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
Hi Mike, You function is discontinuous at -0.8, so you can expect everything :-}! But this is not the only problem. The algorithm for optimize never gets there. In general there exists no universal method to find the global maximum of a function (unless it satisfies certain conditions). You can always get stuck with a local maximum. One possibility may be to divide the interval [-1,1] into many small intervals, compute the function in every interval and decide where approximately your maximum is. Then you can use optimize for a smaller interval. In your case, searching in [-1,0] gives correct result. But this also may fail for a function which is not continuous. Regards, Moshe. --- Mike Lawrence <Mike.Lawrence at dal.ca> wrote:> 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. >
Given a univariate problem where the maximum must be between -1 and 1, I would test with a grid of points, then refine that if necessary. Patrick Burns patrick at burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and "A Guide for the Unwilling S User") Mike Lawrence wrote:>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. > > > >
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.