CUJ 8 Feb 2014_Hari

Inverse Theory Applications in

Petrophysics

K.C. Hari Kumar

ONGC, Baroda. 2013

25

Interpretation

Issues and Challenges

ONGC a Wealth Creator

1

2

Inverse Theory Application

3

4

SVD and Anatomy of Inverse Problem

Presentation Overview

Regularization Methods

Data Inversion

Role of numerical methods in Petrophysics

4ONGC a Wealth Creator

The down-hole measurements (N number) make up the data

used for volumetric parameter estimation

data, d d d d = [d1, d2, dN]T Earth model in terms of mineral volumes is to be retrieved

model parameters, m m m m = [m1, m2, mM]T Physical theory or quantitative model to predict the data is the

forward problem

Inverse theory aims at estimating an earth model from the data

Quantitative Model

Quantitative Model

est pre

est obs

estimates predictions

observationsestimates

Forward Theory

Inverse Theory

6Play of Errors

ddddpre ddddobs

mmmmest mmmmtrueObservational errors

Error propagation

Understanding the effects of observational error is central to

Inverse Theory

Tool response at a depth point di is a composite function of an array

of formation properties and other parameters (gi): = , , 1 T (di), the log values at depth di represent a measurement averaged

over an earth volume surrounding the point of measurement and the

collective signal from various parts of the borehole.

The surrounding formation is represented by the convolution integral

of the form:

7

Down-hole measurements

! = " " " #! ; %, &, 3600

0

0

%, &, %& 2

The kernel K (di; x, r, ) accounts for the geometrical effects included in integration to ensure that the signal belongs to the formation in the

vicinity of the depth point: x, r, are the cylindrical coordinates and g(x, r, ) is the geophysical property distribution of the probe sensitive volume.

The integral equation is converted into a set of m linear

algebraic equations in n unknowns (mn or m n)represented as +, = - .

8

Well Log Data Inversion

A is the response matrix derived of the parameter values of

each tool for 100% of each formation component, x the

volume vector of formation components and b the data vector

constituted of the different tool measurements.

ToolsA x b

Quartz Calcite Kaolinite Muscovite Water Volumes Tool Data

RHOB 2.65 2.71 2.41 2.79 1 0.41 2.196NPHI -0.06 -0.02 0.37 0.25 1 0.07 0.318

DT 55.5 48 120 55 189 0.22 102.515GR 12 0 105 270 0 0.05 41.520

VOLSUM 1 1 1 1 1 0.25 1.000

9The Inverse Problem: x = A-1b

The linear equations Ax = b are solved under constrained

conditions of m = n subject to appropriate handling of the ill-

conditioned system. One of the earliest demonstrations of the

method may be seen with Doveton (1986). Sensitivity of A-1

makes the solution unstable for noise and round off error.

Names x A-1 b

Volumes Inverse Operator Tool Data

Quartz 0.41 -4.519 -10.642 0.022 0.011 10.968 2.196Calcite 0.07 3.827 9.567 -0.030 -0.014 -7.745 0.318

Kaolinite 0.22 2.174 0.494 0.023 -0.002 -6.976 102.515Muscovite 0.05 -0.645 0.281 -0.010 0.004 2.225 41.520

Water 0.25 -0.838 0.300 -0.005 0.000 2.528 1.000

Geophysical inverse problems are ill-posed as the solutions

are either non-unique or unstable or both.

Hadamard (1902) Existence, Uniqueness and Instability

In general the model that one seeks is a continuous function

of the space variables with infinitely many degrees of freedom.

On the other hand, the data space is discrete and always of

finite dimension because any real experiment can only result

in a finite number of measurements.

A simple count of variables shows that the mapping from the

data to a model cannot be unique; or equivalently, there must

be elements of the model space that have no influence on the

data.

This lack of uniqueness is apparent even for problems

involving idealized, noise-free measurements. The problem

only becomes worse when the uncertainties of real

measurements are taken into account.

No guarantee that Ax = b contains enough information for

unique estimate x and ascertaining of the sufficiency of

information devolves upon the discrete inverse theory.

10

The Challenge arising from Ax = b

11

Model Space and Data Space

Earth is a model space of infinite degree of freedom

Physics of the experiment decides the model and finite data

Earth model and the linear Ax = b formulation: A is also known

as the Sensitivity matrix which contains measurements (100%

volumes of minerals) corresponding to the end members.

Volume vector has the obvious constraint that the sum is equal

to 1.

Tool vector consists of the different measurements or data

inverted.

12

./0120/ 34567 +

88892:;64?

5@22: 9/=526 -

Linear model from earth

13

Generalized Inverse = Least Squares

A x bVariance

1.5%

77.97 79.73 70.91 82.09 29.42 0.41 66.67 1

-14.16 -4.72 87.32 59.00 235.99 0.09 66.67 1

38.14 32.98 82.46 37.79 129.88 0.22 66.67 1

17.05 0.00 149.19 383.63 0.00 0.07 66.67 1

66.67 66.67 66.67 66.67 66.67 0.21 66.67 1

Variance reduced to unity

Given the assumptions of normal distribution of errors which

are uncorrelated, the L-2 norm is employed to characterize the

solution vector x. The Generalized inverse gives the minimum

length solution, m > n and becomes the maximum likelihood

solution when m = n.

Method fails when (ATA) has no inverse

14

x

0.0538 -0.0424 -0.0276 0.0084 0.0103 12377.58 0.41

-0.0424 0.0340 0.0204 -0.0061 -0.0077 11644.29 0.09

-0.0276 0.0204 0.0175 -0.0056 -0.0063 30436.01 0.22

0.0084 -0.0061 -0.0056 0.0018 0.0020 41945.12 0.07

0.0103 -0.0077 -0.0063 0.0020 0.0023 30796.89 0.21

The limits or the uncertainty of the model parameters is

estimated from A = B+C+ DE . When B =1, the Hessianinverse operator +C+ DE is the model variance-covariancematrix with the diagonal elements as the variance.

Uncertainty in the estimated model parameters xi is obtained

as the square root of the variance converted into percentage.

