Re: [R] optim() not finding optimal values
If you are going to make this program available for general use you want to take every precaution to make it bulletproof. This is a fairly informative data set. The model will undoubtedly be used on far less informative data. While the model looks pretty simple it is very challenging from a numerical point of view. I took a moment to code it up in AD Model Builder. The true minimum is 1619.480495 So I think Ravi has finally arrived pretty close to the answer. One way of judging the difficulty of a model is to look at the eigenvalues of the Hessian at the minimum. They are 3.122884668e-09 1.410866202e-08 1866282.520 1.330233652e+13 so the condition number is around 1.e+21. One begins to see why these models are challenging. The model as formulated represents the state of the art in fisheries models circa 1985. A lot of progress has been made since that time. Using B_t for the biomass and C_t for the catch the equation in the code is B_{t+1} = B_t + r *B_t*(1-B_t/K) -C_t (1) First notice that for (1) to make sense the following conditions must be satisfied B_t > 0 for all t r > 0 K>0 Strictly speaking it is not necessary that B_t<=K but if B_t>K and r is large then B_{t+1} could be <0. So formulation (1) gives Murphys law a good chance. How to fix it. Notice that (1) is really a rough approximation to the solution of a differential equation B'(t) = r *B(t)*(1-B(t)/K) -C (2) where in (2) C is a constant catch rate. To fix (1) we use a semi-implicit differencing scheme. Because it is useful to allow smaller step sizes than one we denote them by d. B_{t+d} = B_t + d* r *B_t*(1-B_{t+d}/K) -d*C_t*B_{t+d}/B_t (1) The idea is that the quantity 1-x with x>0 will be replaced by 1/(1+x). Expanding 2 and solving for B_{t+d} yields B_{t+d} = (1+d*r) B_t / (1+d*r*B_t/K +d*C_t/B_t) (3) So long as r>0, K>0 C_t>0 then starting from an initial value B_0 > 0 ensures that B_t> 0 for all t>0. We can let d=1/nsteps where nsteps is the number of steps in the approximate integration for each year which can be increased until the solution is judged to be close enough to the exact solution from (2) Notice that in (3) as C_t --> infinity B_{t+d} --> 0 So that you can never catch more fish than you have. I coded up this version of the model in AD Model Builder and fit it to the data. It is now much more resistant to bad starting values for the parameters etc. If anyone wants the tpl file for the model in ADMB they can contact me off list. __ 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.
Re: [R] optim() not finding optimal values
Oops, I my previous email, the second line in the `SPsse.log' function should have been: par <- exp(par) # log-transformation Ravi. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Ravi Varadhan Sent: Monday, June 28, 2010 9:48 AM To: 'Rubén Roa'; 'Derek Ogle'; 'R' Subject: Re: [R] optim() not finding optimal values Ruben, Transforming the parameters is also a good idea, but the obvious caveat is that the transformation must be feasible. The log-transformation is only feasible for positive parameter domain. This happens to be the case for OP's problem. In fact, the log-transform does better than ratio scaling in terms of improving rate of convergence for this particular model/data combination. SPsse.log <- function(par,B,CPE,SSE.only=TRUE) { n <- length(B) # get number of years of data par <- tan(par) # log-transformation B0 <- par["B0"]# isolate B0 parameter K <- par["K"] # isolate K parameter q <- par["q"] # isolate q parameter r <- par["r"] # isolate r parameter predB <- numeric(n) predB[1] <- B0 for (i in 2:n) predB[i] <- predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] predCPE <- q*predB sse <- sum((CPE-predCPE)^2) if (SSE.only) sse else list(sse=sse,predB=predB,predCPE=predCPE) } > SPoptim3 <- optim(log(parsbox),SPsse.log,B=d$catch,CPE=d$cpe, method="BFGS") > SPoptim3 $par B0 K q r 13.5035475 13.9634098 -8.8142230 -0.9030033 $value [1] 1619.481 $counts function gradient 546 $convergence [1] 0 $message NULL > exp(SPoptim3$par) # transforming to original scale B0Kqr 7.320086e+05 1.159396e+06 1.486044e-04 4.053505e-01 > I don't think there is a need to set B0=K since convergence is pretty good with scaling and/or with log-transformation. Best, Ravi. