INTRODUÇÃO À GEOESTATÍSTICA 7-Introduction to Stochastic ... · Geostatistical Models:...

7-Introduction to Stochastic

Simulation

INTRODUÇÃO À

GEOESTATÍSTICA

Amílcar Soares

CERENA-IST

[email protected]

Objective of Geostatistical

Simulation Models : Integration of the

concept of uncertainty in high resolution

static models of the oil reservoirs.

Applications: simulation of lithofacies,

simulation of petrophysical properties,

seismic inversion, geostatistical history

matching.

Geostatistical Models: Estimation or Simulation ?

Is the average value sufficient to represent the knowledge of a given

variable: oil in place, porosity, permeability of a reservoir?

…depends on the variability/variance of the variable; depends on our ignorance about it

Is the average “image” of porosity sufficient to represent the

knowledge of such variable? ..and the knowledge of the top of a layer

of a sedimentary structure?

Estimation Model

Simulation gives a set of equiprobable images of the physical phenomenon

with the same variability and spatial continuity as revealed by the

experimental data.

Simulation

appropriated for the characterization of very heterogeneous

properties (high variability)

- Porosity, Permeability

- Reserves

-Fractures

-…

Stochastic Simulation of Heterogeneous Variables

Main purpose: Integration of the uncertainty in the oil reservoir models

for decision making processes

The Concept of Uncertainty

Porosity,

permeability,..

Response of the dynamic

model of reserves

Estimated model of the most

probable values

Provides only the dynamic response of the most likely model which is not

necessarily the most likely response

It provides no measure of uncertainty, extreme values , risk, etc...

Only one response: pressure

curves, water cut, ..

...

The Concept of Spatial Uncertainty

For any spatial location x0, uncertainty measures can be taken from the local

histograms calculated with the N simulated images of a petrophysical property,

...

The Concept of Spatial Uncertainty

Spatial uncertainty measures the joint probability of a set of points

of a given area A1 located at x0, to be lower than a threshold, etc..

This can be taken from the local histograms calculated with the N

simulated images of a petrophysical property.

The Concept of Uncertainty

Porosity,

permeability,..

Response of the dynamic

model of reserves

Quantifying the uncertainty of internal properties in terms of reserves,

risk analysis, ...Uncertainty is measured in the responses of a dynamic

simulator (non-linear function of petrophysical properties).

Simulated Model

Uncertainty of Reserves

Simulation of one random variable

Sampling of a random variable from its probability distribution

How to get 1000 values from these distributions ?

The Monte Carlo method

“the use of random numbers to solve deterministic systems or stochastic

models”

1- Inverse transform sampling

2- Acceptance / Rejection method

(Rejection Sampling)

Simulation of one random variable

FZ(z) is a function continuous and monotonically increasing:

1 > FZ( z ) > 0 z1 < z2 then FZ( z1 ) < Fz( z2 )

1 - Generate U from U( 0,1 )

2- z1 = F-1z(U )

0

1

FZ1(z1)

u

z1=F-1Z1(u)

0

1

FZ1(z1)

u

z1=F-1Z1(u)

a b

1- Inverse transform sampling

Monte Carlo simulation


2- Acceptance / Rejection method

(Rejection Sampling)

Specify a function (x) that upper bounds the density function f: t(x) >= f(x) for all x.

1

xdxfxdxtc

x1 is accepted

x2 is accepted

x1 is rejected

x2 is rejected

x1 x2

f(x)

t(x)

r(x) = t(x) / c is a density function

1- Generate x

2- Generate u from U( 0,1 )

3- u <= f(x) / t(x) x is accepted

Otherwise x is rejected

Algorithms to generate continuous random variables

Uniform

Inverse method: Generate U U( 0,1 ) Returns X = F-1( U )

1 - Generates U U( 0,1 )

2 - Returns X = a + ( b - a ) . U

Exponential

1 - Generates U U( 0,1 )