15

Parameter Covariance Matrix A = B+C+ DEModel Uncertainty % & Correlation of xi x

23.19 -0.04 -0.03 0.01 0.01 0.41

-0.04 18.45 0.02 -0.01 -0.01 0.09

-0.03 0.02 13.23 -0.01 -0.01 0.22

0.01 -0.01 -0.01 4.25 0.00 0.07

0.01 -0.01 -0.01 0.00 4.80 0.21

Tool QUAR SM1 SMEC ILLI KAOL CHLO W x bSTD

Error

RHOB 2.65 2.35 2.12 2.53 2.42 2.77 1.09 0.35 2.3153 0.035

SGR 5 150 180 12 44 2.04 0 0.05 51.333 0.770

TH/K 3 10 12 3.5 14 16 0 0.2 7.085 0.106

TNPH 0.04 0.4 0.44 0.3 0.37 0.52 1 0.09 0.3513 0.005

PEF 2 2 2.04 3.45 1.83 6.37 0.8 0.11 2.3254 0.035

DT 53.5 60 60 87 77 100 189 0.08 80.705 1.211

SUM 1 1 1 1 1 1 1 0.12 1 0.015

16

Example 2

Model Uncertainty % & Correlation of xi x46.82 0.06 -0.01 -0.29 -0.13 0.11 0.05 0.35

0.06 24.75 -0.04 -0.09 -0.04 0.03 0.02 0.05

-0.01 -0.04 16.86 0.02 0.01 -0.01 0.00 0.20

-0.29 -0.09 0.02 63.24 0.18 -0.14 -0.07 0.09

-0.13 -0.04 0.01 0.18 28.18 -0.06 -0.03 0.11

0.11 0.03 -0.01 -0.14 -0.06 22.56 0.03 0.08

0.05 0.02 0.00 -0.07 -0.03 0.03 11.29 0.12

Anatomy of Inversion

The ill-posed character of the discrete problem i.e. the

structure and numerical behavior of the response matrix A

and the modes of linear transformation can be understood

with the help of singular value decomposition (SVD).

The operator matrix A can be resolved into three component

matrices in such a way that -

18

Singular Value Decomposition

F = G8 = H I!! J!K

!=1

U and V are the orthogonal matrices of left and right

singular vectors representing the model space and the data

space while is a diagonal matrix of singular values i whichare amplitudes of the mode of transformation.

, = FD- = H I@L

M

J

19

SVD of Operator A

A = ,

2.650.0655.5121

2.710.024801

2.410.371201051

2.79 10.25 1 55 189270 1 1 1= UVT

Left Singular Vectors: U -0.01336 0.00499 0.94988 -0.09353 -0.29796-0.00216 0.00364 -0.12703 0.75599 -0.64212-0.54151 0.84058 -0.01229 -0.00658 0.00126-0.84056 -0.54164 -0.00878 -0.00058 0.00081-0.00566 0.00454 0.28525 0.64783 0.70633

Singular Value Spectrum: 319.69 0 0 0 0

0 201.053 0 0 00 0 3.281 0 0

0 0 0 0.185 0

0 0 0 0 0.047

Right Singular Vectors (Transpose) VT-0.12569 0.1998 0.61649 -0.09546 0.74506-0.08144 0.20077 0.69249 0.34282 -0.59665-0.47946 0.21892 0.04011 -0.80287 -0.27565-0.80321 -0.49734 -0.0431 0.31647 0.07408-0.32021 0.79025 -0.37004 0.35862 0.0862

Important insights offered by the SVD in the present case are:

Condition number can be derived as the ratio of the largest and

the least singular values to be 6834.

Ill-conditioning indicated by Cond(A) finds support also from

the elements of the left and right singular vectors u, and v,

which tend to be more oscillatory i.e. show more sign changes

as the index i increases or the corresponding singular values

decreases.

A small singular value i as compared to 1 = ||A||2, pointtowards the existence of a certain linear combination of the

columns of A spanned by the elements of the right singular

vector vi such that ||Avi||2= i is small. Similar argument holdsfor the left singular vector ui and the rows of A. Small singular

values therefore are indicative that the operator is nearly rank

deficient and that the associated ui and vi are effectively the

numerical null vectors of AT and A respectively.

20

Implications of the SVD

The accumulative contribution ratio for the possible truncation

stages is expressed as:

21

Impact of Truncation

TU = UM MModes i

Modes% Cumulative

Normalizedi %

1 319.69 60.98 60.98 100.02 201.053 38.35 99.33 62.93 3.281 0.63 99.96 1.04 0.185 0.04 99.99 0.15 0.047 0.01 100.00 0.0

U V -0.013 0.005 0.95 -0.094 -0.298 -0.126 0.2 0.616 -0.095 0.745-0.002 0.004 -0.127 0.756 -0.642 -0.081 0.201 0.692 0.343 -0.597-0.542 0.841 -0.012 -0.007 0.001 -0.479 0.219 0.04 -0.803 -0.276-0.841 -0.542 -0.009 -0.001 0.001 -0.803 -0.497 -0.043 0.316 0.074-0.006 0.005 0.285 0.648 0.706 -0.32 0.79 -0.37 0.359 0.086 99.33 % of the total variance is contributed by the 1st and 2nd modes

u1 and v1 have no sign changes at all and sign changes increases with

other columns. 3 = 1.02% of 1while and 4 and 5 are abysmally lower and suggests a rank 3 approximation

The minimum norm solution x obtained is [0.23, 0.24, 0.21,

0.06, 0.26] Reconstructed kernel A3 = A with condition number

97 against the 6834 for A. 22

Rank 3 Approximation

F = G8@ % = 8WG@LOriginal A Rank 3 Reconstruction A3 x

2.65 2.71 2.41 2.79 1 2.66 2.71 2.39 2.80 1.01 0.232-0.06 -0.02 0.37 0.25 1 -0.02 -0.09 0.47 0.21 0.95 0.23655.5 48 120 55 189 55.50 48.00 120.00 55.00 189.00 0.21412 0 105 270 0 12.00 0.00 105.00 270.00 0.00 0.0601 1 1 1 1 0.99 0.98 1.10 0.96 0.95 0.261

Data Vector b for x Solution x: Truncated SVD Solutions x b Rank 5 Rank 4 Rank 3 Rank 2

0.41 2.196 0.41 0.24 0.23 0.100.07 0.318 0.07 0.21 0.24 0.090.22 102.515 0.22 0.28 0.21 0.200.05 41.520 0.05 0.03 0.06 0.070.25 1.000 0.25 0.23 0.26 0.34

Diagonal elements of the resolution matrix equal to 1 is

indicative of good resolution Unique solution

Non-zero elements other than the diagonal ones in a row

indicate that the predicted value of the parameter is a weighted

average of the observed data.

