1. Why need probabilistic approach? Rain probability How does that affect our behaviour? ?

2. Uncertainties in Engineering Natural Hazards Natural Hazards Material Properties Material Properties Design Models Design Models Construction Errors Construction Errors

3. Absolute Safety Not Guaranteed Engineers need to: model, analyze, update uncertainties model, analyze, update uncertainties evaluate probability of failure evaluate probability of failure

4. Questions What is acceptable failure probability? -stadium vs shed

5. Questions Should one want to be conservative if a perfectly safe system is possible? -overbooking in airlines -parking permits

6. Questions Should one minimize risk if money is not a problem? -system consideration – e.g.dam

7. Trade-off Decision Analysis Risk vs. consequence Risk vs. consequence System risk System risk

8. Formal analysis of uncertainties and probability Not all problems can be solved by analysis of data Not all problems can be solved by analysis of data Set Theory Set Theory Sample space: collection of all possibilities Sample space: collection of all possibilities Sample point: each possibility Sample point: each possibility Event: subset of sample space Event: subset of sample space Probability Theory Probability Theory

9. Union: either E 1 or E 2 occur E 1 ∪ E 2 Intersection: both E 1 and E 2 occur E 1 ∩ E 2 or E 1 E 2

10. Examples A B No communication between A and B = E 1 E 2 C B A ∪ No communication between A and B = E 3 ∪ E 1 E 2 1 2 E 1 = road 1 closed E 2 = road 2 closed E 3 = road 3 closed 1 2 3

11. Example - pair of footings 1 2 E 1 = 1 settles Ē 1 = 1 does not settle E 2 = 2 settles Ē 2 = 2 does not settle ∪ Settlement occurs = E 1 ∪ E 2 Ē ∪ Ē Tilting occurs = E 1 Ē 2 ∪ Ē 1 E 2

12. de Morgan’s rule E 1 = pipe 1 breaks E 2 = pipe 2 breaks 1 2 Water Supply E = failure in water supply = E 1 ∪ E 2 no failure in water supply = Ē = E 1 ∪ E 2

13. Event of “no failure” Extension to n events

14. de Morgan #2

15. Basis of Probability Estimation a) Subjective assumption e.g. P(Q) = 1/2 b) Relative frequency e.g. P(Q)=502/1000 c) Bayesian (a)+(b) judgment + limited observation

16. Probability of Union in general E1E1 E2E2

17. Using de Morgan’s rule P (intersection) conditional probability

18. or

19. Statistical independence if E 1 and E 2 are s.i. or s.i.

20. Example: E 2 = flood in 廣西 on June E 3 = flood in 哈爾濱 on June E 1 = flood in 廣東 on June P(E 1 ) = 0.1; P(E 2 )=0.1; P(E 3 ) = 0.1 E 1 and E 2 are not s.i. E 1 and E 3 are s.i.

21. if E 1 and E 2 are s.i. if all are s.i.

22. s.i. and m.e. (mutually exclusive) if E 1 and E 2 are m.e. if E 1 and E 2 are s.i.

23. B C A 2 1 3 P(E 1 )=2/5 P(E 2 )=3/4 P(E 3 )=2/3 P(E 3 |E 2 )=4/5 P(E 1 |E 2 E 3 )=1/2 a)P(go from A to B through C) E 1 : ① is open P 2.15

24. b) P(go from A to B) 1/23/5

25. T.O.T (Theorem of Total Probabilities) Bayes theorem P(A) = P(A|E 1 )P(E 1 )+P(A|E 2 )P(E 2 )+…+P(A|E n )P(E n ) E i ’s are m.e. and c.e.

26. good enough for construction positive 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

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

28. After another successful independent test, P(G)?

29. What if the two tests were performed at the same time? 0.7 0.3 0.7 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

30. Random variables A device to: a) formalize description of event b) facilitate computation of probability

31. PMF P X (x) F X (x) PDF f X (x) F X (x) CDF

32. Main descriptors of R.V. The PMF or PDF completely define the r.v. Descriptors give partial information about the r.v.