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Rubén Roa Sent: Monday, June 28, 2010 2:24 AM To: Derek Ogle; R (r-help@R-project.org) Subject: Re: [R] optim() not finding optimal values Derek, As a general strategy, and as an alternative to parscale when using optim, you can just estimate the logarithm of your parameters. So in optim the par argument would contain the logarithm of your parameters, whereas in the model itself you would write exp(par) in the place of the parameter. The purpose of this is to bring all parameters to a similar scale to aid the numerical algorithm in finding the optimum over several dimensions. Due to the functional invariance property of maximum likelihood estimates your transformed pars back to the original scale are also the MLEs of the pars in your model. If you were using ADMB you'd get the standard error of the pars in the original scale simply by declaring them sd_report number class. With optim, you would get the standard error of pars in the original scale post-hoc by using Taylor series (a.k.a. Delta method) which in this case is very simple since the transformation is just the exponential. In relation to your model/data combination, since you have only 17 years of data and just one series of cpue, and this is a rather common case, you may want to give the choice to set B0=K, i.e. equilibrium conditions at the start, in your function, to reduce the dimensionality of your profile likelihood approximation thus helping the optimizer. HTH Dr. Rubén Roa-Ureta AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g 48395 Sukarrieta (Bizkaia) SPAIN > -Mensaje original- > De: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org] En nombre de Derek Ogle > Enviado el: sábado, 26 de junio de 2010 22:28 > Para: R (r-help@R-project.org) > Asunto: [R] optim() not finding optimal values > > I am trying to use optim() to minimize a sum-of-squared > deviations function based upon four parameters. The basic > function is defined as ... > > SPsse <- function(par,B,CPE,SSE.only=TRUE) { > n <- length(B) # get number of > years of data > B0 <- par["B0"]# isolate B0 parameter > K <- par["K"] # isolate K parameter > q <- par["q"] # isolate q parameter > r <- par["r"] # isolate r parameter > predB <- numeric(n) > predB[1] <- B0 > for (i in 2:n) p
Re: [R] optim() not finding optimal values
Ruben, Transforming the parameters is also a good idea, but the obvious caveat is that the transformation must be feasible. The log-transformation is only feasible for positive parameter domain. This happens to be the case for OP's problem. In fact, the log-transform does better than ratio scaling in terms of improving rate of convergence for this particular model/data combination. SPsse.log <- function(par,B,CPE,SSE.only=TRUE) { n <- length(B) # get number of years of data par <- tan(par) # log-transformation B0 <- par["B0"]# isolate B0 parameter K <- par["K"] # isolate K parameter q <- par["q"] # isolate q parameter r <- par["r"] # isolate r parameter predB <- numeric(n) predB[1] <- B0 for (i in 2:n) predB[i] <- predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] predCPE <- q*predB sse <- sum((CPE-predCPE)^2) if (SSE.only) sse else list(sse=sse,predB=predB,predCPE=predCPE) } > SPoptim3 <- optim(log(parsbox),SPsse.log,B=d$catch,CPE=d$cpe, method="BFGS") > SPoptim3 $par B0 K q r 13.5035475 13.9634098 -8.8142230 -0.9030033 $value [1] 1619.481 $counts function gradient 546 $convergence [1] 0 $message NULL > exp(SPoptim3$par) # transforming to original scale B0Kqr 7.320086e+05 1.159396e+06 1.486044e-04 4.053505e-01 > I don't think there is a need to set B0=K since convergence is pretty good with scaling and/or with log-transformation. Best, Ravi. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Rubén Roa Sent: Monday, June 28, 2010 2:24 AM To: Derek Ogle; R (r-help@R-project.org) Subject: Re: [R] optim() not finding optimal values Derek, As a general strategy, and as an alternative to parscale when using optim, you can just estimate the logarithm of your parameters. So in optim the par argument would contain the logarithm of your parameters, whereas in the model itself you would write exp(par) in the place of the parameter. The purpose of this is to bring all parameters to a similar scale to aid the numerical algorithm in finding the optimum over several dimensions. Due to the functional invariance property of maximum likelihood estimates your transformed pars back to the original scale are also the MLEs of the pars in your model. If you were using ADMB you'd get the standard error of the pars in the original scale simply by declaring them sd_report number class. With optim, you would get the standard error of pars in the original scale post-hoc by using Taylor series (a.k.a. Delta method) which in this case is very simple since the transformation is just the exponential. In relation to your model/data combination, since you have only 17 years of data and just one series of cpue, and this is a rather common case, you may want to give the choice to set B0=K, i.e. equilibrium conditions at the start, in your function, to reduce the dimensionality of your profile likelihood approximation thus helping the optimizer. HTH Dr. Rubén Roa-Ureta AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g 48395 Sukarrieta (Bizkaia) SPAIN > -Mensaje original- > De: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org] En nombre de Derek Ogle > Enviado el: sábado, 26 de junio de 2010 22:28 > Para: R (r-help@R-project.org) > Asunto: [R] optim() not finding optimal values > > I am trying to use optim() to minimize a sum-of-squared > deviations function based upon four parameters. The basic > function is defined as ... > > SPsse <- function(par,B,CPE,SSE.only=TRUE) { > n <- length(B) # get number of > years of data > B0 <- par["B0"]# isolate B0 parameter > K <- par["K"] # isolate K parameter > q <- par["q"] # isolate q parameter > r <- par["r"] # isolate r parameter > predB <- numeric(n) > predB[1] <- B0 > for (i in 2:n) predB[i] <- > predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] > predCPE <- q*predB > sse <- sum((CPE-predCPE)^2) > if (SSE.only) sse > else list(sse=sse,predB=predB,predCPE=predCPE) > } > > My call to optim() looks like this > > # the data > d <- data.frame(catch= > c(9,113300,155860,181128,198584,198395,139040,109969,71896 > ,59314,62300,65343,76990,88606,118016
Re: [R] optim() not finding optimal values
Derek, As a general strategy, and as an alternative to parscale when using optim, you can just estimate the logarithm of your parameters. So in optim the par argument would contain the logarithm of your parameters, whereas in the model itself you would write exp(par) in the place of the parameter. The purpose of this is to bring all parameters to a similar scale to aid the numerical algorithm in finding the optimum over several dimensions. Due to the functional invariance property of maximum likelihood estimates your transformed pars back to the original scale are also the MLEs of the pars in your model. If you were using ADMB you'd get the standard error of the pars in the original scale simply by declaring them sd_report number class. With optim, you would get the standard error of pars in the original scale post-hoc by using Taylor series (a.k.a. Delta method) which in this case is very simple since the transformation is just the exponential. In relation to your model/data combination, since you have only 17 years of data and just one series of cpue, and this is a rather common case, you may want to give the choice to set B0=K, i.e. equilibrium conditions at the start, in your function, to reduce the dimensionality of your profile likelihood approximation thus helping the optimizer. HTH Dr. Rubén Roa-Ureta AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g 48395 Sukarrieta (Bizkaia) SPAIN > -Mensaje original- > De: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org] En nombre de Derek Ogle > Enviado el: sábado, 26 de junio de 2010 22:28 > Para: R (r-help@R-project.org) > Asunto: [R] optim() not finding optimal values > > I am trying to use optim() to minimize a sum-of-squared > deviations function based upon four parameters. The basic > function is defined as ... > > SPsse <- function(par,B,CPE,SSE.only=TRUE) { > n <- length(B) # get number of > years of data > B0 <- par["B0"]# isolate B0 parameter > K <- par["K"] # isolate K parameter > q <- par["q"] # isolate q parameter > r <- par["r"] # isolate r parameter > predB <- numeric(n) > predB[1] <- B0 > for (i in 2:n) predB[i] <- > predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] > predCPE <- q*predB > sse <- sum((CPE-predCPE)^2) > if (SSE.only) sse > else list(sse=sse,predB=predB,predCPE=predCPE) > } > > My call to optim() looks like this > > # the data > d <- data.frame(catch= > c(9,113300,155860,181128,198584,198395,139040,109969,71896 > ,59314,62300,65343,76990,88606,118016,108250,108674), > cpe=c(109.1,112.4,110.5,99.1,84.5,95.7,74.1,70.2,63.1,66.4,60. > 5,89.9,117.0,93.0,116.6,90.0,105.1)) > > pars <- c(80,100,0.0001,0.17) # put > all parameters into one vector > names(pars) <- c("B0","K","q","r") # > name the parameters > ( SPoptim <- optim(pars,SPsse,B=d$catch,CPE=d$cpe) )# run optim() > > > This produces parameter estimates, however, that are not at > the minimum value of the SPsse function. For example, these > parameter estimates produce a smaller SPsse, > > parsbox <- c(732506,1160771,0.0001484,0.4049) > names(parsbox) <- c("B0","K","q","r") > ( res2 <- SPsse(parsbox,d$catch,d$cpe,SSE.only=FALSE) ) > > Setting the starting values near the parameters shown in > parsbox even resulted in a movement away from (to a larger > SSE) those parameter values. > > ( SPoptim2 <- optim(parsbox,SPsse,B=d$catch,CPE=d$cpe) )# > run optim() > > > This "issue" most likely has to do with my lack of > understanding of optimization routines but I'm thinking that > it may have to do with the optimization method used, > tolerance levels in the optim algorithm, or the shape of the > surface being minimized. > > Ultimately I was hoping to provide an alternative method to > fisheries biologists who use Excel's solver routine. > > If anyone can offer any help or insight into my problem here > I would be greatly appreciative. Thank you in advance. > > __ > 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. > __ 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.
