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Lec. 0312735: Urban Systems Modeling component reliability

12735: Urban Systems Modeling

instructor: Matteo Pozzi

1

component risk analysis

Lec. 03


outline

2

‐ risk analysis for components‐ uncertain demand and uncertain capacity;‐ multivariate normal distribution: definition;‐ properties;‐ transformation to standard normal space;‐ design point;‐ multivariate log‐normal distribution;‐ First Order Reliability Method; ‐ sensitivity analysis.


example of components, general framework

3

A structure, say a bridge.A road segment.An electrical component.A pump in a water system.

A component is modeled by a set of random variables, describing loads, demands, capacity, resistance, features affecting the behavior. These variables are modeled by a joint distribution: The functioning of the component is described by a binary variable: the safe, of functioning state, and the failure.

90%

Task: computing the probability of failure, P failure

→


PART I

approaches to component risk analysis

4


reliability with uncertain load and resistance: method I

5

, the load: ; ,, the resistance: ; ,

independence: s

limit state function: 0 → safecondition0 → failure

Pfind

0 1 2 3 4 5 6 70

0.5

r, s

p(s)

, p(

r)

p(s)p(r)

varalways true

The difference between two normal rv.s is a normal rv. [to be proved later]

; ,

P 0 0

Φ ; ,

varvar

0 Φ

standard normal cdf

Φ

if resistance R is known

think of as the “residual capacity”


reliability with uncertain load and resistance: method II

6

1

2exp

12

1

2exp

12

02

46

0

5

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

x1 = sx2 = r

∝ ,

00

→

11

with correlated random variables:

→ 2

joint prob. in vector notation,

00.20.4

g

p(g)

P 0


example of reliability problem

7

1015 KN

00

11

00

→ 5KN

2 2.11KN

11

4 0.90.9 2.25 KN

vector notation:2.0KN1.5KN

30%

52.11 2.37 Φ 0.89%

reliability index [defined later]: probability of damage:

: multivariate normal variable [defined later]


-1 -0.5 0 0.5 1

0

0.02

0.04

0.06

0.08

sr

Pf

-1 -0.5 0 0.5 10

2

4

6

8

10

sr

example of reliability, changing correlation

8

varying the correlation coefficient:

∈ 1; 1

x 2 = r

12 = -0.99

5 10 15 205

10

15

20

x 2 = r

12 = -0.3

5 10 15 205

10

15

20

x1 = s

x 2 = r

12 = 0

5 10 15 205

10

15

20

x1 = s

x 2 = r

12 = 0.6

5 10 15 205

10

15

20

, 0

, 0

, 0

, 0

: multivariate normal variable [defined later]


reliability with uncertain load and resistance: method III

9

02

46

0

5

0

0.1

0.2

0.3

x1 = s

x2 = r

p(x 1,x

2)

x1 = s

x 2 = r

0 2 4 60

1

2

3

4

5

6

7

02

46

0

5

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

x1 = sx2 = r

p(x 1,x

2),

g(x1

,x2)

, , 0

, 0

∝ ,,

,

failure domain

safe domain

classical reliability problem: solve an integral.


reliability with uncertain load and resistance: method III

10

x1 = s

x 2 = r

0 2 4 60

1

2

3

4

5

6

7, , 0

, 0

x1 = s

x 2 = r

0 2 4 60

1

2

3

4

5

6

7

, 0

, , 0

x1

x2

0 x1

x2

0

u1

u2

0

transformation:

→

,

1

→orthogonal to 0

standard normal var.:

transformation to standard normal space:

then it is easy…


reliability with uncertain load and resistance: method IV

11

x1 = s

x 2 = r

0 2 4 60

1

2

3

4

5

6

7, , 0

x1

x2

0

, ,

,

|

s

conditional cumulative distribution

prob. that resistance is lower than load, for

load equal to .


reliability with uncertain load and resistance: method IV

12

, ,

,

|

s

alternative formulation:

1 | 1 1

conditional cumulative distribution

prob. that resistance is lower than load, for

load equal to .

prob. that load is higher than resistance,

for resistance equal to .