33. Mean value Define  = E(X) = expected value of X or mean value of X a measure of central tendency

34. Measure of spread Standard deviation  X  X dimensionless % range

35. Expected value of function recall

36. Recall After some algebra,

37. f X (x) x Normal distribution  = 5 X : N ( ,  ) N (5, 2)

38. Effect of varying parameters (  &  ) f X (x) x A B C   for C  for B

39. S: N (0,1) Standard normal distribution f X (x) x

40. a

41. Page 380 Table of Standard Normal Probability

42. Example: retaining wall x F Suppose X = N(200,30)

43. If the retaining wall is designed such that the reliability against sliding is 99%, How much friction should be provided? 2.33

44. Lognormal distribution Parameter   f X (x) x

45. Parameters  for   0.3,

46. Probability for Log-normal distribution If a is x m, then  is not needed.

47. Other distributions Exponential distribution Triangular distribution Uniform distribution Rayleigh distribution p.224-225: table of common distribution

48. Exponential distribution x f X (x) x  0

49. Beta distribution x f X (x) q = 2.0 ; r = 6.0 a = 2.0 b = 12 probability

50. Standard beta PDF q = 1.0 ; r = 4.0 q = r = 3.0 q = 4.0 ; r = 2.0 q = r = 1.0 x f X (x) (a = 0, b = 1)

51. Bernoulli sequence Discrete repeated trials 2 outcomes for each trial s.i. between trials Probability of occurrence same for all trials S F p = probability of a success

52. S F x = number of success p = probability of a success P ( x success in n trials) = P ( X = x | n, p) Binomial distribution

53. Examples Number of flooded years Number of failed specimens Number of polluted days

54. Example: Given: probability of flood each year = 0.1 Over a 5 year period P ( at most 1 flood year) = P (X =0) + P(X=1) = 0.9 5 + 0.328 = 0.919

55. P (flooding during 5 years) = P (X  1) = 1 – P( X = 0) = 1- 0.9 5 = 0.41

56. For Bernoulli sequence Model No. of success  binomial distribution Time to first success  geometric distribution E(T) =1/p = return period

57. 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

58. P (1 st exceedence of 68.6 m/s in 100th year) = 0.99 99  0.01 = 0.0037 P (no exceedence of 68.6 m/s within a service life of 100 years) = 0.99 100 = 0.366 P (no exceedence of 68.6 m/s within the return period of design) = 0.366

59. 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 (1 st exceedence of 70.6 m/s in 100th year) = 0.995 99  0.005 = 0.003

60. P (no exceedence of 70.6 m/s within return period of design) = 0.995 200 = 0.367 P (no exceedence of 70.6 m/s within a service life of 100 years) = 0.995 100 = 0.606 > 0.366

61. 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

62. N (72,8) Example V 100 0.01 Design for return period of 100 years: p = 1/100 = 0.01  V 100 = 90.6 mph Annual max wind velocity Frequency

63. Suppose we design it for 100 mph, what is the corresponding return period? T  4300 years Alternative design criteria 1

64. P f = P (exceedence within 100 years) = 1- P (no exceedence within 100 years) =1- (1-0.000233) 100 = 0.023 Probability of failure

65. P ( x occurrences in n trials) = x = 0, 1, 2, … Poisson distribution

66. 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)

67. (a) P( X  1 in 15 miles ) = ? No. of service stations

68. (b) P( none of the next 3 stations have gasoline) binomial No. of stations with gasoline

69. (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) No. of station in 15 miles binomial Poisson

70. xP( S| X = x )P( X = x )P( S| X = x ) P( X = x ) 01e -1.5 = 0.2230.223 10.21.5 e -1.5 = 0.3350.067 20.2 2 1.5 2 /2! e -1.5 = 0.2510.010 30.2 3 1.5 3 /3! e -1.5 = 0.1260.001 40.2 4 1.5 4 /4! e -1.5 = 0.0470.00007 Total = 0.301

71. Alternative approach Mean rate of service station = 0.1 per mile Probability of gas at a station = 0.8  Mean rate of “wet” station = 0.1  0.8 = 0.08 per mile Occurrence of “wet” station is also Poisson  P (S) = P ( no wet station in 15 mile)

72. Time to next occurrence in Poisson process Time to next occurrence = T is a continuous r.v. = P (X = 0 in time t) Recall for an exponential distribution

73.  T follows an exponential distribution with parameter =  E(T) =1/ If = 0.1 per year, E(T) = 10 years

74. Bernoulli Sequence Poisson Process IntervalDiscreteContinuous No. of occurrenceBinomialPoisson Time to next occurrenceGeometricExponential Time to kth occurrenceNegative binomialGamma Comparison of two families of occurrence models

