search for: extremum

Dear All,   I am working on a time series of hourly river flow measurements from 2000 - 2003  and have been trying to compute the extremum index from the original series as well as a new series from  sampling with replacement . The extremum Index produced from the original data series looks  fine but sadly, a strange value of index is given from the sampling. The flow measurements are in range of 0.1 to 35m3/s and are stored in data f...

Greetings, R-Wizards: I'm trying to find an extremum subject to a nonlinear constraint. (Yes, I have perused the archives but have found nothing positive.) The details of the problem are these: In a paper published some years ago in Technometrics, ("Confidence bands for cumulative distribution functions of continuous random variables" Tec...

Hello, I have a x,y,z file.Z is not corresponding to a simple analytical function of x and y (see below). I am trying to find the minimum location of this surface and the z value corresponding to this location by a spline interpolation or from a polynomial fit. I tried with the akima package but as the location of the point I am looking for (the minimum) is outside of the convex hull it's

...l 5129 7 xlevels 5129 4 call 5129 5 terms 5129 5 model 5129 7 formula 5129 5 anova 1026 2303 1040 1 1033 2 lm 254 1026 1 5129 18 gwRZ.coded.min.par 1038 3 0.1207451127917877 89.70226077469248 284.9345951599792 254 -------------- next part -------------- # Surfaceplots durch vorgegebenen Punkt "extremum" "surfacesinopt" <- function(x.lm, extremum, var.names=NULL, data=NULL, n=20, conf.level=0.95, sing leplot=T, extcol=2, titletext= NULL, var.labs=NULL, pred.var.lab=NULL, charcex=1 , chartype=10,maximize=FALSE) { if (is.null(data)) data <- x.lm$model if ((!is.character(var...

...ion? or use dput to share x and y outputs of UnivarMixingDistribution function and I can look for the issue. Also You might have multiple maximas has same maximum y-values (for instance, y = sin(x)) and this definition is relevant to defined interval. For instance again, we can talk about a global extremum for a function like y = ax^2+bx + c. We know we will have only a single extremum point (maxima or minima). For higher order functions, we can talk about only local maximas. So, I assume you want to obtain maximum one of this local maximas, right? If yes, why don?t we find the maximum y-value and c...

...> outputs of UnivarMixingDistribution function and I can look for the issue. > > > > Also You might have multiple maximas has same maximum y-values (for > instance, y = sin(x)) and this definition is relevant to defined interval. > For instance again, we can talk about a global extremum for a function like > y = ax^2+bx + c. We know we will have only a single extremum point (maxima > or minima). For higher order functions, we can talk about only local > maximas. So, I assume you want to obtain maximum one of this local maximas, > right? > > > > If yes, why...

...produciple example, and let?s focus on the problematic part. I assure you that your problem will be solved faster. I could install the distr package at the end, but I?m getting another error while running your code. But I don?t believe the question is related to this package. Let me explain about extremum points although most of you know about it. Let?s assume we have a y(x) function. An extremum (max/min) point is defined as where dy/dx = 0. If second order derivative of y is less than 0, it?s a maximum, if greater than 0, it?s a minimum point. If zero, it?s undefined. So, at the end, we need x an...

...and let?s focus on the problematic part. I assure you that your problem will be solved faster. > > I could install the distr package at the end, but I?m getting another error while running your code. But I don?t believe the question is related to this package. > > Let me explain about extremum points although most of you know about it. Let?s assume we have a y(x) function. An extremum (max/min) point is defined as where dy/dx = 0. If second order derivative of y is less than 0, it?s a maximum, if greater than 0, it?s a minimum point. If zero, it?s undefined. So, at the end, we need x an...

Hi, Thanks for your mail, and time It is not working for some arguments, when mean value is like >6. case mc0 <- c(0.08844446,0.1744455,0.1379778,0.1209769,0.1573065,0. 1134463,0.2074027) rv <-UnivarMixingDistribution(Norm(486.4255, 53.24133), Norm(664.0713, 3.674773), Norm(669.0484, 4.101381),

..._2) Note that it is the same non-linear transformation f() that is applied to both x_1 and x_2. So my first question is how can I do it in R? A more general question is this: suppose I have a utility function U(a_i, f()), where f() is say a spline. Is there a general optimizer that could find an extremum of such U()? If not, how easy it would be to hack up something like this? Would it become easier if U() depended on f() only, i.e. no a_i terms? Thanks, Vadim -------------------------------------------------- DISCLAIMER \ This e-mail, and any attachments thereto, is intend ... [[dropped]]

...quot; - "Error in numericDeriv(form[[3]], names(ind), env) : \n Valeur manquante ou infinie obtenue au cours du calcul du mod?le\n") - "Error in nlsModel(formula, mf, start, wts) : \n singular gradient matrix at initial parameter estimates\n" So it seems that I reach a local extremum each time. I know that most of the problem comes from the choice of the initial values of the parameters Asym_inf, Asym_inf, n1, n2, tmid1, tmid2, scal1and scal2. My question is how could I estimate those initial values so that the nls fitting works. Thanks in advance -- Nathalie YAUSCHEW-R...

Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01, x1 is the quantile of normal distribution (0.0032,x) with probability of 0.7, and the changing value should be x. Initial value for x is 0.0207. I am using the following codes, but it does not work. fr <- function(x) { x1<-qnorm(0.7,0.0032,x) x2=0.01 x1-x2 } xsd <- optim(0.0207, fr,