Considering the 1strow for example, [0.44, 0.44, 0.21, -0.06, -

0.06], the first element of the predicted data vector bp = [b1P, b2

P,

b3P, b4

P, b5P] viz., b1

P will be predicted as a weighted average of

the observed data d = [d1, d2, d3, d4, d5]. With d = [2.1936,

0.3225, 102.12, 40.38, 0.99], bp = [2.128, 0.326, 103.141,

39.572, 0.99]

Note: 0.44*2.196+0.44*2.196+0.21*2.196-0.06*2.196-0.06*2.196 = 2.12823

Model Resolution (V5V5T = I5 )

Model Resolution V4V4T (Rank 4) Model Resolution V3V3T (Rank 3)0.44 0.44 0.21 -0.06 -0.06 0.44 0.48 0.13 -0.02 -0.030.44 0.64 -0.16 0.04 0.05 0.48 0.53 0.11 -0.06 -0.070.21 -0.16 0.92 0.02 0.02 0.13 0.11 0.28 0.27 0.31-0.06 0.04 0.02 0.99 -0.01 -0.02 -0.06 0.27 0.89 -0.12-0.06 0.05 0.02 -0.01 0.99 -0.03 -0.07 0.31 -0.12 0.86

Data resolution matrix presents the information density i.e.

which data contribute independent information to the

solution. Value of 1 for the diagonal element indicates

information independent of other observations. U3U3T I3

shows that there exists a data null space and the model fit for

the data shall be poor.

These results are in agreement with the ill-conditioned

behavior of the original design matrix A which had a lesser

number of linearly independent vectors than those indicated

by the rank of 5.

24

Data Resolution

Data Resolution U4U4T(Rank 4) Data Resolution U3U3T(Rank 3)0.91 -0.19 0.00 0.00 0.21 0.90 -0.12 0.00 0.00 0.27-0.19 0.59 0.00 0.00 0.45 -0.12 0.02 0.01 0.00 -0.040.00 0.00 1.00 0.00 0.00 0.00 0.01 1.00 0.00 0.000.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.000.21 0.45 0.00 0.00 0.50 0.27 -0.04 0.00 0.00 0.08

The three energetic model factors of the activated space explains

99.96% of the variance in the data. The first function u1 and v1contribute 66.98% followed by u2 and v2 38.35 and u3 and v3adding 0.63%. Vk represents the optimized distribution of row

vectors of the model space and Uk represents the optimized

distribution of column vectors of the data space.

Singular value decomposition thus enables an objective ranking

of uncorrelated modes of variability or latent variables which

helps to differentiate the data from noise.25

SVD Conclusions

Modes iModes

%Cumulative

Normalized

i %

1 319.69 60.98 60.98 100.0

2 201.053 38.35 99.33 62.9

3 3.281 0.63 99.96 1.0

4 0.185 0.04 99.99 0.1

5 0.047 0.01 100.00 0.0

The original kernel A5x5 may be viewed as an A3x3 data kernel

perturbed by a noise matrix of norm . Such a perturbation shallchange the zero singular values by and therefore any singularvalues < stand the chance of being contributed by the noise.

R(A) = 5 needs to be contrasted with the scenario R(A,) = 5 byexamining the norm of the data error possible with the kernel A.

For any of the variables in the kernel A i.e. columns, the standard

error is 1.5% and for the set of observations considered viz., RHOB,

NPHI, DT, GR and VOL_SUM, the likely norm of the error may be

computed from a likely data vector, say, [2.196, 0.318, 102.515,

41.52, 1] for which the errors will be [0.033, 0.005, 1.538, 0.623,

0.015], the norm of the error shall be 1.66.

If we consider the balanced uncertainties given by proprietors of

certain software tools, the noise vector shall be [0.027, 0.015, 2.25, 6,

1.5] but the value for GR (=6) is almost unpredictable and inclusion

may lead to a high value of error norm. Excluding GR, we obtain 2.7 and hence by any count the low singular values we seen above, 4= 0.185 and 5 = 0.047 cannot be treated as causing a genuineaddition to the range of the problem.

26

Impact of Noise

The truncated solutions, theoretically free from noise, differ

much from the true model xtrue used for deriving the data

vector b. Rank-4 and Rank-3 solutions are nearly the same

but significantly different from xtrue.

Here a question arises: The linear formulation has

sufficient information content as to permit the retrieval

of the true model?

Each column of A represented an end-member configuration

realizable as data in 2 decimal digits like 2.65, 0.35 etc. In

other words, the forward problem is expressed in terms of 2

decimal digits multiplied by a fractional volume vector to yield

data of maximum accuracy up to 1 decimal digit only. Given

such a forward problem, can the measurements be of

information content as to retrieve the volume vector in 2

decimal digits?

27

Truth of the linear formulation Ax = b

Rank-4 and Rank-3 operators have not contributed much to

the stability of the solution as may be noted from the above

data.

Constraints are needed to ensure positive values and to make

the solution agree with prior information about the solution x,

volume vector.

Situation demands one or the other of the known regularization

methods depending upon their suitability adjudged domain

wise

28

Round-off Error

b Rank-5x5

b* x5* -L2Rank-4

-L2Rank-3

-L2x4 x4* x3 x3*2.196 0.41 2.22 0.53

0.186

0.240 0.253

0.125

0.232 0.253

0.0050.318 0.07 0.3 -0.05 0.206 0.165 0.236 0.165

102.515 0.22 103.51 0.28 0.282 0.385 0.214 0.38541.52 0.05 42.5 0.02 0.033 -0.003 0.060 -0.003

1 0.25 1 0.22 0.231 0.188 0.261 0.188

Application of regularization methods is problem specific. Advantage

of one method over the other to address instability due to rank

deficiency in specific kind of problems is missing in the literature.

Tikhonov regularization can be found to be quite popular in its

application

Common for all these regularization methods is that they replace the

ill-posed problem with a nearby well-posed problem which is much

less sensitive to perturbations Per Christian Hansen

Given the rank deficient problem of comparatively small size kernel

we have, Tikhonov regularization is one of the best options

29

Model Space

Equations of the kind, Ax = b do not yield the right numerical

solution when the data b contains noise. If it is known that the

given data satisfies an error estimate L L X , Tikhonovstates that an approximate solution can be found by

minimizing the regularization functional:

30

Tikhonov Regularization

C ,, - = , = +, -E YBB Z B Y, YBB [\], ^ _L is typically the identity matrix or a well conditioned discrete

approximation to some derivative operator. In operator

notation:

, = ++C Z ` DE+C-, a ^ _

31

Tikhonov continued

% = 8bccccd

!1!12 Z 0 00 !

