R help - Jun 2010 - 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(90000,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674),
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(800000,1000000,0.0001,0.17)                   # 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.

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.list at 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 
>
(90000,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674
> ),  
> 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(800000,1000000,0.0001,0.17)                   # 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 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.

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
          B0            K            q            r 
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: rvaradhan at jhmi.edu


----- Original Message -----
From: Derek Ogle <DOgle at northland.edu>
Date: Saturday, June 26, 2010 4:28 pm
Subject: [R] optim() not finding optimal values
To: "R (r-help at R-project.org)" <r-help at 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(90000,113300,155860,181128,198584,198395,139040,109969,71896,59314,62300,65343,76990,88606,118016,108250,108674),
>
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(800000,1000000,0.0001,0.17)                   # 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 at r-project.org mailing list
>  
>  PLEASE do read the posting guide 
>  and provide commented, minimal, self-contained, reproducible code.

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. Rub?n Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN


> -----Mensaje original-----
> De: r-help-bounces at r-project.org 
> [mailto:r-help-bounces at r-project.org] En nombre de Derek Ogle
> Enviado el: s?bado, 26 de junio de 2010 22:28
> Para: R (r-help at 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(90000,113300,155860,181128,198584,198395,139040,109969,71896
> ,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(800000,1000000,0.0001,0.17)                   # 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 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.
>

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.