5 10 15 200

0.5

1

r, s

F(s)

, F(

r)

5 10 15 20

0

0.05

0.1

0.15

0.2

0.25

r, s

p(s)

, p(

r)

p(s)p(r)


towards multivariate normal distribution

13

1

2exp

12

1

2exp

12

00

1

2exp

12

1

2exp

12

2,number of dimension determinant

≜ ; ,multivariate normal distribution

parameters

00


PART II

Gaussian model for component risk analysis

14


multivariate normal distribution

15

; ,1

2exp

12pdf:

-50

510

15

-5

0

5

10

150

0.005

0.01

0.015

0.02

0.025

0.03

x1

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

x2

p(x 1,x

2)

⋮

⋮

…⋱ ⋮

⋮ ⋱…

⋱ ⋮…

vector of random variables

mean vector

covariance matrix

∀ : E

∀ : var∀ , :

⟶


multivariate normal distribution in log. scale

16

log ; ,12

pdf: ⟶

-50

510

15

-5

0

5

10

15

10-15

10-10

10-5

x1

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

x2

p(x 1,x

2)

⋮

⋮

⋮ ⋮ ⋮

⋮ ⋮

⋮ ⋮ ⋮

maximum:←

gradient:

Hessian matrix:

uniform curvature, always negative definite. So MVN is log‐concave.

Simplest case: the standard MVN:

log ; ,12

12


multivariate normal distribution: contour plot

17

pdf: ⟶

contour line:const.

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

-5 0 5 10 15-5

0

5

10

15

log ; ,12

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0

-5 0 5 10 15-5

0

5

10

15

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = -0.9

-5 0 5 10 15-5

0

5

10

15

lines are ellipses centered in the mean

examples, changing the correlation only


multivariate normal distribution: eigenvalues

18

pdf: ⟶

contour line:const.

eigen‐value decomposition:

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

-5 0 5 10 15-5

0

5

10

15

: eigenvector matrix …

Eigenvectors form an ortho‐normal base:

10

: eigenvalue matrix0 0

0 ⋱ 00 0

eigen‐problem:

i. e. 1

log ; ,12



19

pdf: ⟶

contour line:const.

⇒

/ ⇒ /

1

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

-5 0 5 10 15-5

0

5

10

15eigen‐value decomposition:

principal components :

in terms of variables , the contour line (surfaces) are circles (spheres):

log ; ,12

isstandard MVN



20

/ 1

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

-5 0 5 10 15-5

0

5

10

15

principal components:

/

inverse relation: original rand. var.s as a function of the components:

basic idea of eigen‐values:to change point of view: from canonical base to an ortho‐normal base …centered in the mean. Now variables looks uncorrelated. Re‐scale using : now variable has also unit‐variance.

Matlab: [m_V,m_L]=eig(m_Sigma)

, eig


example of: eigen‐value decomposition

21

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.4

-5 0 5 10 15-5

0

5

10

15covariance matrix:

4 2.42.4 9 → 0.37 0.93

0.93 0.37

Λ 9.97 00 3.03

4 2.42.4 9

0.37 0.930.93 0.37

9.97 00 3.03

0.37 0.930.93 0.37

3.16 1.74

11

Length of ellipse’s principal axes:

eigenvector matrix:

eigenvalue matrix:

you may check that:


covariance matrix

22

properties: symmetry: ,

positive definitiveness: ∀ ∈ : 0

11

…⋱ ⋮

⋮ ⋱…

⋱ ⋮… 1

:correlation matrix 00

… 0⋱ ⋮

⋮ ⋱0 …

⋱ 00

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0

-5 0 5 10 15-5

0

5

10

15

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = 0.6

-5 0 5 10 15-5

0

5

10

15

x1

x 2

1 = 4 , 2 = 6 , 1 = 2 , 2 = 3 , 12 = -0.9

-5 0 5 10 15-5

0

5

10

15


properties of MN: marginalization

23

; ,1

2exp

12

⋮

⋮

…⋱ ⋮

⋮ ⋱…

⋱ ⋮…

\ ; ,

; ,

parameters

rand. vars.

= 99% = 50%

= -30%

,same

same marginal

samemarginal

marginal is normal


properties of MN: marginalization [cont.]

24

partition:

⋮ ⋮

…⋱ ⋮

⋮ ⋱…

⋱ ⋮…

; ,marginal probability:

; ,1

2exp

12

Marginalization may be computationally expensive in general.But if a vector of rand. vars. is jointly normal, any subset is jointly normal as well, and parameters can be directly read in those of the joint set.


properties of MN: conditional

25

; ,1

2exp

12

; | , |

|

|

After observing , the conditional distribution of is still normal:

if uncorrelated , :

||

For jointly normal rand. vars., uncorrelation and independence are equivalent.

the reduction of variance does not depend on the value observed


properties of MN: conditional [cont.]

