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I’ve been waiting for hours for my code to finish running. I imagine there must be a smarter way to do this, after all I don’t need to visit every point when I have a monotonous function around a minimum point.
My problem is related to the queue theory, I have the average time in the queue, the c and the exit rate.
lambda <-seq(0,ceiling(mu*PDV),0.0001)
n.lambda<-seq_along(lambda)
diferenca <-NULL
diferenca1<-300
for (i in n.lambda) {
try(wq<-queueing::QueueingModel(
queueing::NewInput.MMC(lambda=lambda[i], mu=mu, c=PDV))$Wq, silent = TRUE)
if(is.na(wq)) next
diferenca<-(wq-Wq)^2
if(diferenca<diferenca1){
lambdaWin <-lambda[i]
}
diferenca1<-diferenca
}
Any suggestions on how to improve it?
If I understand correctly, your problem is to find the minimum point of a function, correct? Is there a reason not to use the function
optimofR? I believe if you edit the question with somefunção_de_interesse, someone here will be able to help you implement a solution usingoptim.– Marcus Nunes
It’s just that my problem is with queue theory.
queueing::QueueingModel(
 queueing::NewInput.MMC(lambda=lambda[i], mu=mu, c=PDV))$LqI have the length of the queue I want to know the lambda that would return it.– Márcio Mocellin
Then I won’t be able to help. It’s not a subject I master.
– Marcus Nunes
@Marcusnunes Thanks. I had heard of codes that partitioned the domain successively to search for minimum points.
– Márcio Mocellin