2 - Returns X = - . ln(U)


Standard Normal

1 - Generate U1 e U2 U( 0,1 )

2 - Returns X1 = sqrt[ -2 . ln(U1) ] . cos(2) . U2

Method 1: Box & Muller (1958)

X2 = sqrt[ -2 . ln(U1) ] . sin(2) . U2

X1 e x2 N( 0,1 )

Method 2: Polar method (Marsaglia & Bray(1964))

1 - Generate U1 e U2 U( 0,1 )

2 - Consider V1 = 2 . U1 - 1 e V2 = 2 . U2 – 1 i=1,2 e W = V12 + V22

3 - If W > 1 returns to step 1. Else: Y = sqrt[ ( -2 . ln(W )) / W ]

X1 = V1 . Y e X2 = V2 . Y X1 e x2 N( 0,1 )

Advanced Reading Box


Lognormal

1 – Generate Y N ( ,2 )

2 – Returns X = eY

Triangular

Note that X triang( 0, 1, ( c – a ) / ( b – a ) ), so X‘ = a + ( b - a )X triang( a, b, c )

1 – Generate U U ( 0,1 )

2 – If U c returns X = sqrt( c . U );

Else: returns X = 1 - sqrt[ ( 1 - c) ( 1 – U ) ]

Advanced Reading Box

Simulation of empirical distributions

Sort decreasingly the n values of Z;

Calculate pi=(i-0.5)/n associated to the sorted Zi

1 – Generate U U (0,1)

2 – Returns the value Z = Zi + ( Zi+1 – zi ) . ( pi – U ) / ( pi - pi+1 )

Ex: n=100 values: 2.1, 0.3, 11.4, ….. After sorting, to each value Zi is associated the following “cumulative probability” pi:

min=Z1=0.3 p1=0.005; Z2=0.6 p2=0.015; Z3=0.65 p3=0.025;…;

max=Z100=11.4 p100=.995

Consider U=0.1 returns Z=0.3+.5x(0.6-0.3)=0.45

Create a cumulative “probability of Z”;

Simulation of one random categorical variable

Lithoclass #1 – 25 %

Lithoclass #2 – 40 % Lithoclass #3 – 35 %

Create a pseudo “cumulative probability” :

.25

.65

1

L1 L2 L3

1 - Generate U U ( 0,1 )

2 - If .25 U > 0 returns X = L1

If .65 U > .25 returns X = L2

If 1 U > .65 returns X= L3

F(x)

X

Simulation of a set of independent random variables

Example 1: optimizing a load and transport operation

Loading the truck

Unloading the cargo

Transporting the cargo

Returning to the initial point

Experimental histograms of

operation times simulations

Total 1 Total2

Unloading the truck

t

f(t)

t3 t3 t3

Total 3

t

f(t)

Simulated

histogram of the

time it takes one

operation cycle

...

Returning to the initial

point t

f(t)

t4 t4 t4 ...

...

Transporting the cargo

t

f(t)

t2 t2 t2 ...

Loading the truck

t

f(t)

t1 t1 t1 ... Sensibility analysis

Simulation of a set of independent random variables

Histogram of “Totals”

Rose plot of fracture orientation

Histogram of the number of

segments in fractures

Example 2 : Simulating fracture networks

Fracture density

Field

dimension

Histogram of fracture length

of of

Simulation of a set of independent variables

1st simulation Superimposing of a 2ª simulation over a

different set of fractures

Simulation of 355 fractures Simulation of 513 fractures


Example 3: Reserves determination

reserves A Th Net

Gross Swi

FVF RF

Cumulative

production

* * * * ( ) * * f 1

1

STOIIP = Stock Tank Oil Initial In Place

GIIP for Gas

A = reservoir area (km2 / acres)

Th = total reservoir thickness (meters)

Net & Gross = total reservoir thickness net & gross (meters)

f = porosity = emptiness ratio

Swi = initial water saturation (ratio)

