[R] optimize() stuck in local plateau ?

Mon Oct 1 16:34:23 CEST 2007

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.

----------------------------------------------------------------------------
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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: rvaradhan at jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

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-----Original Message-----
From: Mike Lawrence [mailto:Mike.Lawrence at dal.ca] 
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: rvaradhan at jhmi.edu
>
> Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/ 
> Varadhan.html
>
>
>
> ---------------------------------------------------------------------- 
> ------
> --------
>
>
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r- 
> 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 at 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