why need probabilistic approach? rain probability how does that affect our behaviour? ?

Why need probabilistic Why need probabilistic approach?approach?

Rain probabilityRain probability

How does that affect our behaviour?How does that affect our behaviour?

? ?

Uncertainties in Uncertainties in EngineeringEngineering

Natural HazardsNatural Hazards

Material PropertiesMaterial Properties

Design ModelsDesign Models

Construction ErrorsConstruction Errors

Absolute Safety Not Absolute Safety Not GuaranteedGuaranteed

Engineers need to:Engineers need to:

model, analyze, update model, analyze, update

uncertaintiesuncertainties

evaluate probability of failureevaluate probability of failure

QuestionsQuestions

What is acceptable failure What is acceptable failure probability?probability?

-- stadium vs shedstadium vs shed

QuestionsQuestions

Should one want to be Should one want to be conservative if a perfectly safe conservative if a perfectly safe system is possible?system is possible?

-- overbooking in airlinesoverbooking in airlines

-- parking permitsparking permits

QuestionsQuestions

Should one minimize risk if Should one minimize risk if money is not a problem?money is not a problem?

-- system consideration – system consideration –

e.g.dame.g.dam

Trade-off Decision AnalysisTrade-off Decision Analysis

Risk vs. consequenceRisk vs. consequence

System riskSystem risk

Formal analysis of Formal analysis of uncertainties and probabilityuncertainties and probability

• Not all problems can be solved by analysis Not all problems can be solved by analysis of dataof data

• Set TheorySet Theory

• Sample spaceSample space: collection of all possibilities: collection of all possibilities

• Sample pointSample point: each possibility: each possibility

• EventEvent: subset of sample space: subset of sample space

• Probability TheoryProbability Theory

Union: Union: eithereither E E1 1 or Eor E2 2 occuroccur

EE11∪∪EE22

Intersection:Intersection:bothboth E E1 1 and Eand E2 2 occuroccur

EE11∩∩ E E22 or E or E11 E E22

ExamplesExamples

A B

No communication between A and B = E1E2

CB

A

No communication between A and B = E3∪∪E1E2

1

2

EE11 = road 1 closed = road 1 closedEE22 = road 2 closed = road 2 closed

EE33 = road 3 closed = road 3 closed1

23

Example - pair of footingsExample - pair of footings

1 2

EE11 = 1 settles = 1 settles

ĒĒ11 = 1 does not = 1 does not

settlesettle

EE22 = 2 settles = 2 settles

ĒĒ22 = 2 does not = 2 does not

settlesettle

Settlement occurs = E1∪∪E2

Tilting occurs = E1ĒĒ2∪∪ĒĒ1E2

de Morgan’s rulede Morgan’s rule

EE1 1 = pipe 1 breaks= pipe 1 breaks

EE22 = pipe 2 breaks = pipe 2 breaks

1 2

Water Supply

E = failure in water supply = EE = failure in water supply = E11∪∪EE22

no failure in water supply = no failure in water supply = ĒĒ = =

EE11∪E∪E22

2121EEEE

nEEEEEE n ....... 2121

Event of “no failure”

Extension to n events

de Morgan #2de Morgan #2

2121 EEEE

nn EEEEEE ........ 2121

Basis of Probability Basis of Probability EstimationEstimation

a)a) Subjective assumption e.g. P(Q) = 1/2Subjective assumption e.g. P(Q) = 1/2

b)b) Relative frequency e.g. Relative frequency e.g.

P(Q)=502/1000P(Q)=502/1000

c)c) Bayesian (a)+(b) Bayesian (a)+(b) judgment + limited observation judgment + limited observation

Probability of UnionProbability of Union

in generalin general

E1 E2

1 2( )P E E

1 2 1 2( ) ( ) ( )P E P E P E E

Using de Morgan’s ruleUsing de Morgan’s rule

)(1)( 321321 EEEPEEEP

)(1 321 EEEP

P (intersection) conditional

probability

( ) ( | ) ( )P AB P A B P B ( ) ( | ) ( )P AB P B A P A

( ) ( | ) ( )P ABC P A BC P BC

( | ) ( )P B C P C

or or

Statistical independence

if E1 and E2 are s.i.

