Homework3EngineeringRiskAnalysis

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Apr 3, 2024

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Problem 1: The eye of a hurricane is 10 km from the east coast of Florida and is moving toward the land at an uncertain angle Θ (see figure below). Available information indicates that the angle is greater than π/6 and smaller than π/3. Assume the angle Θ is equally likely to be anywhere within this range. Also assume the eye of the hurricane moves in a straight line. (a) Determine the range of the coordinate X where the eye of the hurricane will hit the land. Determination of the range of the coordinate X where the eye of the hurricane will hit the land by using trigonometry. Let's denote: • D: the distance from the eye of the hurricane to the coast. • Θ: the angle between the path of the hurricane and the line perpendicular to the coast. • X: the horizontal coordinate along the coast where the hurricane makes landfall. We know that: ! " < Θ < ! # We will use the trigonometric relationship: tan(Θ) = X / D Þ X = D * tan(Θ) Finally, we will calculate the range of X by plugging in the minimum and maximum values for Θ: - ࠵? $%& = 10 × tan ( ! " ) = 10 × √# # ≈ 5,77 - ࠵? $() = 10 × tan ( ! # ) = 10 × √3 ≈ 17,32 To conclude, we can say that the range of the coordinate X where the eye of the hurricane will hit the land is approximately 5,77 km to 17,32 km (along the east coast of Florida). (b) Using simple geometric principles and the uniform likelihood of Θ within its range, derive an expression for the CDF, ࠵? ࠵? (x), of X. Make a sketch of the CDF. Since it's a uniform distribution, we can find its value by calculating the length of the range: f(Θ) = + ! " , ! # = + ! # = " ! Now, we can find the CDF of X: FX(x), using the probability distribution function f(Θ): FX(x) = P (X ≤ x) = P (D × tan(Θ) ≤ x) Our goal is to find the probability that X is less than or equal to x. Let's solve for Θ: tan(Θ) ≤ ) - ⟹ Θ ≤ Arctan ( ) - )

Now, we can find the probability: FX(x) = ∫ ࠵?(Θ) dΘ ./0123 ( $ % ) ! # ⟹ FX(x) = ∫ " ! dΘ ./0123 ( $ % ) ! # Now, integrate with respect to Θ: FX(x) = " ! [Arctan 6 ) - 7 − ! " ] To conclude, the CDF of X is: FX(x) = " ! [Arctan 6 ) - 7 − ! " ] for ! " ≤ Arctan ( ) - ) ≤ ! # Sketch of CDF: c) Derive an expression for the PDF, fX (x), of the coordinate X. Make a sketch of the PDF. Note that ࠵? ࠵?࠵? ࠵?࠵?࠵? ,࠵? ( ࠵? ࠵? ) = ࠵? ࠵? (࠵?;( ࠵? ࠵? ) ࠵? ) . We know that: FX(x) = " ! [Arctan 6 ) - 7 − ! " ] for ! " ≤ Arctan ( ) - ) ≤ ! # Now, we'll differentiate this CDF with respect to x to find the PDF: fX(x) = ࠵? ࠵?࠵? [FX(x)] ⟹ fX(x) = ࠵? ࠵?࠵? [ " ! [Arctan 6 ) - 7 − ! " ] ]

Now, we can use the given derivative of arctan: ࠵? ࠵?࠵? Arctan ( ࠵? ࠵? ) = ࠵? ࠵? (࠵?;( ࠵? ࠵? ) ࠵? ) . ⟹ ࠵? ࠵?࠵? (Arctan ( ) - )) = + -(+;( $ % ) ) Now, differentiate the CDF: fX(x) = " ! × + -(+;( $ % ) ) ⟹ fX(x) = 6 10࠵?(1 + ( ࠵? 10 ) < Sketch: (d) Determine the median, mean and standard deviation of X. You may use MATLAB for numerical integration. • The mean of X, denoted as μ, can be calculated using the PDF fX(x) as follows: μ = ∫(x × fX(x)) dx over the en\re range of X • Median: The median is the value of X for which the cumula\ve distribu\on func\on (CDF) FX(x) is equal to 0.5: FX (median) = 0.5 • Standard Devia\on: The standard devia\on of X, denoted as σ, can be calculated as follows: σ = G[∫((࠵? − µ) < × fX(x)) dx] over the en\re range of X

