Application of Kriging in The Oil Industry: East (KM) Distance (KM)

Application of Kriging in the oil industry
To demonstrate the principle of Punctual Kriging we will estimate the elevation of the oil table point
(p) from the three known elevation measured in three observation well as shown in Figure 1.
100
3
2
(d) in m 2
North (km)
80
60
40
20
0
0
4
East (km)
Figure 1
10
Distance (km)
15
20
Figure 2
The coordinate of the known and unknown wells as well as the distance between them is given in table
1. A prior analysis has produced the variogram in Figure 2, which is linear with a slope of 4.0 m 2/km.
Values of the semivariance corresponding to distances between the points are also given in Table 1;
these may be read directly off the variogram or calculated from the slope.
Point
1
2
3
p
East (km)
3.0
6.3
2.0
3.0
Point
1
2
3
1
2
3
North (km)
Elevation (m)
4.0
120.0
3.4
103.0
1.3
142.0
3.0
?
Distance between points
1
2
3
p
0
3.3541
2.832
1.0
0
4.7854
3.3242
1.9723
Semivariances
1
2
3
p
0
13.4164 11.3280
4.0
19.1416 13.2968
7.8892
The equations that must be solved to find the weights Wi in this example are:
W1 * (0) + W2 * (13.4164) + W3 * (11.3280) + = 4.0
W1 * (13.4164) + W2 * (0) + W3 * (19.1416) + = 13.2968
W1 * (11.3280) + W2 * (19.1416) + W3 * (0) + = 7.8892
W1 + W3 + W3 = 1
And in matrix form,
13.4164 11.328
0
13.4164
0
19.1416
11.328 19.1416
0
1
1
1
1 W1 4.0
1 W3 13.2968
=
1 W3 7.8892

0 1.0
13.4164 11.328
W1 0
W 13.4164
0
19.1416
3 =
W3 11.328 19.1416
0

1
1
1
1
1
1
4.0
13.2968
7.8892
1.0
0.0296
0.0368
0.1860 4.0 0.6042
W1 0.0664
W
0.0393 0.0097
0.4171 13.2968 0.0896
3 =
=
W3
0.0465
0.3968 7.8892 0.3062

10.0919 1.0 0.6705

Symmetric
The estimated elevation of point (p) is given by:
Z e ( p) = W1Z1 + W2 Z 2 + W3 Z 3 = 0.6042*120 + 0.0896*103 + 0.3062*142 = 125.2128 m

The error variance of the estimated elevation is given by:
z2 = W1 (d1 p ) + W2 (d 2 p ) + W3 ( d 3 p ) +
= 0.6042*4 + 0.0896*13.2968 + 0.3062*7.8892 = 6.0239 m2 (i.e. standard error = 2.4544 m)

That is Z(p) = 125.2128 4.9087 meters, with 95% probability
Homework:
1. What happen if point (p) coincide with point (2)
2. Suppose that you were to estimate the elevation of the same point (p) but from the following, what
do you expect in terms of the standard error?
Point
1
2A
3
p
East (km)
3.0
3.8
2.0
3.0
North (km)
4.0
2.4
1.3
3.0
Elevation (m)
120.0
115.0
142.0
?
Universal Kriging
Let us extend our previous example so that we include two more wells as given in the table below.
5
North (km)
2
P
4
2
3
1
0
0
East (km)
Point
1
2
3
4
5
p
Point
1
2
3
4
5
1
2
3
1
0
1
0
East (km)
North (km)
3.0
4.0
6.3
3.4
2.0
1.3
3.8
2.4
1
3.0
3
3.0
Distance between points
2
3
4
3.3541
2.832
1.7888
0
4.7854
2.69
0
2.11
0
Semivariances
2
3
4
13.4164 11.3280
7.1552
0
19.1416
10.76
0
8.44
0.
Elevation (m)
120.0
103.0
142.0
115.0
148.0
?
5
2.236
5.315
1.9723
2.863
p
1.0
3.3242
1.9723
1.00
2.00
5
2.2360
21.26
7.8892
11.452
0
p
4.0
13.2968
7.8892
4.0
8.0
The simplest example consists of kriging a point, assuming the drift is linear and that the
semivariogram of the residuals from the drift is linear as well.
The following is an example of how to estimate both the drift and the regionalized variable by
universal kriging:
( d 11)
( d 21)
( d 31)
( d 41)
( d 51)
1
X
1
Y1
Where,
Xp, Yp
1, 2
Xi, Yi
( d 12 )
( d 22 )
( d 32 )
( d 42 )
( d 52 )
1
X2
Y2
( d 13)
( d 23 )
( d 33 )
( d 43 )
( d 53 )
1
X3
Y3
( d 14 )
( d 24 )
( d 34 )
( d 44 )
( d 54 )
1
X4
Y4
( d 15 )
( d 25 )
( d 35 )
( d 45 )
( d 55 )
1
X5
Y5
1
1
1
1
X1
X2
X3
X4
1
0
X5
0
0
0
0
0
Y1 W1 ( d 1 p )
Y2 W2 ( d 2 p )
Y3 W3 ( d 3 p )

Y4 W4 ( d 4 p )
=
Y5 W5 ( d 5 p )

0 1
0 1 X p

0 2 Y p
are the coordinates of kriged location

are the unknown drift parameters
are the coordinates of known point (i)
To simplify the calculation we can shift the origin to point (p), this will alter the all the coordinates but
not the distances, and therefore the kriging weight will be the same. The following simultaneous
equations have to be solved:
13.4164 11.328 7.1552 2.236
0
13.4164
0
19.1416 10.76
21.26
11.328 19.1416
0
8.44
7.8892
10.76
8.44
0
11.452
7.1552
2.236
21.26
7.8892 11.452
0
1
1
1
1
1
0
3.3
1.0
0.8
2.0
0.4
1.7
0.6
0
1.0
1
1
1
1
1
0
0
0
1 W1 4.0
0.4 W2 13.2968
1.0 1.7 W3 7.8892

0.8 0.6 W4 4.0
=
2.0
0 W5 8.0

0
0 1
0
0 1 0

0
0 2 0
0
3.3
Solving the equation gives a set of eight coefficients, the first five of that are the kriging weights:
W1 0.4119
W 0.0137
2
W3 0.0934

W4 = 0.4126
W5 0.0957

0.7245
0.0660
1
2 0.0229
The estimate of the oil table elevation at the well location (p) is given by:
Zp(e) = 0.4119*(120) 0.0137 *(103) + 0.0934 *(142) + 0.4126 *(115) + 0.0967 *(148) = 123.0403 m
Which is only slightly different than the results obtained from the previous example without assuming
the drift. The estimation error variance can be calculated as follows:
0.4119
0.0137
0.0934
0.4126
2
=
[ 4.8 13.2968 7.8892 4.0 8.0 1
0.0957
0.7245
Which is almost 60% of the previous result.
= 3.9018 m2

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Application of Kriging in the oil industry

And in matrix form,

10.0919 1.0 0.6705

The estimated elevation of point (p) is given by:

Z e ( p) = W1Z1 + W2 Z 2 + W3 Z 3 = 0.6042120 + 0.0896103 + 0.3062*142 = 125.2128 m

= 0.60424 + 0.089613.2968 + 0.3062*7.8892 = 6.0239 m2 (i.e. standard error = 2.4544 m)

are the coordinates of kriged location

0.8 0.6 W4 4.0

Which is almost 60% of the previous result.

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