2 1 2( ) ( )P E E P E

1 2 1( ) ( )P E E P Eor

s.i.

Example:

E2 = flood in 廣西 on June E3 = flood in 哈爾濱 on June

E1 = flood in 廣東 on June

P(E1) = 0.1; P(E2)=0.1; P(E3) = 0.1

121 3.0| EPEEP E1 and E2 are not s.i.

1.0| 31 EEPE1 and E3 are s.i.

1EP

1 2 1 2 2( ) ( ) ( )P E E P E E P E

1( )P Eif E1 and E2 are s.i.

1 2 3 1 2 3( ) ( ) ( ) ( )P E E E P E P E P Eif all are s.i.

s.i. and m.e. (mutually exclusive)

if E1 and E2 are m.e.

1 2( ) 0P E E

1 2 1( ) ( )P E E P E

if E1 and E2 are s.i.

B

C

A

2

1

3

P(E1)=2/5

P(E2)=3/4

P(E3)=2/3

P(E3|E2)=4/5

P(E1|E2E3)=1/2

a) P(go from A to B through C)

32EEP 2 3( ) ( )P E P E

5

3

4

3

5

4

3 2 2( ) ( )P E E P E

E1 : ① is openP 2.15

b)

P(go from A to B)

132 EEEP

132132 EEEPEPEEP

1 2 3 2 3

3 2( | ) ( )

5 5P E E E P E E

7.0 1/2 3/5

T.O.T (Theorem of Total Probabilities)

Bayes theorem

AP

EPEAPAEP jj

j

||

P(A) = P(A|E1)P(E1)+P(A|E2)P(E2)+…+P(A|En)P(En)

Ei’s are m.e. and c.e.

7.0GP 3.0GP

good enough for construction

2.0|;8.0| GTPGTP

9.0|;1.0| GTPGTPpositive

E 2.30 aggregate for construction

engineer's judgment based on geology and experience

crude test

reliability (or quality) is as follows:

not a perfect test

After 1 successful test, what is P(G)?

TP

GPGTPTGP

||

3.01.07.08.0

7.08.0

95.0

( | ) ( )

( | ) ( ) ( | ) ( )

P T G P G

P T G P G P T G P G

After another successful independent test, P(G)?

GPGTPGPGTP

GPGTPTGP

||

||

22

22

05.01.095.08.0

95.08.0

993.0

What if the two tests were performed at the same time?

)()|()()|(

)()|()|(

2121

2121

GPGTTPGPGTTP

GPGTTPTTGP

1 2

1 2 1 2

( | ) ( | ) ( )

( | ) ( | ) ( ) ( | ) ( | ) ( )

P T G P T G P G

P T G P T G P G P T G P T G P G

0.7

0.30.7

3.1.1.7.8.8.

7.8.8.

993.0

P(G)

UST

HKU

0.7

0.3

After 1 test

0.95

0.77

After 2 tests

0.993

0.965

… 5

1.0000

0.9999

Random variables

A device to:

a) formalize description of event

b) facilitate computation of

probability

PMF PX(x) FX(x)

PDF fX(x) FX(x)

CDF

( )( ) X

X

dF xf x

dx

Main descriptors of R.V.

The PMF or PDF completely define the r.v.

Descriptors give partial information about the r.v.

