similar to: lsoda

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

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:

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())

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

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 =

>From my book on corpus linguistics with R: # (10) Imagine you have two vectors a and b such that a<-c("d", "d", "j", "f", "e", "g", "f", "f", "i", "g") b<-c("a", "g", "d", "f", "g", "a", "f", "a",

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

Dear Hank, Last question first: really, only you can say for sure if 4e-281 and 5e-11 are small enough; it depends on the units you measure your state variables in. However, this strategy cannot get the state variables to exactly 0. Obviously, you could get closer to 0.0 faster by setting the derivatives even larger in absolute value. You may run into problems with the solver when the

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 =

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),

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

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 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 <-

The information at http://cran.r-project.org/doc/manuals/R-admin.html#The-MinGW-compilers and http://www.murdoch-sutherland.com/Rtools/ is slightly inconsistent about the compiler used to build Windows binary packages available through cran. The 'candidate' package of the recommended MinGW-5.0.0.exe installs g++/g77 3.4.4 (as does the updated installer MinGW-5.0.2.exe). "An

My students are working with several SPSS dataset provided by the European Social Survey. If you register your name, you can download it too. This is the 2004 data, for example: http://ess.nsd.uib.no/ess/round2/ I cannot give you the European Survey dataset, but you can download it for free if you like, and then you could run these commands to re-produce this weird pattern described below.

A couple of minor points about the odesolve package (which I am otherwise enjoying very much): 1. "scalar" is misspelled as "scaler" in the definitions of the rtol and atol parameters 2. it is possible to crash R by doing something dumb, e.g failing to read the documentation carefully enough and (a) returning only a vector of derivatives and not a list of (derivatives,

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, 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 <-

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

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()