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Using MATLAB, we get: • μ ≈10.5 km • Median ≈ 10 km • σ ≈ 3.2 km (e) Determine the probability of the event E = {10 < X ≤ 15} (km). P(E) = =>(+?), =>(+@) =>(+A,#), =>(++,?) With Matlab we find: P(E) = 0,37 Problem 2: The seismic fragility of a building, denoted g(x), is defined as the conditional failure probability of the building for a given intensity measure of the ground acceleration. Here the peak ground acceleration (PGA) is selected as the intensity measure, and g(x) is g(x) = P (Failure | PGA = x)

For a particular class of buildings, the fragility function is given by ࠵?(࠵?) = K ࠵? , ࠵? ≤ ࠵?. ࠵? ࠵?. ࠵?(࠵? − ࠵?. ࠵?) , ࠵?. ࠵? < ࠵? < ࠵?. ࠵? ࠵? , ࠵? ≥ ࠵?. ࠵? where x is measured in units of gravity acceleration. Suppose the PGA of an earthquake has the exponential distribution with mean 0.05 units of gravity acceleration. (a) Determine the probability of failure of the building during an earthquake. • μ (PGA of the earthquake) = 0,05 ⟹ + C = 0,05 ⟹ ࠵? = 20 So, we have f(x) = 20 ࠵? ,<@) • P (failure of the building) = ∫ ࠵?(࠵?) ∙ ࠵?(࠵?)࠵?࠵? D @ = ∫ 0 ∙ ࠵?(࠵?)࠵?࠵? @,+ @ + ∫ 2,5(࠵? − 0,1) ∙ ࠵?(࠵?)࠵?࠵? @,? @,+ + ∫ ࠵?(࠵?)࠵?࠵? D @,? = ∫ 50࠵? ࠵? ,<@) − 5࠵? ,<@ ࠵?࠵? @,? @,+ + ∫ 20࠵? ,<@) ࠵?࠵? D @,? = # E ࠵? ,< - ++ E ࠵? ,+@ - + F ࠵? ,< + + F ࠵? ,+@ + ࠵? ,+@ = 0,017 (b) If the building is known to have failed, what is the probability density function of the PGA of the earthquake that caused the failure. f(x|Failure) = G()) ∙I()) J(G(%KLMN) . We have 3 possibilites • If x ≤ 0,1: f(x|Failure) = G())∙@ J(G(%KLMN) = 0 • If 0,1 < x < 0,5: f(x|Failure) = <@ N *)+$ ∙<,? (),@,+) @,@+A • If x ≥ 0,5 : f(x|Failure) = <@ N *)+$ @,@+A (c) If the building survived an earthquake, what is the probability density function of the PGA of that earthquake. f(x|Survived) = =(O) ∙(+, P(O)) +, Q(=2RST/U) . We have 3 possibilites • If x ≤ 0,1: f(x|Survived) = <@ U *)+, @,VE# = 0 • If 0,1 < x < 0,5: f(x|Survived) = <@ U *)+, ∙(+,<,? (O,@,+)) @,VE# • If x ≥ 0,5 : f(x|Survived) = =(O)∙(+,+) @,VE# = 0 (d) Plot and compare the probability density functions of the PGA for three earthquake ground motions: (a) a randomly selected earthquake, (b) an earthquake that caused the building to fail, (c) an earthquake during which the building survived. Comment on the differences between the

distributions. You may use the following integral: [ ࠵? ࠵?࠵?࠵?(−࠵?࠵?)࠵?࠵? = − ࠵? + ࠵?࠵? ࠵? ࠵? ࠵?࠵?࠵? (−࠵?࠵?) Plot : On this plot we can notice that the buildings always survived when x ≤ 0,1 and on the contrary they always failed when x ≥ 0,5 Finally we see that the random earthquake curve is quit similiar to the curve where the building survi ved.

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