26

; | , |

00.2

0.40.6

0.81

0

0.5

10

5

10

15

20

p(x2)

x1

p(x1,x2=0.45)

p(x1=0.35,x2)

x2

p(x1)

p(x 1,x

2)

00.5

1

0

0.5

10

5

10

x1x2

p(x 2

x 1)

00.5

1

0

0.5

10

2

4

6

x1x2

p(x 1

x 2)


example of marginalization/conditional

27

65104

4 1.21.2 9

2.4 2.46 3.6

2.4 62.4 3.6

16 3.23.2 4

mean vector

covariance matrix

vector of random variables

; ,

; ,

549 3.63.6 4

marginalization

conditionalsuppose to observe 7

12 ; | , |

|9 3.63.6 4

1.2 62.4 3.2

4 2.42.4 16

1.2 2.46 3.2

6.7 2.22.2 2.3

|54

1.2 62.4 3.2

4 2.42.4 16

712

610

5.84.8

reduction of uncertainty


recap: transformation of random variables in 1‐d

28

,,

≜

x

p x(x) ,

Fx(x

)

0

1

x

z

z = f(x)

x = f -1(z) = g(z)

pz(z) , Fz(z)

z

px(x)

Fx(x)

pz(z)

Fz(z)

random variable

transformation

: monotonically increasing

new random variable

inverse

conservation of probability

′


transformation of multivariate rand. vars.

29

≜

…

⋮ ⋱ ⋮

…

, , … ,

⋮

vector of rv.s joint probability

dim: 1 : ⟶

dim.:

≜

…

⋮ ⋱ ⋮

…

invertible map

Jacobian

, , … ,

find:

Jacobian of the inverse map

when

inverse map:

determinant of the Jacobian general formula:

1


00.5

11.5 0

0.51

1.50

0.5

1

1.5

2

2.5

y2y1

transformation of multivariate rand. vars. [cont.]

30

0 0.5 10

0.2

0.4

0.6

0.8

1

x1

x 2

0 0.5 1 1.50

0.2

0.4

0.6

0.8

1

1.2

y1

y 2

00.5

1 00.5

10

0.5

1

x2x1

area( )

area( )2.2

area( )

area( )45%

equal probability (volume)

e.g.: uniform


sign of the determinant of the Jacobian

31

x1

x2

y1

y2

y1

y2

the map preservesorientation

0

the map inverts orientation

0

we are only interested in the ratio between areas,

hence we take the absolute value of the

determinant.

http://noirbabes.com/precode/2012/08/26/1352/


linear transformation of mult. rand. vars.

32

0 0.5 10

0.2

0.4

0.6

0.8

1

x1

x 2

0 0.2 0.4 0.6 0.80

0.2

0.4

0.6

0.8

1

1.2

y1

y 20.41

linear transformation:

inverse transformation:

1

1

for a linear transformation, the Jacobian (and consequently its determinant) is uniform.

→


example of linear transformations:

33

0 0.5 10

0.2

0.4

0.6

0.8

1

x1

x 2

0 0.2 0.40

0.5

1

1.5

y1

y 2

0.5 00 1.5

0.75

0 0.5 1-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

y1

y 2

cos /3 sin /3sin /3 cos /3

1

.65 .35.35 1.3

0.73

-0.4 -0.2 0 0.2 0.4 0.6

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

y1y 2

diagonal(no rotation)

general

pure rotation

1

1.33 1.37


linear transformation of jointly normal rand. vars.

34

; ,1

2exp

12

; ,

∝ exp12

exp12

∝ ; ,

linear invertible transformation:

A linear combination of jointly normal rand. var.s is also jointly normal. This is true in general, also for a transformation to a smaller space, e.g. from vector to scalar (proved by marginalization).

this proves that:

same as for mean vector and covariance matrix of every

multivariate rand. var.s


summary on jointly normal rand. vars.

35

; ,

→ ; ,

‐ the joint probability is completely defined by mean vector and covariance matrix, which are the parameters of the distribution.

‐ the conditional distribution, given any subset of variable, is also jointly normal.‐ each subset of is jointly normally distributed, and marginalization is computationally

trivial (just copy part of and ).‐ note: if the marginal probability of each variable is normal, this does not imply that

the set of variables is jointly normal.‐ any linear transformation of the variables is jointly normal:

ADVANCED‐ the variables can be easily mapped into the «standard normal space».


linear transformation of jointly normal rand. vars.

36

; ,

⋯

; ,

sum: 1 1 2

difference: 1 1 2

example: 10 7 3 5 10 7 3 5

many loads, many resistances:

10 4 3 6 8 2

10 4 3 6 8 2 0

resistances loads

limit state function

from linear transformation rule

distribution of :


sum of two random variables in the general case

37

, , , ,

, ,

0

joint probability

if independency

convolution integral,hopefully it can be solved

for specific distributions , .

, ,

→

second moment representation:

difference:

always true


reliability for normal vars., with linear limit state func.

