similar to: lsoda with arbitrary zero thresholds (with psuedo-solution)

I have been trying to make use of the odesolve library on my university's Linux grid - currently R version 2.0.1 is installed and the system runs 64-bit Scientific Linux based on Redhat. I cannot seem to get lsoda working when I define the model as a shared C library. For example, the following snippet uses the mymod.c example bundled with the package: ### START rm(list=ls())

I am running R 1.2.3 with ESS5.1.18 with Windows 98. I am trying to use lsoda in the odesolve apckage and am having problems. Question: The return value of the function of the system of ode's has to be a list that includes first, the ode's and second, "a vector (possibly with a `names' attribute) of global values that are required at each point in `times'." I

Hi R used with C-code experts, I had a look at the archives and did not find anything on this, so hopefully I am not doubling up. I have previously used the following approach where I needed some very small/large numbers (using Brobdingnag): surfacewithdiff <- function(t, y, p) { const=p["const"] kay =p["kay"] psii=p["psii"]

Hi, I would like to solve a system of coupled ordinary differential equations, where there is a delay (time lag) term. I would like to use the "lsoda" function "odesolve" package. However, I am not sure how to specify the delay term using the syntax allowed by odesolve. Here is an example of the kind of problem that I am trying to solve: > library(odesolve)

Hi, I am doing a project on the simulation of glucose metabolism based on a pharmacokinetic modeling in which we have 4 differential equations. I did this in R by using the odesolve package. It works very well, but I have two questions: Here is the odemodel function _________________________________________________ Ogtt.Odemodel <- function(t, y, p) { absx <- c(-60, -45, -30,

Dear list - Does anyone have any ideas / comments about why I am receiving the following warning when I run lsoda: 1: lsoda-- at t (=r1), too much accuracy requested in: lsoda(start, times, model, parms) 2: for precision of machine.. see tolsf (=r2) in: lsoda(start, times, model, parms) I have tried changing both rtol and atol but without success. I saw the thread in the

I'm trying to use odesolve for integrating various series of coupled 1st order differential equations (derived from a system of enzymatic catalysis and copied below, apologies for the excessively long set of parameters). The thing that confuses me is that, whilst I can run the function rk4: out <- rk4(y=y,times=times,func=func, parms=parms) and the results look not unreasonable:

Dear List, Wonder if you have some thoughts on the following question using lsoda in desolve: I have the following data and function: require(deSolve) times <- c(0:24) tin  <- 0.5 D <- 400 V    <- 26.3 k <-0.056 k12  <- 0.197118 k21  <- 0.022665 yini <- c(dy1 = 0,dy2 = 0)  events <- data.frame(var = "dy1",time = c(10,15),value = c(200,100),method =

Hi - I am getting different results when I run the numerical integrator function lsoda (odesolve package) on a Mac and a PC. I am trying to simulating a system of 10 ODE's with two exogenous pulsed inputs to the system, and have had reasonably good success with many model parameter sets. Under some parameter sets, however, the simulations fail on the Mac (see error message below). The

Hi All, I am running R 2.2.0 on Mac OS 10.4.2, dual G5 processors with 8 Gig RAM. I am running a simulation with lsoda that requires ~378 s to complete one set of time intervals. I need to optimize the parameters, and so need to considerably speed up the simulation. I have tried to figure out how to change the appropriate memory allocation and have search R help and Introductory

I'm solving the differential equation dy/dx = xy-1 with y(0) = sqrt(pi/2). This can be used in computing the tail of the normal distribution. (The actual solution is y(x) = exp(x^2/2) * Integral_x_inf {exp(-t^2/2) dt} = Integral_0_inf {exp (-xt - t^2/2) dt}. For large x, y ~ 1/x, starting around x~2.) I'm testing both lsoda and rk4 from the package odesolve. rk4 is accurate using step

HI R Gurus, > This is my first foray into using c-code with R, so ... > I had a look at the archives and did not find anything on this, so > hopefully I am not doubling up. > I have previously used the following approach where I needed some very small numbers/large (using Brobdingnag): surfacewithdiff <- function(t, y, p) { const=p["const"] kay

R help list subscribers, I am a new user of R. I am attempting to use R to explore a set of equations specifying the dynamics of a three trophic level food chain. I have put together this code for the function that is to be evaluted by LSODA. My equations Rprime, Cprime, and Pprime are meant to describe the actual equation of the derivative. When I run LSODA, I do not get the output that

Hello, I wanted to test the odesolve package and tried to use compiled C-code. But when I do: erg <- lsoda(y, times, "mond", parms, rtol, atol, tcrit=NULL, jacfunc=NULL, verbose=FALSE, dllname="mond", hmin=0, hmax=Inf) I get the error message: Error in lsoda(y, times, "mond", parms, rtol, atol, tcrit = NULL, jacfunc =

Hello, I am an R newbie, trying to use lsoda to solve standard Lotka-Volterra competition equations. My question is: how do I pass a parameter that varies with time, like say, phix <- 0.7 + runif(tmax) in the example below. # defining function lotvol <- function(t,n,p){ x <- n[1]; y <- n[2] rx <- p["rx"]; ry <- p["ry"] Kx <-

Thanks to Dieter Menne and Spencer Graves I started to get my way through lsoda() Now I need to use it in with nls() to assess parameters I have a go with a basic example dy/dt = K1*conc I try to assess the value of K1 from a simulated data set with a K1 close to 2. Here is (I think) the best code that I've done so far even though it crashes when I call nls()

Dear Colleagues, Our group is also working on implementing the use of R for pharmacokinetic compartmental analysis. Perhaps I have missed something, but > fit <- nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=0.5, k2=0.5)), + data=C1.lsoda, + start=list(K1=0.3, k2=0.7), + trace=T + ) Error in eval(as.name(varName), data) : Object

Hello there, Suppose you want to solve the following system of ODE's (a simple Lotka-Volterra predator prey model) dP/dt = beta*P*V - mu*P dV/dt = r*V - beta*P*V where P and V are the numbers of predators and prey. Now, this is easy to do, but suppose you have a system of equations like this, dP1/dt = beta1*P1*V1 - mu1*P1 dP2/dt = beta2*P2*V2 - mu2*P2 dV1/dt = r1*V1 - beta1*P1*V1

Dear R users: I encountered difficulties in michaelis-menten equation. I found that when I use right model definiens, I got wrong Km vlaue, and I got right Km value when i use wrong model definiens. The value of Vd and Vmax are correct in these two models. #-----right model definiens-------- PKindex<-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),

Dear All,   I could use a bit of help here, this function is hard to figure out (for me at least) I have the following so far:   PKindex<-data.frame(Subject=c(1),time=c(1,2,3,4,6,10,12),conc=c(32,28,25,22,18,14,11)) Dose<-200 Tinf <-0.5   defun<- function(time, y, parms) {  dCpdt <- -parms["kel"] * y[1]  list(dCpdt)  } modfun <- function(time,kel, Vd) {   out <-