HW5_solution

CEE 373 Homework #5 Due date: October 6 at 5pm Total: 100 points 1. (20 points) The following questions pertain to a probabilistic model for the ran- dom variable, X , which represents the force acting on a structural column. Suppose X is assumed to have a Normal (i.e., Gaussian) distribution, param- eterized by a mean of µ X = 15 . 0kips and a standard deviation of σ X = 1 . 25 kips. (a) Using the Standard Normal Table, find the probability that X will be be- tween 14 and 16 kips. (b) Using the Standard Normal Table, find the probability that X will be greater 18 kips. (c) Suppose the steel manufacturer you are considering tells you they can fabri- cate columns with the capacity to carry loads (or forces) within 95% of the average forces acting on this column. What is the range in kips for the load capacity for these columns? (d) Now, suppose you need to design the column such that its probability of failure, under random loading, is less than 5 × 10 − 3 . The load at which the column fails is called its buckling load , or x fail . Using the Standard Normal Table, what is the minimum buckling load the column can have, and still meet this design requirement? Hint: You will want P ( X ≥ x fail ) = 5 × 10 − 3 . (e) ( Computer Needed. ) Confirm your answers for parts (a) - (d) using Python (or your other favorite coding language). If you are using Python, you can reference the functions introduced in lecture, which are also documented at https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats. norm.html#scipy.stats.norm . Solution: (a) First, express the problem in terms of the equivalent standard normal random variable, Z . We have that Z = X − µ X σ X and thus if X ∈ [14 , 16] this is equivalent to Z being in the range Z ∈ 14 − 15 1 . 25 , 16 − 15 1 . 25 = [ − 0 . 8 , 0 . 8] Page 1 of 12

CEE 373 Homework #5 The probability of this is P [ − 0 . 8 < Z ≤ 0 . 8] =Φ(0 . 8) − Φ( − 0 . 8) =0 . 7881 − 0 . 2119 =0 . 5762 (b) Similarly, we have that the range X ∈ [18 , ∞ ) is equivalent in the standard normal variable Z to Z ∈ 18 − 15 1 . 25 , ∞ = [2 . 4 , ∞ ) Consequently we have that P [2 . 4 < Z < ∞ ) =Φ( ∞ ) − Φ(2 . 4) =1 − 0 . 9918 =0 . 0082 (c) The average force acting on this column is µ X = 15 . 0 kips. The question asks for loads within 95% of µ X . Knowing that X is normally distributed, we know that 95% of the probability falls within 2 standard deviations σ X of the mean for Normal distributions (because of the 68-95-99.7 rule). Therefore, the range of loads the manufacturers’ columns can carry is [ µ X − 2 σ X , µ X + 2 σ X ] = [15 − 2(1 . 25) , 15 + 2(1 . 25)] = [12 . 5 kips, 17 . 5 kips ] (d) First, look for the value of z such that P [ Z > z ] = 5 × 10 − 3 Equivalently, P [ Z ≤ z ] = 1 − 5 × 10 − 3 = 0 . 995 From the tables, this value is approximately z = 2 . 575. The corresponding value of x is then x = µ X + σ X z = 15 + 1 . 25 × 2 . 575 = 18 . 21 Thus, we have that P [ X ≥ 18 . 21] = 5 × 10 − 3 . (e) The solution code can be found here: https://colab.research.google. com/drive/1fMi4hcGmK_Wa57fJCLkgoXrG5sRkURwf?usp=sharing . Page 2 of 12

CEE 373 Homework #5 • σ lnR = r ln ( σ 2 R µ 2 R + 1 ) = 0.3246 • µ lnR = ln( µ R ) − 1 2 σ 2 lnR = 6.7497 And for the uniform distribution, you need to calculate a and b using the equations for µ X and σ X provided in our common distributions handout. The resulting parameters for the uniform distribution are: • a = µ R − σ R √ 12 2 • b = 2 µ R − a A plot of all three distributions using these parameters is shown below: (b) To calculate the exceedance probabilities, you can use the equations for the CDF, F R ( r ) for each distribution provided in the handout. P ( R > r ) = 1 − F R ( r ). The resulting exceedance probabilities are: Distribution P ( R > r ) = 0 . 5 P ( R > r ) = 0 . 1 P ( R > r ) = 0 . 01 P ( R > r ) = 0 . 001 Normal 900 1284 . 5 1597 . 9 1827 . 1 Lognormal 853 . 8 1294 . 3 1816 . 8 2328 . 0 Uniform 900 1315 . 7 1409 . 2 1418 . 6 (c) As the probability of exceedance gets smaller, the choice of distribution type has a greater impact. The mean and standard deviation are good descriptors of the behavior of a probability distribution near its central values. The distributions thus give similar results in this region (i.e., for a 0.5 probability of exceedance). But as we consider outcomes in the tails of the distributions, the differences between the distributions become much more pronounced, and the choice of distribution model becomes much more significant. Page 4 of 12