!2 Z 00 0 !Z1!Z12 Z ef

fffg

&

GL

hiijiik 8

bccccd

!1!12 Z 0 00 !

!2 Z 00 0 !Z1!Z12 Z ef

fffg

&

G lmIno&!pm qKJm&rm FZlsiitiiu

hiijiik

0, FZlFZ: 8bcccccd 1! 0 00 1

! 00 0 1

! efffffg&

G wmKm&on!pm qKJm&rm FZsiitiiu

Application of a small value for such as =1, in fact gave asolution closer to the truncated A3 solution. Minimum norm

solution happens for = 0 and this is not an expected behaviorand may be the consequence of design or the use of

hypothetical data.

Petro-physicists are in need of the mathematicians help to

interpret the results

32

Tikhonov Example-1

b xtrue =0.001 =1 =5 2.196 0.41 0.36 0.05 0.22 0.19 0.19 0.220.318 0.07 0.11 -0.04 0.22 -0.15 0.19 -0.12

102.515 0.22 0.24 -0.02 0.22 0.00 0.21 0.0141.52 0.05 0.05 0.00 0.06 -0.01 0.06 -0.01

1 0.25 0.24 0.01 0.27 -0.02 0.29 -0.04L2 norm 0.54 0.51 0.07 0.47 0.24 0.45 0.25

The L-curve is a plot of the size

of the regularized solution

versus the size of the

corresponding residual for all

valid regularization parameters.

L-curve does not depict a

distinct elbow but the point of

maximum curvature can be

identified to be between = 4and = 1.

33

Tikhonov Example-2

b% = FWL% = 0

% = x F% L % = y F% L % = 1 F% L % = F% L

2.401 -1.29 0.20 -0.034 0.24 -0.046 0.24 -0.096 0.22 -0.2060.412 1.44 0.00 -0.040 0.26 -0.032 0.25 -0.022 0.23 -0.007

115.260 1.01 0.34 0.000 0.27 0.000 0.24 -0.001 0.23 -0.00336.749 -0.20 -0.01 0.000 0.02 0.000 0.03 0.001 0.04 0.004

1.0 0.04 0.26 0.082 0.29 0.090 0.31 0.079 0.33 0.047F% L 4.8 0.31 0.0096 0.29 0.0111 0.28 0.0160 0.26 0.0446

IIAx-bII2

L - Curve

I

I

x

I

I

2

L= 5 = 0.046

L= 4

= 0.185

L= 1

L= 3

= 3.28

Considering the few solutions which are likely to be valid, for = 4 =0.18, =1 and = 2, it can be seen that for the log data vector used,the porosity values %xhad been on the increase as increased i.e.0.29, 0.31, 0.32 which raises a question mark on the validity of the

regularization as the log values cannot be of so high a porosity value.

34

Validity of the Regularized solution

b % = y F% L % = 1 F% L % = 2 F% L

2.401 0.24 -0.046 0.24 -0.096 0.23 -0.1490.412 0.26 -0.032 0.25 -0.022 0.24 -0.014

115.260 0.27 0.000 0.24 -0.001 0.24 -0.00236.749 0.02 0.000 0.03 0.001 0.03 0.002

1.0 0.29 0.090 0.31 0.079 0.32 0.064F% L 0.29 0.0111 0.28 0.0160 0.27 0.0265x51 =1-

(x11+..x41) 0.21 0.24 0.26With the unity constraint imposed and the porosity value is derived as

x51 =1-(x11+..x41), the value for porosity appears to be a distorted valueexceeding 0.20, viz., 0.24 when = 1 and 0.26 when = 2. For a matrixdensity of 2.65, the porosity could have been only around 0.15.

35

Tikhonov Regularization with prior

z% = o& {!K F% L Z |% %1626 , = BYCY Z +C+ DE BYCY,}~\~ Z +C

Prior = 0.23 = 0.30 = 1.0 = 5 = 15xprior x ax-b x ax-b x ax-b x ax-b x ax-b0.46 0.40 -0.033 0.43 -0.033 0.46 -0.028 0.48 0.027 0.48 0.0490.07 0.11 -0.057 0.09 -0.056 0.07 -0.054 0.08 -0.061 0.09 -0.0640.22 0.32 0.000 0.29 0.000 0.25 -0.001 0.25 -0.012 0.25 -0.0980.05 0.00 0.000 0.00 0.000 0.02 0.000 0.02 0.005 0.02 0.0460.20 0.27 0.085 0.27 0.088 0.29 0.096 0.28 0.113 0.28 0.119

Norms ||x||2 ||ax-b||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||20.31 0.3418 0.0116 0.3511 0.0121 0.3672 0.0129 0.3775 0.0174 0.3810 0.0325x 1.09 1.09 1.10 1.11 1.12

x51had been on the increase as increased and the unity constraint wasnot adhered to in the above exercise of regularization. With the xprior

initially assumed for deriving b, the minimum norm solution and

minimum residual norm happens for = 0

Unity constraint may be enforced by deriving x51 = 1-

SUM(x11+x21+ x31+x41) or by altering the algorithm to solve for

only the (n-1) elements and deriving the nth element as 1

minus sum of (n-1) elements.

36

Prior with Unity Constraint

Volumes = 0.23 = 0.30 = 1 = 5 = 15 No Priorx11 0.40 0.43 0.46 0.48 0.48 0.24x21 0.11 0.09 0.07 0.08 0.09 0.25x31 0.32 0.29 0.25 0.25 0.25 0.24x41 0.00 0.00 0.02 0.02 0.02 0.03

x51 =1-(x11+..x41) 0.18 0.19 0.19 0.17 0.16 0.23

With the unity constraint imposed, the porosity values have

become reasonable but when the same is contrasted with the

no-prior solution, it becomes apparent that the linear

formulation without prior values cannot return even a

reasonably true model

No distinct elbow is seen and in the following slide an

alternative prior vector is used to study the issue further

37

L-Curve in the above case

L - Curve

I

I

x

I

I

2

IIAx-bII2

In this case it is seen that the prior without the unity

constraint fails to retrieve a reasonable solution

With = 1 and the sum of volumes =1 constraint, the prior isalmost reproduced.

