 S 1 ) + 2 (S 3 S 2 ) + 3 (S 3 S 3 ) + = (S 3 R )
1 + 2 + 3 =1

Because the samples in blocks are not positioned regularly, so the relation of
among points is replaced by relation of with estimated plane according to:

1
(S1S1 ) (R,R ) or (S1S2 ) (R, A )
n

1
1 (R , R ) + 2 (R , A ) + 3 (R , V ) + = (R , R )
n
1
1 ( A, R ) + 2 ( A, A) + 3 ( A,V ) + = ( A, R )
m
1 (V , R ) + 2 (V , A) + 3 (V ,V ) + = (V , R )
1 + 2 + 3 =1
On the other hand, we need to notice that the extension of deposit (V) is larger than
range a, so (V, R ) = (R , V ) = (V, A ) = (A, V ) = (V, V ) = C 0 + C = K and then = 0.

Equation of Kriging system can be simplified to be:

1
1 ( R , R ) + 2 ( R , A ) + 3 K = (R , R )
n
1
1 ( A , R ) + 2 ( A, A) + 3 K = ( A, R )
m
1 + 2 + 3 =1

The relation of among planes which used to solve the equation can be obtained
by numerical calculation via integration as explained previously for the
differentiation of auxiliary function F.
Example of kriging application on mineral deposit (Royle, 1971)
sample grade

A slice of mineral deposit is known

having block size 100 100 ft.
On each block, the sample was taken
randomly (random stratified grid).
Variogram of data samples gives
Matheron model with parameters:

C = 16.50%; C0 = 3.80%; = 0.23;

a = 240 ft; z = 4.27%

To correct the samples values by

considering block grades around them,
kriging is indispensable.
The calculation is performed with kriged estimated block

minimum neighborhood (within

searching radius) is 5 samples.
Estimator: z* = 1.z1+ 2.z2+ 3.z3
with: 3 = 1 - 1 - 2
z1 = sample grade in the middle of block
z2 = samples grades of 5 to 8 (in blocks surrounding)
z3 = z = average grades of all samples

Variance of estimation depends on the number of samples involved in estimation:

Conto di within
Nb of sample tengahblock aureol
Aureoles varians
Variance Simpangan baku
Standard deviation

1 8 3,68 1,9
1 7 3,99 2,0
1 6 4,25 2,1
0 6 8,43 2,9

Previous map/figure shows sample values (written as large font) and kriged values
below it (written as italic small font)
Histogram of original grades:
Minimum = 0.02; Maximum = 26.40; Average = 4.24; N = 85; Variance = 14.637;
Standard dev. = 3.826; Skewness = 2.820; Kurtosis = 15.090; Number of class = 12;
Interval = 2.5

NO. BATAS FREKUENSI FREKUENSI FREKUENSI

KELAS ATAS RELATIF KUMULATIF
1 2.5000 27. 31.76 31.76
2 5.0000 34. 40.00 71.76
3 7.5000 13. 15.29 87.06
4 10.0000 6. 7.06 94.12
5 12.5000 3. 3.53 97.65
6 15.0000 0. 0.00 97.65
7 17.5000 1. 1.18 98.82
8 20.0000 0. 0.00 98.82
9 22.5000 0. 0.00 98.82
10 25.0000 0. 0.00 98.82
11 27.5000 1. 1.18 100.00
12 30.0000 0. 0.00 100.00
Histogram of sample grades after kriging:
Minimum = 1.57; Maximum = 15.51; Average = 4.50; N = 78; Variance = 5.504;
Standard dev. = 2.346; Skewness = 1.835; Kurtosis = 8.045; Number of class = 12;
Interval = 2.5

NO. BATAS FREKUENSI FREKUENSI FREKUENSI

KELAS ATAS RELATIF KUMULATIF
1 2.5000 13. 16.67 16.67
2 5.0000 41. 52.56 69.23
3 7.5000 15. 19.23 88.46
4 10.0000 7. 8.97 97.44
5 12.5000 1. 1.28 98.72
6 15.0000 0. 0.00 98.72
7 17.5000 1. 1.28 100.00
8 20.0000 0. 0.00 100.00
9 22.5000 0. 0.00 100.00
10 25.0000 0. 0.00 100.00
11 27.5000 0. 1.18 100.00
12 30.0000 0. 0.00 100.00
 According to the configuration of blocks and searching radius of kriging with 5
samples in minimum, then there are 78 kriged blocks from the total of 88 blocks.
The cut-off grade is defined as 3.0%.

