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Lecture 10

# Lecture 10 - Lecture 10 Root Finding Methods and Interfaces...

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Lecture 10: Root Finding Methods and Interfaces Problem: Estimate alpha and beta for each wind turbine. Here is one student's approach that uses a stepper method to find the value of alpha given the values of both the mean and the median. This method is not considered good although the Java code works just fine. Not only is the stepper inefficient, but the ratio mean/median is not a sensitive barometer for finding alpha. For your projects it is hoped you will use a better method. The method is invoked as: MyFunctions.findAlpha(mean, median) If it is defined in the class MyFunctions Before using the method you would need to calculate both the mean and median for each array of 1000 wind speed measurements. The method would also need to be placed inside a Java class. What's Wrong: This approach uses the less sensitive mean/median ratio: r 1 = μ median = ! 1 + 1 " # \$ % & ( ln(2) 1/ " (and stepper methods though simple, are notoriously inefficient.) Notice β cancels out when we compute this ratio using the first two of the following identities. The Weibull mean is given by: μ = ! " 1 + 1 # \$ % & ( ) The Weibull median is given by: Median = ! ln 2 ( ) 1/ " The Weibull variance is given by: ! 2 = " 2 # 1 + 2 \$ % & ( ) * + # 1 + 1 \$ % & ( ) * 2 % & ( ) * The most serious problem is that around the physically meaningful value of α = 2, the ratio r 1 is not very sensitive.

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You can see this lack of sensitivity in the following graph. Maple Code: > rhs := GAMMA(1 + 1/alpha)/ln(2)^(1/alpha); > y := 'y': plot( rhs, alpha = 0.25..5.001, y = 0..8, thickness=2, font = [TIMES,BOLD,18], ytickmarks = 10, gridlines=true); For larger values of alpha, this method will not be very sensitive as the curve is very flat! A table can summarize the same info. Example: Suppose you find that the mean divided by the median is 1.2 What would be a good estimate for alpha? Look's like 1.35 should be close. Alpha RHS [1., 1.442695041] [1.1, 1.346455653] [1.2, 1.276662828] [1.3, 1.224376348] [1.4, 1.184175973] [1.5, 1.152607899] [1.6, 1.127380096] [1.7, 1.106919306] [1.8, 1.090114490] [1.9, 1.076161730] Find α if the ratio mean/median = 1.2
[2.0, 1.064467020] [2.1, 1.054583356] [2.2, 1.046168919] [2.3, 1.038958573] [2.4, 1.032744044] [2.5, 1.027359844] [2.6, 1.022673135] [2.7, 1.018576281] [2.8, 1.014981328] [2.9, 1.011815848] [3.0, 1.009019767] You might try the following stepper method. Parameter Finding Method 1 : Estimate alpha by finding the root of the equation: ! 1 + 1 " # \$ % & ( ln(2) 1/ " ) μ median = 0 Again, this is not a good method because the ratio is not sensitive for the range in which we expect to find α and will not earn full marks . Warning: The following stepper method is not recommended, but the Java is good. STEPPER METHOD The fact that the red curve falls monotonically, suggests the following strategy. Start at a small value of alpha and progressively move to the right in small steps of say δ = 0.0001 until the error ratio – mean/median changes sign from positive to negative. The value where this occurs provides an estimate for alpha.

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Nevertheless, here is the findAlpha method using this rather insensitive approach. public static double findAlpha( double mean, double median ){ double arg, den, num; //1. Calculate ratio(1) by dividing the mean by the median.
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