Can such use of priors have any meaning or lead to a valid

technique when Ax= b lacks the information?

38

Alternate prior

Prior2

= 0.6 = 1 = 2 = 3 = 5x ax-b x ax-b x ax-b x ax-b x ax-b

0.35 0.55 0.011 0.36 -0.030 0.26 -0.180 0.21 -0.33 0.17 -0.530.15 0.00 -0.071 0.16 -0.043 0.21 -0.013 0.19 0.01 0.15 0.040.22 0.28 0.000 0.25 -0.001 0.24 -0.005 0.23 -0.01 0.23 -0.040.08 0.00 0.000 0.02 0.000 0.03 0.003 0.04 0.01 0.04 0.010.20 0.27 0.103 0.30 0.096 0.32 0.054 0.34 0.01 0.37 -0.05||x||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||2 ||x||2 ||ax-b||20.24 0.4536 0.0159 0.3094 0.0120 0.2690 0.0356 0.2521 0.1088 0.2378 0.2831

x51 =1-(x11+..x41) 0.17 0.20 0.27 0.33 0.42x 1.10 1.10 1.05 1.01 0.95

L-Curve depicts a distinct elbow and = 1 becomes anobvious choice as a regularization parameter

Interpretation has to be specific to the problem

39

L-Curve in the above case

L - Curve

I

I

x

I

I

2

IIAx-bII2

The SVD solution for x is be expressed as:

40

Discrete Picard Condition

% = FZL = H I!L !

K!=1

J! Discrete Picard condition states that the numerator (I@L)

must decay faster than the denominator (i). But the

literature presents contradictory opinions on the trust

that can be placed on the discrete Picard condition. Due

to noise in A or other reasons of design of the inverse

model DPC is found to get violated.

May be it is not applicable to the small size problem

discussed above but why it is so remains to be discussed

in applied mathematics literature.

The DPC for A and A scaled to remove the units AUF

41

Discrete Picard Condition

Discrete Picard condition is apparently not satisfied. Is this

relevant to the small size inversion problems?

A AUF

FourierCoefficientsI@L

SingularValues

I@L

FourierCoefficient

sI@LSingularValues

I@L

90.45 319.69 0.28 118.73 506.15 0.2363.70 201.05 0.32 82.20 259.24 0.320.71 3.28 0.22 37.13 140.34 0.260.02 0.18 0.09 1.67 14.06 0.120.01 0.05 0.23 0.67 3.12 0.21

Data of a 7x7 example is used here to illustrate the failure of

the DPC. But what it means physically to the problem?

42

Picard Plot-1

Kernel A Weight Matrix Modified A = Aw2.65 2.71 2.41 2.79 1 0.555 0 0 0 0 1.47 2.71 0.01 0.00 0.01-0.06 -0.02 0.37 0.25 1 0 1.00 0 0 0 -0.03 -0.02 0.00 0.00 0.0155.5 48 120 55 189 0 0 0.0051 0 0 30.83 48.00 0.62 0.04 1.8912 0 105 270 0 0 0 0 0.0007 0 6.67 0.00 0.54 0.20 0.001 1 1 1 1 0 0 0 0 0.01 0.56 1.00 0.01 0.00 0.01

43

Impact of Weights

Aw = UVTU VT

-0.054 -0.043 0.949 -0.128 -0.280 57.295 0.000 0.000 0.000 0.000 -0.545 -0.838 -0.011 -0.001 -0.0330.001 -0.003 -0.106 -0.990 0.093 0.000 5.630 0.000 0.000 0.000 0.834 -0.543 0.088 0.033 -0.021-0.996 -0.061 -0.060 0.006 -0.004 0.000 0.000 0.102 0.000 0.000 0.000 0.039 -0.014 0.075 -0.996-0.064 0.997 0.041 -0.007 0.001 0.000 0.000 0.000 0.004 0.000 0.085 -0.042 -0.914 -0.394 -0.019-0.020 -0.014 0.288 0.059 0.955 0.000 0.000 0.000 0.000 0.00027 0.006 -0.002 -0.395 0.915 0.074

1. Instantly it strikes that the scaled matrix presents a far

greater condition number and ill-conditioning than the

original matrix.

2. Vector u1 and v1 depict increased oscillation

Number of energetic modes has been reduced when

contrasted with SVD of A.

Plot of the singular values as shown has a distinct elbow

which calls for a rank 2 approximation in contrast to the rank

3 approximation of A.

Number of manifest variables has become 2 instead of 3 in A

and the singular values right of the elbow are discarded as of

noise.

44

Modes and variances

Modes iModes

% CumulativeNormalized

i %

1 57.295 90.899 90.899 100 12 5.6303 8.9326 99.832 9.83 103 0.1016 0.1612 99.993 0.18 5644 0.004 0.0064 100 0.01 142355 0.0003 0.0004 100 0.00 215394

A number of issues arise against the background of the above discussion:1. Where can we find the inversion theory applicable for the

small scale problems of the above kind?

2. Applied Mathematics literature is silent about diagnosis in

respect of (a) information content of the linear formulation

(b) SVD analysis of the small size problems (c) Discussions

on the weights and scaling of the problems (d)

Regularization (e) Discrete Picard condition

3. Can the iterative procedures do any magic when direct

solutions are not valid?

4. Levenberg-Marquardt algorithm can offer something extra

in an iterative process?

45

Diagnosis for a valid solution

Where = F@F & F@F is the Hessian with which weightedleast squares method is sought to be implemented.

H will be more ill-conditioned than A and hence a positive

constant times the diagonal elements of H are added toensure greater eigen values *diag (H).

Direction of error decides the tuning of up and down tominimize the error and Marquardts innovation had helped to

take a large step in the direction with low curvature and a

small step when the update reaches a direction with high

curvature.

