Create a process from random vectors and processes

The objective is to create a process defined from a random vector and a process.

We consider the following limit state function, defined as the difference between a degrading resistance r(t) = R - bt and a time-varying load S(t):

\begin{align*}
g(t)= r(t) - S(t) = R - bt - S(t)
\end{align*}

We propose the following probabilistic model: - R is the initial resistance, and R \sim \mathcal{N}(\mu_R, \sigma_R); - b is the deterioration rate of the resistance; it is deterministic; - S(t) is the time-varying stress, which is modeled by a stationary Gaussian process of mean value \mu_S, standard deviation \sigma_S and a squared exponential covariance model; - t is the time, varying in [0,T].

First, import the python modules:

from openturns import *
from openturns.viewer import View
from math import *

1. Create the gaussian process (\omega, t) \rightarrow S(\omega,t)

Create the mesh which is a regular grid on [0,T], with T=50, by step =1:

b = 0.01
t0 = 0.0
step = 1
tfin = 50
n = round((tfin-t0)/step)
myMesh = RegularGrid(t0, step, n)

Create the squared exeponential covariance model:

C(s,t) = \sigma^2e^{-\frac{1}{2} \left( \dfrac{s-t}{l} \right)^2}

where the scale parameter is l=\frac{10}{\sqrt{2}} and the amplitude \sigma = 1.

l = 10/sqrt(2)
myCovKernel = SquaredExponential([l])
print('cov model = ', myCovKernel)

Out:

cov model =  SquaredExponential(scale=[7.07107], amplitude=[1])

Create the gaussian process S(t):

S_proc = GaussianProcess(myCovKernel, myMesh)

2. Create the process (\omega, t) \rightarrow R(\omega)-bt

First, create the random variable R \sim \mathcal{N}(\mu_R, \sigma_R), with \mu_R = 5 and \sigma_R = 0.3:

muR = 5
sigR = 0.3
R = Normal(muR, sigR)

The create the Dirac random variable B = b:

B = Dirac(b)

Then create the process (\omega, t) \rightarrow R(\omega)-bt using the FunctionalBasisProcess class and the functional basis \phi_1 : t \rightarrow 1 and \phi_2: -t \rightarrow t :

R(\omega)-bt = R(\omega)\phi_1(t) + B(\omega) \phi_2(t)

with (R,B) independent.

const_func = SymbolicFunction(['t'], ['1'])
linear_func = SymbolicFunction(['t'], ['-t'])
myBasis = Basis([const_func, linear_func])

coef = ComposedDistribution([R, B])

R_proc = FunctionalBasisProcess(coef, myBasis, myMesh)

3. Create the process Z: (\omega, t) \rightarrow R(\omega)-bt + S(\omega, t)

First, aggregate both processes into one process of dimension 2: (R_{proc}, S_{proc})

myRS_proc = AggregatedProcess([R_proc, S_proc])

Then create the spatial field function that acts only on the values of the process, keeping the mesh unchanged, using the ValueFunction class. We define the function g on \mathbb{R}^2 by:

g(x,y) = x-y

in order to define the spatial field function g_{dyn} that acts on fields, defined by:

\forall t\in [0,T], g_{dyn}(X(\omega, t), Y(\omega, t)) = X(\omega, t) - Y(\omega, t)

g = SymbolicFunction(['x1', 'x2'], ['x1-x2'])
gDyn = ValueFunction(g, myMesh)

Now you have to create the final process Z thanks to g_{dyn}:

Z_proc = CompositeProcess(gDyn, myRS_proc)

4. Draw some realizations of the process

N = 10
sampleZ_proc = Z_proc.getSample(N)
graph = sampleZ_proc.drawMarginal(0)
graph.setTitle(r'Some realizations of $Z(\omega, t)$')
view = View(graph)
Some realizations of $Z(\omega, t)$

5. Evaluate the probability that Z(\omega, t) \in \mathcal{D}

We define the domaine \mathcal{D} = [2,4] and the event Z(\omega, t) \in \mathcal{D}:

domain = Interval([2], [4])
print('D = ', domain)
event = ProcessEvent(Z_proc, domain)

Out:

D =  [2, 4]

We use the Monte Carlo sampling to evaluate the probability:

MC_algo = ProbabilitySimulationAlgorithm(event)
MC_algo.setMaximumOuterSampling(1000000)
MC_algo.setBlockSize(100)
MC_algo.setMaximumCoefficientOfVariation(0.01)
MC_algo.run()

result = MC_algo.getResult()

proba = result.getProbabilityEstimate()
print('Probability = ', proba)
variance = result.getVarianceEstimate()
print('Variance Estimate = ', variance)
IC90_low = proba - result.getConfidenceLength(0.90)/2
IC90_upp = proba + result.getConfidenceLength(0.90)/2
print('IC (90%) = [', IC90_low, ', ', IC90_upp, ']')
view.ShowAll()

Out:

Probability =  0.7557575757575757
Variance Estimate =  5.6497333296231344e-05
IC (90%) = [ 0.7433940814993385 ,  0.768121070015813 ]

Total running time of the script: ( 0 minutes 0.186 seconds)

Gallery generated by Sphinx-Gallery