Re: [R] optim() not finding optimal values
Ravi, Thank you very much for the pointer to parscale. This is extremely useful -- in this and some other problems that I am working on. Thanks again for the valuable help. > -Original Message- > From: Ravi Varadhan [mailto:rvarad...@jhmi.edu] > Sent: Saturday, June 26, 2010 11:52 PM > To: Ravi Varadhan > Cc: Derek Ogle; R (r-help@R-project.org) > Subject: Re: [R] optim() not finding optimal values > > A slightly better scaling is the following: > > par.scale <- c(1.e06, 1.e06, 1.e-05, 1) # "q" is scaled differently > > > SPoptim <- optim(pars, SPsse, B=d$catch, CPE=d$cpe, > control=list(maxit=1500, parscale=par.scale)) > > SPoptim > $par > B0Kqr > 7.320899e+05 1.159939e+06 1.485560e-04 4.051735e-01 > > $value > [1] 1619.482 > > $counts > function gradient > 585 NA > > $convergence > [1] 0 > > $message > NULL > > > Note that the Nelder-Mead converges in half the number of iterations > compared to that under previous scaling. > > Ravi. > > > Ravi Varadhan, Ph.D. > Assistant Professor, > Division of Geriatric Medicine and Gerontology > School of Medicine > Johns Hopkins University > > Ph. (410) 502-2619 > email: rvarad...@jhmi.edu > > > - Original Message - > From: Ravi Varadhan > Date: Sunday, June 27, 2010 0:42 am > Subject: Re: [R] optim() not finding optimal values > To: Derek Ogle > Cc: "R (r-help@R-project.org)" > > > > Derek, > > > > The problem is that your function is poorly scaled. You can see > > that the parameters vary over 10 orders of magnitude (from 1e-04 to > > 1e06). You can get good convergence once you properly scale your > > function. Here is how you do it: > > > > par.scale <- c(1.e06, 1.e06, 1.e-06, 1.0) > > > > SPoptim <- optim(pars, SPsse, B=d$catch, CPE=d$cpe, > > control=list(maxit=1500, parscale=par.scale)) > > > > > SPoptim > > $par > >B0Kqr > > 7.329553e+05 1.160097e+06 1.484375e-04 4.050476e-01 > > > > $value > > [1] 1619.487 > > > > $counts > > function gradient > > 1401 NA > > > > $convergence > > [1] 0 > > > > $message > > NULL > > > > > > Hope this helps, > > Ravi. > > > > ____________________ > > > > Ravi Varadhan, Ph.D. > > Assistant Professor, > > Division of Geriatric Medicine and Gerontology > > School of Medicine > > Johns Hopkins University > > > > Ph. (410) 502-2619 > > email: rvarad...@jhmi.edu > > > > > > - Original Message - > > From: Derek Ogle > > Date: Saturday, June 26, 2010 4:28 pm > > Subject: [R] optim() not finding optimal values > > To: "R (r-help@R-project.org)" > > > > > > > I am trying to use optim() to minimize a sum-of-squared deviations > > > > > function based upon four parameters. The basic function is > defined > > as > > > ... > > > > > > SPsse <- function(par,B,CPE,SSE.only=TRUE) { > > >n <- length(B) # get number of > years > > of > > > data > > >B0 <- par["B0"]# isolate B0 > parameter > > >K <- par["K"] # isolate K > parameter > > >q <- par["q"] # isolate q > parameter > > >r <- par["r"] # isolate r > parameter > > >predB <- numeric(n) > > >predB[1] <- B0 > > >for (i in 2:n) predB[i] <- predB[i-1]+r*predB[i-1]*(1-predB[i- > 1]/K)-B[i-1] > > >predCPE <- q*predB > > >sse <- sum((CPE-predCPE)^2) > > >if (SSE.only) sse > > > else list(sse=sse,predB=predB,predCPE=predCPE) > > > } > > > > > > My call to optim() looks like this > > > > > > # the data > > > d <- data.frame(catch= > > > > > > c(9,113300,155860,181128,198584,198395,139040,109969,71896,59314,62 > 300,65343,76990,88606,118016,108250,108674), > > > > > > cpe=c(109.1,112.4,110.5,99.1,84.5,95.7,74.1,70.2,63.1,66.4,60.5,89.9,11 > 7.0,93.0,116.6,90.0,105.1)) &