38

0 → safecondition0 → failure

P 0 0

linear limit state function:

distribution of : ; ,

02

46

0

5

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

x1 = sx2 = r

p(x 1,x

2),

g(x1

,x2)

0 Φ

reliability index :

probability of failure :


why do we assume a MVN model?

39

-5 0 5 10

0

0.1

0.2

0.3

0.4

p(x)

pnorm. appr.

-5 0 5 100

0.5

1

g

F(g)

=44% appr. =27%

consider ~ [not necessarily ]

does and exist (can be computed)?

consider linearly related to : :

can we compute and var ?

why we need ~ ?

so that 0 → safecondition0 → failure

Because we get ~ and we can easily compute ℙ 0


PART III

transformation of the Gaussian model

40


transformation to standard normal space

41

; ,

Cholesky decomposition:Given any matrix (positive‐definite),

chol is a lower triangular matrix so that .

; ,

→

Given and , find and so that:

0 0⋮ ⋱ 0

…/

Eigenvalue analysis:Given any matrix , is orthonormal matrix, is a diagonal matrix so that

.

not the same map. Cholesky is simpler.

/

Matlab: m_L=chol(m_Sigma,'lower')

1‐d:

standard normalization:


example of transformation to standard normal space

42

65104

4 1.21.2 9

2.4 2.46 3.6

2.4 62.4 3.6

16 3.23.2 4

; ,

chol2.00 00.60 2.94

0 00 0

1.20 1.801.20 0.98

3.36 00 1.26

/

0.57 1.660.31 0.97

0.43 0.851.68 2.27

0.04 0.060.81 0.80

1.48 3.721.04 1.28

0.53 0.290.38 0.22

0.04 0.748 0.0142 0.184

0.068 0.270.040 0.11

0.236 0.1660.174 0.060

0.50 00.10 0.34

0 00 0

0.12 0.180.40 0.26

0.28 00 0.79

1.221.77

0.052.76

3.001.091.32

0.53

inverse relation:

Cholesky

Cholesky

Eigenvalue

Eigenvalue


density in the standard normal space

43

; ,1

2exp

12

1

2exp

12

1

2

; 0,1

0

polar coord.

u1

u 2

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

-20

2

-20

2

0

0.05

0.1

0.15

0.2

u1

p(u1=0,u2)p(u1,u2=0)

u2

p(u 1,u

2)

0 1 2 3

0.6

0.8

1

(

)

0 1 2 3

0

0.1

0.2

0.3

0.4

()

pdf

cdf

Maximum density in the origin, fast decay in radial direction. Radial symmetry: density only dependents on .


rotation in the standard normal space

44

; ,1

2exp

12

1

2exp

12

1

2

; 0,1

0

polar coord.

u1

u 2

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

0 1 2 3

0.6

0.8

1

(

)

0 1 2 3

0

0.1

0.2

0.3

0.4

()

pdf

→ →

; , ; ,

new reference system:…

new coordinates:

distribution in new coordinates

for an ortho‐normal system:

the distribution is invariant respect to rotation

cdf


properties of the standard normal space

45

; ,1

2exp

12

1

2exp

12

1

2

; 0,1

0

polar coord.

‐ for each number of dimensions , there is just one standard normal space;‐ the distribution is invariant respect to rotation;‐ the origin, , is the mean vector and it is the (only) mode (i.e. maximum);‐ each variable is scaled to (zero mean and) unit standard deviation;‐ each variable is independent from the others; ‐ all marginal distributions are the same: ∀ , (standard normal);‐ the density at one point ( ) depends only by the distance from the origin ( ) and the

number of dimensions ( ).


reliability in the standard normal space

46

limit state functionin the standard normal space linear limit state functions stay linear:

transformation from standard normal to physical space:

as expected, reliability in standard normal space and in the physical space are equivalent:

0 0probability of failure :


design point in the standard normal space

47

∗

0

design point

If 0 :

∗ argmax

design point: the “most dangerous” condition:

∈ FailureDomain

∝ exp12

in standard normal space:

∗ argmin

the design point is the point in the failure domain closest to the origin.

∗ argmin

design point:it belongs to the failure domain (it is on the edge safe/failure);it has a high probability (the highest in the failure domain);it can be found by solving a constrained optimization problem;for linear limit state functions, the solution is very simple.

origin in the safe domain, i.e.low probability of failure.

log12


design point in the standard normal space [cont.]

