The golden search method: Question 1

Transcription

1 1. Golde Sectio Search for the Mode of a Fuctio The golde search method: Questio 1 Suppose the last pair of poits at which we have a fuctio evaluatio is x(), y(). The accordig to the method, If f(x()) f(y()), the mode lies i [x(-1), y()], where y() x ( ) y( 1) x( 1), with 5 1 y( 1) x( ) y( ) x( 1) ; x( ) x( 1) (1 )( y( 1) x( 1)) y( ) x( 1) ( y( 1) x( 1)) 5 1 x( ) x( 1) y( ) x( 1) 5 1 ; Which meas the subiterval [x(-1), y()] is divided i golde sectio by x(). If f(x()) f(y()), the mode lies i [x(), y(-1)], where by the same computatio, y( 1) x( ) ( y( 1) x( 1)) ; y( 1) y( ) (1 )( y( 1) x( 1)) ; y( 1) y( ) y( 1) x( ) Which meas the subiterval [x(), y(-1)] is divided i golde sectio by y(). Programmig: Programmig Task Program usig golde sectio search is listed i page _ (1) Whe f(x)=f(y), accordig to the golde search method, we have the mode lies i [a,y] ad i [x,b], which a<x<y<b. Therefore the mode lies i [x,y]. I the program, whe f(x)=f(y), the ew iterval is defied to be [x,y]. I this way, we restrict the rage for mode to a smaller iterval ad could possibly achieve the required precisio i smaller umber of iteratios. () It is preferable to use (1). Because by questio 1, wheever we get a ew iterval, there is already a poit with a fuctio evaluatio iside this ew iterval ad divides the ew iterval i golde sectio. Therefore equatio () eeds both x ad y to be evaluated which is time cosumig. (3) I the situatio where the mode s ( suppose, WLOG, this mode is a uique maximum) positio were at a ed-poit of the origial iterval, the, the other ed-poit of the origial iterval is also a mode ( a uique miimum) for this iterval. The program searches for mode1 ad mode withi required precisio, oe of which is x for a uique maximum ad the other oe is x for a uique miimum. The suppose WLOG mode1<mode, ad if (f(mode1)-f(a))(f(mode)-f(mode1))>0 ad (f(b)-f(mode))(f(mode)- f(mode1))>0, which idicates the fuctio is mootoically icreasig or decreasig, the the program prits the fuctio is mootoically icreasig/decreasig with modes at both eds. The calculatio of oe of these mode is as follows:

2 WLOG, suppose the mode is at the lower boudary a. Deote a as x(0), b as y(0), ad the ratio 5-1 by μ the we have y(1)= μy(0)+(1- μ)x(0) with the ew iterval [x(0), y(1)] by iductio, we have y( ) y(0) (1 ) x(0) with the required iterval [x(0), y()] where y()-x(0) tol. tol is the required precisio. Here, y ( ) x (0) ( y (0) x (0)) ad adjustig the value of to reach the require precisio. To show it works, try (x-1)^+1 with origial boudary [1,5] with required precisio 0.01, the the program shows The fuctio is mootoically icreasig i the give iterval, with modes at both eds. f ( x) 1 x x 4x 4, Questio df x 16x 0 x dx is the mode of the fuctio f(x). The program for questio is listed i page _, With a chose accuracy of 0.001, the result is: 1 iteratios to get the mode= , correct to Computatioal cost: Questio 3 For a real life problem, the alterative algorithm would be more time cosumig ad more expesive. This is because the alterative algorithm does ot have the special property of golde search method described i questio 1 (i.e. for each ew subiterval, we eed to locate two poits ad make two fuctio evaluatios for the alterative algorithm where as described i questio 1(ii), oly oe poit eed to be located ad oe fuctio evaluatio to be made). To compare the umber of umerical operatios required for both algorithm, we could take a extreme example where the mode is at the lower boudary a. Deote a as x(0), b as y(0), ad the ratio 5-1 & 3 by μ the we have y(1)= μy(0)+(1- μ)x(0) with the ew iterval [x(0), y(1)] by iductio, we have y( ) y(0) (1 ) x(0) with the required iterval [x(0), y()] where y()-x(0) tol. tol is the required precisio.

3 Here, y ( ) x (0) ( y (0) x (0)) For both algorithm to reach the same tol, Take log o both sides to get 5 1 1log log For each iteratios, the alterative algorithm does double umber of umerical operatios tha the golde sectio search method (excludig the case where f(x)=f(y)). Therefore roughly, Number of umerical operatios of golde sectio search method = Number of umerical operatios of alterative algorithm. Questio 4 Whe the value of the fuctio evaluatio f(x) does ot exceed the realmax value i MATLAB, the umerical accuracy that is attaiable could be as accurate as wated. However, whe value of f(x) ad f(y) both exceed the realmax, the computer judge the two values to be equal ad the accuracy would be affected. Here use the example of f(x)=abs(ta(x)). The program goldeforq4.m geerates the mode for abs(ta(x)) accordig to the differet iitial iterval ad accuracy. Questio 6 (i) (ii) Iterval used for ier iteratios is [6.5, 16] because the stylus goes from outer groove to the ier groove, ad so x chages from 6.5 to 16. Iterval used for outer loop is [17.5, 30.5] because of the triagular iequality. abs(l-x)<d<abs(l+x) with x rages from 6.5 to 16 so 17.5<d<30.5

4 Figure1: the outer iteratio is uimodal figure : the ier iteratio is uimodal Here figure 1 shows how the graph for ier iteratio looks for differet values of d. the graph moves dowward as d icreases. Figure shows how the graph for outer iteratio looks (iii) Figure 3: how the required accuracy for ier iteratio affects the value of optimum agle

5 Figure 4: how the required accuracy for outer iteratio affects the value of optimum agle