75. Significance of correlation coefficient  = +1.0  = -1.0 y: strength x: Length Glass y: elongation x: Length Steel

76.  = 0 0<  <1.0 y: ID No x: height y: weight x: height

77. Functions of Random Variable (R.V.) In general, Y = g(X) Y = g(X 1, X 2,…, X n ) If we know distribution of X  distribution of Y? M = 4X + 10 X 5 22 M 1) X b  b = Xtan  2) cost of delay = aX 2 3) where X – length of delay

78. Consider M = 4X + 10 Observation: the distribution of y depends on (1) Distribution of X (2) g(X) X 654 0.6 0.2 M 343026 0.6 0.2 X654 M343026 wider distribution

79. YXX X Y X Eq. 4.6 E4.1 See E4.1 in text for details X X where Y X For monotonic function g(X)

80. if X 1 and X 2 are Poisson with mean rates 1 and 2 respectively  Z is Poisson with z = 1 + 2 (see E4.5 on p. 175) if X is N( ,  )  Y is N(a  + b, a  ) if X is N( ,  )  Y is N(a , a  ) if X is LN(,  )  Y is LN(ln a +,  ) Summary of Common Results

81. Summary of Common Results (Cont’d) if X 1 and X 2 are N(  ,   ) and N(  ,   ) respectively  Z is N(  z,  z ) where if X 1 and X 2 are s.i.   12 = 0 if X i = N(  i,  i ) ; i = 1 to n  Z is N(  z,  z ) where

82. E4.8 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 = 14.1  S =  0.3 2 + 0.8 2 + 0.7 2 =1.1 a) P(S > 18) = 1 –  = 1 –  (3.55) = 0.000193 Assume D, L, W are s.i.

83. E4.8 (Cont’d) P(failure) = P(R < S) = P(R – S < 0) = P(X < 0) R = N(  R,  R  R ) where  R = 1.5  S = 1.5 x 14.1 = 21.15  R = 0.15  R = N(21.15, 3.17)  X = -14.1 + 21.15 = 7.05  X =  1.1 2 + 3.17 2 = 3.36 P(F) =  =  (-2.1) = 0.018 X = R – S  R – design capacity 1.5 – design safety factor, SF

84. If the target is P(F) = 0.001   R = ? and assume  R = 0.15 Recall:  X =  R - 14.1  X =  (0.15  R ) 2 + 1.1 2  =  -1 (0.001) = – 3.09   0.0225  R 2 +1.21 = 20.8 – 2.95  R + 0.105  R 2   0.0825  R 2 – 2.95  R +19.59 = 0  R = 8.812 or 26.9 Set: P(F) =  = 0.001 Since  R should be larger than 21.5  R = 26.9 Check  R = 8.812 P(F) =  =  (3.09)  1.0

85. 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 = 2  W = 200  T = 2  W = 40 Total weight = T = W 1 + W 2  Normal with  T = 200  T = = if s.i. and  1 =  2 = 20 2. ? T

86. E4.10 Sand Footing P S Sand Property – MFooting Property – B and I Assume P, B, I, and M are s.i. and log-normal with parameters P, B, I, M and  P,  B,  I,  M, respectively

87. Central Limit Theorem S will approach a normal distribution regardless the individual probability distribution of X i if N is large enough

88. First order approximation: g(  X ) g’(  X )  (X –  X ) X g(X) XX X Taylor Series Approximation  x -  x Var(X) …

89. Observe validity of linear approx depends on: 1) Function g is almost linear, i.e. small curvature 2)  x is small, i.e. distribution of X is narrow

90. Uses 1. Easy calculations 2. Compare relative contributions of uncertainties – allocation of resource 3. Combine individual contributions of uncertainties

91. Reliability Computation Suppose R denotes resistance or capacity S denotes load or demand Satisfactory Performance = {S < R} P S = P(S < R) and P f = 1 - P S Case 1: If R, S are normal where  Z =  S –  R and  Z =  S 2 +  R 2 Case 2: If R, S are lognormal where Z = S – R and  Z =  S 2 +  R 2

92. Case 3: If R is discrete, S is continuous Example on Case 3 S = N(5, 1) r P(R = r) 567 0.1 0.3 0.6

93. Reliability – Based Design Observe for Case 1 in which R and S are both Normal If    P S   Reliability    = Reliability index =  -1 (P S )  Design  R =  S +  S 2 +  R 2 

94. Example: S = N(5, 2) R = N(  R, 1)  R = ? Require P f = 0.001 or P S = 0.999  =  -1 (0.999) = 3.1  Design  R = 5 + 3.1  2 2 + 1 2 = 11.93