48

∗

∗

0

design point

0limit state function

∗ ∗ ∗

∗ 0

∗ 0 ⟶ ∗

coordinates of the design point

conditions to find design point∗ ∥ ∗ 0

∗ ∗

∗ 0 ⟶ ∗

vector of norm 1


reliability using design point, in stand. normal space

49

∗ coordinates of the design point

∗

0

design point

∗ Φ

once you have found the design point, you can measure how far the failure domain is from the origin, and compute the probability of failure.

reliability index

[check that this is consistent with previous result]

Summary:Transform your belief in the standard normal space → ,

and define the new limit that function .Find the design point ∗, measure how far is it from the origin to get the reliability index.


design point in the physical space [normal variables]

50

x1 = s

x 2 = r

1 = 10 , 2 = 15 , 1 = 2 , 2 = 1.5 , 12 = 0.6

5 10 15 205

10

15

20

u1

u 2

-4 -2 0 2 4-4

-3

-2

-1

0

1

2

3

4

∗ ∗

∗

∗

∗ argmax∶

∗ argmax∶

The design point is the “most dangerous” scenario: it is a (incipient) failure condition, and it is the scenario with highest probability in the failure domain.If the map is linear, the design point in the physical coordinates is ∗ ∗ .If it is not linear, previous equation can be only approximate.

standard normal space

physical space


0 2 4 610

-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Pf

reliability index

51

≜ Φ Φ

2

0 50%

1 15.6%

2 2.28%

3 1.35%

4 3.1710‐5

5 2.8710‐7

6 0.9910‐9

it gives the order of magnitude of the probability of failure:

example, problem:

; ,

1 ΦΦ 1

IF the limit state function is linear in standard normal space, than the reliability index is the distance between the origin and the design point.IF NOT, it is not necessary.

[actually, if the set 0 is an hyper plane, and is regular, continuous]


log‐normal multivariate distribution

52

; , ln ; ,exp

log

exp 0 00 ⋱ 00 0 exp

exp

0 00 ⋱ 00 0

1∏

log ; ,

1

2 ∏exp

12 log log ≜ ln ; ,

Jacobian determinant of the Jacobian

log‐normal density

log scale linear scale


moments to parameters for log‐norm. mult. distr.

53

ln ; ,

ln 1 ≅ for small

ln ≅ ln for small

⋮

11

…⋱ ⋮

⋮ ⋱…

⋱ ⋮… 1

00

… 0⋱ ⋮

⋮ ⋱0 …

⋱ 00

⋮

moments of :

; ,

log

11

…⋱ ⋮

⋮ ⋱…

⋱ ⋮… 1

00

… 0⋱ ⋮

⋮ ⋱0 …

⋱ 00

parameters:

ln 1 ≅ for small

as for 1‐drelations:

linear scale

log scale


properties of log‐normal multivariate distribution

54

z1

z 2

1 = 1 , 2 = 1.5 , 1 = 0.5 , 2 = 0.8 , x1x2 = 0.6

0 2 4 6 8 100

1

2

3

4

5

6

7

8

9

10

0 2 4 6 8 100

5

100

0.02

0.04

0.06

0.08

1 = 1 , 2 = 1.5 , 1 = 0.5 , 2 = 0.8 , x1x2 = 0.6

z1

z2

p(z 1,z

2)

completely defined by mean vector and covariance matrix;marginal and conditional distributions of any subset of rand. vars. are lognormal;product functions are jointly lognormal; uncorrelation implies independence.


r‐s problem with log‐normal rand. var.s

55

ln ; ,

limit state function: log log log 0 → safecondition0 → failure

00 → 0

safeconditionif → 1 → log 0 → log log 0

failureif → 1 → log 0 → log log 0

consider the load to resistance ratio:

load and resistance:

; ,log

log log

1 1

in the logarithm scale, the formulation is equivalent to that with normal rand. vars.


reliability problem with log‐normal rand. var.s

56

ln ; ,

limit state function log log log

; ,

log

… …

⋮vector of random variables

failure ↔ 1

≜ log

we can re‐shape the problem as that of a linear limit state function on a jointly normal distributed variable:

failure ↔ 72 4 3 7 ∙ 2 ∙ 43 18.7 1 3 0.5

log 2.9372 43 1

example:

with , >0


PART IV

general approach and FORM

57


general reliability problem

58

givenlimit state functionjoint probability

compute

x1

x 2

0 2 4 6 8 100

2

4

6

8

10

When many random variables are involved in the problem (high dimensional space) , it is expensive to compute the integral.

The integral can be solved numerically, by counting along a grid (but that method is ineffective because of the course of dimensionality).

No analytical solutions are generally available.

Approximate solutions are provided by reliability methods (FORM: first order reliability method).Or by simulations (Monte Carlo).