Equivalence is drawn between the stochastic inverse A+s and

the LM algorithm46

Levenberg-Marquardt algorithm

%W = % Z aq D %W = % Z a[!o] D = AT AAT Z n2

m2 I1 = ATA Z n2

m2 I1 AT

47

Example for Uncertainty vs QUAR SM1 SMEC ILLI KAOL CHLO W b b*0.015 \

RHOB 2.65 2.35 2.12 2.53 2.42 2.77 1.09 2.3287 0.034931 638.77SGR 5 150 180 12 44 2.04 0 32.082 0.48123 334.89TH/K 3 10 12 3.5 14 16 0 6.455 0.096825 192.39TNPH 0.04 0.4 0.44 0.3 0.37 0.52 1 0.3338 0.005007 89.15PEF 2 2 2.04 3.45 1.83 6.37 0.8 2.1498 0.032247 64.75DT 53.5 60 60 87 77 100 189 83.29 1.24935 3.82SUM 1 1 1 1 1 1 1 1 0.015 1.12

=0 =1 7 =2 =3 =3.82=6 =5

xUncert.

% xUncert.

% xUncert.

% xUncert.

% xUncert.

% xUncert.

%0.4 47.06 0.37 26.54 0.35 11.91 0.34 6.58 0.33 4.55 0.33 3.04

0.05 24.81 0.04 20.20 0.05 15.80 0.05 12.26 0.06 9.86 0.07 7.260.07 16.84 0.07 15.58 0.06 13.01 0.06 10.25 0.05 8.28 0.05 6.110.09 63.58 0.13 35.51 0.16 15.30 0.17 7.89 0.17 5.14 0.17 3.170.2 28.33 0.22 15.87 0.23 6.97 0.23 3.77 0.24 2.65 0.24 1.91

0.05 22.68 0.04 12.68 0.03 5.50 0.02 2.90 0.02 1.97 0.02 1.340.14 11.33 0.13 6.37 0.13 2.82 0.13 1.55 0.12 1.09 0.12 0.78

48

Model & Data Resolution for = 0, 2 R = 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.000.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.000.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.000.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.000.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.000.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.000.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00

= ++

=20.78 -0.04 -0.01 0.29 0.13 -0.10 -0.05 0.85 -0.01 0.00 0.00 0.02 -0.05 0.19-0.04 0.86 0.11 0.09 0.03 -0.03 -0.01 -0.01 1.00 0.00 0.01 0.00 -0.03 0.02-0.01 0.11 0.91 -0.01 0.00 0.01 0.00 0.00 0.00 1.00 0.02 0.00 -0.03 0.020.29 0.09 -0.01 0.61 -0.17 0.14 0.07 0.00 0.01 0.02 0.91 0.00 0.17 -0.110.13 0.03 0.00 -0.17 0.92 0.06 0.03 0.02 0.00 0.00 0.00 1.00 0.00 -0.02-0.10 -0.03 0.01 0.14 0.06 0.95 -0.02 -0.05 -0.03 -0.03 0.17 0.00 0.66 0.28-0.05 -0.01 0.00 0.07 0.03 -0.02 0.99 0.19 0.03 0.02 -0.11 -0.02 0.28 0.61

Given the uncertainty and resolution as above can the linear

formulation Ax = b under discussion contain sufficient

information as to facilitate a retrieval of a reasonably true

solution?

If the above deductions are incorrect, how the efficiency of a

linear formulation and method of inversion can be analysed for

efficiency?

How the small scale problems can be better understood with

the help of mathematical theory?

49

Model & Data resolution for = 5

0.71 0.00 -0.06 0.36 0.16 -0.13 -0.07 0.66 0.00 0.02 -0.10 0.04 0.05 0.330.00 0.62 0.30 0.09 0.03 -0.04 -0.01 0.00 1.00 0.00 0.02 0.00 -0.04 0.03-0.06 0.30 0.75 -0.01 0.01 0.00 -0.01 0.02 0.00 0.99 0.03 0.00 -0.05 0.010.36 0.09 -0.01 0.51 -0.22 0.17 0.09 -0.10 0.02 0.03 0.82 0.02 0.29 -0.090.16 0.03 0.01 -0.22 0.90 0.08 0.04 0.04 0.00 0.00 0.02 0.99 -0.01 -0.03-0.13 -0.04 0.00 0.17 0.08 0.94 -0.03 0.05 -0.04 -0.05 0.29 -0.01 0.48 0.28-0.07 -0.01 -0.01 0.09 0.04 -0.03 0.98 0.33 0.03 0.01 -0.09 -0.03 0.28 0.47

Lagrangian to be minimized is defined here as:

Y = Z Z = F% L Z , , 2 Z Sx

S=

1 1 0 0 00 1 1 0 0000000

100110

011 Differentiating the Lagrangian with respect to x and

equating to zero, the solution x can be obtained as:

% = F@F Z q Z T@r DF@L Z qT 50

Constrained Optimization

No appreciable improvement in the solution characteristics by

the use of the derivative operator

With the parameter choice of 1 =1 and 2 =1, the errors to thetune of 1.5% do not alter the regularized solution computed

against a specific and precise solution prior vector (xref)

chosen.

But the departure of the regularized solution from the solution

prior vector and the relation of both to the true solution vector

becomes a subtle issue that needs more detailed deliberations.

51

b1 %W 1 2 xref x = x1 Ax-b1 b2= b1+e %W x2 b3= b2+e %W x32.2518 0.46

1 1

0.46 0.31 -0.019 2.2856 0.30 0.31 2.3199 0.30 0.320.2829 0.05 0.05 0.19 0.023 0.2871 0.17 0.19 0.2915 0.17 0.1997.36 0.2 0.2 0.16 0.000 98.8204 0.31 0.17 100.3027 0.31 0.1745.42 0.07 0.07 0.09 0.000 46.1013 0.04 0.09 46.7928 0.04 0.09

1 0.22 0.22 0.24 0.001 1 0.19 0.25 1.0000 0.19 0.25L2-norm 0.55 0.55 0.48 0.03 0.50 0.48 0.50 0.49

Solutions by the Lagrange multiplier method

Here the precise solution prior vector has been changed to see

the impact on the solution and it can be found that the

regularized solution did not significantly respond to the prior

solution vector used.

But when a log vector is applied to the same scenario, the

solution showed departure from the unity constraint.

