The local maximum (x* = 0.106) found by the golden-section search does lie in the interval [x_1, x_2]. So, it is consistent with the explanation on the help page.
---------------------------------------------------------------------------- ------- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html ---------------------------------------------------------------------------- -------- -----Original Message----- From: Mike Lawrence [mailto:[EMAIL PROTECTED] Sent: Monday, October 01, 2007 10:27 AM To: Ravi Varadhan Cc: 'Rhelp' Subject: Re: [R] optimize() stuck in local plateau ? I may be misunderstanding, but my example seems to violate the "local minimum inside [x_1,x_2] will be found as solution, even when f is constant in there" rule since the search in my example continues on towards +1. On 1-Oct-07, at 10:40 AM, Ravi Varadhan wrote: > Please read the help for optimize() carefully. The following > excerpted from > there should help explain your problem: > > "The first evaluation of f is always at x_1 = a + (1-phi)(b-a) > where (a,b) = > (lower, upper) and phi = (sqrt 5 - 1)/2 = 0.61803.. is the golden > section > ratio. Almost always, the second evaluation is at x_2 = a + phi(b- > a). Note > that a local minimum inside [x_1,x_2] will be found as solution, > even when f > is constant in there, see the last example." > > In your case, >> x_1 = a + (1-phi)*(b-a) >> x_1 > [1] -0.236068 >> x_2 = a + phi*(b-a) >> x_2 > [1] 0.236068 >> > > Since your function is constant in (x_1, x_2), you get your > solution in that > interval. > > Try a different interval, and you'll get your answer: > >> optimize(my.func,interval=c(-1,0),maximum=TRUE) > [1] -0.618034 0.618034 > [1] -0.381966 0.000000 > [1] -0.763932 0.763932 > [1] -0.8090170 0.4045085 > [1] -0.7016261 0.7016261 > [1] -0.7401333 0.7401333 > [1] -0.781153 0.781153 > [1] -0.791796 0.791796 > [1] -0.7983739 0.7983739 > [1] -0.8024392 0.4012196 > [1] -0.7958614 0.7958614 > [1] -0.7999267 0.7999267 > [1] -0.8008864 0.4004432 > [1] -0.7993336 0.7993336 > [1] -0.8002933 0.4001466 > [1] -0.7997001 0.7997001 > [1] -0.8000667 0.4000334 > [1] -0.7998402 0.7998402 > [1] -0.7999802 0.7999802 > [1] -0.8000209 0.4000104 > [1] -0.7999802 0.7999802 > $maximum > [1] -0.7999802 > > $objective > [1] 0.7999802 > > > Best, > Ravi. > > > ---------------------------------------------------------------------- > ------ > ------- > > Ravi Varadhan, Ph.D. > > Assistant Professor, The Center on Aging and Health > > Division of Geriatric Medicine and Gerontology > > Johns Hopkins University > > Ph: (410) 502-2619 > > Fax: (410) 614-9625 > > Email: [EMAIL PROTECTED] > > Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/ > Varadhan.html > > > > ---------------------------------------------------------------------- > ------ > -------- > > > -----Original Message----- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] > project.org] On > Behalf Of Mike Lawrence > Sent: Monday, October 01, 2007 1:29 AM > To: Rhelp > Subject: [R] optimize() stuck in local plateau ? > > Hi all, > > Consider the following function: > > #### > my.func = function(x){ > y=ifelse(x>-.5,0,ifelse(x< -.8,abs(x)/2,abs(x))) > print(c(x,y)) #print what was tested and what the result is > return(y) > } > curve(my.func,from=-1,1) > #### > > When I attempt to find the maximum of this function, which should be > -.8, I find that optimize gets stuck in the plateau area and doesn't > bother testing the more interesting bits of the function: > > #### > optimize(my.func,interval=c(-1,1),maximum=TRUE) > #### > > I really don't understand why the search moves to the positive/ > constant area of the function and neglects the more negative area of > the function. On step #4, after finding that there is no difference > between tests at -.23, .23 & .52, shouldn't the algorithm try -.52? > In fact, it seems to me that it would make sense to try -.52 on step > 3, so that we've tested one negative, one positive (found no > difference), now one negative again. Thoughts? > > Of course I could define my interval more reasonably for this > particular function, but this is in fact simply one of a class of > functions I'm exploring, none of which have known formal descriptions > as above (I'm exploring a large number of 'black boxes'). I do know > that the maximum must occur between -1 and 1 for all however. Please > advise on how I might use optimize more usefully. > > Mike > > -- > Mike Lawrence > Graduate Student, Department of Psychology, Dalhousie University > > Website: http://memetic.ca > > Public calendar: http://icalx.com/public/informavore/Public > > "The road to wisdom? Well, it's plain and simple to express: > Err and err and err again, but less and less and less." > - Piet Hein > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public "The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less." - Piet Hein ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