Mean value

Define = E(X)

= expected value of X or mean value of X

( )i X ii

x P xa measure of central tendency a measure of central tendency

Measure of spread

Standard deviation X

X dimensionless %

range

( )Var X

X

X

Expected value of function

( ) ( )XE X xf x dx

recall

( ) ( ) ( )XE g X g X f x dx

Recall2( ) [ ( )] ( )XVar X x E X f x dx

2 2( ) ( ) ( )Var X E X E X

After some algebra,

2 ( )Xx f x dx

0.00

0.05

0.10

0.15

0.20

0.25

-5 0 5 10 15 20

fX(x)

x

Normal distribution

= 5

X : N (, )

N (5, 2)

x

Effect of varying parameters ( & )

fX(x)

x

A

B

C

for C

for B

S: N (0,1)

Standard normal distribution

0.00

0.10

0.20

0.30

0.40

0.50

-6 -4 -2 0 2 4 6

fX(x)

x

(0) 0.5 (1) 0.8413

(2) 0.97725 ( 1) 1 (1) 0.1587

(4) 1 0.00003167 0.9999683

-5 -4 -3 -2 -1 0 1 2 3 4 5

21

21( )

2

a sa e ds

a

Page 380 Table of Standard Normal Probability

Example: retaining wall

x

F

Suppose X = N(200,30)

(230 260)

260 200 230 200

30 30

(2) (1)

0.977 0.841 0.136

P x

If the retaining wall is designed such that the reliability against sliding is 99%,

How much friction should be provided?

1

( ) 99%

200 2000.99

30 30

200(0.99)

30

P x F

F

F

1200 30 (0.99) 270F

2.33

Lognormal distribution

Parameter

0 2 4 6 8

fX(x)

x

21ln

2

2

2 22

ln 1 ln 1

Parameters

for 0.3,

ln mx

Probability for Log-normal distribution

ln( )

aP X a

If a is xm, then is not needed.

ln ln( )

b aP a X b

Other distributions

Exponential distribution Triangular distribution Uniform distribution Rayleigh distribution

p.224-225: table of common distribution

Exponential distribution

x

fX(x)

( ) xXf x e x 0

1( )E X

21

( )Var X

100%X

Beta distribution1 1

1

( ) ( ) ( )( )

( ) ( ) ( )

q r

X q r

q r x a b xf x

q r b a

a x b

0.0

0.1

0.2

0.3

0 2 4 6 8 10 12x

fX(x)q = 2.0 ; r = 6.0

a = 2.0 b = 12

probability

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

Standard beta PDF

q = 1.0 ; r = 4.0

q = r = 3.0 q = 4.0 ; r = 2.0

q = r = 1.0

x

fX(x)

(a = 0, b = 1)

Bernoulli sequence

Discrete repeated trials 2 outcomes for each trial s.i. between trials Probability of occurrence same for all

trials

SF

p = probability of a success

SF

x = number of successp = probability of a success

P ( x success in n trials)

= P ( X = x | n, p) (1 )

( 0,1,2,..., )

x n xnp p

x

x n

Binomial distribution

Examples

Number of flooded years

Number of failed specimens

Number of polluted days

Example:

Given: probability of flood each year = 0.1

Over a 5 year period

1 45( 1) 0.1 (0.9) 0.328

1P X

P ( at most 1 flood year) = P (X =0) + P(X=1)

= 0.95 + 0.328

= 0.919

P (flooding during 5 years)

= P (X 1)

= 1 – P( X = 0)

= 1- 0.95

= 0.41

For Bernoulli sequence Model

No. of success binomial distribution

Time to first success geometric

distribution

E(T) =1/p = return period

Significance of return period in design

Suppose a bldg is expected to last 100 years and if it is designed against 100 year-wind of 68.6 m/s

P (exceedence of 68.6 m/s each year) = 1/100 = 0.01

P (exceedence of 68.6 m/s in 100th year) = 0.01

Service life

design return period

P (1st exceedence of 68.6 m/s in 100th year)

= 0.99990.01 = 0.0037

P (no exceedence of 68.6 m/s within a service life of 100 years)

= 0.99100 = 0.366

P (no exceedence of 68.6 m/s within the return period of design) = 0.366

If it is designed against a 200 year-wind of 70.6 m/s

P (exceedence of 70.6 m/s each year) = 1/200 = 0.005

P (1st exceedence of 70.6 m/s in 100th year)

= 0.995990.005 = 0.003

P (no exceedence of 70.6 m/s within return period of design)

= 0.995200 = 0.367

P (no exceedence of 70.6 m/s within a service life of 100 years)

= 0.995100 = 0.606 > 0.366

How to determine the design wind speed for a given return period?