For reliability methods and for simulations it is convenient to formulate the problem in the standard normal space.


u1

u 2

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

general reliability problem in stand. norm. space

59

givenlimit state functionjoint probability

compute

find transformation to the standard normal space:

: ; ,

≜

x1

x 2

0 2 4 6 8 100

2

4

6

8

10

00


going to the standard normal space

60

It can be easily done for any jointly normal distribution (e.g. using Cholesky).Also for any jointly log‐normal distribution (taking the log, and using Cholesky).It can also be done for any distribution (Rosenblatt transformation), but it may complicate.

why we prefer this space:‐ once here, distribution is very simple.‐ variables are uniform, in the “same scale”, uncorrelated.‐ you can easily generate samples from the distribution,‐ you can approximate the solution finding the design point (FORM)

u1

u 2

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3


First Order Reliability Method (FORM)

61

‐ find design point:∗ argmax

∶argmin∶

‐ compute approximate reliability index:

≅ ∗ Φ

in standard normal space

∗

0

design point

approximate limit state (linear approximation at the design point)


-5 0 5-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

p(u)

, g

(u)

FORM in 1‐d

62

4 1 10

1 5

non linear limit state function:

derivative (gradient in higher dim.)

design point: ∗ : ∗ 0

0 0 0 is in the safe domain

≅ ∗ Φ


approximation:

find design point ⟶ find zeroNewton–Raphson method

start at approximate (Taylor)

≅∗ 0

∗ ≅ ′

′

repeat

until convergence

∗

safe failure


-5 0 5-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

p(u)

, g

(u)

FORM in 1‐d [cont.]

63

4 1 10

1 5

non linear limit state function:

derivative (gradient in higher dim.)

design point: ∗ : ∗ 0

safe failure

0 0 0 is in the safe domain

≅ ∗ Φ


approximation:

find design point ⟶ find zeroNewton–Raphson method

start at approximate (Taylor)

≅∗ 0

∗ ≅ ′

′

repeat

until convergence


FORM for more than 1 variable

64

…

≅

∗ ≅∙

∗ ≅∙

probability

; , [generally non‐linear]


∙ ∙

linear approximation around [Taylor]

gradient

design point:

distant from the origin:

[standard normal space]


FORM iterative method

65

∗

∙

select

repeat, from 0

until convergence at ∗

set

Φ

compute

1


example of FORM

66

-50

5

-5

0

5

-3

-2

-1

0

1

2

u1u2

g(u 1,u

2)

-50

5

-505

-3

-2

-1

0

1

2

u1u2

g(u 1,u

2)

0 0


example of FORM

67

-5

0

5

-505

-3

-2

-1

0

1

2

u1u2

g(u 1,u

2)

-50

5

-505

-3

-2

-1

0

1

2

u1u2

g(u 1,u

2)


example of FORM

68

-8 -6 -4 -2 0 2 4 6 8-8

-6

-4

-2

0

2

4

6

8

u1

u 2

# iter.1 02 5.33333 4.07864 3.99915 3.99746 3.99717 3.9970

3.21 10

4.50 10 5%Monte Carlo:


gradient following a transformation

69

…

⋮ ⋱ ⋮

…

⋮

…

⋮ ⋱ ⋮

…

⋮

∙: ⟶

invertible map inverse map: limit state function

Suppose we have the Jacobian → , and the gradient → , we can compute the gradient → as chain rule, multiplying Jacobian and gradient .

proof


Jacobian of a composed map

70

: ⟶

invertible maps

Suppose we have the Jacobian → , and that for → , we can compute the Jacobian → as chain rule, multiplying the two matrices .

: ⟶

…

⋮ ⋱ ⋮

…

…

⋮ ⋱ ⋮

…

…

⋮ ⋱ ⋮

…

∙

proof


example of reliability problem by FORM

71

ln ; , 1.41.2

exp

; ,

70% 70%

20%⇢ .49 .098

.098 .49

→ 1 130

exp 00 exp

00

0.70 00.14 0.69

; , ∙ ∙

∙ exp

exp exp


standard normal space

failure ↔

normal space


example of reliability problem by FORM

72

-5 0 5-5

0

5

u1

u 1-5

05 -5

05

-200

-150

-100

-50

0

50

100

150

200

u2u1

g(u 1,u

2)

limit state function is not linear in the standard normal space

0.97% (from integration and Monte Carlo)

design point


reliability problem by FORM: iterative scheme

73



k = 1

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

0 04.64 3.20

‐3.30 ‐2.28

22.62

∙0

5.64

50%

9 10



74



k = 2

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

4.64 3.203.70 1.78

‐81.20 ‐39.17

‐132

5.64

4.11

9 10

2 10∙



75



k = 3

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

3.70 1.782.91 0.93

‐40.58 ‐13.02

‐43.2

4.11

3.06

2 10

1.1%∙



76



k = 4

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

2.91 0.932.52 0.71

‐23.14 ‐6.50

‐10.6

3.06

2.61

0.11%

0.45%∙



77



k = 5

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

2.52 0.712.44 0.73

‐17.61 ‐5.26

‐1.29

2.61

2.544

0.45%

0.548%∙



78



k = 6

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

2.44 0.732.42 0.77

‐16.71 ‐5.28

‐0.036

2.544

2.542

0.548%

0.551%∙



79



k = 7

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)