CEE 373 Homework #5 3. (20 points) You are installing active dampers in a new high-rise building, to limit the deflections of the building during high winds. Your job is to select the number of dampers to install in the building. The building’s damping system will fail if at least half of its dampers fail. Assume that all dampers have an identical probability of failure, p f , in a wind- storm and that the failure of one damper is independent of the failure of any other one. Hint: a damper has a binary outcome when exposed to high winds: failure or no failure. This will help you select which distribution to use. Note: To see what a damper is, visit this video: https: // youtu. be/ 9FJ7oFtqxpQ? si= y4dJzmLmbV9QeEJh (a) If p f = 0 . 1, would a two-damper system or a four-damper system be safer? You can consider a safe system to be a system that has a lower probability of failure. (b) If p f = 0 . 5, is the two-damper system or the four-damper system safer? (c) ( Computer recommended. ) Now let’s imagine p f = 0 . 5 still, but you are con- sidering installing 20-dampers (this is a big high-rise). What is the probabil- ity of failure of the system in this case? If you approximate this probability of failure with the normal distribution, do you get a similar value? (d) What assumptions did you make to calculate the probabilities of failure in parts (a) - (c)? Solution: Note: if you used a different definition of ”at least” for this problem, we will accept both interpretations (either greater than or greater than or equal to). (a) We are looking for the probability of failure for a damper system. Each damper can either fail or not fail (a single experiment of the Bernoulli Distribution). The combined failures of all dampers is a Bernoulli sequence (multiple exper- iments or dampers). We are interested in a certain number of dampers failing in order for the system to fail. In order for a two-damper system to fail, at least half of its dampers must fail (e.g., 2 dampers). Therefore, P ( fail ) = P ( x = 2). In this case, the parameters for the binomial distribution are: • n = 2 dampers • p = p f = 0 . 1 • x = 2 dampers To calculate P ( fail ) for the two-damper system, we plug these parameters into the binomial PDF: p X ( x = 2) = 2 2 p 2 (1 − 0 . 1) 2 − 2 = 0 . 01 Page 5 of 12

CEE 373 Homework #5 • µ = np = 10 dampers • σ X = p np (1 − p ) = 2 . 236 In this case, you are looking for P (10 < X ≤ 20) The probability of failure using the normal CDF is thus (evaluated on com- puter): P ( fail ) = F X ( x = 20) − F X ( x = 10) = 0 . 5 Same as above, because the upper bound of this distribution has negligible probability (i.e., when x > 20), this is really similar to: P ( fail ) = 1 − F X ( x = 10) = 0 . 4999999 (The value of x here can be flexible inside of 10 to 11, because this is just approximation.) The probabilities we calculated are shown below: The binomial and normal values are not similar. If there were more dampers, these would become more similar. (d) The results above rely on the assumption of independent damper failures. This assumption may not be valid (e.g., the dampers are manufactured and serviced using similar procedures, and they may all fail if the power fails during a windstorm, etc.). These potential common cause failures may cause us to question our results. A less critical assumption is that the system failure is a single trial at a single instant in time. Four dampers might provide extra safety because the failure of the first damper may serve as a warning of trouble for the other dampers, allowing the operators to perform inspections on the other dampers. Page 7 of 12