Problems:
1. How many parts of 78 kriged blocks which have original sample grades
>3.0%?
2. How many parts of 78 kriged blocks which have kriged grades >3.0%?
3. Please remark the blocks with kriged grades >3.0% as mineable area!

References
Royle, A.G., A Practical Introduction to Geostatistics. Course Notes of the
University of Leeds, Dept. of Mining and Mineral Sciences, Leeds, 1971.
 sample grade
The same mineral deposit is
calculated again using Kriging by
assuming that all drillholes exactly
located in the middle of grid/block.
The kriging result can be shown in the
figure beside.

Kriged blocks with max and

min samples of 9 and 6
samples respectively

kriged estimated block

 sample grade
sample grade

kriged estimated block

kriged estimated block
Evaluation:
Estimation variance ( K2 ) is lower
than before.
Depending on the drillhole
configuration and the number of N
drillholes, the estimated values are
different.
The figure beside shows the
properties of estimation variance
and estimated values Z* in relation
to the number of drillholes N with 2
different patterns.
It seems that 5 - 6 drillholes are
quite enough for estimation.

The effect of pattern and number

of samples to the kriging variance
and average values
Point/Punctual Kriging Method

In general samples points are distributed irregularly, so in order to arrange the

isoline map, the interpolation for creating a regular grid is indispensable.
Interpolation can be performed by some methods i.e. NNP (Nearest Neighboring
Polygon) and IDW (Inverse Distance Weighted, ID, IDS, or ID3).
From some discussions about estimation methods, Kriging has been known
producing best and reliable estimator.
In order to solve the estimation in point, the same kriging system previously is
used. In this case, we use only variogram, because only the relation among points
is considered.
As for example (see the next figure): if there are 3 points xi to estimate the 4th
point x0, the Kriging system written as:
1 ( x1 x1 ) + 2 ( x1 x 2 ) + 3 ( x1 x3 ) + = ( x1 x0 )
1 ( x 2 x1 ) + 2 (x 2 x 2 ) + 3 ( x 2 x3 ) + = ( x 2 x0 )
1 ( x3 x1 ) + 2 (x 3 x 2 ) + 3 (x 3 x3 ) + = (x 3 x0 )
1 + 2 + 3 =1
 Kriging system solution and data configuration:

( x1 x1 ) = ( x 2 x 2 ) = (x 3 x3 ) = 0 ,0
X1

( x1 x 2 ) = ( x 2 x1 ) = C 0 + C ( x1 x 2 )

( x1 x 3 ) = ( x 3 x1 ) = C 0 + C ( x1 x 3 )

X2
( x 2 x3 ) = (x 3 x 2 ) = C0 + C ( x 2 x3 )

X0

( x1 x0 ) = C 0 + C ( x1 x0 )

( x 2 x0 ) = C 0 + C ( x 2 x0 )

X3
( x 3 x0 ) = C 0 + C ( x 3 x0 )

Estimation/kriging variance is simplified to be:

= + j (x j x0 )
n
2
K
j =1
 Point Kriging method has properties that if estimation process gives a point x i = x0 ,
then in this point we will have z * (x0 ) = z (xi ) .
This matter is explained in the simple case below:
A mineral deposit shows variogram of Matheron model with parameters C0 = 0,
C = 1, a = 60 m, assume that we have 3 points as:

x1 x3 x2
I-------------------I--------------------I
20 m x0 20 m

( x1 x1 ) = ( x2 x 2 ) = (x3 x3 ) = 0 ,0 1 0 ,0 + 2 0 ,852 + 3 0 ,481 + = 0 ,481

40 1 0 ,852 + 2 0 ,0 + 3 0 ,481 + = 0 ,481
( x 1 x 2 ) = ( x 2 x1 ) = C 0 + C ( x1 x 2 ) = = 0 ,852
60
1 0 ,481 + 2 0 ,481 + 3 0 ,0 + = 0 ,0
20 + 3
( x 1 x 3 ) = ( x 3 x 1 ) = C 0 + C ( x1 x 3 ) = = 0 ,481 1 + 2 = 1,0
60

20
( x 2 x3 ) = ( x3 x 2 ) = C0 + C ( x 2 x3 ) = = 0 ,481 Solution:
60
( x3 x0 ) = ( x0 x0 ) = 0 ,0 3 = 1 1 = 2 = 0
K2 = 0
Example of point kriging application by Delfiner & Delhomme (1973): Optimum
interpolation by kriging, in Davies & McCullagh (Eds.), Display and Analysis of
Spatial Data.

Linear variogram of rainfall Rainfall measurement points (mm),

data in Wadi Kadjemeur, where contours constructed based on
Afrika Tengah manual interpolation

Contour s constructed based on Contours constructed based on point kriging

polynomial order-2 method
Exercise for Homework!

500 mN
x4

400
x3

300
V
?
x2
200

x1
100

0 100 200 300 400 500 mE

There are 4 drillholes of x1, x2, x3, and x4 which located randomly in the middle of
regular grid where the grid spacing is 100100 m (see figure). Those drillholes
were derived from exploration of primary Au deposit. The Au grades for each
known points are: z(x1) = 10 ppm, z(x2) = 20 ppm, z(x3) = 5 ppm, and z(x4) = 15
ppm. If the variogram model for Au grades distribution is Spherical with range =
200 m, nugget variance (C0) = 5 ppm2, and sill (C) = 25 ppm2, then:
Continued.Exercise for Homework! Collected due to Final Test day!

1. Please estimate the Au grade for Block V or z(V)* by using Ordinary

Block Kriging, by assuming the searching radius for estimation is 250 m
from the central of Block V! The average variogram between data points
and estimated block are:
(x 1 , V ) = 25 ppm 2 (x 3 , V ) = 20 ppm 2
(x 2 , V ) = 30 ppm 2 (x 4 , V ) = 30 ppm 2
2. Please calculate kriging variance from the estimation result! The average
variogram within block is: (V, V ) = 15 ppm . Also calculate the relative standard
2

deviation of estimation (error) on Block V! Note: relative to the estimated

grade in Block V.

3. Please compare the estimated grade of Au in Block V by using

conventional estimation methods such as IDS and NNP with searching
radius of estimation is 250 m from the central of Block V!
 Parameters of 3D kriging application using KT3D (in GSLIB, 1998)
Parameters for KT3D
*****************
START OF PARAMETERS:
../data/Pb_Sediments.txt -file with data
1 2 3 4 0 -columns for X, Y, Z, var, sec var
0 999 -trimming limits
0 -option: 0=grid, 1=cross, 2=jackknife
jack.dat -file with jackknife data
1 2 3 4 0 -columns for X,Y,Z,vr and sec var
1 -debugging level: 0,1,2,3
3dkrig-pb-sed.dbg -file for debugging output
3dkrig-pb-sed.out -file for kriged output
248 32270 15 -nx,xmn,xsiz
210 49550 15 -ny,ymn,ysiz
114 2440 15 -nz,zmn,zsiz
10 10 10 -x,y and z block discretization
3 15 -min, max data for kriging
0 -max per octant (0-> not used)
300 300 50 -maximum search radii
0 0 0 -angles for search ellipsoid
1 0.10438 -0=SK,1=OK,2=non-st SK,3=exdrift
0 0 0 0 0 0 0 0 0 -drift: x,y,z,xx,yy,zz,xy,xz,zy
0 -0, variable; 1, estimate trend
grid.dat -gridded file with drift/mean
0 -column number in gridded file
1 0.06 -nst, nugget effect
1 0.055 0 0 0 -it,cc,ang1,ang2,ang3
200 200 200 -a_hmax, a_hmin, a_vert
Spatial distribution of coal qualities estimated using Ordinary Kriging (OK)