Re: [R] optim() not finding optimal values
A slightly better scaling is the following: par.scale <- c(1.e06, 1.e06, 1.e-05, 1) # "q" is scaled differently > SPoptim <- optim(pars, SPsse, B=d$catch, CPE=d$cpe, control=list(maxit=1500, > parscale=par.scale)) > SPoptim $par B0Kqr 7.320899e+05 1.159939e+06 1.485560e-04 4.051735e-01 $value [1] 1619.482 $counts function gradient 585 NA $convergence [1] 0 $message NULL Note that the Nelder-Mead converges in half the number of iterations compared to that under previous scaling. Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu - Original Message - From: Ravi Varadhan Date: Sunday, June 27, 2010 0:42 am Subject: Re: [R] optim() not finding optimal values To: Derek Ogle Cc: "R (r-help@R-project.org)" > Derek, > > The problem is that your function is poorly scaled. You can see > that the parameters vary over 10 orders of magnitude (from 1e-04 to > 1e06). You can get good convergence once you properly scale your > function. Here is how you do it: > > par.scale <- c(1.e06, 1.e06, 1.e-06, 1.0) > > SPoptim <- optim(pars, SPsse, B=d$catch, CPE=d$cpe, > control=list(maxit=1500, parscale=par.scale)) > > > SPoptim > $par >B0Kqr > 7.329553e+05 1.160097e+06 1.484375e-04 4.050476e-01 > > $value > [1] 1619.487 > > $counts > function gradient > 1401 NA > > $convergence > [1] 0 > > $message > NULL > > > Hope this helps, > Ravi. > > > > Ravi Varadhan, Ph.D. > Assistant Professor, > Division of Geriatric Medicine and Gerontology > School of Medicine > Johns Hopkins University > > Ph. (410) 502-2619 > email: rvarad...@jhmi.edu > > > - Original Message - > From: Derek Ogle > Date: Saturday, June 26, 2010 4:28 pm > Subject: [R] optim() not finding optimal values > To: "R (r-help@R-project.org)" > > > > I am trying to use optim() to minimize a sum-of-squared deviations > > > function based upon four parameters. The basic function is defined > as > > ... > > > > SPsse <- function(par,B,CPE,SSE.only=TRUE) { > >n <- length(B) # get number of years > of > > data > >B0 <- par["B0"]# isolate B0 parameter > >K <- par["K"] # isolate K parameter > >q <- par["q"] # isolate q parameter > >r <- par["r"] # isolate r parameter > >predB <- numeric(n) > >predB[1] <- B0 > >for (i in 2:n) predB[i] <- > predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] > >predCPE <- q*predB > >sse <- sum((CPE-predCPE)^2) > >if (SSE.only) sse > > else list(sse=sse,predB=predB,predCPE=predCPE) > > } > > > > My call to optim() looks like this > > > > # the data > > d <- data.frame(catch= > > > c(9,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674), > > > > > cpe=c(109.1,112.4,110.5,99.1,84.5,95.7,74.1,70.2,63.1,66.4,60.5,89.9,117.0,93.0,116.6,90.0,105.1)) > > > > pars <- c(80,100,0.0001,0.17) # put all > > > parameters into one vector > > names(pars) <- c("B0","K","q","r") # name the > parameters > > ( SPoptim <- optim(pars,SPsse,B=d$catch,CPE=d$cpe) )# run optim() > > > > > > This produces parameter estimates, however, that are not at the > > minimum value of the SPsse function. For example, these parameter > > > estimates produce a smaller SPsse, > > > > parsbox <- c(732506,1160771,0.0001484,0.4049) > > names(parsbox) <- c("B0","K","q","r") > > ( res2 <- SPsse(parsbox,d$catch,d$cpe,SSE.only=FALSE) ) > > > > Setting the starting values near the parameters shown in parsbox > even > > resulted in a movement away from (to a larger SSE) those