Get histogram of annual max. wind velocity

Fit probability model Calculate wind speed for a design

return period

N (72,8)Example

V100

0.01

Design for return period of 100 years:

p = 1/100 = 0.01

100( ) 0.99P V V

100 720.99

8

V

V100 = 90.6 mph

Annual max wind velocity

Frequency

Suppose we design it for 100 mph, what is the corresponding return period?

( 100)

100 721

8

1 (3.5)

0.000233

P V

T 4300 years

Alternative design criteria 1

Pf = P (exceedence within 100 years)

= 1- P (no exceedence within 100 years)

=1- (1-0.000233)100 = 0.023

Probability of failure

1x n xn t t

x n n

P ( x occurrences in n trials)

= limn

( )

!

xtt

ex

x = 0, 1, 2, …

Poisson distributionPoisson distribution

P 3.42

Service stations along highway are located according to a Poisson process

Average of 1 station in 10 miles = 0.1 /mile

P(no gasoline available in a service station)

( ) 0.2P G

(a) P( X 1 in 15 miles ) = ?

0 1

0 1.5 1 1.5

( 0) ( 1)

( ) ( )

0! 1!

(1.5) (1.5)

0! 1!0.223 0.335

0.558

t t

P X P X

t e t e

e e

No. of service stations

(b) P( none of the next 3 stations have gasoline)

0 3

( 0 | 3, )

( 0 | 3,0.8)

3(0.8) (0.2)

0

0.008

P Y p

P Y

binomial

No. of stations with gasoline

(c) A driver noticed the fuel gauge reads empty; he can go another 15 miles from experience.

P (stranded on highway without gasoline) = ?

P (S)( | 0) ( 0) ( | 1) ( 1)

( | 2) ( 2) ......

P S X P X P S X P X

P S X P X

No. of station in 15 miles

binomial Poisson

x P( S| X = x ) P( X = x ) P( S| X = x ) P( X = x )

0 1 e-1.5 = 0.223 0.223

1 0.2 1.5 e-1.5 = 0.335 0.067

2 0.22 1.52/2! e-1.5 = 0.251 0.010

3 0.23 1.53/3! e-1.5 = 0.126 0.001

4 0.24 1.54/4! e-1.5 = 0.047 0.00007

Total = 0.301

Alternative approach

Mean rate of service station = 0.1 per mile

Probability of gas at a station = 0.8

Mean rate of “wet” station = 0.10.8 = 0.08 per mile

Occurrence of “wet” station is also Poisson

P (S) = P ( no wet station in 15 mile)0

0.08 15 1.2(0.08 15)0.301

0!e e

Time to next occurrence in Poisson process

Time to next occurrence = T is a continuous r.v.

( ) ( ) ( ) 1 ( )T T

d d df t F t P T t P T t

dt dt dt

( )P T t = P (X = 0 in time t)te

( ) 1 t tT

df t e e

dt

Recall for an exponential distribution

( ) xXf x e

T follows an exponential distribution with parameter =

E(T) =1/

If = 0.1 per year, E(T) = 10 years

Bernoulli Sequence

Poisson Process

Interval Discrete Continuous

No. of occurrence Binomial Poisson

Time to next occurrence Geometric Exponential

Time to kth occurrence Negative binomial Gamma

Comparison of two families of occurrence models

Significance of correlation coefficient

= +1.0 = -1.0

y: strength

x: Length

GlassGlass

y: elongation

x: Length

SteelSteel

= 0 0< <1.0

y: ID No

x: height

y: weight

x: height

Functions of Random Variable Functions of Random Variable (R.V.)(R.V.)