When the data vector used is different from the inverse crime

scenario the solution is perturbed suggesting instability of the

solution despite the LMM implementation

52

b1 %W 1 2 xref x = V1 Ax-b1 xref x Log xref %:2W x Ax-b12.2518 0.46

1 1

0.50 0.32 -0.007 0.40 0.29 2.468 0.40 -2.46 0.34 0.1440.2829 0.05 0.10 0.19 0.022 0.20 0.22 0.4782 0.20 2.54 0.26 0.07097.36 0.2 0.15 0.15 0.000 0.10 0.15 110.3399 0.10 1.09 0.14 12.98045.42 0.07 0.03 0.09 0.000 0.08 0.10 32.0829 0.08 -0.19 0.05 -13.336

1 0.22 0.22 0.25 0.006 0.22 0.25 1 0.22 0.03 0.31 0.104L2-norm 0.55 0.55 0.48 0.02 0.55 0.48 0.55 0.55

Changed Prior Vector

Tikhonov solution vectors are different many elements

differing drastically from the constrained optimization output.

Here the issue comes up:

How to choose between different optimization methods?

Under what circumstances one regularization method is

preferred over the other?

53

Tikhnov versus LMM solutions

L = 0 L = 0.04678 L = 0.18459 L = 0.001 L= 0.01 L=1b1 xtikh = x+ xtikh Ax-b xtikh Ax-b xtikh Ax-b xtikh Ax-b xtikh Ax-b

2.2518 0.46 0.25 -0.01 0.24 0.00 0.39 0.0006 0.29 0.002 0.23 -0.060.2829 0.05 0.23 0.02 0.24 0.02 0.10 0.0031 0.19 0.010 0.23 0.0397.36 0.20 0.23 0.00 0.22 0.00 0.22 0.0002 0.25 0.000 0.21 0.0045.42 0.07 0.07 0.00 0.07 0.00 0.06 -0.0008 0.06 -0.001 0.08 0.00

1 0.22 0.22 0.00 0.22 0.00 0.21 -0.0028 0.21 -0.006 0.23 -0.02L2-norm 0.55 0.47 0.023 0.47 0.016 0.51 0.00 0.48 0.01 0.46 0.07

(L2)2 0.31 0.22 0.00 0.22 0.00 0.26 0.00 0.23 0.00 0.21 0.01

1.Where does the problem lie, A is a smoothing operator?

When A is a smoothing operator, the forward solution of

equation Ax = b dampens the variability in the vector

space x when transforming it to the vector space b.

Smoothing operators under inversion acts like amplifiers.

The smoothing can also dampen the signals in the input

data b below the noise levels, effectively leading to a loss of

information, with no means of getting it back in an

inversion operation.

2. Is it a problem with the accuracy of the data vector b?

Can enhanced instrumentation and accuracy to b bring in

more quality or efficiency to the data inversion process?

54

Diagnosis for a valid solution x from Ax = b

3. Ill-conditioning of the problem is also a result of the

discretization of the continuous functions in which information

is lost. As the resolution of the discretised function increases,

each component of the solution vector x will have less and less

influence on the data b.

In the inversion problem, the influence of x can be viewed as

the information content of x in each data point bi. Therefore,

as the influence of x decreases, each data point bi will contain

less information about the higher resolution elements of x.

As the information content decreases, it is subjected to higher

and higher amplification to reconstruct the unknowns. This

amplification works on the error in the observations and

kernel as well, creating situations where the error of the

reconstructed values are very much larger than the values

themselves. Thus, a trade-off between the solution resolution

and error exists.

55

Diagnosis for a valid solution x from Ax = b

1. Small scale inverse problems of the kind encountered in

well log data inversion (Petrophysics) call for adequate

theory to explain the usefulness of the method.

2. Interpretation of the inverse theory elements and SVD are

domain specific and Petrophysics remains an untouched

area.

3. Areas like use of the regularization method, L-curve and

applicability of the discrete Picard condition towards the

diagnosis of a valid solution remains to be explored.

4. Efficiency of Tikhonov regularization and Levenberg-

Marquardt algorithm to be studied against the linear earth

model used to describe the measurements.

5. Possibility of developing better alternatives and the relative

merits over the deterministic approach are to be reckoned.

6. Quantification of the uncertainty of model parameters is an

essential requirement.

56

Conclusions

Condition number is 4108157 and it is obvious that the

residual cannot be of any indication about the quality of the

solution.

Efficacy of iterative methods also comes into question as the

Picard coefficients are unfavorably distributed against small

singular values.

58

SVD Analysis Example

UT*b iUT*b/

i

Variance(I)^2

% Variance IU

T*b/iI%

EnergyRatio to

max value20.1974 4502.54 0.004 20272866.45 0.95 0.004 0.004 0.012-7.1581 1069.07 -0.007 1142910.66 0.05 0.007 0.006 0.0184.5046 108.86 0.041 11850.72 0.00 0.041 0.037 0.111-0.4225 10.58 -0.040 112.02 0.00 0.040 0.036 0.1081.8436 4.97 0.371 24.66 0.00 0.371 0.335 1.0000.3201 1.07 0.299 1.14 0.00 0.299 0.270 0.8060.0004 0.00 0.344 0.00 0.00 0.344 0.311 0.928

Scaling can be resorted to reduce the condition number and

application of a weight matrix derived as the inverse of the error

estimates on data (1.5%) gives a weighted operator of condition

number 427708.

As we are working in an inverse crime scenario where the data

vector has been artificially created from the model, we get the

model x used above as the minimum norm solution.

Adding a noise of mean zero and standard deviation 1 to the

data vector b, the weighted least square solutions depicts

residuals varying from 1.13 to 3.14 for the same solution vector.

In fact, the prior is dominating the solution as the linear

operator is lacking in structure to illuminate the model space

with information retrieved from data space.

No other explanation can be thought of for the Picard

coefficients weighted towards the ill-conditioned model space.

59

SVD Analysis contd.

Inverse Crime Operations Axi = bi and xi = A+bi

60

Data vector at a depth

Depth Q C K M W A Transpose b N t 1 0.46 0.05 0.20 0.07 0.22 2.65 -0.06 55.5 12 1 2.252 0.283 97.36 45.422 0.48 0.03 0.18 0.07 0.24 2.71 -0.02 48 0 1 2.222 0.295 98.89 43.563 0.50 0.03 0.16 0.05 0.26 2.41 0.37 120 105 1 2.191 0.301 100.28 36.34 0.50 0.04 0.16 0.05 0.25 2.79 0.25 55 270 1 2.209 0.291 98.87 36.35 0.52 0.03 0.15 0.03 0.27 1 1 189 0 1 2.175 0.301 100.98 30.09

x1 x2 x3 x4 x5 b1 b2 b3 b4 b50.46 0.48 0.50 0.50 0.52 2.2518 2.2224 2.1914 2.2085 2.17450.05 0.03 0.03 0.04 0.03 0.2829 0.2947 0.3011 0.2909 0.30120.20 0.18 0.16 0.16 0.15 97.36 98.89 100.28 98.87 100.980.07 0.07 0.05 0.05 0.03 45.42 43.56 36.3 36.3 30.090.22 0.24 0.26 0.25 0.27 1 1 1 1 1

The coefficient matrix is retrieved from observed tool data by

using the inverse of the volume matrix.