2.42 0.772.42 0.79

‐16.59 ‐5.40

‐3.19x10‐3

2.542

2.5415

0.551%

0.552%∙



80

k = 8

u1

u 1

-5 0 5-5

0

5

-50

5 -5 0 5

-200

-100

0

100

200

u2u1

g(u 1,u

2)



2.42 0.792.41 0.80

‐16.53 ‐5.48

‐1.05x10‐3

2.5415

2.5414

0.552%

0.552%

0.97%compare with Monte Carlo result:

∙


FORM importance measures in the stand. norm. space

81

∗

0

01

∗

0

∑ 1

: irrelevant is more important that

importance measure:

0: isa′load’

0: isa′capacity′

0is a capacity

∗

∗ ∗ ∗ ∗ ∙ ∗

∗ ∙

≜∗

∗

∗

∗

∗ ≜ ∗

∗ ∙

∗

design point:

linearized limit state function: direction of the design point:


FORM importance measures in the stand. norm. space

82

∗

0

01

∗

0

∑ 1

: irrelevant is more important that

importance measure:

0: isa′load′

0: isa′capacity′

0is a capacity

∗ ≜∗

∗

∗

∗

∗

∗ ∗ ∗ ⇢

linearized limit state function:

importance measure

; ,stand. norm. space:

direction of the design point:

gives the importance of variable in the problem.


FORM importance measures in physical space

83

≅ ∗ ∗ ∙ ∗

≅ ∗ ∗ ∙ ∗ ∗ ∙

∗ ≜ ∗

∗ ∗ ∗

∗ ∗ ∗ ∙

we assume correlation matrix , because we are not interested in the correlation, to define the importance singular variables.

∗ ≜ ∗ ≜

linearized map

linearized inverse map

linearized limit state function

contribution to the uncertainty (variance) of : 00

… 0⋱ ⋮

⋮ ⋱0 …

⋱ 00

constants

stand. dev. of :

normalized vector gives the importance measures of variables in .

1


example of importance measures in physical space

84

∗ 2.41 0.80 →∗

∗ 95 32 %

∗ ∗ ∙ ∗

∗∗ 00 ∗ ∙ 15.32 0

1.14 5.67

∗ 3.09 2.09∗ 21.88 8.12

∗

∗ ∗ 65.3 013.3 179 10

S 00

4.12 00 3.37

4.12 00 3.37

65.3 13.30 179

9532 10 24

19 % → 7763 %

design point:

Normal space

Standard normal space

Physical space

exp 1 exp 2

∗ exp ∗

∗


PART V

further remarks on component reliability

85


-4 -2 0 2 40

0.5

1

x

p x(x)

-4 -2 0 2 4

0

0.2

0.4

0.6

0.8

1

u

xx = Fx-1(w)

w = Fx(x)

0 0.5 1

0

0.2

0.4

0.6

0.8

1

pw(w)

w

general transformation to the standard normal space

86

11; ∈ 0,1

a) every rand. var. distributed by: , can be transformed into a uniform

rand. var. by transformation .


general transformation to the standard normal space

87

11; ∈ 0,1

a) every rand. var. distributed by: , can be transformed into a uniform

rand. var. by transformation .

b) uniform rand. var. can be transformed into rand. var. , distributed by , , through transformation

1

hence can be derived by through . In particular, can be mapped into the standard normal distribution by transformation Φ .

multivariate case:

independentrand. vars.

∀ :Φ

dependentrand. vars.

ΦΦ | |

Φ | … | …⋮

Rosenblatttransform.

-4 -2 0 2 40

0.5

1

u , x

p u(u),

p x(x)

-4 -2 0 2 4

0

0.2

0.4

0.6

0.8

1

u

xu = -1(w)

w = (u)x = Fx

-1(w)

w = Fx(x)

0 0.5 1

0

0.2

0.4

0.6

0.8

1

pw(w)

w


properties of limit state function, to use FORM

88

-4 -3 -2 -1 0 1 2 3 4-0.5

0

0.5

1

u

p(u)

, g

(u)

∗

∗

‐ only the sign of the limit state function is relevant. Functions and , so that sign sign , are equivalent.

∗

∗

1

‐ a linear limit state function is convenient, because local data , at any point define the all boundary 0.