In general, Y = g(X)

Y = g(X1, X2,…, Xn)

If we know distribution of X distribution of Y?

M = 4X + 10

X5

22

M1)

X

b

b = Xtan 2)

cost of delay = aX23) where X – length of delay

Consider M = 4X + 10Consider M = 4X + 10

Observation: the distribution of y depends on

(1) Distribution of X

(2) g(X)

X654

0.6

0.2 0.2

M343026

0.6

0.2 0.2

X654 M343026

wider distribution

2

11

2

1

2

2 3 2

1 1 2 3 3exp 2

2 22 2

1e 0,1

2

, 0,1

y x x

y

dgf y f g f y

dy

y

N

xif x N y N

Y X X

X YX

Eq. 4.6Eq. 4.6

E4.1

See E4.1 in text for details

3

2

xy

,

3, 2x N

1 2 3 2dx

g x ydy

X

21

21

2

x

xf x e

X

where

1

1y x

dgf y f g

dy

Y X

For monotonic For monotonic function g(X)function g(X)

if X1 and X2 are Poisson with mean rates 1 and 2 respectively Z is Poisson with z = 1 + 2

(see E4.5 on p. 175)

1 2Z X X

if X is N(,) Y is N(a + b, a)Y aX b ( 0)a

if X is N(,) Y is N(a, a)

if X is LN(,) Y is LN(ln a + , )Y aX

( 0)a

Summary of Common ResultsSummary of Common Results

Summary of Common Results Summary of Common Results (Cont’d)(Cont’d)

if X1 and X2 are N(,) and N(,) respectively Z is N(z,z)where

if X1 and X2 are s.i. 12 = 0

1 1 2 2Z a X a X

2 2 2 21 1 2 2 1 2 12 1 22Z a a a a

if Xi = N(i,i) ; i = 1 to n Z is N(z,z)where1 1 2 2

... n n

Z a X a X

a X

1 1 2 2 ...Z n na a a

2 2Z i i

i

a correlation terms

1 1 2 2Z a a

E4.8E4.8

S = D + L + W

Column with capacity, R

D = 4.2 , D = 0.3 , D = 7%L = 6.5 , L = 0.8 , L = 12%W = 3.4 , W = 0.7 , W = 21%

S = total load = D + L + W

S = D + L + W = 14.1S = 0.32 + 0.82 + 0.72 =1.1

a) P(S > 18) = 1 –

= 1 – (3.55) = 0.000193

18 14.1

1.1

Assume D, L, W Assume D, L, W are s.i.are s.i.

E4.8 (Cont’d)E4.8 (Cont’d)P(failure) = P(R < S)

= P(R – S < 0)= P(X < 0)

R = N(R, RR)where R = 1.5S

= 1.5 x 14.1 = 21.15R = 0.15 R = N(21.15, 3.17)

X = -14.1 + 21.15 = 7.05X = 1.12 + 3.172 = 3.36

P(F) = = (-2.1) = 0.0180 7.05

3.36

X = R – SX = R – S

RR – design capacity – design capacity

1.5 – design safety factor, SF1.5 – design safety factor, SF

If the target is P(F) = 0.001 R = ? and assume R = 0.15

Recall:X = R - 14.1

X = (0.15R)2 + 1.12

= -1(0.001) = – 3.09

0.0225R2+1.21 = 20.8 – 2.95R + 0.105R

2

0.0825R2 – 2.95R +19.59 = 0

R = 8.812 or 26.9

2 2

14.1

0.15 1.1R

R

2

22 2 14.1(0.15 ) 1.1 4.56 0.324

3.09R

R R

Set: P(F) = = 0.001 2 2

14.1

0.15 1.1R

R

Since R should be larger than 21.5R = 26.9

Check R = 8.812

P(F) = = (3.09) 1.0

8.812 14.1

1.7197

W = weight of a truck = N(100, 20)We are interested in the total weight of 2

trucks

or

1.