CEE 373 Homework #5 4. (20 points) Let X = the time between two arrivals of a car at a stoplight. The average time between intervals based on another stoplight in town is 2.5 minutes. (a) What is the standard deviation for the average time between two arrivals? (b) What is the probability that the arrival time is between 1 and 2 minutes? (c) What is the probability that the arrival time is less than the average? (d) What is the median arrival time? Is this greater or less than the mean arrival time? Why? Solution: (a) Given that we are interested in the time between events, this should indicate to us that X has an exponential PDF. Therefore, because the mean and standard deviation for exponential PDF’s are the same, we get sigma X = µ X = 2 . 5 minutes. Another way to solve this is solve for λ . Based on the equation for µ for a exponential random variable, λ = 1 µ X = 0 . 4. Plugging this value in for λ , we get σ X = q 1 λ 2 = 2 . 5 minutes. (b) To calculate P (1 ≤ x ≤ 2) we can use the CDF of the exponential PDF: P (1 ≤ x ≤ 2) = F X (2) − F X (1) = (1 − e − 0 . 4 ∗ 2 ) − (1 − e − 0 . 4 ∗ 1 ) = 0 . 2209 (c) To calculate P ( x ≤ µ X ), we can still use the CDF: P ( x ≤ µ X ) = F X ( µ X ) = (1 − e − 0 . 4 ∗ 2 . 5 ) = 0 . 632 (d) The median arrival time, ˜ x , is the arrival with 50% of the probability less than this value or F X (˜ x ) = 0 . 5. Therefore: 1 − e − 0 . 4 ∗ ˜ x = 0 . 5 Solving for ˜ x , we get ln (1 − 0 . 5) − 0 . 4 = 1 . 732 minutes. This is less than the mean because the exponential distribution is skewed right. Therefore the mean is pulled towards the tail, which is on the right of the distribution, so it’s greater than the median. In addition, F X ( µ X ) > F X (˜ x ), indicating there is more area to the left of µ X than ˜ x , indicating µ X is larger. 5. (20 points) [ Computer required. ] This problem will ask you about a series of dis- tributions we covered in lecture. You will use a combination of code you developed Page 8 of 12

CEE 373 Homework #5 Note: if you used a different definition of ”at least” for this problem, we will accept both interpretations (either greater than or greater than or equal to). (a) If we decide earthquake forces as x , X ∼ Gamma ( α, β ). In the question, µ = α β = 2000 σ = α β 2 = 800 2 From these values, α = 6 . 25 and β = 0 . 003125. Therefore, the gamma distribution is: f ( x ) = β α Γ( α ) x α − 1 e − βx = 0 . 003125 6 . 25 Γ(6 . 25) x 6 . 25 − 1 e − 0 . 003125 x Plots of PDF and CDF: The probability can be calculated like this. P ( x < 3000) = F (3000) = 0 . 888 (b) p = P ( x < 3000) = F (3000) = 0 . 888 , q = 1 − p = 0 . 112 Three buildings will be damaged: P ( three buildings damaged ) = 5 C 3 ( p 3 ) q 2 = 0 . 088 Four buildings will be damaged: P ( four buildings damaged ) = 5 C 4 ( p 4 ) q 1 = 0 . 348 Five buildings will be damaged: P ( five buildings damaged ) = 5 C 5 ( p 5 ) q 0 = 0 . 552 At least three buildings will be damaged: P ( at least three buildings damaged ) = 0 . 988 Page 10 of 12

CEE 373 Homework #5 (c) This is an exponential distribution (g(t)) and its CDF (G(t)). g ( t ) = 0 . 1 e − 0 . 1 t G ( t ) = 1 − e − 0 . 1 t Here, g(t) is the exponential distribution (T ∼ Exponential(0.1)). The two probability can be calculated using G(x). P ( no x > 3000 during 25 years ) = 1 − G (25) = 0 . 082 (1) (d) If we set n as years when the building will collapse, the geometric distribution (N ∼ Geom(1/40)) can be used: p = 1 40 q = 39 40 f ( n ) = pq n − 1 Therefore, the probability that the building will collapse after 40 years is: P ( n = 40 yrs. ) = ( 1 40 )( 39 40 ) 40 − 1 = 0 . 009 Plots of PMF and CDF: (e) When calculating parameters( α , β ): α = 2000 1000 3 = 8 β = 10 The replacement cost ratio follows Beta distribution ( C ∼ Beta ( α, β )). f ( x ) = Γ( α + β ) Γ( α )Γ( β ) x α − 1 (1 − x ) β − 1 = Γ(8 + 10) Γ(8)Γ(10) x 8 − 1 (1 − x ) 10 − 1 Page 11 of 12