FVF = Oil & Gas formation factor (non-dimensional)

RF = Recovery factor (ratio)

A*Th = GRV = Gross rock volume


PDFs of input variables:

Simulation of two dependent random variables

When two random variables show some statistical dependence, one

whish to generate both by reproducing the marginal distributions F(Z1 )

and F(Z2 ) but also the joint distribution of probabilities F( Z1, Z2 )

To reproduce the joint distribution K and Φ

(high values of K correspond to high values

of Φ ) they can not be simulated

independently

..let us suppose that one whish to simulate pairs of porosity e permeability

values that show a positive correlation

Principle of Sequential Simulation

Applying Bayes law in sequential steps: F( Z1, Z2 ) = F( Z2 | Z1 ) . F(Z1)

Simulating two values z1 e z2 from a joint distribution

F( Z1,Z2 ):

1st Generate z1 from pdf F(Z1).

2nd Generate z2 from conditional distribution function F( Z2 | Z1=z1)

Simulation of two dependent random variables


Simulating two variables from bivariate distribution:

Permeability and porosity

K

f

z1 Fz1

K z1

u

F(f|K=z1)

f z2

u

1- simulate the first

value of K from the

marginal distribution

of permeability by

using a Monte Carlo

algorithm

2- From the joint

distributtion F(Φ,K),

extract the conditional

distribution: F(Φ|K=z1)

3- Simulate the porosity

value Φ =z2 from the

conditional distribution:

F(Φ|K=z1), by using a Monte

Carlo algorithm


Simulating fracture networks with dependent variables

Orientation

lenght

Bi-histogram obtained

from an outcrop

F(Z1,Z2,….,ZN)=F(Z1).F(Z2|Z1).F(Z3|Z1,Z2,Z3)… F(ZN|Z1,…ZN-1)

Generate (simulate) a set of values z1,…,zN from the

distribution law F(Z1,Z2,…,ZN), which can be made from

sequential steps:

1º- We generate a value z1 from the marginal law F(Z1)

2º- Than we simulate a value z2 from the conditional

distribution law F(Z2|Z1=z1)

Simulate value zN from F (ZN |Z1=z1,Z2=z2,…,ZN-1=zN-1)

Simulating N random dependant variables

Example: Simulation of values Zi on a regular grid using 5 known samples

Sequential simulation of spatial phenomena


Z(x),=1,7 ?

x8

z

f(z)

z(x8)

x8

( II )

( I )

How to estimate the probability conditioned distribution functions

F(Z2|Z1). F(Z3|Z1, Z2 )... F(ZN|Z1, Z2, , ZN-1) ?

x8

Z(x),=1,8

?

x8

z

f(z)

z(x9)

x9

( IV)

( III )

x8

Example : Sequential simulation of 3 lithofacies: t1, t2 andt3

?

Prob x0 t1 =60%

Prob x0 t2=25%

Prob x0 t3 =15%

1-

60%

85%

100%

2-

1- Random choice of one node on the regular grid,

x0 .

2- Estimate the local probabilities

3- Create the Pseudo local cumulative

law of distribution:

If r<.6 then x0 t1

If r>=.6 and r<..85 then x0 t2

If r>=.85 then x0 t3

Monte Carlo: Generating r from U(0,1)

60%

85%

100%

3- Simulating T1, T2 e T3 in x0: 3-

3- The simulated value is added to the “experimental” values.

Again a random node is chosen from the regular grid to be simulated:

? Prob x0 t1 =5%

Prob x0 t2=75%

Prob x0 t3 =20%

5%

80%

100% If r<.05 then x0 Lf1

If r>=.05 and r<..80 then x0

Lf2

If r>=...8 then x0 Lf3

4- This procedure continues

until all nodes are simulated.

INTRODUÇÃO À GEOESTATÍSTICA 7-Introduction to Stochastic ... · Geostatistical Models:...

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