‐ the reliability problem is defined by boundary: 0 [and sign]

the reliability does not depend on the slope of the gradient, or on the “magnitude” of :

‐ to use FORM (Newton method), we require function to be continuous and differentiable: ∃ .


convergence of Newton‐Rapshon method

89

‐ The method may not converge in some conditions:

‐ To overcome this problem, one may pose a maximum size in the steps taken by the algorithm: .

see Wolfe Conditions and Armijo rule: http://en.wikipedia.org/wiki/Wolfe_conditions

-2 0 2 4 6 8-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

u

p(u)

, g

(u)


SORM

90

∗

0

design point

limit state function is approximated taking curvature into account

Second Order Reliability Method: it approximates the limit state function with a quadratic form around the design point.

≅ ∗ ∗ ∗ 12

∗ ∗ ∗

Hessian matrix

It is more accurate than FORM, but computationally more expensive because it requires to obtain curvatures ( ) at the design point.It is a further step after FORM:‐ find the design point ∗

‐ compute curvature at ∗.


bounds for FORM and SORM

91

0

By definition, the design point is the closest to the origin. Hence the all region∶ ∗ belongs to the safe domain. Let us define ∗ ≜ ∗ .

∗

design point

P ∗ ∗ ∗

χ ∗

upper bound to the probability of failure:Cumulative Chi‐squared

distribution, with degrees of freedom.

∗ 1 χ ∗

0 0.5 1 1.5 2 2.5 310

-3

10-2

10-1

100

*

Pf

FORMupper bound

20[no lower bound]

FORM: ≅ ∗


note about design point and reliability index

92

∗

0

design point

≜ Φ Φdefinition of reliability index:

distance of the design point from the origin: ∗ ≜ ∗

FORM approximation: ≅ ∗

0

∗design point

∗ argmax∶

argmax∶

∗ ∗

for a non‐linear map , the design point in the stand. norm. space is not necessary mapped into the max. of the physical space in the failure domain. analogy: the mode of the (e.g. uni‐variate) normal distr. is not mapped into the mode of the log‐normal.


refereces

93

on wikipedia:Cholesky decomposition ‐ Eigenvalues and eigenvectors ‐ Gradient ‐ Jacobian ‐Positive‐definite matrix ‐ Multivariate normal distribution ‐ Newton's method ‐Chi‐squared distribution ‐ Wolfe Condition ‐ Chain rule

Barber, B. (2012). Bayesian Reasoning and Machine Learning. Cambridge UP. Downloadable from http://web4.cs.ucl.ac.uk/staff/D.Barber/pmwiki/pmwiki.php?n=Brml.HomePage. Section 8.4 on “Multivariate Gaussian”.

Der Kiureghian, A. (2005) "First and Second‐Order Reliability Methods", in book: E. Nikolaidis, D.M. Ghiocel, S. Singhal (Eds), The Engineering design reliability handbook, CRC Press LLC.

Ditlevsen, O. and H.O. Madsen. (1996). Structural reliability methods. J. Wiley & Sons, New York, NY. Downloadable from http://www.web.mek.dtu.dk/staff/Od/books/OD‐HOM‐StrucRelMeth‐Ed2.3.7‐June‐September.pdf. Sections 2.1‐3, 4.1‐2, 5.

Faber, M. (2009) “Risk and Safety in Engineering, lecture notes”, Lectures 5‐6, available at http://www.ibk.ethz.ch/emeritus/fa/education/ws_safety/Non_printable_script.pdf

Sørensen, J.D. (2004) "Notes in Structural Reliability Theory And Risk Analysis", notes 3‐5, avail. at http://www.waterbouw.tudelft.nl/fileadmin/Faculteit/CiTG/Over_de_faculteit/Afdelingen/Afdeling_Waterbouwkunde/sectie_waterbouwkunde/people/personal/gelder/publications/citations/doc/citatie215.pdf


MW Matlab ‐ commands

94

M=zeros(n,m) : it defines matrix ‘M’, of size (nm), with all entries zero . length(v) : number of entries in vector ‘v’. M=diag(v) : it makes diagonal matrix ‘M’, putting vector ‘v’ on the diagonal. L=chol(M,'lower') : compute lower triangular matrix ‘L’, from Cholesky decomp. of ‘M’. M1.*M2: when matrices (or vectors) ‘M1’ and ‘M2’ have the same dimension, it makes

matrices ‘M3’ as element‐by‐element product of ‘M1’ and ‘M2’. Similar allowed operations are: M1./M2 , 1./M1 , M1.^2 .

instructor matteo pozzi - cmufaculty.ce.cmu.edu/pozzi/files/2015/09/12735_15_lec_03.pdf · 12735:...

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