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reliability_TNO_RP8.md

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RP8 reliability problem

set_id problem_id
-1 1

challenge set_1

Overview

Category Value
Type DACC
Number of random variables 6
Failure probability, $P_\mathrm{f}$ 7.84e-4
Reliability index, $\beta=-\Phi^{-1}(P_\mathrm{f})$ 3.16
Number of performance functions 1
Continuity $\geq C^1$
Reference Xu2018

Performance function

$$g({\bf X}) = X_1 + 2X_2 + 2X_3 + X_4 - 5X_5 - 5X_6$$

Random variables

The parametrization of distributions follows that of in.

Variable Description Distribution (\theta_1) (\theta_2) Mean Std
$X_1$ NA Lognormal 4.783 0.09975 120.0 12.0
$X_2$ NA Lognormal 4.783 0.09975 120.0 12.0
$X_3$ NA Lognormal 4.783 0.09975 120.0 12.0
$X_4$ NA Lognormal 4.783 0.09975 120.0 12.0
$X_5$ NA Lognormal 4.783 0.09975 50.0 10.0
$X_6$ NA Lognormal 4.783 0.09975 40.0 8.0

The random variables are mutually independent.

from sampling import linesampling as ls
from sampling import dists as dists
from reliability.tnochallenge import problem

Instantiate the object with all the data identifying the model

RP8 = problem('RP8')
print(RP8)
Name: RP8
Link: https://rprepo.readthedocs.io/en/latest/reliability_problems.html#sec-rp-8

# The available information about the problem can be seen in one go with
RP8.info
{'description': 'Tutorial set problem. Linear performance function with six lognormal variables.',
 'name': 'RP8',
 'kind': 'Tutorial set',
 'latexFormula': '$$g({\\bf X}) = X_1 + 2X_2 + 2X_3 + X_4 - 5X_5 - 5X_6$$',
 'latexPF': '$7.84\\ 10^{-4}$',
 'targetBeta': 3.16,
 'Nv': 6,
 'Ng': 1,
 'Distributions': ['lognormal'],
 'Inputs': {'X[1:4]': {'dist': 'lognormal', 'mean': 120, 'std': 12},
  'X[5]': {'dist': 'lognormal', 'mean': 50, 'std': 10},
  'X[6]': {'dist': 'lognormal', 'mean': 40, 'std': 8}},
 'index': 1}

# We will be using two piece of information for the problem:
# 1. The (independent) copula function of the input random vairables
C8 = RP8.inputs() # Pi-copula of the problem
C8.marginals()
# returns 
# (X1 ~ Lognormal(t1=4.78252, t2=0.0997513),
#  X2 ~ Lognormal(t1=4.78252, t2=0.0997513),
#  X3 ~ Lognormal(t1=4.78252, t2=0.0997513),
#  X4 ~ Lognormal(t1=4.78252, t2=0.0997513),
#  X5 ~ Lognormal(t1=3.89241, t2=0.198042),
#  X6 ~ Lognormal(t1=3.66927, t2=0.198042))
# A sample of the copula function can be obtained as follows:
x,z,u = C8.sample(N=1,seed=30) # shape of x: (6, N), void produces N=100 samples  
print(x) 
# prints:  
# [array([120.57857706]), array([125.37349755]), array([105.56579493]), array([125.6978498]), array([47.80327458]), array([38.43929189])]
# The copula can be used to perform transformation from the physical space X to the standard normal space Z
print(C8.map2std(x))
# returns:
# [array([-1.21348165]), array([1.96858688]), array([0.54451422]), array([-1.36668842]), array([1.30673409]), array([0.14566481])]
# 2. The performance function of the problem:
g8 = RP8.g()
# the performance can be evaluated on a sample from the copula as follows:
g8(*x)
# returns:
# array([-233.33485107])

Let's dive into the solution of the reliability problem

Line sampling requires an important direction to work. Let's compute it using the gradient in the standard normal space.

# Let us create a new problem object to remove the previous evaluation from the total count of model's evaluations.
RP8 = problem('RP8')
alpha = ls.initialiseAlpha(RP8,C8,gradient=True)
print(alpha)
[-0.1637928660244218, -0.3275857320489999, -0.3275857320489999, -0.16379286602457807, 0.6679596573890065, 0.5343677259112521]

LS = ls.LineSampling(lines=20,alpha=alpha,linegrid=[2,3,4,5])

Perfom the analysis with line sampling

pF, b, dp, LSdata, LSdata2, cvar, PFLine = LS.failureProbability(C8,RP8,additional=4,seed=7)
print('failure probability:  [%.2e, %.2e]'%(pF[0],pF[1]))
print('reliability index:    [%g, %g]'%(b[0],b[1]))
print('coeff. of variation:  %g'%cvar)
print('total number of runs: %i'%RP8.evaluations())
failure probability:  [6.70e-04, 8.84e-04]
reliability index:    [3.12664, 3.2073]
coeff. of variation:  0.16861
total number of runs: 327

Verifying LS results

By doing the analysis on the line through the origin of the standard space, we can compute the Euclidean norm of the closest point to the origin.

cl,cr,data = LS.doLineZero(C8,RP8,additional=9)
print('Euclidean norm of design point: [%g, %g]'%(cl,cr))
Euclidean norm of design point: [3.27344, 3.2793]

# This norm appears to be smaller than any other nors computed by the line sampling anlysis
# The smallest norm found on the line samples 
print('Smallest norm: %g'%dp[0])
Smallest norm: 3.31325

# which is greater than the upper bound of the Euclidean norm 
print('Smallest norm on lines %g > %g Euclidean norm of design point'%(dp[0],cr))
# So the analysis ends here, as no better important direction was found during the analysis. 
# Of course increasing the number of lines may increase the possibility to find a better direction.
Smallest norm on lines 3.31325 > 3.2793 Euclidean norm of design point

Let's plot the results

LS.plotLines(LSdata,figsize=(16,8))

png

LS.plotLines2(LSdata2,figsize=(16,8))

png

LS.plotLines3([LSdata,LSdata2],figsize=(16,8))

png

LS.plot([LSdata],space='X')

png

LS.plot([LSdata],space='Z')

png