61

Kernel A from the data

Inverse of Volume Matrix Observed data Coefficient Matrix-22.5 54 -106.5 42 34 2.252 0.283 97.36 45.42 1 2.65 -0.06 55.5 12.00 127.5 -96 93.5 42 -66 2.222 0.295 98.89 43.56 1 2.71 -0.02 48 0.00 127.5 4 -6.5 -58 34 2.191 0.301 100.28 36.3 1 2.41 0.37 120 105.00 1-22.5 4 43.5 42 -66 2.209 0.291 98.87 36.3 1 2.79 0.25 55 270.00 127.5 -96 193.5 -58 -66 2.175 0.301 100.98 30.09 1 1 1 189 0.00 1

A x Ax = b Error2.65 2.71 2.41 2.79 1 0.41 2.196 0.0329-0.06 -0.02 0.37 0.25 1 0.07 0.318 0.004855.5 48 120 55 189 0.22 102.515 1.537712 0 105 270 0 0.05 41.52 0.62281 1 1 1 1 0.25 1 0.015

62

Additional Examples-1

Unit Free Operator AUF bUF Error80.449 82.271 73.163 84.699 30.358 66.66667 1-12.579 -4.193 77.568 52.411 209.644 66.66667 136.092 31.215 78.037 35.767 122.909 66.66667 119.268 0.000 168.593 433.526 0.000 66.66667 166.667 66.667 66.667 66.667 66.667 66.66667 1

U VT-0.25 0.15 0.684 -0.133 0.656 506.15 0.00 0.00 0.00 0.00 -0.11 -0.079 -0.427 -0.875 -0.184

-0.229 0.716 -0.502 -0.38 0.195 0.00 259.24 0.00 0.00 0.00 0.121 0.166 0.218 -0.325 0.897-0.185 0.483 0.093 0.844 -0.107 0.00 0.00 140.34 0.00 0.00 0.662 0.665 0.119 -0.14 -0.292-0.896 -0.393 -0.2 0.046 -0.033 0.00 0.00 0.00 14.06 0.00 0.145 -0.455 0.786 -0.311 -0.239-0.221 0.277 0.482 -0.352 -0.721 0.00 0.00 0.00 0.00 3.12 -0.717 0.563 0.371 -0.113 -0.138

Tikhonov Regularization for AUF

Scaling did show a significant effect on the singular value spectrum

of the operator. Rationale followed in achieving a good scaling is

that the uncertainties in all the elements of A are of the same order.

Quality of the scaling done may be understood by looking at the

norm of the columns of the operator.

63

b xtrue L=0 e L =1 e L=2 e L = 3 e2.196 0.41 0.41 0.00 0.40 0.015 0.38 0.026 0.37 0.040.318 0.07 0.07 0.00 0.08 -0.014 0.09 -0.024 0.10 -0.03

102.515 0.22 0.21 0.00 0.22 -0.012 0.23 -0.017 0.23 -0.0241.52 0.05 0.05 0.00 0.05 0.000 0.05 0.001 0.05 0.00

1 0.25 0.25 0.00 0.25 0.001 0.25 0.003 0.24 0.01L2 norm 0.53 0.01 0.53 0.023 0.52 0.04 0.52 0.05

b xtrue L=5 e L =10 e L=30 e L = 50 e2.196 0.41 0.36 0.053 0.33 0.079 0.29 0.119 0.28 0.130.318 0.07 0.11 -0.045 0.14 -0.066 0.17 -0.101 0.18 -0.11

102.515 0.22 0.24 -0.029 0.25 -0.040 0.26 -0.052 0.26 -0.0541.52 0.05 0.05 0.005 0.04 0.008 0.04 0.011 0.04 0.01

1 0.25 0.24 0.008 0.24 0.012 0.23 0.017 0.23 0.02L2 norm 0.51 0.08 0.50 0.11 0.49 0.17 0.48 0.18

How to interpret the change in the norm?

64

Interpreting the norm of the columns

Operator Norms of the columnsA 57 48 159 276 189

AUF 113 110 224 451 254

A: L-Curve Parameters Regularized Solution[x]T ||Ax-b|| ||x|| ||e|| = x11 x21 x31 x41 x51

0.0468 0.01 0.48 0.22 0.24 0.22 0.24 0.05 0.251 0.06 0.47 0.24 0.22 0.22 0.22 0.06 0.273 0.16 0.46 0.25 0.20 0.20 0.21 0.06 0.285 0.23 0.45 0.25 0.19 0.19 0.21 0.06 0.29

10 0.34 0.44 0.27 0.17 0.16 0.21 0.06 0.3020 0.46 0.43 0.28 0.15 0.14 0.21 0.07 0.31

65

Tikhonov in both the cases

A is good in minimizing the residual norm while AUF minimizes

the residual solution x error norm. Norm of x is almost thesame in both. Tikhonov regularization gives approximate

solution x as the unique minimizer of the quadratic cost

function F% L Z % % i.e. two options are available (contd)

66

Tikhonov in both the cases

(a) Define an upper bound for the solution error norm andminimize the residual

(b) (b)Limit the residue by choice and minimize the error norm =||x-x||.

AUF: L-Curve Parameters Regularized Solution[x]T ||Ax-b|| ||x|| ||e|| = x11 x21 x31 x41 x515 0.22 0.51 0.07 0.36 0.11 0.24 0.05 0.2410 0.29 0.50 0.11 0.33 0.14 0.25 0.04 0.2420 0.41 0.49 0.14 0.30 0.16 0.26 0.04 0.2350 0.70 0.48 0.18 0.28 0.18 0.26 0.04 0.23

100 1.06 0.48 0.20 0.26 0.20 0.26 0.04 0.23

67


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CUJ 8 Feb 2014_Hari