Seam thickness Ash content Total sulphur

T T
Seam 0m T Seam 0m
Seam 0m

-20 m -20 m
(m) -20 m (%) (%)
3.5 10 2.5

R
Seam
3.15 9.1 2.25
R
Seam 2.8 Seam
R 8.2 2

Q Q
Seam
1.75
Seam
2.45 7.3
-75 m Q -75 m
Seam
2.1 6.4 1.5
-75 m
P 1.75
P 1.25
Seam
5.5
-90 m P -90 m Seam -90 m
Seam
1.4 4.6 1

1.05 3.7 0.75

-115 m -115 m 0.5
0.7
-115 m 2.8
0.35 0.25
1.9
0 0
1
 Spatial distribution of Ni+Co grades estimated using Ordinary Kriging (OK)
Graphic Edit: Section XOZ 11
X (m)
42.
-6100. -6090. -6080. -6070. -6060. -6050. 42.
VII-3 VII-20 VII-24 VII-33 VII >=3.4 32.
40. 40.
VI-102

3.08125
VI-91

32.
II-47 VI VI-71
VI-28 VI-39
VI-17

SW
II-36

2.7625
NE
II-25
II-14

35. II
VII

35. 2.44375 22.

22.
Z (m)

Z (m)
2.125

.
1.80625

-6906.
-6916.
.
-6095
VI-91 VII-33

-6926.
.
1.4875
VI VII-24

-6085
30. 30.
VI-39 VII-3 VII-20

-6936.
VI-71

.
-6075

-6946.
VI-102

1.16875

.
VI-28

-6065
VI-17

-6055
II-47
II-36
0.85
25. II-14 a) 25. 0.53125 Isatis
0.2125
II-25

II
<0
-6100. -6090. -6080. -6070. -6060. -6050. N/A
X (m)
X (m)
-6100. -6090. -6080. -6070. -6060. -6050.
VII-3 VII >=0.338897 Fence diagram in SW-NE
40. 40.
VII-20 VII-24 VII-33

0.322302
VI-102
VI-71 VI-91

direction
II-47 VI-39 VI
VI-17 VI-28

II-25
II-36
0.305708
0.289113
II-14

35. 35.
VII
II
Z (m)

0.272519

Z (m)
VI-91 VII-33
0.255925
30.
VI-39 VI
VI-71
VI-102
VII-3 VII-20
VII-24

30. 0.23933
VI-17
VI-28
0.222736
0.206141
Sections of: (a)
II-47
II-36

25. II-14
II-25
b) 25. 0.189547
0.172953
II
<0.16189
-6100. -6090. -6080. -6070. -6060. -6050. N/A
estimated Ni+Co grades,
X (m)
X (m) (b) kriging standard
-6100. -6090. -6080. -6070. -6060. -6050.
40. II-47
VI-28
VI-39 VI VI-71 VI-91
VI-102 VII-3 VII-20 VII-24 VII-33 VII

40.
>=3.4
3.08125 deviation, and (c) profile
2.7625
VI-17
II-36

of original grades from

II-25
II-14
2.44375
35. 35.
VII

2.125
II
Z (m)

Z (m)

1.80625
1.4875
VI-91 VII-33
VI VII-24

test pit
VI-39

30. 30.
VII-3 VII-20
VI-71

1.16875
VI-102
VI-28
VI-17

II-36
II-47
0.85
25. 25. 0.53125
II
II-14
II-25 c) 0.2125
<0
-6100. -6090. -6080. -6070. -6060. -6050. N/A
X (m)
Application of Kriging using SGeMS (2004)
 Kriged Pb in
HSZ (%)

Level 3025 m

1000 m
 Krig.Var. of Pb
in HSZ

Level 3025 m

1000 m