Approximate methods for calculating
probability of failure
• Working with normal distributions is
appealing
• First-order second-moment method (FOSM)
• Most probable point
• First order reliability method (FORM)
• Section 4.1 in Choi, Grandhi, & Canfield
The normal distribution is attractive
• It has the nice property that linear functions of
normal variables are normally distributed.
• Also, the sum of many random variables tends to be
normally distributed.
• Probability of failure varies over many orders of
magnitude.
• Reliability index, which is the number of standard
deviations away from the mean solves this problem.
   1 ( Pf ) Pf   (  )
 is the normal CDF
Approximation about mean
• Predecessor of FORM called first-order
second-moment method (FOSM)
g ( X )  g (  X )   g (  X )T  X   X 
Then, easy to show
1/2
 n  g 

2
 g  g (  X )  g       xi 
 i 1  xi 

(How can you use that to get rid of unimportant variables?)
The reliability index and probability of failure are
2
g

Pf   (  )
g
 is the normal CDF
Beam under central load example
Example 4.2
• Probability of exceeding plastic moment
capacity g (P, L,W , T )  WT  PL / 4
P
L
W (plastic section
modulus)
T (yield stress)
mean
10kN
8m
0.0001m^3
600,000 kN/m^2
Standard deviation
2kN
0.1m
0.00002m^3
100,000kN/m^2
• All normal
Reliability index for example
• Using the linear approximation get
g ( P, L,W , T )  WT  PL / 4
 g  g (  X )  0.0001 600, 000  10  8 / 4  60  20  40kNm (safety factor of 3!)
g
 W  0.0001
T
g
  L / 4  2
P
g
 T  600, 000
W
g
  P / 4  2.5
L
1/2
 n  g 2 
 g       x2i 
 i 1  xi 


(0.0001100, 000) 2  (600, 000  0.00002) 2   2  2   (2.5  0.1) 2  16.13kNm
2

g
 2.48,
g
Pf   (2.48)  0.0066
• Example 4.2 of CGC shows that if we change to g=T0.25PL/W we get 3.48 (0.00025)
• MCS with 10 million samples gave 2.85 (0.00218)
Top Hat question
• In the beam example, the error in estimating
the reliability index was due to
– Linear approximation
– g is not normal
– both
Most probable point (MPP)
• The error due to the linear approximation is
exacerbated due to the fact that the expansion
may be about a point that is far from the failure
region (due to the safety margin).
• Hasofer and Lind suggested remedying this
problem by finding the most probable point and
linearizing about it.
• The joint distribution of all the random variables
assigns a probability density to every point in the
random space. The point with the highest density
on the line g=0 is the MPP.
Response minus capacity illustration
r=randn(1000,1)*1.25+10;
c=randn(1000,1)*1.5+13;
f=@(x) x;
fplot(f,[5,20])
hold on
plot(r,c,'ro')
xlabel('r')
ylabel('c')
20
c
15
10
5
5
10
15
r
20
Recipe for finding MPP with
independent normal variables
• Transform into standard normal variables
(zero mean and unity standard deviation)
ui 
xi   xi
x
i
• Find the point on g=0 of minimum distance to
origin. The point will be the MPP and the
distance to the origin will be the reliability
index based on linear approximation there.
  min U TU 
1/2
U g (U )  0
First order reliability method (FORM)
• Limit state g(X). Failure when g<0.
• Linear approximation of limit state together
with assumption that random variables are
normal.
• Approximate around most probable point.
• Then limit state is also normal variable.
• Reliability index is the distance of the mean of
g from zero measured in standard deviations.
Visual
Linear Example
•
•
•
•
For linear example g  R  C, R  N (10,1.25 )
 10
c  13
Then u  r1.25
u 
g  1.25u  1.5u  3
1.5
The failure boundary g  0  u  (5 / 6)u  2
Distance from origin
 2  u12   5u1 / 6  2 
C  N(13,1.52 )
2
1
2
1
2
2
1
3
2
2
1
0
61
u2  1.1528(r  c  11.27)
60
Pf   (  )  0.0621
u1 
u2
d 2
25
10
 2u1  u1   0
du1
18
3
-1
-2
  1.537
-3
-4
-5
-3
-2
-1
0
1
u1
2
3
4
Check by MCS
•
•
•
•
r=randn(1000,1)*1.25+10;
c=randn(1000,1)*1.5+13;
>> s=0.5*(sign(r-c)+1);pf=sum(s)/1000=0.0550
Six repetitions gave: 0.0660, 0.0640,
0.0670,0.0450, 0.0550, 0.0640
• With million samples got 0.06258
• So even with a million samples, accurate only
to two digits.
Problems FORM
• Repeat the linear example of Slide 12 with a
slightly different limit state g=R2-C2 with both
FORM and MCS.
• Do the beam problem of slides 4 and 5 with
FORM. Note that you are now in 4
dimensional random space.
General case
• If random variables are normal but correlated, a
linear transformation will transform them to
independent variables.
• If random variables are not normal, can be
transformed to normal with similar probability of
failure. See Section 4.1.5 of CGC (It is called the
Rosenblatt transformation)
• Murray Rosenblatt, Remarks on a Multivariate
Transformation, Ann. Math. Statist. Volume 23,
Number 3 (1952), 470-472.
Approximate transformation
• If we want the transformed variable u to have the
same CDF as the original x at a certain value of x we
would require (u )  F ( x)
• Unfortunately we cannot enforce that everywhere.
• If we focus on MPP we get the following
transformation (Eqs. 4.38, 4.39), which must be used
iteratively, starting from some guess.
u
X* replaced
by xMPP in
equations.
x  x'
 x' 
 x'

  1  F ( x MPP ) 

f ( x MPP )
 x '  x MPP   1  F ( x MPP )   x '