[R] optimize() stuck in local plateau ?

[Ravi Varadhan]

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

 

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-----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

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