412 CHAP. 8 NUMERICAL OPTIMIZATION. The Fibonacci Search

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1 412 CHAP 8 NUMERICAL OPTIMIZATION The Fibonacci Search Fibonacci Search In the golden ratio search two function evaluations are made at the first iteration and then only one function evaluation is made for each subsequent iteration The value of

2 SEC 81 MINIMIZATION OF A FUNCTION OF ONE VARIABLE 413 r remains constant on each subinterval and the search is terminated at the kth subinterval, provided that b k a k or f b k ) f a k ) satisfies predefined tolerances The Fibonacci search method differs from the golden ratio method in that the value of r is not constant on each subinterval Additionally, the number of subintervals iterations) is predetermined and based on the specified tolerances The Fibonacci search is based on the sequence of Fibonacci numbers {F k } k=0 defined by the equations 5) 6) F 0 = 0, F 1 = 1 F n = F n 1 + F n 2 for n = 2, 3, Thus the Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, Assume we are given a function f x) that is unimodal on the interval [a 0, b 0 ]As in the golden ratio search a value r 0 1/2 < r 0 < 1) is selected so that both of the interior points c 0 and d 0 will be used in the next subinterval and there will be only one new function evaluation Without loss of generality assume that f c 0 )> f d 0 ) It follows that a 1 = a 0, b 1 = d 0, and d 1 = c 0 see Figure 84) If there is to be only one new function evaluation, then we select r 1 1/2 < r 1 < 1) for the subinterval [a 1, b 1 ], such that d 0 c 0 = b 1 d 1 2r 0 1)b 0 a 0 ) = 1 r 1 )b 1 a 1 ) 2r 0 1)b 0 a 0 ) = 1 r 1 )r 0 b 0 a 0 )) 2r 0 1 = 1 r 1 )r 0 r 1 = 1 r 0 r 0 Substituting r 0 = F n 1 /F n, n 4, into this last equation yields r 1 = 1 F n 1 F n F n 1 F n = F n F n 1 F n 1 = F n 2 F n 1 since, by equation 6), F n = F n 1 + F n 2 Reasoning inductively, it follows that the Fibonacci search can be begun with r 0 = F n 1 /F n and continued using r k = F n 1 k /F n k for k = 1, 2,, n 3 Note that r n 3 = F 2 /F 3 = 1/2, thus no new points can be added at this stage Therefore, there are a total of n 3) + 1 = n 2 steps in this process

3 414 CHAP 8 NUMERICAL OPTIMIZATION 1-r 0 2r 0-1 a 0 c 0 d 0 b 0 r 1 1-r 1 a 1 c 1 d 1 b 1 Figure 84 The Fibonacci search intervals [a 0, b 0 ] and [a 1, b 1 ] The k + 1)st subinterval is obtained by reducing the length of the kth subinterval by a factor of r k = F n 1 k /F n k The length of the last subinterval is F n 1 F n 2 F 2 b 0 a 0 ) = F 2 b 0 a 0 ) F n F n 1 F 3 F n = 1 b 0 a 0 ) = b 0 a 0 F n F n If the absissa of the minimum is to be found with a tolerance of ɛ, then we need to find the smallest value of n such that b 0 a 0 7) <ɛ or F n > b 0 a 0 F n ɛ The interior points c k and d k of the kth subinterval [a k, b k ] are found, as needed, using the formulas c k = a k + 1 F ) n k 1 8) b k a k ) F n k d k = a k + F n k 1 9) b k a k ) F n k Note the value of n used in formulas 8) and 9) is found using inequality 7) Each iteration requires the determination of two new interior points, one from the previous iteration and the second from formula 8) or 9) When r 0 = F 2 /F 3 = 1/2, the two interior points will be concurrent in the middle of the interval To distinguish the two interior points a small distinguishability constant, e, is introducted Thus when formula 8) or 9) is used, the coefficients of b k a k ) are 1/2 e or 1/2 + e, respectively

4 SEC 81 MINIMIZATION OF A FUNCTION OF ONE VARIABLE 415 Example 83 Find the minimum of the function f x) = x 2 sinx) on the interval [0, 1] using the Fibonacci search method Use a tolerance of ɛ = 10 4 and the distinguishability constant e = 001 The smallest Fibonacci number satisfying F n > b 0 a 0 ɛ = 1 0 = 10,000, 10 4 is F 21 = 10,946 Thus n = 21 Let a 0 = 0 and b 0 = 1 Formulas 8) and 9) yield c 0 = F ) ) F 21 d 0 = 0 + F ) F 21 We set a 1 = a 0, b 1 = d 0, and d 1 = c 0, since f ) = and f ) = f d 0 ) f c 0 )) The new subinterval containing the abscissa of the minimum of f is [a 1, b 1 ]=[0, ] Now use formula 8) to calculate the interior point c 1 : c 1 = a F F 21 1 ) b 1 a 1 ) = F ) ) F Now compute and compare f c 1 ) and f d 1 ) to determine the new subinterval [a 2, b 2 ], and continue the iteration process Some of the computations are shown in Table 83 At the seventeenth iteration the interval has been narrowed down to [a 17, b 17 ] = [ , ], where c 17 = , d 17 = , and f d 17 ) f c 17 ) Thus [a 18, b 18 ]=[ , ] and d 18 = At this stage the multiplier is r 18 = 1 F 2 /F 3 = 1 1/2 = 1/2 and the distinguishability constant e = 001 is used to calculate c 18 : c 18 = a )b 18 a 18 ) = ) Since f d 18 ) f c 18 ),thefinal subinterval is [a 19, b 19 ]=[ , ] This interval has width We choose to report the abscissa of the minimum as the midpoint of this interval Therefore, the minimum value is f ) = Both the Fibonacci and golden ratio search methods can be applied in cases where f x) is not differentiable It should be noted that when n is small the Fibonacci method is more efficient than the golden ratio method However, for n large the two methods are almost identical

5 416 CHAP 8 NUMERICAL OPTIMIZATION Table 83 Fibonacci Search for the Minimum of f x) = x 2 sinx) k a k c k d k b k y = fx) y = fx) a p 0 p b a p p 0 b Figure 85 Using f x) to find the minimum value of the unimodal function f x) on the interval [a, b]

6 Numerical Methods Using Matlab, 4 th Edition, 2004 John H Mathews and Kurtis K Fink ISBN: Prentice-Hall Inc Upper Saddle River, New Jersey, USA

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