Re: [R] optim() not finding optimal values
Derek, The problem is that your function is poorly scaled. You can see that the parameters vary over 10 orders of magnitude (from 1e-04 to 1e06). You can get good convergence once you properly scale your function. Here is how you do it: par.scale <- c(1.e06, 1.e06, 1.e-06, 1.0) SPoptim <- optim(pars, SPsse, B=d$catch, CPE=d$cpe, control=list(maxit=1500, parscale=par.scale)) > SPoptim $par B0Kqr 7.329553e+05 1.160097e+06 1.484375e-04 4.050476e-01 $value [1] 1619.487 $counts function gradient 1401 NA $convergence [1] 0 $message NULL Hope this helps, Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu - Original Message - From: Derek Ogle Date: Saturday, June 26, 2010 4:28 pm Subject: [R] optim() not finding optimal values To: "R (r-help@R-project.org)" > I am trying to use optim() to minimize a sum-of-squared deviations > function based upon four parameters. The basic function is defined as > ... > > SPsse <- function(par,B,CPE,SSE.only=TRUE) { >n <- length(B) # get number of years of > data >B0 <- par["B0"]# isolate B0 parameter >K <- par["K"] # isolate K parameter >q <- par["q"] # isolate q parameter >r <- par["r"] # isolate r parameter >predB <- numeric(n) >predB[1] <- B0 >for (i in 2:n) predB[i] <- predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] >predCPE <- q*predB >sse <- sum((CPE-predCPE)^2) >if (SSE.only) sse > else list(sse=sse,predB=predB,predCPE=predCPE) > } > > My call to optim() looks like this > > # the data > d <- data.frame(catch= > c(9,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674), > > cpe=c(109.1,112.4,110.5,99.1,84.5,95.7,74.1,70.2,63.1,66.4,60.5,89.9,117.0,93.0,116.6,90.0,105.1)) > > pars <- c(80,100,0.0001,0.17) # put all > parameters into one vector > names(pars) <- c("B0","K","q","r") # name the parameters > ( SPoptim <- optim(pars,SPsse,B=d$catch,CPE=d$cpe) )# run optim() > > > This produces parameter estimates, however, that are not at the > minimum value of the SPsse function. For example, these parameter > estimates produce a smaller SPsse, > > parsbox <- c(732506,1160771,0.0001484,0.4049) > names(parsbox) <- c("B0","K","q","r") > ( res2 <- SPsse(parsbox,d$catch,d$cpe,SSE.only=FALSE) ) > > Setting the starting values near the parameters shown in parsbox even > resulted in a movement away from (to a larger SSE) those parameter values. > > ( SPoptim2 <- optim(parsbox,SPsse,B=d$catch,CPE=d$cpe) )# run optim() > > > This "issue" most likely has to do with my lack of understanding of > optimization routines but I'm thinking that it may have to do with the > optimization method used, tolerance levels in the optim algorithm, or > the shape of the surface being minimized. > > Ultimately I was hoping to provide an alternative method to fisheries > biologists who use Excel's solver routine. > > If anyone can offer any help or insight into my problem here I would > be greatly appreciative. Thank you in advance. > > __ > R-help@r-project.org mailing list > > PLEASE do read the posting guide > and provide commented, minimal, self-contained, reproducible code. __ 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.
