Your code is not reproducible nor minimal, but why don't you put a
command print(acceptprob) in and see if you are getting reasonable
values. If these values are extremely low it shouldn't surprise you
that your loop takes a long time to run.
More generally, read up on the use of print() and browser() as debugging tools.
Michael
On Mon, Dec 5, 2011 at 3:47 PM, Gyanendra Pokharel
<gyanendra.pokha...@gmail.com> wrote:
> I forgot to upload the R-code in last email, so heare is one
>
> epiann <- function(T0 = 1, N=1000, ainit=1, binit=1,rho = 0.99, amean = 3,
> bmean=1.6, avar =.1, bvar=.1, f){
>
> moving <- 1
> count <- 0
> Temp <- T0
> aout <- ainit
> bout <- binit
> while(moving > 0){
> moving <- 0
> for (i in 1:N) {
> aprop <- rnorm (1,amean, avar)
> bprop <- rnorm (1,bmean, bvar)
> if (aprop > 0 & bprop > 0){
> acceptprob <- min(1,exp((f(aout, bout) -
> f(aprop,bprop))/Temp))
> u <- runif(1)
> if(u<acceptprob){
> moving <- moving +1
> aout <- aprop
> bout <- bprop
> }
> else{aprob <- aout
> bprob <- bout}
> }
> }
> Temp <- Temp*rho
> count <- count +1
>
> }
> fmin <- f(aout,bout)
> return(c(aout, bout,fmin, count) )
>
> }
> out<- epiann(f = loglikelihood)
>
> On Mon, Dec 5, 2011 at 3:46 PM, Gyanendra Pokharel <
> gyanendra.pokha...@gmail.com> wrote:
>
>> Hi all,
>> I have the following code,
>> When I run the code, it never terminate this is because of the while loop
>> i am using. In general, if you need a loop for which you don't know in
>> advance how many iterations there will be, you can use the `while'
>> statement so here too i don't know the number how many iterations are
>> there. So Can some one suggest me whats going on?
>> I am using the Metropolis simulated annealing algorithm
>> Best
>>
>
> [[alternative HTML version deleted]]
>
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> and provide commented, minimal, self-contained, reproducible code.
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