Total weight = T = 2W

Normal with

T = 2W = 200

T = 2W = 40

Total weight = T = W1 + W2

Normal with T = 200

T =

=

=

if s.i. and 1 = 2 = 20

2 21 220 20 2

2 220 2020 2

2.

?

T

E4.10E4.10

Sand Footing

P

S

Sand Property – M Footing Property – B and I

PBIS

M

Assume P, B, I, and M are s.i. and log-normal with parameters P, B, I, M and P, B, I, M, respectively

c.o.v.

P B I

M

1.0 6.0 0.6

32.0

0.10 0

0.10 0.15

Central Limit TheoremCentral Limit Theorem

S will approach a normal distribution regardless the individual probability distribution of Xi if N is large enough

0.5

0.7

0.4

0.2

0.6

0.3

0.1

-1 1

P N = 1 S = X1

0.5

0.7

0.4

0.2

0.6

0.3

0.1

-2 0 2

P N = 2 S = X1 + X2

0.5

0.7

0.4

0.2

0.6

0.3

0.1

2 -4 -2 0 4 8 6 10

P N = 10 S = X1 + X2 +…+X10

P

0.5

0.7

0.4

0.2

0.6

0.3

0.1

18 16 14 12 10 8 6 4 2 0 -2 -4

N = 20 S = X1 + X2 +…+ X20

First order approximation:

2 2

X X X

X

X

X X

dgE g X g E X g

dX

dg dgVar g X Var X Var X

dX dX

g(X)

g’(X) (X – X)

X

g(X)

X X

Taylor Series ApproximationTaylor Series Approximation

X X X

2X

g(X) g( ) g ( ) X-

"...

2 X

gX

x - x

X X

dgg(X) g( ) X-

dXX

Var(X)

…

is known and are known valuesX

X

dgg X g

dX

2 2

X X X

X

X

X X

dgE g X g E X g

dX

dg dgVar g X Var X Var X

dX dX

Observe validity of linear approx Observe validity of linear approx depends on:depends on:

1) Function g is almost linear, i.e. small curvature

2) x is small, i.e. distribution of X is narrow

UsesUses

1. Easy calculations

2. Compare relative contributions of uncertainties – allocation of resource

3. Combine individual contributions of uncertainties

Reliability ComputationReliability ComputationSuppose R denotes resistance or capacity

S denotes load or demandSatisfactory Performance = {S < R}

PS = P(S < R) and Pf = 1 - PS

Case 1: If R, S are normal

where Z = S – R and Z = S2 + R

2

0 0

0 Z

Z

P S R P S R P Z

Case 2: If R, S are lognormal

where Z = S – R and Z = S2 + R

2

ln11 1 Z

Z

SP S R P P Z

R

Case 3: If R is discrete, S is continuous

i ii

S i R ii

P S R P S R R r P R r

F r P r

Example on Case 3

S = N(5, 1)

5 5 6 6 7 7P S R P S P R P S P R P S P R

5 5 6 5 7 50.1 0.3 0.6

1 1 1

0 0.1 1 0.3 2 0.6

0.5 0.1 0.84 0.3 0.98 0.6

0.889

r

P(R = r)

5 6 7

0.1

0.3

0.6

Reliability – Based DesignReliability – Based Design

Observe for Case 1 in which R and S are both Normal

2 2

R SS

S R

P

If PS Reliability

= Reliability index = -1(PS)

Design Design RR = = SS + + SS22 + + RR

22

Example:S = N(5, 2)R = N(R, 1)

R = ?

Require Pf = 0.001 or PS = 0.999 = -1(0.999) = 3.1

Design R = 5 + 3.122 + 12 = 11.93