Re: [R] optim() not finding optimal values
Your function is very irregular, so the optim is likely to return local minima rather than global minima. Try different methods (SANN, CG, BFGS) and see if you get the result you need. As with all numerical optimsation, I would check the sensitivity of the results to starting values. Nikhil Kaza Asst. Professor, City and Regional Planning University of North Carolina nikhil.l...@gmail.com On Jun 26, 2010, at 4:27 PM, Derek Ogle wrote: I am trying to use optim() to minimize a sum-of-squared deviations function based upon four parameters. The basic function is defined as ... SPsse <- function(par,B,CPE,SSE.only=TRUE) { n <- length(B) # get number of years of data B0 <- par["B0"]# isolate B0 parameter K <- par["K"] # isolate K parameter q <- par["q"] # isolate q parameter r <- par["r"] # isolate r parameter predB <- numeric(n) predB[1] <- B0 for (i in 2:n) predB[i] <- predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)- B[i-1] predCPE <- q*predB sse <- sum((CPE-predCPE)^2) if (SSE.only) sse else list(sse=sse,predB=predB,predCPE=predCPE) } My call to optim() looks like this # the data d <- data.frame(catch= c (9,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674 ), cpe = c (109.1,112.4,110.5,99.1,84.5,95.7,74.1,70.2,63.1,66.4,60.5,89.9,117.0,93.0,116.6,90.0,105.1 )) pars <- c(80,100,0.0001,0.17) # put all parameters into one vector names(pars) <- c("B0","K","q","r") # name the parameters ( SPoptim <- optim(pars,SPsse,B=d$catch,CPE=d$cpe) )# run optim() This produces parameter estimates, however, that are not at the minimum value of the SPsse function. For example, these parameter estimates produce a smaller SPsse, parsbox <- c(732506,1160771,0.0001484,0.4049) names(parsbox) <- c("B0","K","q","r") ( res2 <- SPsse(parsbox,d$catch,d$cpe,SSE.only=FALSE) ) Setting the starting values near the parameters shown in parsbox even resulted in a movement away from (to a larger SSE) those parameter values. ( SPoptim2 <- optim(parsbox,SPsse,B=d$catch,CPE=d$cpe) )# run optim() This "issue" most likely has to do with my lack of understanding of optimization routines but I'm thinking that it may have to do with the optimization method used, tolerance levels in the optim algorithm, or the shape of the surface being minimized. Ultimately I was hoping to provide an alternative method to fisheries biologists who use Excel's solver routine. If anyone can offer any help or insight into my problem here I would be greatly appreciative. Thank you in advance. __ 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. __ 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.
[R] optim() not finding optimal values
I am trying to use optim() to minimize a sum-of-squared deviations function based upon four parameters. The basic function is defined as ... SPsse <- function(par,B,CPE,SSE.only=TRUE) { n <- length(B) # get number of years of data B0 <- par["B0"]# isolate B0 parameter K <- par["K"] # isolate K parameter q <- par["q"] # isolate q parameter r <- par["r"] # isolate r parameter predB <- numeric(n) predB[1] <- B0 for (i in 2:n) predB[i] <- predB[i-1]+r*predB[i-1]*(1-predB[i-1]/K)-B[i-1] predCPE <- q*predB sse <- sum((CPE-predCPE)^2) if (SSE.only) sse else list(sse=sse,predB=predB,predCPE=predCPE) } My call to optim() looks like this # the data d <- data.frame(catch= c(9,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674), cpe=c(109.1,112.4,110.5,99.1,84.5,95.7,74.1,70.2,63.1,66.4,60.5,89.9,117.0,93.0,116.6,90.0,105.1)) pars <- c(80,100,0.0001,0.17) # put all parameters into one vector names(pars) <- c("B0","K","q","r") # name the parameters ( SPoptim <- optim(pars,SPsse,B=d$catch,CPE=d$cpe) )# run optim() This produces parameter estimates, however, that are not at the minimum value of the SPsse function. For example, these parameter estimates produce a smaller SPsse, parsbox <- c(732506,1160771,0.0001484,0.4049) names(parsbox) <- c("B0","K","q","r") ( res2 <- SPsse(parsbox,d$catch,d$cpe,SSE.only=FALSE) ) Setting the starting values near the parameters shown in parsbox even resulted in a movement away from (to a larger SSE) those parameter values. ( SPoptim2 <- optim(parsbox,SPsse,B=d$catch,CPE=d$cpe) )# run optim() This "issue" most likely has to do with my lack of understanding of optimization routines but I'm thinking that it may have to do with the optimization method used, tolerance levels in the optim algorithm, or the shape of the surface being minimized. Ultimately I was hoping to provide an alternative method to fisheries biologists who use Excel's solver routine. If anyone can offer any help or insight into my problem here I would be greatly appreciative. Thank you in advance. __ 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.